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I.For salt A Solubility of A before heating = mass of A x 100 Volume of water added => 40 x 100 = 400g/100g Water 10 (Theoretical)Solubility of A before heating = 400 g Less (From graph ) Solubility of A after heating at 70oC = 48g Mass of crystals that can not dissolve at70oC = 352 g (From graph ) Solubility of A after heating at 70oC = 48g Less (From graph ) Solubility of A after cooling to 10oC = 31g Mass of crystals that crystallize out on cooling to10oC = 17 g Mass of crystals that can not dissolve at70oC = 352 g Add Mass of crystals that crystallize out on cooling to10oC = 17 g Total mass of A that does not dissolve/crystallize/precipitate = 369 g I.For salt B Solubility of B before heating = mass of B x 100 Volume of water added => 60 x 100 = 600g/100g Water 10 (Theoretical)Solubility of B before heating = 600 g Less (From graph ) Solubility of B after heating at 70oC = 138g Mass of crystals that cannot dissolve at70oC = 462 g (From graph ) Solubility of B after heating at 70oC = 138g Less (From graph ) Solubility of B after cooling to 10oC = 21g Mass of crystals that crystallize out on cooling to10oC = 117 g Mass of crystals that cannot dissolve at70oC = 462 g Add Mass of crystals that crystallize out on cooling to10oC = 117 g Total mass of A that does not dissolve/crystallize/precipitate = 579 g
(f)State the assumption made in (e)above Solubility of one salt has no effect on the solubility of the other 8. When 5.0 g of potassium chlorate (V) was put in 10cm3 of water and heated, the solid dissolves. When the solution was cooled , the temperature at which crystals reappear was noted. Another 10cm3 of water was added and the mixture heated to dissolve then cooled for the crystals to reappear .The table below shows the the results obtained Total volume of water added(cm3) 10.0 20.0 30.0 40.0 50.0 Mass of KClO3 5.0 5.0 5.0 5.0 5.0 Temperature at which crystals appear 80.0 65.0 55.0 45.0 30.0 Solubility of KclO3 50.0 25.0 16.6667 12.5 10.0 (a)Complete the table to show the solubility of KclO3 at different temperatures.
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When 5.0 g of potassium chlorate (V) was put in 10cm3 of water and heated, the solid dissolves. When the solution was cooled , the temperature at which crystals reappear was noted. Another 10cm3 of water was added and the mixture heated to dissolve then cooled for the crystals to reappear .The table below shows the the results obtained Total volume of water added(cm3) 10.0 20.0 30.0 40.0 50.0 Mass of KClO3 5.0 5.0 5.0 5.0 5.0 Temperature at which crystals appear 80.0 65.0 55.0 45.0 30.0 Solubility of KclO3 50.0 25.0 16.6667 12.5 10.0 (a)Complete the table to show the solubility of KclO3 at different temperatures. (b)Plot a graph of mass of KClO3 per 100g water against temperature at which crystals form. (c)From the graph, show and determine ; (i)the solubility of KClO3 at I. 50oC From a well plotted graph = 14.5 g KClO3/100g water II. 35oC From a well plotted graph = 9.0 g KclO3/100g water (ii)the temperature at which the solubility is: I.10g/100g water From a well plotted graph = 38.0 oC II.45g/100g water From a well plotted graph = 77.5 oC (d)Explain the shape of the graph. Solubility of KClO3 increase with increase in temperature/more KclO3dissolve as temperature rises. (e)What happens when 100g per 100g water is cooled to 35.0 oC Solubility before heating = 100.0
(From the graph) Solubility after cooling = 9.0 Mass of salt precipitated/crystallization = 91.0 g 9. 25.0cm3 of water dissolved various masses of ammonium chloride crystals at different temperatures as shown in the table below.
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Solubility of KClO3 increase with increase in temperature/more KclO3dissolve as temperature rises. (e)What happens when 100g per 100g water is cooled to 35.0 oC Solubility before heating = 100.0
(From the graph) Solubility after cooling = 9.0 Mass of salt precipitated/crystallization = 91.0 g 9. 25.0cm3 of water dissolved various masses of ammonium chloride crystals at different temperatures as shown in the table below. Mass of ammonium chloride(grams) 4.0 4.5 5.5 6.5 9.0 Temperature at which solid dissolved(oC) 30.0 50.0 70.0 90.0 120.0 Solubility of NH4Cl 16.0 18.0 22.0 26.0 36.0 (a)Complete the table (b)Plot a solubility curve (c)What happens when a saturated solution of ammonium chloride is cooled from 80oC to 40oC. (From the graph )Solubility at 80oC = 24.0 g Less (From the graph )Solubility at 40oC = 16.8 g Mass of crystallized/precipitated = 7.2 g 20. Solubility and solubility curves are therefore used (i) to know the effect of temperature on the solubility of a salt (ii)to fractional crystallize two soluble salts by applying their differences in solubility at different temperatures. (iii)determine the mass of crystal that is obtained from crystallization. 21.Natural fractional crystallization takes place in Kenya/East Africa at: (i) Lake Magadi during extraction of soda ash(Sodium carbonate) from Trona(sodium sesquicarbonate) (ii) Ngomeni near Malindi at the Indian Ocean Coastline during the extraction of common salt(sodium chloride). 22.Extraction of soda ash from Lake Magadi in Kenya Rain water drains underground in the great rift valley and percolate underground where it is heated geothermically. The hot water dissolves underground soluble sodium compounds and comes out on the surface as alkaline springs which are found around the edges of Lake Magadi in Kenya.
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21.Natural fractional crystallization takes place in Kenya/East Africa at: (i) Lake Magadi during extraction of soda ash(Sodium carbonate) from Trona(sodium sesquicarbonate) (ii) Ngomeni near Malindi at the Indian Ocean Coastline during the extraction of common salt(sodium chloride). 22.Extraction of soda ash from Lake Magadi in Kenya Rain water drains underground in the great rift valley and percolate underground where it is heated geothermically. The hot water dissolves underground soluble sodium compounds and comes out on the surface as alkaline springs which are found around the edges of Lake Magadi in Kenya. Temperatures around the lake are very high (30-40oC) during the day. The solubility of trona decrease with increase in temperature therefore solid crystals of trona grows on top of the lake (upto or more than 30metres thick) A bucket dredger mines the trona which is then crushed ,mixed with lake liquor and pumped to washery plant where it is further refined to a green granular product called CRS. The CRS is then heated to chemically decompose trona to soda ash(Sodium carbonate) Chemical equation 2Na2CO3.NaHCO3.2H2O(s) -> 3Na2CO3 (s) + CO2(g) + 5H2O(l) Soda ash(Sodium carbonate) is then stored .It is called Magadi Soda. Magadi Soda is used : (i) make glass (ii) for making soapless detergents (iii) softening hard water. (iv) Common salt is colledcted at night because its solubility decreases with decrease in temperature. It is used as salt lick/feed for animals. Summary flow diagram showing the extraction of Soda ash from Trona Sodium chloride and Trona dissolved in the sea Natural fractional crystallization Crystals of Trona (Day time) Crystals of sodium chloride(At night) Dredging /scooping/ digging Crushing Furnace (Heating) Carbon(IV) oxide Soda ash
23.Extraction of common salt from Indian Ocean at Ngomeni in Kenya Oceans are salty.They contain a variety of dissolved salts (about 77% being sodium chloride). During high tide ,water is collected into shallow pods and allowed to crystallize as evaporation takes place.The pods are constructed in series to increase the rate of evaporation.
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It is used as salt lick/feed for animals. Summary flow diagram showing the extraction of Soda ash from Trona Sodium chloride and Trona dissolved in the sea Natural fractional crystallization Crystals of Trona (Day time) Crystals of sodium chloride(At night) Dredging /scooping/ digging Crushing Furnace (Heating) Carbon(IV) oxide Soda ash
23.Extraction of common salt from Indian Ocean at Ngomeni in Kenya Oceans are salty.They contain a variety of dissolved salts (about 77% being sodium chloride). During high tide ,water is collected into shallow pods and allowed to crystallize as evaporation takes place.The pods are constructed in series to increase the rate of evaporation. At the final pod ,the crystals are scapped together,piled in a heap and washed with brine (concentrated sodium chloride). It contains MgCl2 and CaCl2 . MgCl2 and CaCl2are hygroscopic. They absorb water from the atmosphere and form a solution. This makes table salt damp/wet on exposure to the atmosphere. 24.Some water form lather easily with soap while others do not. Water which form lather easily with soap is said to be “soft” Water which do not form lather easily with soap is said to be “hard” Hardness of water is caused by the presence of Ca2+ and Mg2+ ions. Ca2+ and Mg2+ ions react with soap to form an insoluble grey /white suspension/precipitate called Scum/ curd. Ca2+ and Mg2+ ions in water come from the water sources passing through rocks containing soluble salts of Ca2+ and Mg2+ e.g. Limestone or gypsum There are two types of water hardness: (a)temporary hardness of water (b)permanent hardness of water (a)temporary hardness of water
Temporary hardness of water is caused by the presence of dissolved calcium hydrogen carbonate/Ca(HCO3)2 and magnesium hydrogen carbonate/Mg(HCO3)2 When rain water dissolve carbon(IV) oxide from the air it forms waek carbonic(IV) acid i.e. CO2(g) + H2O(l) -> H2CO3(aq) When carbonic(IV) acid passes through limestone/dolomite rocks it reacts to form soluble salts i.e.
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Ca2+ and Mg2+ ions in water come from the water sources passing through rocks containing soluble salts of Ca2+ and Mg2+ e.g. Limestone or gypsum There are two types of water hardness: (a)temporary hardness of water (b)permanent hardness of water (a)temporary hardness of water
Temporary hardness of water is caused by the presence of dissolved calcium hydrogen carbonate/Ca(HCO3)2 and magnesium hydrogen carbonate/Mg(HCO3)2 When rain water dissolve carbon(IV) oxide from the air it forms waek carbonic(IV) acid i.e. CO2(g) + H2O(l) -> H2CO3(aq) When carbonic(IV) acid passes through limestone/dolomite rocks it reacts to form soluble salts i.e. In limestone areas; H2CO3(aq) + CaCO3(s) -> Ca(HCO3)2 (aq) In dolomite areas; H2CO3(aq) + MgCO3(s) -> Mg(HCO3)2 (aq) (b)permanent hardness of water Permanent hardness of water is caused by the presence of dissolved calcium sulphate(VI)/CaSO4 and magnesium sulphate(VI)/Mg SO4 Permanent hardness of water is caused by water dissolving CaSO4 and MgSO4 from ground rocks. Hardness of water can be removed by the following methods: (a)Removing temporary hardness of water (i)Boiling/heating. Boiling decomposes insoluble calcium hydrogen carbonate/Ca(HCO3)2 and magnesium hydrogen carbonate/Mg(HCO3)2 to insoluble CaCO3 and MgCO3 that precipitate away. i.e Chemical equation Ca(HCO3)2(aq) -> CaCO3 (s) + CO2(g) + H2O(l) Mg(HCO3)2(aq) -> MgCO3 (s) + CO2(g) + H2O(l) (ii)Adding sodium carbonate (IV) /Washing soda. Since boiling is expensive on a large scale ,a calculated amount of sodium carbonate decahydrate /Na2CO3.10H2O precipitates insoluble Ca2+(aq) and Mg2+(aq) ions as carbonates to remove both temporary and permanent hardness of water .This a double decomposition reaction where two soluble salts form an insoluble and soluble salt. i.e.
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i.e Chemical equation Ca(HCO3)2(aq) -> CaCO3 (s) + CO2(g) + H2O(l) Mg(HCO3)2(aq) -> MgCO3 (s) + CO2(g) + H2O(l) (ii)Adding sodium carbonate (IV) /Washing soda. Since boiling is expensive on a large scale ,a calculated amount of sodium carbonate decahydrate /Na2CO3.10H2O precipitates insoluble Ca2+(aq) and Mg2+(aq) ions as carbonates to remove both temporary and permanent hardness of water .This a double decomposition reaction where two soluble salts form an insoluble and soluble salt. i.e. (i)with temporary hard water Chemical equation Na2CO3 (aq) + Ca(HCO3) 2 (aq) -> NaHCO3(aq) + CaCO3 (s) Na2CO3 (aq) + Mg(HCO3) 2 (aq) -> NaHCO3(aq) + MgCO3 (s) Ionic equation CO32 (aq) + Ca2+ (aq) -> CaCO3 (s)
CO32 (aq) + Mg2+ (aq) -> MgCO3 (s) (ii)with permanent hard water Chemical equation Na2CO3 (aq) + MgSO4 (aq) -> Na2SO4 (aq) + MgCO3 (s) Na2CO3 (aq) + CaSO4 (aq) -> Na2SO4 (aq) + MgCO3 (s) Ionic equation CO32 (aq) + Ca2+ (aq) -> CaCO3 (s) CO32 (aq) + Mg2+ (aq) -> MgCO3 (s) (iii)Adding calcium (II)hydroxide/Lime water Lime water/calcium hydroxide removes only temporary hardness of water from by precipitating insoluble calcium carbonate(IV). Chemical equation Ca(OH)2 (aq) + Ca(HCO3) 2 (aq) -> 2H2O(l) + 2CaCO3 (s) Excess of Lime water/calcium hydroxide should not be used because it dissolves again to form soluble calcium hydrogen carbonate(IV) causing the hardness again.
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i.e. (i)with temporary hard water Chemical equation Na2CO3 (aq) + Ca(HCO3) 2 (aq) -> NaHCO3(aq) + CaCO3 (s) Na2CO3 (aq) + Mg(HCO3) 2 (aq) -> NaHCO3(aq) + MgCO3 (s) Ionic equation CO32 (aq) + Ca2+ (aq) -> CaCO3 (s)
CO32 (aq) + Mg2+ (aq) -> MgCO3 (s) (ii)with permanent hard water Chemical equation Na2CO3 (aq) + MgSO4 (aq) -> Na2SO4 (aq) + MgCO3 (s) Na2CO3 (aq) + CaSO4 (aq) -> Na2SO4 (aq) + MgCO3 (s) Ionic equation CO32 (aq) + Ca2+ (aq) -> CaCO3 (s) CO32 (aq) + Mg2+ (aq) -> MgCO3 (s) (iii)Adding calcium (II)hydroxide/Lime water Lime water/calcium hydroxide removes only temporary hardness of water from by precipitating insoluble calcium carbonate(IV). Chemical equation Ca(OH)2 (aq) + Ca(HCO3) 2 (aq) -> 2H2O(l) + 2CaCO3 (s) Excess of Lime water/calcium hydroxide should not be used because it dissolves again to form soluble calcium hydrogen carbonate(IV) causing the hardness again. (iv)Adding aqueous ammonia Aqueous ammonia removes temporary hardness of water by precipitating insoluble calcium carbonate(IV) and magnesium carbonate(IV) Chemical equation 2NH3 (aq) + Ca(HCO3) 2 (aq) -> (NH4) 2CO3(aq) + CaCO3 (s) 2NH3 (aq) + Mg(HCO3) 2 (aq) -> (NH4) 2CO3(aq) + MgCO3 (s) (v)Use of ion-exchange permutit This method involves packing a chamber with a resin made of insoluble complex of sodium salt called sodium permutit. The sodium permutit releases sodium ions that are exchanged with Mg2+ and Ca2+ ions in hard water making the water to be soft. i.e.
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(iv)Adding aqueous ammonia Aqueous ammonia removes temporary hardness of water by precipitating insoluble calcium carbonate(IV) and magnesium carbonate(IV) Chemical equation 2NH3 (aq) + Ca(HCO3) 2 (aq) -> (NH4) 2CO3(aq) + CaCO3 (s) 2NH3 (aq) + Mg(HCO3) 2 (aq) -> (NH4) 2CO3(aq) + MgCO3 (s) (v)Use of ion-exchange permutit This method involves packing a chamber with a resin made of insoluble complex of sodium salt called sodium permutit. The sodium permutit releases sodium ions that are exchanged with Mg2+ and Ca2+ ions in hard water making the water to be soft. i.e. Na2X(aq) + Ca2+ (aq) -> Na+ (aq) + CaX(s) Na2X(aq) + Mg2+ (aq) -> Na+ (aq) + MgX(s) Hard water containing Mg2+ and Ca2+
Ion exchange resin as Sodium permutit ------- Na+ ions replace Mg2+ and Ca2+ to make the water soft. When all the Na+ ions in the resin is fully exchanged with Ca2+ and Ng2+ ions in the permutit column ,it is said to be exhausted. Brine /concentrated sodium chloride solution is passed through the permutit column to regenerated /recharge the column again. Hard water containing Mg2+ and Ca2+ Ion exchange resin as Sodium permutit
------- Na+ ions replace Mg2+ and Ca2+ to make the water soft. (vi)Deionization /demineralization This is an advanced ion exchange method of producing deionized water .Deionized water is extremely pure water made only of hydrogen and oxygen only without any dissolved substances. Deionization involve using the resins that remove all the cations by using: (i)A cation exchanger which remove /absorb all the cations present in water and leave only H+ ions. (ii)An anion exchanger which remove /absorb all the anions present in water and leave only OH- ions. The H+(aq) and OH- (aq) neutralize each other to form pure water.
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Deionization involve using the resins that remove all the cations by using: (i)A cation exchanger which remove /absorb all the cations present in water and leave only H+ ions. (ii)An anion exchanger which remove /absorb all the anions present in water and leave only OH- ions. The H+(aq) and OH- (aq) neutralize each other to form pure water. Chemical equation H+(aq) + OH- (aq) -> H2O(l) When exhausted the cation exchanger is regenerated by adding H+(aq) from sulphuric(VI)acid/hydrochloric acid. When exhausted the anion exchanger is regenerated by adding OH-(aq) from sodium hydroxide. Advantages of hard water Hard water has the following advantages: (i)Ca2+(aq) in hard water are useful in bone and teeth formation (ii) is good for brewing beer (iii)contains minerals that cause it to have better /sweet taste (iv)animals like snails and coral polyps use calcium to make their shells and coral reefs respectively. (v)processing mineral water Disadvantages of hard water
Hardness of water: (i)waste a lot of soap during washing before lather is formed. (ii)causes stains/blemishes/marks on clothes/garments (iii)causes fur on electric appliances like kettle ,boilers and pipes form decomposition of carbonates on heating .This reduces their efficiency hence more/higher cost of power/electricity. Sample revision questions In an experiment, soap solution was added to three separate samples of water. The table below shows the volumes of soap solution required to form lather with 1000cm3 of each sample of water before and after boiling. Sample I Sample II Sample III Volume of soap before water is boiled (cm3) 27.0 3.0 10.0 Volume of soap after water is boiled(cm3) 27.0 3.0 3.0 a) Which water sample is likely to be soft? Explain. (2mks) Sample II: Uses little sample of soap . c) Name the change in the volume of soap solution used in sample III (1mk) On heating the sample water become soft bcause it is temporary hard.
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Explain. (2mks) Sample II: Uses little sample of soap . c) Name the change in the volume of soap solution used in sample III (1mk) On heating the sample water become soft bcause it is temporary hard. 2.Study the scheme below and use it to aanswer the questions that follow:
(a)Write the formula of: (i)Cation in solution K Al3+ (ii)white ppt L Al(OH)3 (iii) colourless solution M [Al(OH)4]- (iv) colourless solution N AlCl3 (v)white ppt P Al(OH)3 (b)Write the ionic equation for the reaction for the formation of: (i)white ppt L Al3+(aq) + 3OH- (aq) -> Al(OH)3(s) (v)white ppt P Al3+(aq) + 3OH- (aq) -> Al(OH)3(s) (c)What property is illustrated in the formation of colourless solution M and N. Amphotellic
19.0.0 ENERGY CHANGES IN CHEMICAL AND PHYSICAL PROCESSES (25 LESSONS) 1.Introduction to Energy changes Energy is the capacity to do work. There are many/various forms of energy like heat, electric, mechanical, and/ or chemical energy.There are two types of energy: (i)Kinetic Energy(KE) ;the energy in motion. (ii)Potential Energy(PE); the stored/internal energy. Energy like matter , is neither created nor destroyed but can be transformed /changed from one form to the other/ is interconvertible. This is the principle of conservation of energy. e.g. Electrical energy into heat through a filament in bulb. Chemical and physical processes take place with absorption or evolution/production of energy mainly in form of heat The study of energy changes that accompany physical/chemical reaction/changes is called Thermochemistry. Physical/chemical reaction/changes that involve energy changes are called thermochemical reactions. The SI unit of energy is the Joule(J).Kilo Joules(kJ)and megaJoules(MJ) are also used. The Joule(J) is defined as the: (i) quantity of energy transferred when a force of one newton acts through a distance of one metre. 2 (ii) quantity of energy transferred when one coulomb of electric charge is passed through a potential difference of one volt.
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The SI unit of energy is the Joule(J).Kilo Joules(kJ)and megaJoules(MJ) are also used. The Joule(J) is defined as the: (i) quantity of energy transferred when a force of one newton acts through a distance of one metre. 2 (ii) quantity of energy transferred when one coulomb of electric charge is passed through a potential difference of one volt. All thermochemical reactions should be carried out at standard conditions of: (i) 298K /25oC temperature (ii)101300Pa/101300N/m2 /760mmHg/1 atmosphere pressure. 2.Exothermic and endothermic processes/reactions Some reactions / processes take place with evolution/production of energy. They are said to be exothermic while others take place with absorption of energy. They are said to be endothermic. Practically exothermic reactions / processes cause a rise in temperature (by a rise in thermometer reading/mercury or alcohol level rise) Practically endothermic reactions / processes cause a fall in temperature (by a fall in thermometer reading/mercury or alcohol level decrease) To demonstrate/illustrate exothermic and endothermic processes/reactions a) Dissolving Potassium nitrate(V)/ammonium chloride crystals Procedure: Measure 20cm3 of water in a beaker. Determine and record its temperature T1.Put about 1.0g of Potassium nitrate(V) crystals into the beaker. Stir the mixture carefully and note the highest temperature rise /fall T2.Repeat the whole procedure by using ammonium chloride in place of Potassium nitrate (V) crystals. Sample results Temperture (oC) Using Potassium nitrate(V) crystals Using Ammonium chloride crystals T2(Final temperature) 21.0 23.0 T1 (Initial temperature) 25.0 26.0 Change in temperature(T2 –T1) 4.0 3.0 Note: (i)Initial(T1) temperature of dissolution of both potassium nitrate(V) crystals and ammonium chloride crystals is higher than the final temperature(T2) (ii) Change in temperature(T2 –T1) is not a mathematical “-4.0” or “-3.0”.
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Determine and record its temperature T1.Put about 1.0g of Potassium nitrate(V) crystals into the beaker. Stir the mixture carefully and note the highest temperature rise /fall T2.Repeat the whole procedure by using ammonium chloride in place of Potassium nitrate (V) crystals. Sample results Temperture (oC) Using Potassium nitrate(V) crystals Using Ammonium chloride crystals T2(Final temperature) 21.0 23.0 T1 (Initial temperature) 25.0 26.0 Change in temperature(T2 –T1) 4.0 3.0 Note: (i)Initial(T1) temperature of dissolution of both potassium nitrate(V) crystals and ammonium chloride crystals is higher than the final temperature(T2) (ii) Change in temperature(T2 –T1) is not a mathematical “-4.0” or “-3.0”. (iii)Dissolution of both potassium nitrate(V) and ammonium chloride crystals is an endothermic process because initial(T1) temperature is higher than the final temperature(T2) thus causes a fall/drop in temperature. b) Dissolving concentrated sulphuric(VI) acid/sodium hydroxide crystals
3 Procedure: Measure 20cm3 of water in a beaker. Determine and record its temperature T1.Carefully put about 1.0g/four pellets of sodium hydroxide crystals into the beaker. Stir the mixture carefully and note the highest temperature rise /fall T2.Repeat the whole procedure by using 2cm3 of concentrated sulphuric(VI) acid in place of sodium hydroxide crystals. CAUTION: (i)Sodium hydroxide crystals are caustic and cause painful blisters on contact with skin. (ii) Concentrated sulphuric (VI) acid is corrosive and cause painful wounds on contact with skin. Sample results Temperture (oC) Using Sodium hydroxide pellets Using Concentrated sulphuric(VI) acid T2(Final temperature) 30.0 32.0 T1 (Initial temperature) 24.0 25.0 Change in temperature(T2 –T1) 6.0 7.0 Note: (i)Initial (T1) temperature of dissolution of both concentrated sulphuric (VI) acid and sodium hydroxide pellets is lower than the final temperature (T2).
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CAUTION: (i)Sodium hydroxide crystals are caustic and cause painful blisters on contact with skin. (ii) Concentrated sulphuric (VI) acid is corrosive and cause painful wounds on contact with skin. Sample results Temperture (oC) Using Sodium hydroxide pellets Using Concentrated sulphuric(VI) acid T2(Final temperature) 30.0 32.0 T1 (Initial temperature) 24.0 25.0 Change in temperature(T2 –T1) 6.0 7.0 Note: (i)Initial (T1) temperature of dissolution of both concentrated sulphuric (VI) acid and sodium hydroxide pellets is lower than the final temperature (T2). (ii)Dissolution of both Sodium hydroxide pellets and concentrated sulphuric (VI) acid is an exothermic process because final (T2) temperature is higher than the initial temperature (T1) thus causes a rise in temperature. The above reactions show heat loss to and heat gain from the surrounding as illustrated by a rise and fall in temperature/thermometer readings. Dissolving both potassium nitrate(V) and ammonium chloride crystals causes heat gain from the surrounding that causes fall in thermometer reading. Dissolving both Sodium hydroxide pellets and concentrated sulphuric (VI) acid causes heat loss to the surrounding that causes rise in thermometer reading. At the same temperature and pressure ,heat absorbed and released is called enthalpy/ heat content denoted H. Energy change is measured from the heat content/enthalpy of the final and initial products. It is denoted ∆H(delta H).i.e.
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At the same temperature and pressure ,heat absorbed and released is called enthalpy/ heat content denoted H. Energy change is measured from the heat content/enthalpy of the final and initial products. It is denoted ∆H(delta H).i.e. Enthalpy/energy/ change in heat content ∆H = Hfinal – Hinitial For chemical reactions: ∆H = Hproducts – Hreactants For exothermic reactions, the heat contents of the reactants is more than/higher than the heat contents of products, therefore the ∆H is negative (-∆H)
4 For endothermic reactions, the heat contents of the reactants is less than/lower than the heat contents of products, therefore the ∆H is negative (+∆H) Graphically, in a sketch energy level diagram: (i)For endothermic reactions the heat content of the reactants should be relatively/slightly lower than the heat content of the products (ii)For exothermic reactions the heat content of the reactants should be relatively/slightly higher than the heat content of the products Sketch energy level diagrams for endothermic dissolution Energy (kJ) H2 KNO3(aq) +∆H = H2 – H1 H1 KNO3(s) Reaction path/coordinate/progress Energy (kJ) H2 NH4Cl (aq) +∆H = H2 – H1 H1 NH4Cl (s) Reaction path/coordinate/progress Sketch energy level diagrams for exothermic dissolution
5 H2 NaOH (s) Energy(kJ) -∆H = H2 – H1 H1 NaOH (aq) Reaction path/coordinate/progress H2 H2SO4 (l) Energy (kJ) -∆H = H2 – H1 H1 H2SO4 (aq) Reaction path/coordinate/progress 3.Energy changes in physical processes Melting/freezing/fusion/solidification and boiling/vaporization/evaporation are the two physical processes. Melting /freezing point of pure substances is fixed /constant. The boiling point of pure substance depend on external atmospheric pressure. Melting/fusion is the physical change of a solid to liquid. Freezing is the physical change of a liquid to solid. Melting/freezing/fusion/solidification are therefore two opposite but same reversible physical processes.
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Melting/fusion is the physical change of a solid to liquid. Freezing is the physical change of a liquid to solid. Melting/freezing/fusion/solidification are therefore two opposite but same reversible physical processes. i.e A (s) ========A(l)
6 Boiling/vaporization/evaporation is the physical change of a liquid to gas/vapour. Condensation/liquidification is the physical change of gas/vapour to liquid. Boiling/vaporization/evaporation and condensation/liquidification are therefore two opposite but same reversible physical processes. i.e B (l) ========B(g) Practically (i) Melting/liquidification/fusion involves heating a solid to weaken the strong bonds holding the solid particles together. Solids are made up of very strong bonds holding the particles very close to each other (Kinetic Theory of matter).On heating these particles gain energy/heat from the surrounding heat source to form a liquid with weaker bonds holding the particles close together but with some degree of freedom. Melting/freezing/fusion is an endothermic (+∆H)process that require/absorb energy from the surrounding. (ii)Freezing/fusion/solidification involves cooling a a liquid to reform /rejoin the very strong bonds to hold the particles very close to each other as solid and thus lose their degree of freedom (Kinetic Theory of matter). Freezing /fusion / solidification is an exothermic (-∆H)process that require particles holding the liquid together to lose energy to the surrounding. (iii)Boiling/vaporization/evaporation involves heating a liquid to completely break/free the bonds holding the liquid particles together. Gaseous particles have high degree of freedom (Kinetic Theory of matter). Boiling /vaporization / evaporation is an endothermic (+∆H) process that require/absorb energy from the surrounding. (iv)Condensation/liquidification is reverse process of boiling /vaporization / evaporation.It involves gaseous particles losing energy to the surrounding to form a liquid.It is an exothermic(+∆H) process. The quantity of energy required to change one mole of a solid to liquid or to form one mole of a solid from liquid at constant temperature is called molar enthalpy/latent heat of fusion. e.g.
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(iv)Condensation/liquidification is reverse process of boiling /vaporization / evaporation.It involves gaseous particles losing energy to the surrounding to form a liquid.It is an exothermic(+∆H) process. The quantity of energy required to change one mole of a solid to liquid or to form one mole of a solid from liquid at constant temperature is called molar enthalpy/latent heat of fusion. e.g. H2O(s) -> H2O(l) ∆H = +6.0kJ mole-1 (endothermic process) H2O(l) -> H2O(s) ∆H = -6.0kJ mole-1 (exothermic process) The quantity of energy required to change one mole of a liquid to gas/vapour or to form one mole of a liquid from gas/vapour at constant temperature is called molar enthalpy/latent heat of vapourization. e.g. H2O(l) -> H2O(g) ∆H = +44.0kJ mole-1 (endothermic process) H2O(g) -> H2O(l) ∆H = -44.0kJ mole-1 (exothermic process)
7 The following experiments illustrate/demonstrate practical determination of melting and boiling a) To determine the boiling point of water Procedure: Measure 20cm3 of tap water into a 50cm3 glass beaker. Determine and record its temperature.Heat the water on a strong Bunsen burner flame and record its temperature after every thirty seconds for four minutes. Sample results Time(seconds) 0 30 60 90 120 150 180 210 240 Temperature(oC) 25.0 45.0 85.0 95.0 96.0 96.0 96.0 97.0 98.0 Questions 1.Plot a graph of temperature against time(y-axis) Sketch graph of temperature against time boiling point 96 oC Temperature(0C) 25oC time(seconds) 2.From the graph show and determine the boiling point of water Note: Water boils at 100oC at sea level/one atmosphere pressure/101300Pa but boils at below 100oC at higher altitudes. The sample results above are from Kiriari Girls High School-Embu County on the slopes of Mt Kenya in Kenya. Water here boils at 96oC.
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Sample results Time(seconds) 0 30 60 90 120 150 180 210 240 Temperature(oC) 25.0 45.0 85.0 95.0 96.0 96.0 96.0 97.0 98.0 Questions 1.Plot a graph of temperature against time(y-axis) Sketch graph of temperature against time boiling point 96 oC Temperature(0C) 25oC time(seconds) 2.From the graph show and determine the boiling point of water Note: Water boils at 100oC at sea level/one atmosphere pressure/101300Pa but boils at below 100oC at higher altitudes. The sample results above are from Kiriari Girls High School-Embu County on the slopes of Mt Kenya in Kenya. Water here boils at 96oC. 8 3.Calculate the molar heat of vaporization of water.(H= 1.0,O= 16.O) Working: Mass of water = density x volume => (20 x 1) /1000 = 0.02kg Quantity of heat produced = mass of water x specific heat capacity of water x temperature change =>0.02kg x 4.2 x ( 96 – 25 ) = 5.964kJ Heat of vaporization of one mole H2O = Quantity of heat Molar mass of H2O =>5.964kJ = 0.3313 kJ mole -1 18 To determine the melting point of candle wax Procedure Weigh exactly 5.0 g of candle wax into a boiling tube. Heat it on a strongly Bunsen burner flame until it completely melts. Insert a thermometer and remove the boiling tube from the flame. Stir continuously. Determine and record the temperature after every 30seconds for four minutes.
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Insert a thermometer and remove the boiling tube from the flame. Stir continuously. Determine and record the temperature after every 30seconds for four minutes. Sample results Time(seconds) 0 30 60 90 120 150 180 210 240 Temperature(oC) 93.0 85.0 78.0 70.0 69.0 69.0 69.0 67.0 65.0 Questions 1.Plot a graph of temperature against time(y-axis) Sketch graph of temperature against time 93 oC Temperature(0C) melting point 69oC time(seconds) 2.From the graph show and determine the melting point of the candle wax
9 4.Energy changes in chemical processes Thermochemical reactions measured at standard conditions of 298K(25oC) and 101300Pa/101300Nm2/ 1 atmospheres/760mmHg/76cmHg produce standard enthalpies denoted ∆Hᶿ. Thermochemical reactions are named from the type of reaction producing the energy change. Below are some thermochemical reactions: (a) Standard enthalpy/heat of reaction ∆Hᶿr (b) Standard enthalpy/heat of combustion ∆Hᶿc (c) Standard enthalpy/heat of displacement ∆Hᶿd (d) Standard enthalpy/heat of neutralization ∆Hᶿn (e) Standard enthalpy/heat of solution/dissolution ∆Hᶿs (f) Standard enthalpy/heat of formation ∆Hᶿf (a)Standard enthalpy/heat of reaction ∆Hᶿr The molar standard enthalpy/heat of reaction may be defined as the energy/heat change when one mole of products is formed at standard conditions A chemical reaction involves the reactants forming products. For the reaction to take place the bonds holding the reactants must be broken so that new bonds of the products are formed. i.e. A-B + C-D -> A-C + B-D Old Bonds broken A-B and C-D on reactants New Bonds formed A-C and B-D on products The energy required to break one mole of a (covalent) bond is called bond dissociation energy.
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For the reaction to take place the bonds holding the reactants must be broken so that new bonds of the products are formed. i.e. A-B + C-D -> A-C + B-D Old Bonds broken A-B and C-D on reactants New Bonds formed A-C and B-D on products The energy required to break one mole of a (covalent) bond is called bond dissociation energy. The SI unit of bond dissociation energy is kJmole-1 The higher the bond dissociation energy the stronger the (covalent)bond Bond dissociation energies of some (covalent)bonds Bond Bond dissociation energy (kJmole-1) Bond dissociation energy (kJmole-1) H-H 431 I-I 151 C-C 436 C-H 413 C=C 612 O-H 463 C = C 836 C-O 358 N = N 945 H-Cl 428 N-H 391 H-Br 366 F-F 158 C-Cl 346 Cl-Cl 239 C-Br 276 Br-Br 193 C-I 338 H-I 299 O=O 497 Si-Si 226 C-F 494
10 The molar enthalpy of reaction can be calculated from the bond dissociation energy by: (i)adding the total bond dissociation energy of the reactants(endothermic process/+∆H) and total bond dissociation energy of the products(exothermic process/-∆H). (ii)subtracting total bond dissociation energy of the reactants from the total bond dissociation energy of the products(exothermic process/-∆H less/minus endothermic process/+∆H). Practice examples/Calculating ∆Hr 1.Calculate ∆Hr from the following reaction: a) H2(g) + Cl2(g) -> 2HCl(g) Working Old bonds broken (endothermic process/+∆H ) = (H-H + Cl-Cl) => (+431 + (+ 239)) = + 670kJ New bonds broken (exothermic process/-∆H ) = (2(H-Cl ) => (- 428 x 2)) = -856kJ ∆Hr =( + 670kJ + -856kJ) = 186 kJ = -93kJ mole-1 2 The above reaction has negative -∆H enthalpy change and is therefore practically exothermic.
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The SI unit of bond dissociation energy is kJmole-1 The higher the bond dissociation energy the stronger the (covalent)bond Bond dissociation energies of some (covalent)bonds Bond Bond dissociation energy (kJmole-1) Bond dissociation energy (kJmole-1) H-H 431 I-I 151 C-C 436 C-H 413 C=C 612 O-H 463 C = C 836 C-O 358 N = N 945 H-Cl 428 N-H 391 H-Br 366 F-F 158 C-Cl 346 Cl-Cl 239 C-Br 276 Br-Br 193 C-I 338 H-I 299 O=O 497 Si-Si 226 C-F 494
10 The molar enthalpy of reaction can be calculated from the bond dissociation energy by: (i)adding the total bond dissociation energy of the reactants(endothermic process/+∆H) and total bond dissociation energy of the products(exothermic process/-∆H). (ii)subtracting total bond dissociation energy of the reactants from the total bond dissociation energy of the products(exothermic process/-∆H less/minus endothermic process/+∆H). Practice examples/Calculating ∆Hr 1.Calculate ∆Hr from the following reaction: a) H2(g) + Cl2(g) -> 2HCl(g) Working Old bonds broken (endothermic process/+∆H ) = (H-H + Cl-Cl) => (+431 + (+ 239)) = + 670kJ New bonds broken (exothermic process/-∆H ) = (2(H-Cl ) => (- 428 x 2)) = -856kJ ∆Hr =( + 670kJ + -856kJ) = 186 kJ = -93kJ mole-1 2 The above reaction has negative -∆H enthalpy change and is therefore practically exothermic. The thermochemical reaction is thus: ½ H2(g) + ½ Cl2(g) -> HCl(g) ∆Hr = -93kJ b) CH4(g) + Cl2(g) -> CH3Cl + HCl(g) Working Old bonds broken (endothermic process/+∆H ) = (4(C-H) + Cl-Cl) => ((4 x +413) + (+ 239)) = + 1891kJ New bonds broken (exothermic process/-∆H ) = (3(C-H + H-Cl + C-Cl) => (( 3 x - 413) + 428 + 346) = -2013 kJ ∆Hr =( + 1891kJ + -2013 kJ) = -122 kJ mole-1 The above reaction has negative -∆H enthalpy change and is therefore practically exothermic.
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(ii)subtracting total bond dissociation energy of the reactants from the total bond dissociation energy of the products(exothermic process/-∆H less/minus endothermic process/+∆H). Practice examples/Calculating ∆Hr 1.Calculate ∆Hr from the following reaction: a) H2(g) + Cl2(g) -> 2HCl(g) Working Old bonds broken (endothermic process/+∆H ) = (H-H + Cl-Cl) => (+431 + (+ 239)) = + 670kJ New bonds broken (exothermic process/-∆H ) = (2(H-Cl ) => (- 428 x 2)) = -856kJ ∆Hr =( + 670kJ + -856kJ) = 186 kJ = -93kJ mole-1 2 The above reaction has negative -∆H enthalpy change and is therefore practically exothermic. The thermochemical reaction is thus: ½ H2(g) + ½ Cl2(g) -> HCl(g) ∆Hr = -93kJ b) CH4(g) + Cl2(g) -> CH3Cl + HCl(g) Working Old bonds broken (endothermic process/+∆H ) = (4(C-H) + Cl-Cl) => ((4 x +413) + (+ 239)) = + 1891kJ New bonds broken (exothermic process/-∆H ) = (3(C-H + H-Cl + C-Cl) => (( 3 x - 413) + 428 + 346) = -2013 kJ ∆Hr =( + 1891kJ + -2013 kJ) = -122 kJ mole-1 The above reaction has negative -∆H enthalpy change and is therefore practically exothermic. The thermochemical reaction is thus: CH4(g) + Cl2(g) -> CH3Cl(g) + HCl(g) ∆H = -122 kJ c) CH2CH2(g) + Cl2(g) -> CH3Cl CH3Cl (g)
11 Working Old bonds broken (endothermic process/+∆H ) = (4(C-H) + Cl-Cl + C=C) => ((4 x +413) + (+ 239) +(612)) = + 2503kJ New bonds broken (exothermic process/-∆H ) = (4(C-H + C-C + 2(C-Cl) ) => (( 3 x - 413) + -436 +2 x 346 = -2367 kJ ∆Hr =( + 2503kJ + -2367 kJ) = +136 kJ mole-1 The above reaction has negative +∆H enthalpy change and is therefore practically endothermic.
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Practice examples/Calculating ∆Hr 1.Calculate ∆Hr from the following reaction: a) H2(g) + Cl2(g) -> 2HCl(g) Working Old bonds broken (endothermic process/+∆H ) = (H-H + Cl-Cl) => (+431 + (+ 239)) = + 670kJ New bonds broken (exothermic process/-∆H ) = (2(H-Cl ) => (- 428 x 2)) = -856kJ ∆Hr =( + 670kJ + -856kJ) = 186 kJ = -93kJ mole-1 2 The above reaction has negative -∆H enthalpy change and is therefore practically exothermic. The thermochemical reaction is thus: ½ H2(g) + ½ Cl2(g) -> HCl(g) ∆Hr = -93kJ b) CH4(g) + Cl2(g) -> CH3Cl + HCl(g) Working Old bonds broken (endothermic process/+∆H ) = (4(C-H) + Cl-Cl) => ((4 x +413) + (+ 239)) = + 1891kJ New bonds broken (exothermic process/-∆H ) = (3(C-H + H-Cl + C-Cl) => (( 3 x - 413) + 428 + 346) = -2013 kJ ∆Hr =( + 1891kJ + -2013 kJ) = -122 kJ mole-1 The above reaction has negative -∆H enthalpy change and is therefore practically exothermic. The thermochemical reaction is thus: CH4(g) + Cl2(g) -> CH3Cl(g) + HCl(g) ∆H = -122 kJ c) CH2CH2(g) + Cl2(g) -> CH3Cl CH3Cl (g)
11 Working Old bonds broken (endothermic process/+∆H ) = (4(C-H) + Cl-Cl + C=C) => ((4 x +413) + (+ 239) +(612)) = + 2503kJ New bonds broken (exothermic process/-∆H ) = (4(C-H + C-C + 2(C-Cl) ) => (( 3 x - 413) + -436 +2 x 346 = -2367 kJ ∆Hr =( + 2503kJ + -2367 kJ) = +136 kJ mole-1 The above reaction has negative +∆H enthalpy change and is therefore practically endothermic. The thermochemical reaction is thus: CH2CH2(g) + Cl2(g) -> CH3Cl CH3Cl (g) ∆H = +136 kJ Note that: (i)a reaction is exothermic if the bond dissociation energy of reactants is more than bond dissociation energy of products.
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The thermochemical reaction is thus: ½ H2(g) + ½ Cl2(g) -> HCl(g) ∆Hr = -93kJ b) CH4(g) + Cl2(g) -> CH3Cl + HCl(g) Working Old bonds broken (endothermic process/+∆H ) = (4(C-H) + Cl-Cl) => ((4 x +413) + (+ 239)) = + 1891kJ New bonds broken (exothermic process/-∆H ) = (3(C-H + H-Cl + C-Cl) => (( 3 x - 413) + 428 + 346) = -2013 kJ ∆Hr =( + 1891kJ + -2013 kJ) = -122 kJ mole-1 The above reaction has negative -∆H enthalpy change and is therefore practically exothermic. The thermochemical reaction is thus: CH4(g) + Cl2(g) -> CH3Cl(g) + HCl(g) ∆H = -122 kJ c) CH2CH2(g) + Cl2(g) -> CH3Cl CH3Cl (g)
11 Working Old bonds broken (endothermic process/+∆H ) = (4(C-H) + Cl-Cl + C=C) => ((4 x +413) + (+ 239) +(612)) = + 2503kJ New bonds broken (exothermic process/-∆H ) = (4(C-H + C-C + 2(C-Cl) ) => (( 3 x - 413) + -436 +2 x 346 = -2367 kJ ∆Hr =( + 2503kJ + -2367 kJ) = +136 kJ mole-1 The above reaction has negative +∆H enthalpy change and is therefore practically endothermic. The thermochemical reaction is thus: CH2CH2(g) + Cl2(g) -> CH3Cl CH3Cl (g) ∆H = +136 kJ Note that: (i)a reaction is exothermic if the bond dissociation energy of reactants is more than bond dissociation energy of products. (ii)a reaction is endothermic if the bond dissociation energy of reactants is less than bond dissociation energy of products.
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The thermochemical reaction is thus: CH4(g) + Cl2(g) -> CH3Cl(g) + HCl(g) ∆H = -122 kJ c) CH2CH2(g) + Cl2(g) -> CH3Cl CH3Cl (g)
11 Working Old bonds broken (endothermic process/+∆H ) = (4(C-H) + Cl-Cl + C=C) => ((4 x +413) + (+ 239) +(612)) = + 2503kJ New bonds broken (exothermic process/-∆H ) = (4(C-H + C-C + 2(C-Cl) ) => (( 3 x - 413) + -436 +2 x 346 = -2367 kJ ∆Hr =( + 2503kJ + -2367 kJ) = +136 kJ mole-1 The above reaction has negative +∆H enthalpy change and is therefore practically endothermic. The thermochemical reaction is thus: CH2CH2(g) + Cl2(g) -> CH3Cl CH3Cl (g) ∆H = +136 kJ Note that: (i)a reaction is exothermic if the bond dissociation energy of reactants is more than bond dissociation energy of products. (ii)a reaction is endothermic if the bond dissociation energy of reactants is less than bond dissociation energy of products. 12 (b)Standard enthalpy/heat of combustion ∆Hᶿc The molar standard enthalpy/heat of combustion(∆Hᶿc) is defined as the energy/heat change when one mole of a substance is burnt in oxygen/excess air at standard conditions. Burning is the reaction of a substance with oxygen/air. It is an exothermic process producing a lot of energy in form of heat. A substance that undergoes burning is called a fuel. A fuel is defined as the combustible substance which burns in air to give heat energy for domestic or industrial use.
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It is an exothermic process producing a lot of energy in form of heat. A substance that undergoes burning is called a fuel. A fuel is defined as the combustible substance which burns in air to give heat energy for domestic or industrial use. A fuel may be solid (e.g coal, wood, charcoal) liquid (e.g petrol, paraffin, ethanol, kerosene) or gas (e.g liquefied petroleum gas/LPG, Water gasCO2/H2, biogas-methane, Natural gas-mixture of hydrocarbons) To determine the molar standard enthalpy/heat of combustion(∆Hᶿc) of ethanol Procedure Put 20cm3 of distilled water into a 50cm3 beaker. Clamp the beaker. Determine the temperature of the water T1.Weigh an empty burner(empty tin with wick). 13 Record its mass M1.Put some ethanol into the burner. Weigh again the burner with the ethanol and record its mass M2. Ignite the burner and place it below the clamped 50cm3 beaker. Heat the water in the beaker for about one minute. Put off the burner. Record the highest temperature rise of the water, T2. Weigh the burner again and record its mass M3 Sample results: Volume of water used 20cm3 Temperature of the water before heating T1 25.0oC Temperature of the water after heating T2 35.0oC Mass of empty burner M1 28.3g Mass of empty burner + ethanol before igniting M2 29.1g Mass of empty burner + ethanol after igniting M3 28.7g Sample calculations: 1.Calculate: (a) ∆T the change in temperature ∆T = T2 – T1 => (35.0oC – 25.0oC) = 10.0oC (b) the mass of ethanol used in burning mass of ethanol used = M2 – M1 => 29.1g – 28.7g = 0.4g (c) the number of moles of ethanol used in burning moles of ethanol = mass used => 0.4 = 0.0087 /8.7 x 10-3 moles molar mass of ethanol 46 2.
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Put off the burner. Record the highest temperature rise of the water, T2. Weigh the burner again and record its mass M3 Sample results: Volume of water used 20cm3 Temperature of the water before heating T1 25.0oC Temperature of the water after heating T2 35.0oC Mass of empty burner M1 28.3g Mass of empty burner + ethanol before igniting M2 29.1g Mass of empty burner + ethanol after igniting M3 28.7g Sample calculations: 1.Calculate: (a) ∆T the change in temperature ∆T = T2 – T1 => (35.0oC – 25.0oC) = 10.0oC (b) the mass of ethanol used in burning mass of ethanol used = M2 – M1 => 29.1g – 28.7g = 0.4g (c) the number of moles of ethanol used in burning moles of ethanol = mass used => 0.4 = 0.0087 /8.7 x 10-3 moles molar mass of ethanol 46 2. Given that the specific heat capacity of water is 4.2 kJ-1kg-1K-1,determine the heat produced during the burning. Heat produced ∆H = mass of water(m) x specific heat capacity (c)x ∆T => 20 x 4.2 x 10 = 840 Joules = 0.84 kJ 1000 3.Calculate the molar heat of combustion of ethanol Molar heat of combustion ∆Hc = Heat produced ∆H Number of moles of fuel => 0.84 kJ = 96.5517 kJmole-1 0.0087 /8.7 x 10-3 moles 4.List two sources of error in the above experiment. (i)Heat loss to the surrounding lowers the practical value of the molar heat of combustion of ethanol. 14 A draught shield tries to minimize the loss by protecting wind from wobbling the flame. (ii) Heat gain by reaction vessels/beaker lowers ∆T and hence ∆Hc 5.Calculate the heating value of the fuel.
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(i)Heat loss to the surrounding lowers the practical value of the molar heat of combustion of ethanol. 14 A draught shield tries to minimize the loss by protecting wind from wobbling the flame. (ii) Heat gain by reaction vessels/beaker lowers ∆T and hence ∆Hc 5.Calculate the heating value of the fuel. Heating value = molar heat of combustion => 96.5517 kJmole-1 = 2.0989 kJg-1 Molar mass of fuel 46 g 6.Explain other factors used to determine the choice of fuel for domestic and industrial use. (i) availability and affordability-some fuels are more available cheaply in rural than in urban areas at a lower cost. (ii)cost of storage and transmission-a fuel should be easy to transport and store safely. e.g LPG is very convenient to store and use. Charcoal and wood are bulky. (iii)environmental effects –Most fuels after burning produce carbon(IV) oxide gas as a byproduct. Carbon(IV) oxide gas is green house gas that causes global warming. Some other fuel produce acidic gases like sulphur(IV) oxide ,and nitrogen(IV) oxide. These gases cause acid rain. Internal combustion engines exhaust produce lead vapour from leaded petrol and diesel. Lead is carcinogenic. (iv)ignition point-The temperature at which a fuel must be heated before it burns in air is the ignition point. Fuels like petrol have very low ignition point, making it highly flammable. Charcoal and wood have very high ignition point. 7.Explain the methods used to reduce pollution from common fuels. (i)Planting trees-Plants absorb excess carbon(IV)oxide for photosynthesis and release oxygen gas to the atmosphere. (ii)using catalytic converters in internal combustion engines that convert harmful/toxic/poisonous gases like carbon(II)oxide and nitrogen(IV)oxide to harmless non-poisonous carbon(IV)oxide, water and nitrogen gas by using platinum-rhodium catalyst along the engine exhaust pipes. Further practice calculations 1.Calculate the heating value of methanol CH3OH given that 0.87g of the fuel burn in air to raise the temperature of 500g of water from 20oC to 27oC.(C12.0,H=1.0 O=16.0).
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(i)Planting trees-Plants absorb excess carbon(IV)oxide for photosynthesis and release oxygen gas to the atmosphere. (ii)using catalytic converters in internal combustion engines that convert harmful/toxic/poisonous gases like carbon(II)oxide and nitrogen(IV)oxide to harmless non-poisonous carbon(IV)oxide, water and nitrogen gas by using platinum-rhodium catalyst along the engine exhaust pipes. Further practice calculations 1.Calculate the heating value of methanol CH3OH given that 0.87g of the fuel burn in air to raise the temperature of 500g of water from 20oC to 27oC.(C12.0,H=1.0 O=16.0). Moles of methanol used = Mass of methanol used => 0.87 g = 0.02718 moles Molar mass of methanol 32 Heat produced ∆H = mass of water(m) x specific heat capacity (c)x ∆T => 500 x 4.2 x 7 = 14700 Joules = 14.7 kJ 1000
15 Molar heat of combustion ∆Hc = Heat produced ∆H Number of moles of fuel => 14.7 kJ = 540.8389 kJmole-1 0.02718 moles Heating value = molar heat of combustion => 540.8389 kJmole-1 = 16.9012 kJg-1 Molar mass of fuel 32 g 2. 1.0 g of carbon burn in excess air to raise the temperature of 400g of water by 18oC.Determine the molar heat of combustion and hence the heating value of carbon(C-12.0,).
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Further practice calculations 1.Calculate the heating value of methanol CH3OH given that 0.87g of the fuel burn in air to raise the temperature of 500g of water from 20oC to 27oC.(C12.0,H=1.0 O=16.0). Moles of methanol used = Mass of methanol used => 0.87 g = 0.02718 moles Molar mass of methanol 32 Heat produced ∆H = mass of water(m) x specific heat capacity (c)x ∆T => 500 x 4.2 x 7 = 14700 Joules = 14.7 kJ 1000
15 Molar heat of combustion ∆Hc = Heat produced ∆H Number of moles of fuel => 14.7 kJ = 540.8389 kJmole-1 0.02718 moles Heating value = molar heat of combustion => 540.8389 kJmole-1 = 16.9012 kJg-1 Molar mass of fuel 32 g 2. 1.0 g of carbon burn in excess air to raise the temperature of 400g of water by 18oC.Determine the molar heat of combustion and hence the heating value of carbon(C-12.0,). Moles of carbon used = Mass of carbon used => 1.0 g = 0.0833 moles Molar mass of carbon 12 Heat produced ∆H = mass of water(m) x specific heat capacity (c)x ∆T => 400 x 4.2 x 18 = 30240 Joules = 30.24 kJ 1000 Molar heat of combustion ∆Hc = Heat produced ∆H Number of moles of fuel => 30.24 kJ = 363.0252 kJmole-1 0.0833 moles Heating value = molar heat of combustion => 363.0252 kJmole-1= 30.2521 kJg-1 Molar mass of fuel 12 g (c)Standard enthalpy/heat of displacement ∆Hᶿd The molar standard enthalpy/heat of displacement ∆Hᶿd is defined as the energy/heat change when one mole of a substance is displaced from its solution.
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Moles of methanol used = Mass of methanol used => 0.87 g = 0.02718 moles Molar mass of methanol 32 Heat produced ∆H = mass of water(m) x specific heat capacity (c)x ∆T => 500 x 4.2 x 7 = 14700 Joules = 14.7 kJ 1000
15 Molar heat of combustion ∆Hc = Heat produced ∆H Number of moles of fuel => 14.7 kJ = 540.8389 kJmole-1 0.02718 moles Heating value = molar heat of combustion => 540.8389 kJmole-1 = 16.9012 kJg-1 Molar mass of fuel 32 g 2. 1.0 g of carbon burn in excess air to raise the temperature of 400g of water by 18oC.Determine the molar heat of combustion and hence the heating value of carbon(C-12.0,). Moles of carbon used = Mass of carbon used => 1.0 g = 0.0833 moles Molar mass of carbon 12 Heat produced ∆H = mass of water(m) x specific heat capacity (c)x ∆T => 400 x 4.2 x 18 = 30240 Joules = 30.24 kJ 1000 Molar heat of combustion ∆Hc = Heat produced ∆H Number of moles of fuel => 30.24 kJ = 363.0252 kJmole-1 0.0833 moles Heating value = molar heat of combustion => 363.0252 kJmole-1= 30.2521 kJg-1 Molar mass of fuel 12 g (c)Standard enthalpy/heat of displacement ∆Hᶿd The molar standard enthalpy/heat of displacement ∆Hᶿd is defined as the energy/heat change when one mole of a substance is displaced from its solution. A displacement reaction takes place when a more reactive element/with less electrode potential Eᶿ / negative Eᶿ /higher in the reactivity/electrochemical series remove/displace another with less reactive element/with higher electrode potential Eᶿ / positive Eᶿ /lower in the reactivity/electrochemical series from its solution.e.g.
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1.0 g of carbon burn in excess air to raise the temperature of 400g of water by 18oC.Determine the molar heat of combustion and hence the heating value of carbon(C-12.0,). Moles of carbon used = Mass of carbon used => 1.0 g = 0.0833 moles Molar mass of carbon 12 Heat produced ∆H = mass of water(m) x specific heat capacity (c)x ∆T => 400 x 4.2 x 18 = 30240 Joules = 30.24 kJ 1000 Molar heat of combustion ∆Hc = Heat produced ∆H Number of moles of fuel => 30.24 kJ = 363.0252 kJmole-1 0.0833 moles Heating value = molar heat of combustion => 363.0252 kJmole-1= 30.2521 kJg-1 Molar mass of fuel 12 g (c)Standard enthalpy/heat of displacement ∆Hᶿd The molar standard enthalpy/heat of displacement ∆Hᶿd is defined as the energy/heat change when one mole of a substance is displaced from its solution. A displacement reaction takes place when a more reactive element/with less electrode potential Eᶿ / negative Eᶿ /higher in the reactivity/electrochemical series remove/displace another with less reactive element/with higher electrode potential Eᶿ / positive Eᶿ /lower in the reactivity/electrochemical series from its solution.e.g. (i)Zn(s) + CuSO4(aq) -> Cu(s) + ZnSO4(aq) Ionically: Zn(s) + Cu2+(aq) -> Cu(s) + Zn2+ (aq) (ii)Fe(s) + CuSO4(aq) -> Cu(s) + FeSO4(aq) Ionically: Fe(s) + Cu2+(aq) -> Cu(s) + Fe2+ (aq) (iii)Pb(s) + CuSO4(aq) -> Cu(s) + PbSO4(s) This reaction stops after some time as insoluble PbSO4(s) coat/cover unreacted lead.
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Moles of carbon used = Mass of carbon used => 1.0 g = 0.0833 moles Molar mass of carbon 12 Heat produced ∆H = mass of water(m) x specific heat capacity (c)x ∆T => 400 x 4.2 x 18 = 30240 Joules = 30.24 kJ 1000 Molar heat of combustion ∆Hc = Heat produced ∆H Number of moles of fuel => 30.24 kJ = 363.0252 kJmole-1 0.0833 moles Heating value = molar heat of combustion => 363.0252 kJmole-1= 30.2521 kJg-1 Molar mass of fuel 12 g (c)Standard enthalpy/heat of displacement ∆Hᶿd The molar standard enthalpy/heat of displacement ∆Hᶿd is defined as the energy/heat change when one mole of a substance is displaced from its solution. A displacement reaction takes place when a more reactive element/with less electrode potential Eᶿ / negative Eᶿ /higher in the reactivity/electrochemical series remove/displace another with less reactive element/with higher electrode potential Eᶿ / positive Eᶿ /lower in the reactivity/electrochemical series from its solution.e.g. (i)Zn(s) + CuSO4(aq) -> Cu(s) + ZnSO4(aq) Ionically: Zn(s) + Cu2+(aq) -> Cu(s) + Zn2+ (aq) (ii)Fe(s) + CuSO4(aq) -> Cu(s) + FeSO4(aq) Ionically: Fe(s) + Cu2+(aq) -> Cu(s) + Fe2+ (aq) (iii)Pb(s) + CuSO4(aq) -> Cu(s) + PbSO4(s) This reaction stops after some time as insoluble PbSO4(s) coat/cover unreacted lead. (iv)Cl2(g) + 2NaBr(aq) -> Br2(aq) + 2NaCl(aq) Ionically: Cl2(g)+ 2Br- (aq) -> Br2(aq) + 2Cl- (aq)
16 Practically, a displacement reaction takes place when a known amount /volume of a solution is added excess of a more reactive metal.
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A displacement reaction takes place when a more reactive element/with less electrode potential Eᶿ / negative Eᶿ /higher in the reactivity/electrochemical series remove/displace another with less reactive element/with higher electrode potential Eᶿ / positive Eᶿ /lower in the reactivity/electrochemical series from its solution.e.g. (i)Zn(s) + CuSO4(aq) -> Cu(s) + ZnSO4(aq) Ionically: Zn(s) + Cu2+(aq) -> Cu(s) + Zn2+ (aq) (ii)Fe(s) + CuSO4(aq) -> Cu(s) + FeSO4(aq) Ionically: Fe(s) + Cu2+(aq) -> Cu(s) + Fe2+ (aq) (iii)Pb(s) + CuSO4(aq) -> Cu(s) + PbSO4(s) This reaction stops after some time as insoluble PbSO4(s) coat/cover unreacted lead. (iv)Cl2(g) + 2NaBr(aq) -> Br2(aq) + 2NaCl(aq) Ionically: Cl2(g)+ 2Br- (aq) -> Br2(aq) + 2Cl- (aq)
16 Practically, a displacement reaction takes place when a known amount /volume of a solution is added excess of a more reactive metal. To determine the molar standard enthalpy/heat of displacement(∆Hᶿd) of copper Procedure Place 20cm3 of 0.2M copper(II)sulphate(VI)solution into a 50cm3 plastic beaker/calorimeter. Determine and record the temperature of the solution T1.Put all the Zinc powder provided into the plastic beaker. Stir the mixture using the thermometer. Determine and record the highest temperature change to the nearest 0.5oC- T2 .
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Determine and record the temperature of the solution T1.Put all the Zinc powder provided into the plastic beaker. Stir the mixture using the thermometer. Determine and record the highest temperature change to the nearest 0.5oC- T2 . Repeat the experiment to complete table 1 below Table 1 Experiment I II Final temperature of solution(T2) 30.0oC 31.0oC Final temperature of solution(T1) 25.0oC 24.0oC Change in temperature(∆T) 5.0 6.0 Questions 1.(a) Calculate: (i)average ∆T Average∆T = change in temperature in experiment I and II =>5.0 + 6.0 = 5.5oC 2 (ii)the number of moles of solution used Moles used = molarity x volume of solution = 0.2 x 20 = 0.004 moles 1000 1000 (iii)the enthalpy change ∆H for the reaction Heat produced ∆H = mass of solution(m) x specific heat capacity (c)x ∆T => 20 x 4.2 x 5.5 = 462 Joules = 0.462 kJ 1000 (iv)State two assumptions made in the above calculations. Density of solution = density of water = 1gcm-3 Specific heat capacity of solution=Specific heat capacity of solution=4.2 kJ-1kg-1K This is because the solution is assumed to be infinite dilute. 2. Calculate the enthalpy change for one mole of displacement of Cu2+ (aq) ions. Molar heat of displacement ∆Hd = Heat produced ∆H
17 Number of moles of fuel => 0.462 kJ = 115.5 kJmole-1 0.004 3.Write an ionic equation for the reaction taking place. Zn(s) + Cu2+(aq) -> Cu(s) + Zn2+(aq) 4.State the observation made during the reaction. Blue colour of copper(II)sulphate(VI) fades/becomes less blue/colourless. Brown solid deposits are formed at the bottom of reaction vessel/ beaker. 5.Illustrate the above reaction using an energy level diagram.
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Blue colour of copper(II)sulphate(VI) fades/becomes less blue/colourless. Brown solid deposits are formed at the bottom of reaction vessel/ beaker. 5.Illustrate the above reaction using an energy level diagram. Zn(s) + Cu2+(aq) Energy ∆H = -115.5 kJmole-1 (kJ) Cu(s) + Zn2+(aq) Reaction progress/path/coordinates 6. Iron is less reactive than Zinc. Explain the effect of using iron instead of Zinc on the standard molar heat of displacement ∆Hd of copper(II)sulphate (VI) solution. No effect. Cu2+ (aq) are displaced from their solution.The element used to displace it does not matter.The reaction however faster if a more reactive metal is used. 7.(a)If the standard molar heat of displacement ∆Hd of copper(II)sulphate (VI) solution is 209kJmole-1 calculate the temperature change if 50cm3 of 0.2M solution was displaced by excess magnesium. Moles used = molarity x volume of solution = 0.2 x 50 = 0.01 moles 1000 1000 Heat produced ∆H = Molar heat of displacement ∆Hd x Number of moles =>209kJmole-1x 0.01 moles = 2.09 kJ ∆T (change in temperature) = Heat produced ∆H Molar heat of displacement ∆Hd x Number of moles =>2.09 kJ = 9.9524Kelvin 0.01 moles (b)Draw an energy level diagram to show the above energy changes
18 Mg(s) + Cu2+(aq) Energy ∆H = -209 kJmole-1 (kJ) Cu(s) + Mg2+(aq) Reaction progress/path/coordinates 8. The enthalpy of displacement ∆Hd of copper(II)sulphate (VI) solution is 12k6kJmole-1.Calculate the molarity of the solution given that 40cm3 of this solution produces 2.204kJ of energy during a displacement reaction with excess iron filings.
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7.(a)If the standard molar heat of displacement ∆Hd of copper(II)sulphate (VI) solution is 209kJmole-1 calculate the temperature change if 50cm3 of 0.2M solution was displaced by excess magnesium. Moles used = molarity x volume of solution = 0.2 x 50 = 0.01 moles 1000 1000 Heat produced ∆H = Molar heat of displacement ∆Hd x Number of moles =>209kJmole-1x 0.01 moles = 2.09 kJ ∆T (change in temperature) = Heat produced ∆H Molar heat of displacement ∆Hd x Number of moles =>2.09 kJ = 9.9524Kelvin 0.01 moles (b)Draw an energy level diagram to show the above energy changes
18 Mg(s) + Cu2+(aq) Energy ∆H = -209 kJmole-1 (kJ) Cu(s) + Mg2+(aq) Reaction progress/path/coordinates 8. The enthalpy of displacement ∆Hd of copper(II)sulphate (VI) solution is 12k6kJmole-1.Calculate the molarity of the solution given that 40cm3 of this solution produces 2.204kJ of energy during a displacement reaction with excess iron filings. Number of moles = Heat produced ∆H Molar heat of displacement ∆Hd =>2.204 kJ = 0.0206moles 126 moles Molarity of the solution = moles x 1000 Volume of solution used = 0.0206moles x 1000 = 0.5167 M 40 9. If the molar heat of displacement of Zinc(II)nitrate(V)by magnesium powder is 25.05kJmole-1 ,calculate the volume of solution which must be added 0.5 moles solution if there was a 3.0K rise in temperature.
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The enthalpy of displacement ∆Hd of copper(II)sulphate (VI) solution is 12k6kJmole-1.Calculate the molarity of the solution given that 40cm3 of this solution produces 2.204kJ of energy during a displacement reaction with excess iron filings. Number of moles = Heat produced ∆H Molar heat of displacement ∆Hd =>2.204 kJ = 0.0206moles 126 moles Molarity of the solution = moles x 1000 Volume of solution used = 0.0206moles x 1000 = 0.5167 M 40 9. If the molar heat of displacement of Zinc(II)nitrate(V)by magnesium powder is 25.05kJmole-1 ,calculate the volume of solution which must be added 0.5 moles solution if there was a 3.0K rise in temperature. Heat produced ∆H = Molar heat of displacement ∆Hd x Number of moles =>25.08kJmole-1x 0.5 moles = 1.254 kJ x 1000 =1254J Mass of solution (m) = Heat produced ∆H specific heat capacity (c)x ∆T => 1254J = 99.5238 g 4.2 x 3 Volume = mass x density = 99.5238 g x 1 = 99.5238cm3 Note: The solution assumes to be too dilute /infinite dilute such that the density and specific heat capacity is assumed to be that of water. Graphical determination of the molar enthalpy of displacement of copper Procedure:
19 Place 20cm3 of 0.2M copper(II)sulphate (VI) solution into a calorimeter/50cm3 of plastic beaker wrapped in cotton wool/tissue paper. Record its temperature at time T= 0. Stir the solution with the thermometer carefully and continue recording the temperature after every 30 seconds . Place all the (1.5g) Zinc powder provided. Stir the solution with the thermometer carefully and continue recording the temperature after every 30 seconds for five minutes. Determine the highest temperature change to the nearest 0.5oC.
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Place all the (1.5g) Zinc powder provided. Stir the solution with the thermometer carefully and continue recording the temperature after every 30 seconds for five minutes. Determine the highest temperature change to the nearest 0.5oC. Sample results Time oC 0.0 30.0 60.0 90.0 120.0 150.0 180.0 210.0 240.0 270.0 Temperature 25.0 25.0 25.0 25.0 25.0 xxx 36.0 35.5 35.0 34.5 Sketch graph of temperature against time 36.5 Extrapolation Temperature point ∆T oC 130 Time(seconds) Questions 1. Show and determine the change in temperature ∆T From a well constructed graph ∆T= T2 –T1 at 150 second by extrapolation ∆T = 36.5 – 25.0 = 11.5oC 2.Calculate the number of moles of copper(II) sulphate(VI)used given the molar heat of displacement of Cu2+ (aq)ions is 125kJmole-1 Heat produced ∆H = mass of solution(m) x specific heat capacity (c)x ∆T => 20 x 4.2 x 11.5 = 966 Joules = 0.966 kJ 1000 Number of moles = Heat produced ∆H Molar heat of displacement ∆Hd =>.966 kJ = 0.007728moles 125 moles 7.728 x 10-3moles 2. What was the concentration of copper(II)sulphate(VI) in moles per litre. 20 Molarity = moles x 1000 => 7.728 x 10-3moles x 1000 = 0.3864M Volume used 20 4.The actual concentration of copper(II)sulphate(VI) solution was 0.4M.Explain the differences between the two. Practical value is lower than theoretical.
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What was the concentration of copper(II)sulphate(VI) in moles per litre. 20 Molarity = moles x 1000 => 7.728 x 10-3moles x 1000 = 0.3864M Volume used 20 4.The actual concentration of copper(II)sulphate(VI) solution was 0.4M.Explain the differences between the two. Practical value is lower than theoretical. Heat/energy loss to the surrounding and that absorbed by the reaction vessel decreases ∆T hence lowering the practical number of moles and molarity against the theoretical value 5.a) In an experiment to determine the molar heat of reaction when magnesium displaces copper ,0.15g of magnesium powder were added to 25.0cm3 of 2.0M copper (II) chloride solution. The temperature of copper (II) chloride solution was 25oC.While that of the mixture was 43oC. i)Other than increase in temperature, state and explain the observations which were made during the reaction.(3mks) ii)Calculate the heat change during the reaction (specific heat capacity of the solution = 4.2jg-1k-1and the density of the solution = 1g/cm3(2mks) iii)Determine the molar heat of displacement of copper by magnesium.(Mg=24.0). iv)Write the ionic equation for the reaction.(1mk) v)Sketch an energy level diagram for the reaction.(2mks) b)Use the reduction potentials given below to explain why a solution containing copper ions should not be stored in a container made of zinc. Zn2+(aq) + 2e -> Zn(s); Eø = -0.76v Cu2+(aq) + 2e -> Cu(s); Eø = +0.34v (2mks) (c)Standard enthalpy/heat of neutralization ∆Hᶿn The molar standard enthalpy/heat of neutralization ∆Hᶿn is defined as the energy/heat change when one mole of a H+ (H3O+)ions react completely with one mole of OH- ions to form one mole of H2O/water.
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i)Other than increase in temperature, state and explain the observations which were made during the reaction.(3mks) ii)Calculate the heat change during the reaction (specific heat capacity of the solution = 4.2jg-1k-1and the density of the solution = 1g/cm3(2mks) iii)Determine the molar heat of displacement of copper by magnesium.(Mg=24.0). iv)Write the ionic equation for the reaction.(1mk) v)Sketch an energy level diagram for the reaction.(2mks) b)Use the reduction potentials given below to explain why a solution containing copper ions should not be stored in a container made of zinc. Zn2+(aq) + 2e -> Zn(s); Eø = -0.76v Cu2+(aq) + 2e -> Cu(s); Eø = +0.34v (2mks) (c)Standard enthalpy/heat of neutralization ∆Hᶿn The molar standard enthalpy/heat of neutralization ∆Hᶿn is defined as the energy/heat change when one mole of a H+ (H3O+)ions react completely with one mole of OH- ions to form one mole of H2O/water. Neutralization is thus a reaction of an acid /H+ (H3O+)ions with a base/alkali/ OH- ions to form salt and water only. 21 Strong acids/bases/alkalis are completely dissociated to many free ions(H+ /H3O+ and OH- ions). Weak acids/bases/alkalis are partially dissociated to few free ions(H+ (H3O+ and OH- ions) and exist more as molecules. Neutralization is an exothermic(-∆H) process.The enrgy produced during neutralization depend on the amount of free ions (H+ H3O+ and OH-)ions existing in the acid/base/alkali reactant: (i)for weak acid-base/alkali neutralization,some of the energy is used to dissociate /ionize the molecule into free H+ H3O+ and OH- ions therefore the overall energy evolved is comparatively lower/lesser/smaller than strong acid / base/ alkali neutralizations.
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21 Strong acids/bases/alkalis are completely dissociated to many free ions(H+ /H3O+ and OH- ions). Weak acids/bases/alkalis are partially dissociated to few free ions(H+ (H3O+ and OH- ions) and exist more as molecules. Neutralization is an exothermic(-∆H) process.The enrgy produced during neutralization depend on the amount of free ions (H+ H3O+ and OH-)ions existing in the acid/base/alkali reactant: (i)for weak acid-base/alkali neutralization,some of the energy is used to dissociate /ionize the molecule into free H+ H3O+ and OH- ions therefore the overall energy evolved is comparatively lower/lesser/smaller than strong acid / base/ alkali neutralizations. (ii) (i)for strong acid/base/alkali neutralization, no energy is used to dissociate /ionize since molecule is wholly/fully dissociated/ionized into free H+ H3O+ and OH- ions.The overall energy evolved is comparatively higher/more than weak acid-base/ alkali neutralizations. For strong acid-base/alkali neutralization, the enthalpy of neutralization is constant at about 57.3kJmole-1 irrespective of the acid-base used. This is because ionically: OH-(aq)+ H+(aq) -> H2O(l) for any wholly dissociated acid/base/alkali Practically ∆Hᶿn can be determined as in the examples below: To determine the molar enthalpy of neutralization ∆Hn of Hydrochloric acid Procedure Place 50cm3 of 2M hydrochloric acid into a calorimeter/200cm3 plastic beaker wrapped in cotton wool/tissue paper. Record its temperature T1.Using a clean measuring cylinder, measure another 50cm3 of 2M sodium hydroxide. Rinse the bulb of the thermometer in distilled water. Determine the temperature of the sodium hydroxide T2.Average T2 andT1 to get the initial temperature of the mixture T3. Carefully add all the alkali into the calorimeter/200cm3 plastic beaker wrapped in cotton wool/tissue paper containing the acid. Stir vigorously the mixture with the thermometer. Determine the highest temperature change to the nearest 0.5oC T4 as the final temperature of the mixture.
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Carefully add all the alkali into the calorimeter/200cm3 plastic beaker wrapped in cotton wool/tissue paper containing the acid. Stir vigorously the mixture with the thermometer. Determine the highest temperature change to the nearest 0.5oC T4 as the final temperature of the mixture. Repeat the experiment to complete table 1. Table I . Sample results Experiment I II Temperature of acid T1 (oC) 22.5 22.5 Temperature of base T2 (oC) 22.0 23.0 Final temperature of solution T4(oC) 35.5 36.0 Initial temperature of solution T3(oC) 22.25 22.75
22 (a)Calculate T6 the average temperature change T6 = 13.25 +13.75 = 13.5 oC 2 (b)Why should the apparatus be very clean? Impurities present in the apparatus reacts with acid /base lowering the overall temperature change and hence ∆Hᶿn. (c)Calculate the: (i)number of moles of the acid used number of moles = molarity x volume => 2 x 50 = 0.1moles 1000 1000 (ii)enthalpy change ∆H of neutralization. ∆H = (m)mass of solution(acid+base) x (c)specific heat capacity of solution x ∆T(T6) => (50 +50) x 4.2 x 13.5 = 5670Joules = 5.67kJ (iii) the molar heat of neutralization the acid. ∆Hn = Enthalpy change ∆H => 5.67kJ = 56.7kJ mole-1 Number of moles 0.1moles (c)Write the ionic equation for the reaction that takes place OH-(aq)+ H+(aq) -> H2O(l) (d)The theoretical enthalpy change is 57.4kJ. Explain the difference with the results above.
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∆H = (m)mass of solution(acid+base) x (c)specific heat capacity of solution x ∆T(T6) => (50 +50) x 4.2 x 13.5 = 5670Joules = 5.67kJ (iii) the molar heat of neutralization the acid. ∆Hn = Enthalpy change ∆H => 5.67kJ = 56.7kJ mole-1 Number of moles 0.1moles (c)Write the ionic equation for the reaction that takes place OH-(aq)+ H+(aq) -> H2O(l) (d)The theoretical enthalpy change is 57.4kJ. Explain the difference with the results above. The theoretical value is higher Heat/energy loss to the surrounding/environment lowers ∆T/T6 and thus ∆Hn Heat/energy is absorbed by the reaction vessel/calorimeter/plastic cup lowers ∆T and hence ∆Hn (e)Compare the ∆Hn of the experiment above with similar experiment repeated with neutralization of a solution of: (i) potassium hydroxide with nitric(V) acid The results would be the same/similar. Both are neutralization reactions of strong acids and bases/alkalis that are fully /wholly dissociated into many free H+ / H3O+ and OH- ions. (ii) ammonia with ethanoic acid Temperature change( T5) 13.25 13.75
23 The results would be lower/∆Hn would be less. Both are neutralization reactions of weak acids and bases/alkalis that are partially /partly dissociated into few free H+ / H3O+ and OH- ions. Some energy is used to ionize the molecule. (f)Draw an energy level diagram to illustrate the energy changes H2 H+ (aq)+OH (aq) Energy (kJ) ∆H = -56.7kJ H1 H2O (l) Reaction path/coordinate/progress Theoretical examples 1.The molar enthalpy of neutralization was experimentary shown to be 51.5kJ per mole of 0.5M hydrochloric acid and 0.5M sodium hydroxide.
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Both are neutralization reactions of weak acids and bases/alkalis that are partially /partly dissociated into few free H+ / H3O+ and OH- ions. Some energy is used to ionize the molecule. (f)Draw an energy level diagram to illustrate the energy changes H2 H+ (aq)+OH (aq) Energy (kJ) ∆H = -56.7kJ H1 H2O (l) Reaction path/coordinate/progress Theoretical examples 1.The molar enthalpy of neutralization was experimentary shown to be 51.5kJ per mole of 0.5M hydrochloric acid and 0.5M sodium hydroxide. If the volume of sodium hydroxide was 20cm3, what was the volume of hydrochloric acid used if the reaction produced a 5.0oC rise in temperature? Working: Moles of sodium hydroxide = molarity x volume => 0.5 M x 20cm3 = 0.01 moles 1000 1000 Enthalpy change ∆H = ∆Hn => 51.5 = 0.515kJ Moles sodium hydroxide 0.01 moles Mass of base + acid = Enthalpy change ∆H in Joules Specific heat capacity x ∆T => 0.515kJ x 1000 = 24.5238g 4.2 x 5 Mass/volume of HCl = Total volume – volume of NaOH =>24.5238 - 20.0 = 4.5238 cm3 3. ∆Hn of potassium hydroxide was practically determined to be 56.7kJmole-1.Calculate the molarity of 50.0 cm3 potassium hydroxide used to neutralize 25.0cm3 of dilute sulphuric(VI) acid raising the temperature of the solution from 10.0oC to 16.5oC.
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If the volume of sodium hydroxide was 20cm3, what was the volume of hydrochloric acid used if the reaction produced a 5.0oC rise in temperature? Working: Moles of sodium hydroxide = molarity x volume => 0.5 M x 20cm3 = 0.01 moles 1000 1000 Enthalpy change ∆H = ∆Hn => 51.5 = 0.515kJ Moles sodium hydroxide 0.01 moles Mass of base + acid = Enthalpy change ∆H in Joules Specific heat capacity x ∆T => 0.515kJ x 1000 = 24.5238g 4.2 x 5 Mass/volume of HCl = Total volume – volume of NaOH =>24.5238 - 20.0 = 4.5238 cm3 3. ∆Hn of potassium hydroxide was practically determined to be 56.7kJmole-1.Calculate the molarity of 50.0 cm3 potassium hydroxide used to neutralize 25.0cm3 of dilute sulphuric(VI) acid raising the temperature of the solution from 10.0oC to 16.5oC. ∆H = (m)mass of solution(acid+base) x (c)specific heat capacity of solution x ∆T
24 => (50 +25) x 4.2 x 6.5 = 2047.5Joules Moles potassium hydroxide =Enthalpy change ∆H ∆Hn 2047.5Joules = 0.0361 moles 56700Joules Molarity of KOH = moles x 1000 => 0.0361 moles x 1000 = 0.722M Volume used 50cm3 3.Determine the specific heat capacity of a solution of a solution mixture of 50.0cm3 of 2M potassium hydroxide neutralizing 50.0cm3 of 2M nitric(V) acid if a 13.25oC rise in temperature is recorded.(1mole of potassium hydroxide produce 55.4kJ of energy) Moles of potassium hydroxide = molarity KOH x volume 1000 => 2 M x 50cm3 = 0.1 moles 1000 Enthalpy change ∆H = ∆Hn x Moles potassium hydroxide => 55.4kJ x 0.1 moles = 5.54kJ x 1000=5540Joules Specific heat capacity = Enthalpy change ∆H in Joules Mass of base + acid x ∆T => 5540 = 4.1811J-1g-1K-1 (50+50) x 13.25 Graphically ∆Hn can be determined as in the example below: Procedure Place 8 test tubes in a test tube rack .Put 5cm3 of 2M sodium hydroxide solution into each test tube.
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Working: Moles of sodium hydroxide = molarity x volume => 0.5 M x 20cm3 = 0.01 moles 1000 1000 Enthalpy change ∆H = ∆Hn => 51.5 = 0.515kJ Moles sodium hydroxide 0.01 moles Mass of base + acid = Enthalpy change ∆H in Joules Specific heat capacity x ∆T => 0.515kJ x 1000 = 24.5238g 4.2 x 5 Mass/volume of HCl = Total volume – volume of NaOH =>24.5238 - 20.0 = 4.5238 cm3 3. ∆Hn of potassium hydroxide was practically determined to be 56.7kJmole-1.Calculate the molarity of 50.0 cm3 potassium hydroxide used to neutralize 25.0cm3 of dilute sulphuric(VI) acid raising the temperature of the solution from 10.0oC to 16.5oC. ∆H = (m)mass of solution(acid+base) x (c)specific heat capacity of solution x ∆T
24 => (50 +25) x 4.2 x 6.5 = 2047.5Joules Moles potassium hydroxide =Enthalpy change ∆H ∆Hn 2047.5Joules = 0.0361 moles 56700Joules Molarity of KOH = moles x 1000 => 0.0361 moles x 1000 = 0.722M Volume used 50cm3 3.Determine the specific heat capacity of a solution of a solution mixture of 50.0cm3 of 2M potassium hydroxide neutralizing 50.0cm3 of 2M nitric(V) acid if a 13.25oC rise in temperature is recorded.(1mole of potassium hydroxide produce 55.4kJ of energy) Moles of potassium hydroxide = molarity KOH x volume 1000 => 2 M x 50cm3 = 0.1 moles 1000 Enthalpy change ∆H = ∆Hn x Moles potassium hydroxide => 55.4kJ x 0.1 moles = 5.54kJ x 1000=5540Joules Specific heat capacity = Enthalpy change ∆H in Joules Mass of base + acid x ∆T => 5540 = 4.1811J-1g-1K-1 (50+50) x 13.25 Graphically ∆Hn can be determined as in the example below: Procedure Place 8 test tubes in a test tube rack .Put 5cm3 of 2M sodium hydroxide solution into each test tube. Measure 25cm3 of 1M hydrochloric acid into 100cm3 plastic beaker.
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∆Hn of potassium hydroxide was practically determined to be 56.7kJmole-1.Calculate the molarity of 50.0 cm3 potassium hydroxide used to neutralize 25.0cm3 of dilute sulphuric(VI) acid raising the temperature of the solution from 10.0oC to 16.5oC. ∆H = (m)mass of solution(acid+base) x (c)specific heat capacity of solution x ∆T
24 => (50 +25) x 4.2 x 6.5 = 2047.5Joules Moles potassium hydroxide =Enthalpy change ∆H ∆Hn 2047.5Joules = 0.0361 moles 56700Joules Molarity of KOH = moles x 1000 => 0.0361 moles x 1000 = 0.722M Volume used 50cm3 3.Determine the specific heat capacity of a solution of a solution mixture of 50.0cm3 of 2M potassium hydroxide neutralizing 50.0cm3 of 2M nitric(V) acid if a 13.25oC rise in temperature is recorded.(1mole of potassium hydroxide produce 55.4kJ of energy) Moles of potassium hydroxide = molarity KOH x volume 1000 => 2 M x 50cm3 = 0.1 moles 1000 Enthalpy change ∆H = ∆Hn x Moles potassium hydroxide => 55.4kJ x 0.1 moles = 5.54kJ x 1000=5540Joules Specific heat capacity = Enthalpy change ∆H in Joules Mass of base + acid x ∆T => 5540 = 4.1811J-1g-1K-1 (50+50) x 13.25 Graphically ∆Hn can be determined as in the example below: Procedure Place 8 test tubes in a test tube rack .Put 5cm3 of 2M sodium hydroxide solution into each test tube. Measure 25cm3 of 1M hydrochloric acid into 100cm3 plastic beaker. Record its initial temperature at volume of base =0.
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∆H = (m)mass of solution(acid+base) x (c)specific heat capacity of solution x ∆T
24 => (50 +25) x 4.2 x 6.5 = 2047.5Joules Moles potassium hydroxide =Enthalpy change ∆H ∆Hn 2047.5Joules = 0.0361 moles 56700Joules Molarity of KOH = moles x 1000 => 0.0361 moles x 1000 = 0.722M Volume used 50cm3 3.Determine the specific heat capacity of a solution of a solution mixture of 50.0cm3 of 2M potassium hydroxide neutralizing 50.0cm3 of 2M nitric(V) acid if a 13.25oC rise in temperature is recorded.(1mole of potassium hydroxide produce 55.4kJ of energy) Moles of potassium hydroxide = molarity KOH x volume 1000 => 2 M x 50cm3 = 0.1 moles 1000 Enthalpy change ∆H = ∆Hn x Moles potassium hydroxide => 55.4kJ x 0.1 moles = 5.54kJ x 1000=5540Joules Specific heat capacity = Enthalpy change ∆H in Joules Mass of base + acid x ∆T => 5540 = 4.1811J-1g-1K-1 (50+50) x 13.25 Graphically ∆Hn can be determined as in the example below: Procedure Place 8 test tubes in a test tube rack .Put 5cm3 of 2M sodium hydroxide solution into each test tube. Measure 25cm3 of 1M hydrochloric acid into 100cm3 plastic beaker. Record its initial temperature at volume of base =0. Put one portion of the base into the beaker containing the acid. Stir carefully with the thermometer and record the highest temperature change to the nearest 0.5oC. Repeat the procedure above with other portions of the base to complete table 1 below Table 1:Sample results.
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(b)Plot a graph of volume of sodium hydroxide against temperature change. 28.7=T2 temperature(oC) ∆T From the graph show and determine : (i)the highest temperature change ∆T ∆T =T2-T1 => highest temperature-T2 (from extrapolating a correctly plotted graph) less lowest temperature at volume of base=0 :T1 =>∆T = 6.7 – 0.0 = 6.70C (ii)the volume of sodium hydroxide used for complete neutralization From a correctly plotted graph – 16.75cm3 (c)Calculate the number of moles of the alkali used Moles NaOH = molarity x volume =>2M x 16.75cm3 = 0.0335 moles 1000 1000 (d)Calculate ∆H for the reaction ∆H = mass of solution(acid+base) x c x ∆T =>(25.0 + 16.75) x 4.2 x 6.7 = 1174.845 J = 1.174845kJ 1000 (e)Calculate the molar enthalpy of neutralization of the alkali. ∆Hn = ∆Hn = 1.174845kJ = 35.0701kJ Number of moles 0.0335 Volume of sodium hydroxide(cm3) 22. oC T1
26 (d)Standard enthalpy/heat of solution ∆Hᶿs The standard enthalpy of solution ∆Hᶿsis defined as the energy change when one mole of a substance is dissolve in excess distilled water to form an infinite dilute solution. An infinite dilute solution is one which is too dilute to be diluted further. Dissolving a solid involves two processes: (i) breaking the crystal of the solid into free ions(cations and anion).This process is the opposite of the formation of the crystal itself. The energy required to form one mole of a crystal structure from its gaseous ions is called Lattice energy/heat/enthalpy of lattice (∆Hl). Lattice energy /heat/enthalpy of lattice (∆Hl) is an endothermic process (+∆Hl).
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Dissolving a solid involves two processes: (i) breaking the crystal of the solid into free ions(cations and anion).This process is the opposite of the formation of the crystal itself. The energy required to form one mole of a crystal structure from its gaseous ions is called Lattice energy/heat/enthalpy of lattice (∆Hl). Lattice energy /heat/enthalpy of lattice (∆Hl) is an endothermic process (+∆Hl). The table below shows some ∆Hl in kJ for the process MX(s) -> M+ (g) + X- (g) Li Na K Ca Mg F +1022 +900 +800 +760 +631 Cl +846 +771 +690 +2237 +2493 Br +800 +733 +670 +2173 +2226 (ii)surrounding the free ions by polar water molecules. This process is called hydration. The energy produced when one mole of ions are completely hydrated is called hydration energy/ heat/enthalpy of hydration(∆Hh).Hydration energy /enthalpy of hydration(∆Hh) is an exothermic process(∆Hh). The table below shows some ∆Hh in kJ for some ions; ion Li+ Na+ K+ Mg2+ Ca2+ F- Cl- Br- ∆Hh -1091 -406 -322 -1920 -1650 -506 -364 -335 The sum of the lattice energy +∆Hl (endothermic) and hydration energy -∆Hh (exothermic) gives the heat of solution-∆Hs ∆Hs = ∆Hl +∆Hh Note Since ∆Hl is an endothermic process and ∆Hh is an exothermic process then ∆Hs is: (i)exothermic if ∆Hl is less than ∆Hh and hence a solid dissolve easily in water. (ii)endothermic if ∆Hl is more than ∆Hh and hence a solid does not dissolve easily in water.
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The energy produced when one mole of ions are completely hydrated is called hydration energy/ heat/enthalpy of hydration(∆Hh).Hydration energy /enthalpy of hydration(∆Hh) is an exothermic process(∆Hh). The table below shows some ∆Hh in kJ for some ions; ion Li+ Na+ K+ Mg2+ Ca2+ F- Cl- Br- ∆Hh -1091 -406 -322 -1920 -1650 -506 -364 -335 The sum of the lattice energy +∆Hl (endothermic) and hydration energy -∆Hh (exothermic) gives the heat of solution-∆Hs ∆Hs = ∆Hl +∆Hh Note Since ∆Hl is an endothermic process and ∆Hh is an exothermic process then ∆Hs is: (i)exothermic if ∆Hl is less than ∆Hh and hence a solid dissolve easily in water. (ii)endothermic if ∆Hl is more than ∆Hh and hence a solid does not dissolve easily in water. (a)Dissolving sodium chloride crystal/s: (i) NaCl ----breaking the crystal into free ions---> Na +(g)+ Cl-(g) ∆Hl =+771 kJ
27 (ii) Hydrating the ions; Na +(g) + aq -> Na(aq) ∆Hh = - 406 kJ Cl-(g) + aq -> Cl-(aq) ∆Hh = - 364 kJ ∆Hs =∆Hh +∆Hs -> (- 406 kJ + - 364 kJ) + +771 kJ = + 1.0 kJmole-1 NaCl does not dissolve easily in water because overall ∆Hs is endothermic Solubility of NaCl therefore increases with increase in temperature.
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The table below shows some ∆Hh in kJ for some ions; ion Li+ Na+ K+ Mg2+ Ca2+ F- Cl- Br- ∆Hh -1091 -406 -322 -1920 -1650 -506 -364 -335 The sum of the lattice energy +∆Hl (endothermic) and hydration energy -∆Hh (exothermic) gives the heat of solution-∆Hs ∆Hs = ∆Hl +∆Hh Note Since ∆Hl is an endothermic process and ∆Hh is an exothermic process then ∆Hs is: (i)exothermic if ∆Hl is less than ∆Hh and hence a solid dissolve easily in water. (ii)endothermic if ∆Hl is more than ∆Hh and hence a solid does not dissolve easily in water. (a)Dissolving sodium chloride crystal/s: (i) NaCl ----breaking the crystal into free ions---> Na +(g)+ Cl-(g) ∆Hl =+771 kJ
27 (ii) Hydrating the ions; Na +(g) + aq -> Na(aq) ∆Hh = - 406 kJ Cl-(g) + aq -> Cl-(aq) ∆Hh = - 364 kJ ∆Hs =∆Hh +∆Hs -> (- 406 kJ + - 364 kJ) + +771 kJ = + 1.0 kJmole-1 NaCl does not dissolve easily in water because overall ∆Hs is endothermic Solubility of NaCl therefore increases with increase in temperature. Increase in temperature increases the energy to break the crystal lattice of NaCl to free Na +(g)+ Cl-(g) (b)Dissolving magnesium chloride crystal/s// MgCl2 (s) ->MgCl2 (aq) (i) MgCl2 --breaking the crystal into free ions-->Mg 2+(g)+ 2Cl-(g) ∆Hl =+2493 kJ (ii) Hydrating the ions; Mg 2+(g) + aq -> Mg 2+(g) (aq) ∆Hh = - 1920 kJ 2Cl-(g) + aq -> 2Cl-(aq) ∆Hh = (- 364 x 2) kJ ∆Hs =∆Hh +∆Hs -> (- 1920 kJ + (- 364 x 2 kJ)) + +2493 kJ = -155.0 kJmole-1 MgCl2 (s) dissolve easily in water because overall ∆Hs is exothermic .
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(ii)endothermic if ∆Hl is more than ∆Hh and hence a solid does not dissolve easily in water. (a)Dissolving sodium chloride crystal/s: (i) NaCl ----breaking the crystal into free ions---> Na +(g)+ Cl-(g) ∆Hl =+771 kJ
27 (ii) Hydrating the ions; Na +(g) + aq -> Na(aq) ∆Hh = - 406 kJ Cl-(g) + aq -> Cl-(aq) ∆Hh = - 364 kJ ∆Hs =∆Hh +∆Hs -> (- 406 kJ + - 364 kJ) + +771 kJ = + 1.0 kJmole-1 NaCl does not dissolve easily in water because overall ∆Hs is endothermic Solubility of NaCl therefore increases with increase in temperature. Increase in temperature increases the energy to break the crystal lattice of NaCl to free Na +(g)+ Cl-(g) (b)Dissolving magnesium chloride crystal/s// MgCl2 (s) ->MgCl2 (aq) (i) MgCl2 --breaking the crystal into free ions-->Mg 2+(g)+ 2Cl-(g) ∆Hl =+2493 kJ (ii) Hydrating the ions; Mg 2+(g) + aq -> Mg 2+(g) (aq) ∆Hh = - 1920 kJ 2Cl-(g) + aq -> 2Cl-(aq) ∆Hh = (- 364 x 2) kJ ∆Hs =∆Hh +∆Hs -> (- 1920 kJ + (- 364 x 2 kJ)) + +2493 kJ = -155.0 kJmole-1 MgCl2 (s) dissolve easily in water because overall ∆Hs is exothermic . Solubility of MgCl2 (s) therefore decreases with increase in temperature.
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(a)Dissolving sodium chloride crystal/s: (i) NaCl ----breaking the crystal into free ions---> Na +(g)+ Cl-(g) ∆Hl =+771 kJ
27 (ii) Hydrating the ions; Na +(g) + aq -> Na(aq) ∆Hh = - 406 kJ Cl-(g) + aq -> Cl-(aq) ∆Hh = - 364 kJ ∆Hs =∆Hh +∆Hs -> (- 406 kJ + - 364 kJ) + +771 kJ = + 1.0 kJmole-1 NaCl does not dissolve easily in water because overall ∆Hs is endothermic Solubility of NaCl therefore increases with increase in temperature. Increase in temperature increases the energy to break the crystal lattice of NaCl to free Na +(g)+ Cl-(g) (b)Dissolving magnesium chloride crystal/s// MgCl2 (s) ->MgCl2 (aq) (i) MgCl2 --breaking the crystal into free ions-->Mg 2+(g)+ 2Cl-(g) ∆Hl =+2493 kJ (ii) Hydrating the ions; Mg 2+(g) + aq -> Mg 2+(g) (aq) ∆Hh = - 1920 kJ 2Cl-(g) + aq -> 2Cl-(aq) ∆Hh = (- 364 x 2) kJ ∆Hs =∆Hh +∆Hs -> (- 1920 kJ + (- 364 x 2 kJ)) + +2493 kJ = -155.0 kJmole-1 MgCl2 (s) dissolve easily in water because overall ∆Hs is exothermic . Solubility of MgCl2 (s) therefore decreases with increase in temperature. (c)Dissolving Calcium floride crystal/s// CaF2 (s) -> CaF2 (aq) (i) CaF2 -->Ca 2+(g)+ 2F-(g) ∆Hl =+760 kJ (ii) Hydrating the ions; Ca 2+(g) + aq -> Ca 2+(g) (aq) ∆Hh = - 1650 kJ 2F-(g) + aq -> 2F-(aq) ∆Hh = (- 506 x 2) kJ ∆Hs =∆Hh +∆Hs -> (- 1650 kJ + (- 506 x 2 kJ)) + +760 kJ = -1902.0 kJmole-1 CaF2 (s) dissolve easily in water because overall ∆Hs is exothermic .
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Increase in temperature increases the energy to break the crystal lattice of NaCl to free Na +(g)+ Cl-(g) (b)Dissolving magnesium chloride crystal/s// MgCl2 (s) ->MgCl2 (aq) (i) MgCl2 --breaking the crystal into free ions-->Mg 2+(g)+ 2Cl-(g) ∆Hl =+2493 kJ (ii) Hydrating the ions; Mg 2+(g) + aq -> Mg 2+(g) (aq) ∆Hh = - 1920 kJ 2Cl-(g) + aq -> 2Cl-(aq) ∆Hh = (- 364 x 2) kJ ∆Hs =∆Hh +∆Hs -> (- 1920 kJ + (- 364 x 2 kJ)) + +2493 kJ = -155.0 kJmole-1 MgCl2 (s) dissolve easily in water because overall ∆Hs is exothermic . Solubility of MgCl2 (s) therefore decreases with increase in temperature. (c)Dissolving Calcium floride crystal/s// CaF2 (s) -> CaF2 (aq) (i) CaF2 -->Ca 2+(g)+ 2F-(g) ∆Hl =+760 kJ (ii) Hydrating the ions; Ca 2+(g) + aq -> Ca 2+(g) (aq) ∆Hh = - 1650 kJ 2F-(g) + aq -> 2F-(aq) ∆Hh = (- 506 x 2) kJ ∆Hs =∆Hh +∆Hs -> (- 1650 kJ + (- 506 x 2 kJ)) + +760 kJ = -1902.0 kJmole-1 CaF2 (s) dissolve easily in water because overall ∆Hs is exothermic . Solubility of CaF2 (s) therefore decreases with increase in temperature.
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Solubility of MgCl2 (s) therefore decreases with increase in temperature. (c)Dissolving Calcium floride crystal/s// CaF2 (s) -> CaF2 (aq) (i) CaF2 -->Ca 2+(g)+ 2F-(g) ∆Hl =+760 kJ (ii) Hydrating the ions; Ca 2+(g) + aq -> Ca 2+(g) (aq) ∆Hh = - 1650 kJ 2F-(g) + aq -> 2F-(aq) ∆Hh = (- 506 x 2) kJ ∆Hs =∆Hh +∆Hs -> (- 1650 kJ + (- 506 x 2 kJ)) + +760 kJ = -1902.0 kJmole-1 CaF2 (s) dissolve easily in water because overall ∆Hs is exothermic . Solubility of CaF2 (s) therefore decreases with increase in temperature. (d)Dissolving magnesium bromide crystal/s// MgBr2 (s) ->MgBr2 (aq) (i) MgCl2 --breaking the crystal into free ions-->Mg 2+(g)+ 2Br-(g) ∆Hl =+2226 kJ (ii) Hydrating the ions; Mg 2+(g) + aq -> Mg 2+(g) (aq) ∆Hh = - 1920 kJ 2Br-(g) + aq -> 2Br-(aq) ∆Hh = (- 335x 2) kJ ∆Hs =∆Hh +∆Hs -> (- 1920 kJ + (- 335 x 2 kJ)) + +2226 kJ = -364.0 kJmole-1 MgBr2 (s) dissolve easily in water because overall ∆Hs is exothermic . Solubility of MgBr2(s) therefore decreases with increase in temperature. 28 Practically the heat of solution can be determined from dissolving known amount /mass/volume of solute in known mass /volume of water/solvent. From the temperature of solvent before and after dissolving the change in temperature(∆T) during dissolution is determined.
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Solubility of MgBr2(s) therefore decreases with increase in temperature. 28 Practically the heat of solution can be determined from dissolving known amount /mass/volume of solute in known mass /volume of water/solvent. From the temperature of solvent before and after dissolving the change in temperature(∆T) during dissolution is determined. To determine the ∆Hs ammonium nitrate Place 100cm3 of distilled water into a plastic beaker/calorimeter. Determine its temperature and record it at time =0 in table I below. Put all the 5.0g of ammonium nitrate (potassium nitrate/ammonium chloride can also be used)provided into the plastic beaker/calorimeter, stir using a thermometer and record the highest temperature change to the nearest 0.5oCafter every ½ minute to complete table I. Continue stirring the mixture throughout the experiment. Sample results: Table I Time (minutes) 0.0 ½ 1 1 ½ 2 2 ½ 3 3 ½ Temperature()oC 22.0 21.0 20.0 19.0 19.0 19.5 20.0 20.5 (a)Plot a graph of temperature against time(x-axis) 22.0=T1 temperature(oC) ∆T (b)From the graph show and determine the highest temperature change ∆T ∆T =T2-T1 => lowest temperature-T2 (from extrapolating a correctly plotted graph) less highest temperature at volume of base=0 :T1 Time (minutes) 18.7. oC T1
29 =>∆T =18.7 – 22.0 = 3.30C (c)Calculate the number of moles of ammonium nitrate(V) used Moles NH4NO3 = mass used => 5.0 = 0.0625 moles Molar mass 80 (d)Calculate ∆H for the reaction ∆H = mass of water x c x ∆T ->100 x 4.2 x 3.3 = +1386 J = +1.386kJ 1000 (e)Calculate the molar enthalpy of dissolution of ammonium nitrate(V).
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Continue stirring the mixture throughout the experiment. Sample results: Table I Time (minutes) 0.0 ½ 1 1 ½ 2 2 ½ 3 3 ½ Temperature()oC 22.0 21.0 20.0 19.0 19.0 19.5 20.0 20.5 (a)Plot a graph of temperature against time(x-axis) 22.0=T1 temperature(oC) ∆T (b)From the graph show and determine the highest temperature change ∆T ∆T =T2-T1 => lowest temperature-T2 (from extrapolating a correctly plotted graph) less highest temperature at volume of base=0 :T1 Time (minutes) 18.7. oC T1
29 =>∆T =18.7 – 22.0 = 3.30C (c)Calculate the number of moles of ammonium nitrate(V) used Moles NH4NO3 = mass used => 5.0 = 0.0625 moles Molar mass 80 (d)Calculate ∆H for the reaction ∆H = mass of water x c x ∆T ->100 x 4.2 x 3.3 = +1386 J = +1.386kJ 1000 (e)Calculate the molar enthalpy of dissolution of ammonium nitrate(V). ∆Hs = ∆H = +1.386kJ = + 22.176kJ mole-1 Number of moles 0.0625 moles (f)What would happen if the distilled water was heated before the experiment was performed. The ammonium nitrate(V)would take less time to dissolves.
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oC T1
29 =>∆T =18.7 – 22.0 = 3.30C (c)Calculate the number of moles of ammonium nitrate(V) used Moles NH4NO3 = mass used => 5.0 = 0.0625 moles Molar mass 80 (d)Calculate ∆H for the reaction ∆H = mass of water x c x ∆T ->100 x 4.2 x 3.3 = +1386 J = +1.386kJ 1000 (e)Calculate the molar enthalpy of dissolution of ammonium nitrate(V). ∆Hs = ∆H = +1.386kJ = + 22.176kJ mole-1 Number of moles 0.0625 moles (f)What would happen if the distilled water was heated before the experiment was performed. The ammonium nitrate(V)would take less time to dissolves. Increase in temperature reduces lattice energy causing endothermic dissolution to be faster (g)Illustrate the process above in an energy level diagram NH4+ (g) + NO3-(g) +∆H NH4+ (aq)+NO3-(aq) Energy(kJ) +∆H ∆H = -22.176kJ NH4NO3-(s) Reaction path /progress/coordinate (h) 100cm3 of distilled water at 25oC was added carefully 3cm3 concentrated sulphuric(VI)acid of density 1.84gcm-3.The temperature of the mixture rose from 250C to 38oC.Calculate the molar heat of solution of sulphuric(VI)acid (S=32.0,H=1.0,0=16.0) Working
30 Molar mass of H2SO4 = 98g Mass of H2SO4= Density x volume => 1.84gcm-3 x 3cm3 = 5.52 g Mass of H2O = Density x volume => 1.00gcm-3 x 100cm3 = 100 g Moles of H2SO4= mass => 5.52 g = 0.0563 moles Molar mass of H2SO4 98g Enthalpy change ∆H= (mass of acid + water) x specific heat capacity of water x ∆T => (100 +5.52 g) x 4.2 x 13oC = 5761.392 J = 5.761392 kJ 1000 ∆Hs of H2SO4= ∆H => 5.761392 kJ = -102.33378kJmoles-1 Moles of H2SO4 0.0563 moles (e)Standard enthalpy/heat of formation ∆Hᶿf The molar enthalpy of formation ∆Hᶿf is defined as the energy change when one mole of a compound is formed from its elements at 298K(25oC) and 101325Pa(one atmosphere)pressure.
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∆Hs = ∆H = +1.386kJ = + 22.176kJ mole-1 Number of moles 0.0625 moles (f)What would happen if the distilled water was heated before the experiment was performed. The ammonium nitrate(V)would take less time to dissolves. Increase in temperature reduces lattice energy causing endothermic dissolution to be faster (g)Illustrate the process above in an energy level diagram NH4+ (g) + NO3-(g) +∆H NH4+ (aq)+NO3-(aq) Energy(kJ) +∆H ∆H = -22.176kJ NH4NO3-(s) Reaction path /progress/coordinate (h) 100cm3 of distilled water at 25oC was added carefully 3cm3 concentrated sulphuric(VI)acid of density 1.84gcm-3.The temperature of the mixture rose from 250C to 38oC.Calculate the molar heat of solution of sulphuric(VI)acid (S=32.0,H=1.0,0=16.0) Working
30 Molar mass of H2SO4 = 98g Mass of H2SO4= Density x volume => 1.84gcm-3 x 3cm3 = 5.52 g Mass of H2O = Density x volume => 1.00gcm-3 x 100cm3 = 100 g Moles of H2SO4= mass => 5.52 g = 0.0563 moles Molar mass of H2SO4 98g Enthalpy change ∆H= (mass of acid + water) x specific heat capacity of water x ∆T => (100 +5.52 g) x 4.2 x 13oC = 5761.392 J = 5.761392 kJ 1000 ∆Hs of H2SO4= ∆H => 5.761392 kJ = -102.33378kJmoles-1 Moles of H2SO4 0.0563 moles (e)Standard enthalpy/heat of formation ∆Hᶿf The molar enthalpy of formation ∆Hᶿf is defined as the energy change when one mole of a compound is formed from its elements at 298K(25oC) and 101325Pa(one atmosphere)pressure. ∆Hᶿf is practically difficult to determine in a school laboratory.
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The ammonium nitrate(V)would take less time to dissolves. Increase in temperature reduces lattice energy causing endothermic dissolution to be faster (g)Illustrate the process above in an energy level diagram NH4+ (g) + NO3-(g) +∆H NH4+ (aq)+NO3-(aq) Energy(kJ) +∆H ∆H = -22.176kJ NH4NO3-(s) Reaction path /progress/coordinate (h) 100cm3 of distilled water at 25oC was added carefully 3cm3 concentrated sulphuric(VI)acid of density 1.84gcm-3.The temperature of the mixture rose from 250C to 38oC.Calculate the molar heat of solution of sulphuric(VI)acid (S=32.0,H=1.0,0=16.0) Working
30 Molar mass of H2SO4 = 98g Mass of H2SO4= Density x volume => 1.84gcm-3 x 3cm3 = 5.52 g Mass of H2O = Density x volume => 1.00gcm-3 x 100cm3 = 100 g Moles of H2SO4= mass => 5.52 g = 0.0563 moles Molar mass of H2SO4 98g Enthalpy change ∆H= (mass of acid + water) x specific heat capacity of water x ∆T => (100 +5.52 g) x 4.2 x 13oC = 5761.392 J = 5.761392 kJ 1000 ∆Hs of H2SO4= ∆H => 5.761392 kJ = -102.33378kJmoles-1 Moles of H2SO4 0.0563 moles (e)Standard enthalpy/heat of formation ∆Hᶿf The molar enthalpy of formation ∆Hᶿf is defined as the energy change when one mole of a compound is formed from its elements at 298K(25oC) and 101325Pa(one atmosphere)pressure. ∆Hᶿf is practically difficult to determine in a school laboratory. It is determined normally determined by applying Hess’ law of constant heat summation.
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Increase in temperature reduces lattice energy causing endothermic dissolution to be faster (g)Illustrate the process above in an energy level diagram NH4+ (g) + NO3-(g) +∆H NH4+ (aq)+NO3-(aq) Energy(kJ) +∆H ∆H = -22.176kJ NH4NO3-(s) Reaction path /progress/coordinate (h) 100cm3 of distilled water at 25oC was added carefully 3cm3 concentrated sulphuric(VI)acid of density 1.84gcm-3.The temperature of the mixture rose from 250C to 38oC.Calculate the molar heat of solution of sulphuric(VI)acid (S=32.0,H=1.0,0=16.0) Working
30 Molar mass of H2SO4 = 98g Mass of H2SO4= Density x volume => 1.84gcm-3 x 3cm3 = 5.52 g Mass of H2O = Density x volume => 1.00gcm-3 x 100cm3 = 100 g Moles of H2SO4= mass => 5.52 g = 0.0563 moles Molar mass of H2SO4 98g Enthalpy change ∆H= (mass of acid + water) x specific heat capacity of water x ∆T => (100 +5.52 g) x 4.2 x 13oC = 5761.392 J = 5.761392 kJ 1000 ∆Hs of H2SO4= ∆H => 5.761392 kJ = -102.33378kJmoles-1 Moles of H2SO4 0.0563 moles (e)Standard enthalpy/heat of formation ∆Hᶿf The molar enthalpy of formation ∆Hᶿf is defined as the energy change when one mole of a compound is formed from its elements at 298K(25oC) and 101325Pa(one atmosphere)pressure. ∆Hᶿf is practically difficult to determine in a school laboratory. It is determined normally determined by applying Hess’ law of constant heat summation. Hess’ law of constant heat summation states that “the total enthalpy/heat/energy change of a reaction is the same regardless of the route taken from reactants to products at the same temperature and pressure”.
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∆Hᶿf is practically difficult to determine in a school laboratory. It is determined normally determined by applying Hess’ law of constant heat summation. Hess’ law of constant heat summation states that “the total enthalpy/heat/energy change of a reaction is the same regardless of the route taken from reactants to products at the same temperature and pressure”. Hess’ law of constant heat summation is as a result of a series of experiments done by the German Scientist Henri Hess(1802-1850). He found that the total energy change from the reactants to products was the same irrespective of the intermediate products between. i.e. A(s) --∆H1-->C(s) = A(s) --∆H2-->B(s)--∆H3-->C(s) Applying Hess’ law of constant heat summation then:
31 A(s) ∆H2 B(s) ∆H1 ∆H3 C(s) The above is called an energy cycle diagram. It can be used to calculate any of the missing energy changes since: (i) ∆H1 =∆H2 + ∆H3 (ii) ∆H2 =∆H1 + -∆H3 (iii) ∆H3 = - ∆H1 + ∆H2 Examples of applying Hess’ law of constant heat summation 1.Calculate the molar enthalpy of formation of methane (CH4) given that ∆Hᶿc of carbon-graphite is -393.5kJmole-1,Hydrogen is -285.7 kJmole-1 and that of methane is 890 kJmole-1 Working Carbon-graphite ,hydrogen and oxygen can react to first form methane. Methane will then burn in the oxygen present to form carbon(IV)oxide and water. Carbon-graphite can burn in the oxygen to form carbon(IV)oxide. Hydrogen can burn in the oxygen to form water.
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Methane will then burn in the oxygen present to form carbon(IV)oxide and water. Carbon-graphite can burn in the oxygen to form carbon(IV)oxide. Hydrogen can burn in the oxygen to form water. C(s)+ 2H2 (g)+2O2 (g) --∆H1--> CH4(g) +2O2(g) --∆H2--> CO2(g)+2H2O(l) C(s)+ 2H2 (g)+2O2 (g) --∆H3--> CO2(g)+2H2O(l) Energy cycle diagram C(s) + 2H2 (g) + 2O2(g) ∆H1=∆Hᶿc =-890.4kJ CH4(g)+2O2(g) ∆H3=∆Hᶿc =-393.5kJ ∆H3=∆Hᶿc =-285.7kJ x 2 ∆H2= ∆Hᶿf= x CO2(g) + 2H2O(l) Substituting: ∆H3 = ∆H1 + ∆H2 -393.5 + (-285.7 x 2) = -890.4kJ + x x = -74.5 kJ
32 Heat of formation ∆Hᶿf CH4 = -74.5 kJmole-1 2. Calculate the molar enthalpy of formation of ethyne (C2H2) given : ∆Hᶿc of carbon-graphite = -394kJmole-1,Hydrogen = -286 kJmole-1 , (C2H2) = -1300 kJmole-1 Working Carbon-graphite ,hydrogen and oxygen can react to first form ethyne. Ethyne will then burn in the oxygen present to form carbon(IV)oxide and water. Carbon-graphite can burn in the oxygen to form carbon(IV)oxide. Hydrogen can burn in the oxygen to form water.
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Ethyne will then burn in the oxygen present to form carbon(IV)oxide and water. Carbon-graphite can burn in the oxygen to form carbon(IV)oxide. Hydrogen can burn in the oxygen to form water. 2C(s)+ H2 (g)+2 ½ O2 (g) --∆H1--> C2 H2 (g) +2 ½ O2(g) --∆H2--> CO2(g)+H2O(l) 2C(s)+ H2 (g)+ 2 ½ O2 (g) --∆H3--> 2CO2(g)+H2O(l) Energy cycle diagram 2C(s) + H2 (g) +2½O2(g) ∆H1=∆Hᶿf =x C2 H2+2½O2(g) ∆H3=∆Hᶿc =-394kJx 2 ∆H3=∆Hᶿc =-286kJ ∆H2= ∆Hᶿc= -1300kJ 2CO2(g) + H2O(l) Substituting: ∆H3 = ∆H1 + ∆H2 ( -394 x 2) + -286 = -1300kJ + x x = +244 kJ Heat of formation ∆Hᶿf CH4 = +244 kJmole-1 3. Calculate the molar enthalpy of formation of carbon(II)oxide (CO) given : ∆Hᶿc of carbon-graphite = -393.5kJmole-1, ∆Hᶿc of carbon(II)oxide (CO)= -283 kJmole-1 Working Carbon-graphite reacts with oxygen first to form carbon (II)oxide (CO). Carbon(II)oxide (CO) then burn in the excess oxygen to form carbon(IV)oxide. Carbon-graphite can burn in excess oxygen to form carbon (IV) oxide.
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Measure 100cm3 of distilled water into a beaker. Determine its temperature T1 .Put all the crystals of the copper(II)sulphate(VI)pentahydrate carefully into the beaker. Stir using a thermometer and determine the highest temperature change T2 Repeat the procedure again to complete table 1. Table 1:Sample results
35 Experiment I II Highest /lowest temperature T2 27.0 29.0 Initial temperature T1 24.0 25.0 Change in temperature ∆T 3.0 4.0 Experiment II Weigh accurately 8.0g of anhydrous copper(II)sulphate(VI). Measure 100cm3 of distilled water into a beaker. Determine its temperature T1 .Put all the crystals of the copper(II)sulphate(VI)pentahydrate carefully into the beaker. Stir using a thermometer and determine the highest temperature change T2 Repeat the procedure again to complete table II.
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Measure 100cm3 of distilled water into a beaker. Determine its temperature T1 .Put all the crystals of the copper(II)sulphate(VI)pentahydrate carefully into the beaker. Stir using a thermometer and determine the highest temperature change T2 Repeat the procedure again to complete table II. Table II :Sample results Experiment I II Highest /lowest temperature T2 26.0 27.0 Initial temperature T1 25.0 25.0 Change in temperature ∆T 1.0 2.0 Questions (a)Calculate the average ∆T in (i)Table I ∆T= T2 -T1 => 3.0 +4.0 = 3.5 oC 2 (ii)Table II ∆T= T2 -T1 => 1.0 +2.0 = 1.5 oC 2 (b)Calculate the number of moles of solid used in: (i)Experiment I Moles of CuSO4.5H2O = Mass => 12.5 = 0.05 moles Molar mass 250 (ii)Experiment II Moles of CuSO4 = Mass => 8.0 = 0.05 moles Molar mass 160 (c)Calculate the enthalpy change for the reaction in: (i)Experiment I Enthalpy change of CuSO4.5H2O= mass of Water(m) x c x ∆T =>100cm3 x 4.2 x 3.5 oC = -1.47kJ 1000 (ii)Experiment II Enthalpy change of CuSO4 = mass of water(m) x c x ∆T
36 =>100cm3 x 4.2 x 1.5 oC = -0.63kJ 1000 (c)Calculate the molar enthalpy of solution CuSO4 .5H2O (s) form the results in (i)experiment I. ∆Hs = CuSO4.5H2O= ∆H => -1.47kJ = 29.4kJ Number of Moles 0.05 moles (ii)experiment II.
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Stir using a thermometer and determine the highest temperature change T2 Repeat the procedure again to complete table II. Table II :Sample results Experiment I II Highest /lowest temperature T2 26.0 27.0 Initial temperature T1 25.0 25.0 Change in temperature ∆T 1.0 2.0 Questions (a)Calculate the average ∆T in (i)Table I ∆T= T2 -T1 => 3.0 +4.0 = 3.5 oC 2 (ii)Table II ∆T= T2 -T1 => 1.0 +2.0 = 1.5 oC 2 (b)Calculate the number of moles of solid used in: (i)Experiment I Moles of CuSO4.5H2O = Mass => 12.5 = 0.05 moles Molar mass 250 (ii)Experiment II Moles of CuSO4 = Mass => 8.0 = 0.05 moles Molar mass 160 (c)Calculate the enthalpy change for the reaction in: (i)Experiment I Enthalpy change of CuSO4.5H2O= mass of Water(m) x c x ∆T =>100cm3 x 4.2 x 3.5 oC = -1.47kJ 1000 (ii)Experiment II Enthalpy change of CuSO4 = mass of water(m) x c x ∆T
36 =>100cm3 x 4.2 x 1.5 oC = -0.63kJ 1000 (c)Calculate the molar enthalpy of solution CuSO4 .5H2O (s) form the results in (i)experiment I. ∆Hs = CuSO4.5H2O= ∆H => -1.47kJ = 29.4kJ Number of Moles 0.05 moles (ii)experiment II. ∆Hs = CuSO4= ∆H => -0.63kJ = 12.6kJ Number of Moles 0.05 moles (d) Using an energy level diagram, calculate the molar enthalpy change for the reaction: CuSO4 .5H2O (s) -> CuSO4(s) + 5H2O(l) Energy cycle diagram CuSO4(s) + (aq) + 5H2O(l) ∆H1=x CuSO4.
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Table II :Sample results Experiment I II Highest /lowest temperature T2 26.0 27.0 Initial temperature T1 25.0 25.0 Change in temperature ∆T 1.0 2.0 Questions (a)Calculate the average ∆T in (i)Table I ∆T= T2 -T1 => 3.0 +4.0 = 3.5 oC 2 (ii)Table II ∆T= T2 -T1 => 1.0 +2.0 = 1.5 oC 2 (b)Calculate the number of moles of solid used in: (i)Experiment I Moles of CuSO4.5H2O = Mass => 12.5 = 0.05 moles Molar mass 250 (ii)Experiment II Moles of CuSO4 = Mass => 8.0 = 0.05 moles Molar mass 160 (c)Calculate the enthalpy change for the reaction in: (i)Experiment I Enthalpy change of CuSO4.5H2O= mass of Water(m) x c x ∆T =>100cm3 x 4.2 x 3.5 oC = -1.47kJ 1000 (ii)Experiment II Enthalpy change of CuSO4 = mass of water(m) x c x ∆T
36 =>100cm3 x 4.2 x 1.5 oC = -0.63kJ 1000 (c)Calculate the molar enthalpy of solution CuSO4 .5H2O (s) form the results in (i)experiment I. ∆Hs = CuSO4.5H2O= ∆H => -1.47kJ = 29.4kJ Number of Moles 0.05 moles (ii)experiment II. ∆Hs = CuSO4= ∆H => -0.63kJ = 12.6kJ Number of Moles 0.05 moles (d) Using an energy level diagram, calculate the molar enthalpy change for the reaction: CuSO4 .5H2O (s) -> CuSO4(s) + 5H2O(l) Energy cycle diagram CuSO4(s) + (aq) + 5H2O(l) ∆H1=x CuSO4. 5H2O (s)+ (aq) ∆H3= =-29.4kJ ∆H2= -12.6kJ CuSO4 .5H2O (aq) ∆H3 = ∆H1 +∆H2 =>-29.4kJ = -12.6kJ + x =>-29.4kJ - (+12.6kJ) = x x = 16.8kJ b)Practical example II Determination of enthalpy of solution of ammonium chloride Theoretical information.
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∆Hs = CuSO4.5H2O= ∆H => -1.47kJ = 29.4kJ Number of Moles 0.05 moles (ii)experiment II. ∆Hs = CuSO4= ∆H => -0.63kJ = 12.6kJ Number of Moles 0.05 moles (d) Using an energy level diagram, calculate the molar enthalpy change for the reaction: CuSO4 .5H2O (s) -> CuSO4(s) + 5H2O(l) Energy cycle diagram CuSO4(s) + (aq) + 5H2O(l) ∆H1=x CuSO4. 5H2O (s)+ (aq) ∆H3= =-29.4kJ ∆H2= -12.6kJ CuSO4 .5H2O (aq) ∆H3 = ∆H1 +∆H2 =>-29.4kJ = -12.6kJ + x =>-29.4kJ - (+12.6kJ) = x x = 16.8kJ b)Practical example II Determination of enthalpy of solution of ammonium chloride Theoretical information. Ammonium chloride dissolves in water to form ammonium chloride solution. Aqueous ammonia can react with excess dilute hydrochloric acid to form ammonium chloride solution. The heat change taking place can be calculated from the heat of reactions:
37 (i) NH3(aq) + HCl(aq) -> NH4Cl(s) (ii) NH4Cl(s) + (aq) -> NH4Cl(aq) (iii) NH3(aq) + HCl(aq) -> NH4Cl(aq) Experiment procedure I Measure 50cm3 of water into a 100cm3 beaker. Record its temperature T1 as initial temperature to the nearest 0.5oC in table I. Add exactly 5.0g of ammonium chloride crystals weighed carefully into the water. Stir and record the highest temperature change T2 as the final temperature change. Repeat the above procedure to complete table I.
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Add exactly 5.0g of ammonium chloride crystals weighed carefully into the water. Stir and record the highest temperature change T2 as the final temperature change. Repeat the above procedure to complete table I. Sample results TableI Experiment I II final temperature(oC) 19.0 20.0 initial temperature(oC) 22.0 22.0 temperature change ∆T(oC) 3.0 2.0 Experiment procedure II Measure 25cm3 of 2M aqueous ammonia into a 100cm3 beaker. Record its temperature T1 as initial temperature to the nearest 0.5oC in table II. Measure 25cm3 of 2M hydrochloric acid solution. Add the acid into the beaker containing aqueous ammonia. Stir and record the highest temperature change T2 as the final temperature change. Repeat the above procedure to complete table II.
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Add the acid into the beaker containing aqueous ammonia. Stir and record the highest temperature change T2 as the final temperature change. Repeat the above procedure to complete table II. Sample results:Table II Experiment I II final temperature(oC) 29.0 29.0 initial temperature(oC) 22.0 22.0 temperature change ∆T(oC) 7.0 7.0 Sample Calculations: (a)Calculate the average ∆T in (i)Table I ∆T= T2 -T1 => -3.0 +-2.0 = 2.5 oC 2 (ii)Table II ∆T= T2 -T1 => 7.0 +7.0 = 7.0 oC 2 (b)Calculate the enthalpy change for the reaction in:
38 (i)Experiment I Enthalpy change ∆H = mass of Water(m) x c x ∆T =>50cm3 x 4.2 x 2.5 oC = +0.525kJ 1000 (ii)Experiment II Enthalpy change of CuSO4 = mass of water(m) x c x ∆T =>25+25cm3 x 4.2 x 7 oC = +1.47kJ 1000 (c)Write the equation for the reaction taking place in: (i)Experiment I NH4Cl(s) + (aq) -> NH4Cl(aq) (ii)Experiment I NH3(aq) + HCl(aq) -> NH4Cl(aq) (d)Calculate the enthalpy change ∆H for the reaction: NH3(g) + HCl(g) -> NH4Cl(s) given that: (i) NH3(g) + (aq) -> NH3(aq) ∆H= -40.3kJ (ii) (aq) + HCl(g) -> HCl(aq) ∆H= -16.45kJ (e)Applying Hess’ Law of constant heat summation: Energy level diagram N2(g) + 1½ H2(g) + ½ Cl2 ∆Hf NH4Cl(s) + aq +0.525kJ=∆H4 (aq) (aq) - 40.3kJ=∆H1 -16.43kJ=∆H2 NH3 (aq) + HCl(aq) -1.47kJ=∆H3 NH4Cl(s) ∆H1 + ∆H2 + ∆H3 = ∆H4 + ∆Hf - 40.3kJ + -16.43kJ + -1.47kJ = +0.525kJ + ∆Hf =>∆Hf = -58.865kJ.
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Stir and record the highest temperature change T2 as the final temperature change. Repeat the above procedure to complete table II. Sample results:Table II Experiment I II final temperature(oC) 29.0 29.0 initial temperature(oC) 22.0 22.0 temperature change ∆T(oC) 7.0 7.0 Sample Calculations: (a)Calculate the average ∆T in (i)Table I ∆T= T2 -T1 => -3.0 +-2.0 = 2.5 oC 2 (ii)Table II ∆T= T2 -T1 => 7.0 +7.0 = 7.0 oC 2 (b)Calculate the enthalpy change for the reaction in:
38 (i)Experiment I Enthalpy change ∆H = mass of Water(m) x c x ∆T =>50cm3 x 4.2 x 2.5 oC = +0.525kJ 1000 (ii)Experiment II Enthalpy change of CuSO4 = mass of water(m) x c x ∆T =>25+25cm3 x 4.2 x 7 oC = +1.47kJ 1000 (c)Write the equation for the reaction taking place in: (i)Experiment I NH4Cl(s) + (aq) -> NH4Cl(aq) (ii)Experiment I NH3(aq) + HCl(aq) -> NH4Cl(aq) (d)Calculate the enthalpy change ∆H for the reaction: NH3(g) + HCl(g) -> NH4Cl(s) given that: (i) NH3(g) + (aq) -> NH3(aq) ∆H= -40.3kJ (ii) (aq) + HCl(g) -> HCl(aq) ∆H= -16.45kJ (e)Applying Hess’ Law of constant heat summation: Energy level diagram N2(g) + 1½ H2(g) + ½ Cl2 ∆Hf NH4Cl(s) + aq +0.525kJ=∆H4 (aq) (aq) - 40.3kJ=∆H1 -16.43kJ=∆H2 NH3 (aq) + HCl(aq) -1.47kJ=∆H3 NH4Cl(s) ∆H1 + ∆H2 + ∆H3 = ∆H4 + ∆Hf - 40.3kJ + -16.43kJ + -1.47kJ = +0.525kJ + ∆Hf =>∆Hf = -58.865kJ. Practice theoretical examples:
39 1.
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Repeat the above procedure to complete table II. Sample results:Table II Experiment I II final temperature(oC) 29.0 29.0 initial temperature(oC) 22.0 22.0 temperature change ∆T(oC) 7.0 7.0 Sample Calculations: (a)Calculate the average ∆T in (i)Table I ∆T= T2 -T1 => -3.0 +-2.0 = 2.5 oC 2 (ii)Table II ∆T= T2 -T1 => 7.0 +7.0 = 7.0 oC 2 (b)Calculate the enthalpy change for the reaction in:
38 (i)Experiment I Enthalpy change ∆H = mass of Water(m) x c x ∆T =>50cm3 x 4.2 x 2.5 oC = +0.525kJ 1000 (ii)Experiment II Enthalpy change of CuSO4 = mass of water(m) x c x ∆T =>25+25cm3 x 4.2 x 7 oC = +1.47kJ 1000 (c)Write the equation for the reaction taking place in: (i)Experiment I NH4Cl(s) + (aq) -> NH4Cl(aq) (ii)Experiment I NH3(aq) + HCl(aq) -> NH4Cl(aq) (d)Calculate the enthalpy change ∆H for the reaction: NH3(g) + HCl(g) -> NH4Cl(s) given that: (i) NH3(g) + (aq) -> NH3(aq) ∆H= -40.3kJ (ii) (aq) + HCl(g) -> HCl(aq) ∆H= -16.45kJ (e)Applying Hess’ Law of constant heat summation: Energy level diagram N2(g) + 1½ H2(g) + ½ Cl2 ∆Hf NH4Cl(s) + aq +0.525kJ=∆H4 (aq) (aq) - 40.3kJ=∆H1 -16.43kJ=∆H2 NH3 (aq) + HCl(aq) -1.47kJ=∆H3 NH4Cl(s) ∆H1 + ∆H2 + ∆H3 = ∆H4 + ∆Hf - 40.3kJ + -16.43kJ + -1.47kJ = +0.525kJ + ∆Hf =>∆Hf = -58.865kJ. Practice theoretical examples:
39 1. Using an energy level diagram calculate the ∆Hs of ammonium chloride crystals given that.
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Using an energy level diagram calculate the ∆Hs of ammonium chloride crystals given that. ∆Hf of NH3 (aq) = -80.54kJ mole-1 ∆Hf of HCl (aq) = -164.46kJ mole-1 ∆Hf of NH4Cl (aq) = -261.7483kJ mole-1 ∆Hs of NH4Cl (aq) = -16.8517kJ mole-1 N2(g) + 1½ H2(g) + ½ Cl2 ∆Hf=-261.7483kJ NH4Cl(s) + aq x=∆Hs (aq) (aq) - 80.54kJ=∆H1 -164.46kJ=∆H2 NH3 (aq) + HCl(aq) 16.8517kJ=∆H3 NH4Cl(s) ∆H1 + ∆H2 + ∆H3 = ∆H4 + ∆Hf - 80.54kJ + -164.46kJ + -16.8517kJ = -261.7483kJ + ∆Hf =>∆Hf = -33.6kJmole-1. Study the energy cycle diagram below and use it to: (a)Identify the energy changes ∆H1 ∆H2 ∆H3 ∆H4 ∆H5 ∆H6
40 ∆H1 - enthalpy/heat of formation of sodium chloride (∆Hf) ∆H2 -enthalpy/heat of atomization of sodium (∆Hat) ∆H3 -enthalpy/heat of ionization/ionization energy of sodium (∆H i) ∆H4 -enthalpy/heat of atomization of chlorine (∆Hat) ∆H5 -enthalpy/heat of electron affinity of chlorine (∆He) ∆H6 enthalpy/heat of lattice/Lattice energy of sodium chloride(∆H l) (b) Calculate ∆H1 given that ∆H2 =+108kJ , ∆H3=+500kJ, ∆H4 =+121kJ ,∆H5 =-364kJ and ∆H6 =-766kJ Working: ∆H1 =∆H2 +∆H3 +∆H4 +∆H5 +∆H6 Substituting: ∆H1= +108kJ + +500kJ + +121kJ +-364kJ + -766kJ ∆H1= -401kJmole-1 (c) Given the that: (i) Ionization energy of sodium = + 500kJmole-1 (ii)∆Hat of sodium = + 110kJmole-1 (iii) Electron affinity of chlorine = - 363kJmole-1 (iv)∆Hat of chlorine = + 120kJmole-1 (v) ∆Hf of sodium chloride = -411kJ , calculate the lattice energy of sodium chloride using an energy cycle diagram. 41 Working: Applying Hess law then: ∆Hf =∆Ha +∆Hi +∆Ha +∆He +∆Hl Substituting: -411= +108kJ + +500kJ + +121kJ +-364kJ + x -411 + -108kJ + -500kJ + -121kJ + +364kJ = x x= -776kJmole-1
2
20.0.0 REACTION RATES AND REVERSIBLE REACTIONS (15 LESSONS) A.THE RATE OF CHEMICAL REACTION (CHEMICAL KINETICS) 1.Introduction The rate of a chemical reaction is the time taken for a given mass/amount of products to be formed.
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∆Hf of NH3 (aq) = -80.54kJ mole-1 ∆Hf of HCl (aq) = -164.46kJ mole-1 ∆Hf of NH4Cl (aq) = -261.7483kJ mole-1 ∆Hs of NH4Cl (aq) = -16.8517kJ mole-1 N2(g) + 1½ H2(g) + ½ Cl2 ∆Hf=-261.7483kJ NH4Cl(s) + aq x=∆Hs (aq) (aq) - 80.54kJ=∆H1 -164.46kJ=∆H2 NH3 (aq) + HCl(aq) 16.8517kJ=∆H3 NH4Cl(s) ∆H1 + ∆H2 + ∆H3 = ∆H4 + ∆Hf - 80.54kJ + -164.46kJ + -16.8517kJ = -261.7483kJ + ∆Hf =>∆Hf = -33.6kJmole-1. Study the energy cycle diagram below and use it to: (a)Identify the energy changes ∆H1 ∆H2 ∆H3 ∆H4 ∆H5 ∆H6
40 ∆H1 - enthalpy/heat of formation of sodium chloride (∆Hf) ∆H2 -enthalpy/heat of atomization of sodium (∆Hat) ∆H3 -enthalpy/heat of ionization/ionization energy of sodium (∆H i) ∆H4 -enthalpy/heat of atomization of chlorine (∆Hat) ∆H5 -enthalpy/heat of electron affinity of chlorine (∆He) ∆H6 enthalpy/heat of lattice/Lattice energy of sodium chloride(∆H l) (b) Calculate ∆H1 given that ∆H2 =+108kJ , ∆H3=+500kJ, ∆H4 =+121kJ ,∆H5 =-364kJ and ∆H6 =-766kJ Working: ∆H1 =∆H2 +∆H3 +∆H4 +∆H5 +∆H6 Substituting: ∆H1= +108kJ + +500kJ + +121kJ +-364kJ + -766kJ ∆H1= -401kJmole-1 (c) Given the that: (i) Ionization energy of sodium = + 500kJmole-1 (ii)∆Hat of sodium = + 110kJmole-1 (iii) Electron affinity of chlorine = - 363kJmole-1 (iv)∆Hat of chlorine = + 120kJmole-1 (v) ∆Hf of sodium chloride = -411kJ , calculate the lattice energy of sodium chloride using an energy cycle diagram. 41 Working: Applying Hess law then: ∆Hf =∆Ha +∆Hi +∆Ha +∆He +∆Hl Substituting: -411= +108kJ + +500kJ + +121kJ +-364kJ + x -411 + -108kJ + -500kJ + -121kJ + +364kJ = x x= -776kJmole-1
2
20.0.0 REACTION RATES AND REVERSIBLE REACTIONS (15 LESSONS) A.THE RATE OF CHEMICAL REACTION (CHEMICAL KINETICS) 1.Introduction The rate of a chemical reaction is the time taken for a given mass/amount of products to be formed. The rate of a chemical reaction is also the time taken for a given mass/amount of reactant to be consumed /used up.
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Study the energy cycle diagram below and use it to: (a)Identify the energy changes ∆H1 ∆H2 ∆H3 ∆H4 ∆H5 ∆H6
40 ∆H1 - enthalpy/heat of formation of sodium chloride (∆Hf) ∆H2 -enthalpy/heat of atomization of sodium (∆Hat) ∆H3 -enthalpy/heat of ionization/ionization energy of sodium (∆H i) ∆H4 -enthalpy/heat of atomization of chlorine (∆Hat) ∆H5 -enthalpy/heat of electron affinity of chlorine (∆He) ∆H6 enthalpy/heat of lattice/Lattice energy of sodium chloride(∆H l) (b) Calculate ∆H1 given that ∆H2 =+108kJ , ∆H3=+500kJ, ∆H4 =+121kJ ,∆H5 =-364kJ and ∆H6 =-766kJ Working: ∆H1 =∆H2 +∆H3 +∆H4 +∆H5 +∆H6 Substituting: ∆H1= +108kJ + +500kJ + +121kJ +-364kJ + -766kJ ∆H1= -401kJmole-1 (c) Given the that: (i) Ionization energy of sodium = + 500kJmole-1 (ii)∆Hat of sodium = + 110kJmole-1 (iii) Electron affinity of chlorine = - 363kJmole-1 (iv)∆Hat of chlorine = + 120kJmole-1 (v) ∆Hf of sodium chloride = -411kJ , calculate the lattice energy of sodium chloride using an energy cycle diagram. 41 Working: Applying Hess law then: ∆Hf =∆Ha +∆Hi +∆Ha +∆He +∆Hl Substituting: -411= +108kJ + +500kJ + +121kJ +-364kJ + x -411 + -108kJ + -500kJ + -121kJ + +364kJ = x x= -776kJmole-1
2
20.0.0 REACTION RATES AND REVERSIBLE REACTIONS (15 LESSONS) A.THE RATE OF CHEMICAL REACTION (CHEMICAL KINETICS) 1.Introduction The rate of a chemical reaction is the time taken for a given mass/amount of products to be formed. The rate of a chemical reaction is also the time taken for a given mass/amount of reactant to be consumed /used up. Some reactions are too slow to be determined.
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41 Working: Applying Hess law then: ∆Hf =∆Ha +∆Hi +∆Ha +∆He +∆Hl Substituting: -411= +108kJ + +500kJ + +121kJ +-364kJ + x -411 + -108kJ + -500kJ + -121kJ + +364kJ = x x= -776kJmole-1
2
20.0.0 REACTION RATES AND REVERSIBLE REACTIONS (15 LESSONS) A.THE RATE OF CHEMICAL REACTION (CHEMICAL KINETICS) 1.Introduction The rate of a chemical reaction is the time taken for a given mass/amount of products to be formed. The rate of a chemical reaction is also the time taken for a given mass/amount of reactant to be consumed /used up. Some reactions are too slow to be determined. e.g rusting ,decomposition of hydrogen peroxide and weathering. Some reactions are too fast and instantaneous e.g. neutralization of acid and bases/alkalis in aqueous solution and double decomposition/precipitation. Other reactions are explosive and very risky to carry out safely e.g. reaction of potassium with water and sodium with dilute acids. The study of the rate of chemical reaction is useful in knowing the factors that influence the reaction so that efficiency and profitability is maximized in industries. Theories of rates of reaction. The rate of a chemical reaction is defined as the rate of change of concentration/amount of reactants in unit time. It is also the rate of formation of given concentration of products in unit time. i.e. Rate of reaction = Change in concentration/amount of reactants Time taken for the change to occur Rate of reaction = Change in concentration/amount of products formed Time taken for the products to form For the above, therefore the rate of a chemical reaction is rate of decreasing reactants to form an increasing product. The SI unit of time is second(s) but minutes and hours are also used. (a)The collision theory The collision theory is an application of the Kinetic Theory of matter which assumes matter is made up of small/tiny/minute particles like ions atoms and molecules. The collision theory proposes that (i)for a reaction to occur, reacting particles must collide. (ii)not all collisions between reacting particles are successful in a reaction.
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(a)The collision theory The collision theory is an application of the Kinetic Theory of matter which assumes matter is made up of small/tiny/minute particles like ions atoms and molecules. The collision theory proposes that (i)for a reaction to occur, reacting particles must collide. (ii)not all collisions between reacting particles are successful in a reaction. Collisions that initiate a chemical reaction are called successful / fruitful/ effective collisions (iii)the speed at which particles collide is called collision frequency. The higher the collision frequency the higher the chances of successful / fruitful/ effective collisions to form products. (iv)the higher the chances of successful collisions, the faster the reaction. (v)the average distance between solid particles from one another is too big for them to meet and collide successfully. (vi)dissolving substances in a solvent ,make the solvent a medium for the reaction to take place. The solute particle distance is reduced as the particle ions are free to move in the solvent medium. (vii)successful collisions take place if the particles colliding have the required energy and right orientation which increases their vibration and intensity of successful / fruitful/ effective collisions to form products. (b)The Activation Energy(Ea) theory
The Enthalpy of activation(∆Ha) /Activation Energy(Ea) is the minimum amount of energy which the reactants must overcome before they react. Activation Energy(Ea) is usually required /needed in bond breaking of the reacting particles. Bond breaking is an endothermic process that require an energy input. The higher the bond energy the slower the reaction to start of. Activation energy does not influence whether a reaction is exothermic or endothermic. The energy level diagrams below shows the activation energy for exothermic and endothermic processes/reactions. Energy level diagram showing the activation energy for exothermic processes /reactions. Activated complex Energy level diagram showing the activation energy for endothermic processes /reactions. Activated complex A A B B Energy kJ Reaction path/coordinate/path Ea A-A B-B A-B A-B A A B B Energy kJ Reaction path/coordinate/path Ea A-A B-B A-B A-B ∆Hr
The activated complex is a mixture of many intermediate possible products which may not exist under normal physical conditions ,but can theoretically exist. Exothermic reaction proceeds without further heating /external energy because it generates its own energy/heat to overcome activation energy.
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Activated complex Energy level diagram showing the activation energy for endothermic processes /reactions. Activated complex A A B B Energy kJ Reaction path/coordinate/path Ea A-A B-B A-B A-B A A B B Energy kJ Reaction path/coordinate/path Ea A-A B-B A-B A-B ∆Hr
The activated complex is a mixture of many intermediate possible products which may not exist under normal physical conditions ,but can theoretically exist. Exothermic reaction proceeds without further heating /external energy because it generates its own energy/heat to overcome activation energy. Endothermic reaction cannot proceed without further heating /external energy because it does not generates its own energy/heat to overcome activation energy. It generally therefore requires continuous supply of more energy/heat to sustain it to completion. 3. Measuring the rate of a chemical reaction. The rate of a chemical reaction can be measure as: (i)Volume of a gas in unit time; - if reaction is producing a gas as one of the products. - if reaction is using a gas as one reactants (ii)Change in mass of reactants/products for solid products/reactants in unit time. (iii)formation of a given mass of precipitate in unit time (iv)a certain mass of reactants to completely form products/diminish. Reactants may be homogenous or heterogenous. -Homogenous reactions involve reactants in the same phase/state e.g. solid-solid,gas-gas,liquid-liquid. -Heterogenous reactions involve reactants in the different phase/state e.g. solid-liquid,gas-liquid,solid-gas. 4. Factors influencing/altering/affecting/determining rate of reaction The following factors alter/influence/affect/determine the rate of a chemical reaction: (a)Concentration (b)Pressure (c) Temperature (d)Surface area
(e)Catalyst a) Influence of concentration on rate of reaction The higher the concentration, the higher the rate of a chemical reaction. An increase in concentration of the reactants reduces the distance between the reacting particles increasing their collision frequency to form products. Practically an increase in concentration reduces the time taken for the reaction to take place. Practical determination of effect of concentration on reaction rate Method 1(a) Reaction of sodium thisulphate with dilute hydrochloric acid Procedure: Measure 20cm3 of 0.05M sodium thisulphate into a 50cm3 glass beaker.
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An increase in concentration of the reactants reduces the distance between the reacting particles increasing their collision frequency to form products. Practically an increase in concentration reduces the time taken for the reaction to take place. Practical determination of effect of concentration on reaction rate Method 1(a) Reaction of sodium thisulphate with dilute hydrochloric acid Procedure: Measure 20cm3 of 0.05M sodium thisulphate into a 50cm3 glass beaker. Place the beaker on a white piece of filter paper with ink mark ‘X’ on it. Measure 20cm3 of 0.1M hydrochloric acid solution using a 50cm3 measuring cylinder. Put the acid into the beaker containing sodium thisulphate. Immediately start off the stop watch/clock. Determine the time taken for the ink mark ‘X’ to become invisible /obscured when viewed from above. Repeat the procedure by measuring different volumes of the acid and adding the volumes of the distilled water to complete table 1. Sample results:Table 1. Volume of acid(cm3) Volume of water(cm3) Volume of sodium thiosulphate(cm3) Time taken for mark ‘X’ to be invisible/obscured(seconds) Reciprocal of time 1 t 20.0 0.0 20.0 20.0 5.0 x 10-2 18.0 2.0 20.0 23.0 4.35 x 10-2 16.0 4.0 20.0 27.0 3.7 x 10-2 14.0 6.0 20.0 32.0 3.13 x 10-2 12.0 8.0 20.0 42.0 2.38 x 10-2 10.0 10.0 20.0 56.0 1.78 x 10-2 For most examining bodies/councils/boards the above results score for: (a) complete table as evidence for all the practical work done and completed. (b) (i)Consistent use of a decimal point on time as evidence of understanding/knowledge of the degree of accuracy of stop watches/clock.
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Sample results:Table 1. Volume of acid(cm3) Volume of water(cm3) Volume of sodium thiosulphate(cm3) Time taken for mark ‘X’ to be invisible/obscured(seconds) Reciprocal of time 1 t 20.0 0.0 20.0 20.0 5.0 x 10-2 18.0 2.0 20.0 23.0 4.35 x 10-2 16.0 4.0 20.0 27.0 3.7 x 10-2 14.0 6.0 20.0 32.0 3.13 x 10-2 12.0 8.0 20.0 42.0 2.38 x 10-2 10.0 10.0 20.0 56.0 1.78 x 10-2 For most examining bodies/councils/boards the above results score for: (a) complete table as evidence for all the practical work done and completed. (b) (i)Consistent use of a decimal point on time as evidence of understanding/knowledge of the degree of accuracy of stop watches/clock. (ii)Consistent use of a minimum of four decimal points on inverse/reciprocal of time as evidence of understanding/knowledge of the degree of accuracy of scientific calculator. (c) accuracy against a school value based on candidate’s teachers-results submitted. (d) correct trend (time increase as more water is added/acid is diluted) in conformity with expected theoretical results. Sample questions 1. On separate graph papers plot a graph of: (i)volume of acid used(x-axis) against time. Label this graph I (ii) volume of acid used(x-axis) against 1/t. Label this graph II 2. Explain the shape of graph I Diluting/adding water is causes a decrease in concentration. Decrease in concentration reduces the rate of reaction by increasing the time taken for reacting particle to collide to form products.
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Label this graph II 2. Explain the shape of graph I Diluting/adding water is causes a decrease in concentration. Decrease in concentration reduces the rate of reaction by increasing the time taken for reacting particle to collide to form products. Sketch sample Graph I Time (seconds)
Sketch sample Graph II 3.From graph II ,determine the time taken for the cross to be obscured/invisible when the volume of the acid is: (i) 13cm3 From a correctly plotted graph 1/t at 13cm3 on the graph => 2.75 x 10-2 t = 1 / 2.75 x 10-2 = 36.3636 seconds (ii) 15cm3 From a correctly plotted graph 1/t at 15cm3 on the graph => 3.35 x 10-2 t = 1 / 3.35 x 10-2 = 29.8507 seconds (iii) 15cm3 From a correctly plotted graph 1/t Sec-1 x 10-2 Volume of acid (cm3) Volume of acid(cm3) Volume of acid(cm3)
1/t at 17cm3 on the graph => 4.0 x 10-2 t = 1 / 4.0 x 10-2 = 25.0 seconds (iv) 19cm3 From a correctly plotted graph 1/t at 19cm3 on the graph => 4.65 x 10-2 t = 1 / 4.65 x 10-2 = 21.5054 seconds 4.From graph II ,determine the volume of the acid used if the time taken for the cross to be obscured/invisible is: (i)25 seconds 1/t => 1/25 = 4.0 x 10-2 Reading from a correctly plotted graph; 4.0 x 10-2 correspond to 17.0 cm3 (ii)30 seconds 1/t => 1/30 = 3.33 x 10-2 Reading from a correctly plotted graph; 3.33 x 10-2 correspond to 14.7 cm3 (iii)40 seconds 1/t => 1/40 = 2.5 x 10-2 Reading from a correctly plotted graph; 2.5 x 10-2 correspond to 12.3 cm3 4.
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Explain the shape of graph I Diluting/adding water is causes a decrease in concentration. Decrease in concentration reduces the rate of reaction by increasing the time taken for reacting particle to collide to form products. Sketch sample Graph I Time (seconds)
Sketch sample Graph II 3.From graph II ,determine the time taken for the cross to be obscured/invisible when the volume of the acid is: (i) 13cm3 From a correctly plotted graph 1/t at 13cm3 on the graph => 2.75 x 10-2 t = 1 / 2.75 x 10-2 = 36.3636 seconds (ii) 15cm3 From a correctly plotted graph 1/t at 15cm3 on the graph => 3.35 x 10-2 t = 1 / 3.35 x 10-2 = 29.8507 seconds (iii) 15cm3 From a correctly plotted graph 1/t Sec-1 x 10-2 Volume of acid (cm3) Volume of acid(cm3) Volume of acid(cm3)
1/t at 17cm3 on the graph => 4.0 x 10-2 t = 1 / 4.0 x 10-2 = 25.0 seconds (iv) 19cm3 From a correctly plotted graph 1/t at 19cm3 on the graph => 4.65 x 10-2 t = 1 / 4.65 x 10-2 = 21.5054 seconds 4.From graph II ,determine the volume of the acid used if the time taken for the cross to be obscured/invisible is: (i)25 seconds 1/t => 1/25 = 4.0 x 10-2 Reading from a correctly plotted graph; 4.0 x 10-2 correspond to 17.0 cm3 (ii)30 seconds 1/t => 1/30 = 3.33 x 10-2 Reading from a correctly plotted graph; 3.33 x 10-2 correspond to 14.7 cm3 (iii)40 seconds 1/t => 1/40 = 2.5 x 10-2 Reading from a correctly plotted graph; 2.5 x 10-2 correspond to 12.3 cm3 4. Write the equation for the reaction taking place Na2S2O3 (aq) + 2HCl(aq) -> 2NaCl (aq)+ SO2 (g) + S(s) + H2O(l) Ionically: S2O32- (aq) + 2H+ (aq) -> SO2 (g) + S(s) + H2O(l) 5.Name the yellow precipitate Colloidal sulphur Method 1(b)
Reaction of sodium thisulphate with dilute hydrochloric acid You are provided with 2.0M Hydrochloric acid 0.4M sodium thiosulphate solution Procedure: Measure 10cm3 of sodium thisulphate into a 50cm3 glass beaker.
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Decrease in concentration reduces the rate of reaction by increasing the time taken for reacting particle to collide to form products. Sketch sample Graph I Time (seconds)
Sketch sample Graph II 3.From graph II ,determine the time taken for the cross to be obscured/invisible when the volume of the acid is: (i) 13cm3 From a correctly plotted graph 1/t at 13cm3 on the graph => 2.75 x 10-2 t = 1 / 2.75 x 10-2 = 36.3636 seconds (ii) 15cm3 From a correctly plotted graph 1/t at 15cm3 on the graph => 3.35 x 10-2 t = 1 / 3.35 x 10-2 = 29.8507 seconds (iii) 15cm3 From a correctly plotted graph 1/t Sec-1 x 10-2 Volume of acid (cm3) Volume of acid(cm3) Volume of acid(cm3)
1/t at 17cm3 on the graph => 4.0 x 10-2 t = 1 / 4.0 x 10-2 = 25.0 seconds (iv) 19cm3 From a correctly plotted graph 1/t at 19cm3 on the graph => 4.65 x 10-2 t = 1 / 4.65 x 10-2 = 21.5054 seconds 4.From graph II ,determine the volume of the acid used if the time taken for the cross to be obscured/invisible is: (i)25 seconds 1/t => 1/25 = 4.0 x 10-2 Reading from a correctly plotted graph; 4.0 x 10-2 correspond to 17.0 cm3 (ii)30 seconds 1/t => 1/30 = 3.33 x 10-2 Reading from a correctly plotted graph; 3.33 x 10-2 correspond to 14.7 cm3 (iii)40 seconds 1/t => 1/40 = 2.5 x 10-2 Reading from a correctly plotted graph; 2.5 x 10-2 correspond to 12.3 cm3 4. Write the equation for the reaction taking place Na2S2O3 (aq) + 2HCl(aq) -> 2NaCl (aq)+ SO2 (g) + S(s) + H2O(l) Ionically: S2O32- (aq) + 2H+ (aq) -> SO2 (g) + S(s) + H2O(l) 5.Name the yellow precipitate Colloidal sulphur Method 1(b)
Reaction of sodium thisulphate with dilute hydrochloric acid You are provided with 2.0M Hydrochloric acid 0.4M sodium thiosulphate solution Procedure: Measure 10cm3 of sodium thisulphate into a 50cm3 glass beaker. Place the beaker on a white piece of filter paper with ink mark ‘X’ on it.
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Sketch sample Graph I Time (seconds)
Sketch sample Graph II 3.From graph II ,determine the time taken for the cross to be obscured/invisible when the volume of the acid is: (i) 13cm3 From a correctly plotted graph 1/t at 13cm3 on the graph => 2.75 x 10-2 t = 1 / 2.75 x 10-2 = 36.3636 seconds (ii) 15cm3 From a correctly plotted graph 1/t at 15cm3 on the graph => 3.35 x 10-2 t = 1 / 3.35 x 10-2 = 29.8507 seconds (iii) 15cm3 From a correctly plotted graph 1/t Sec-1 x 10-2 Volume of acid (cm3) Volume of acid(cm3) Volume of acid(cm3)
1/t at 17cm3 on the graph => 4.0 x 10-2 t = 1 / 4.0 x 10-2 = 25.0 seconds (iv) 19cm3 From a correctly plotted graph 1/t at 19cm3 on the graph => 4.65 x 10-2 t = 1 / 4.65 x 10-2 = 21.5054 seconds 4.From graph II ,determine the volume of the acid used if the time taken for the cross to be obscured/invisible is: (i)25 seconds 1/t => 1/25 = 4.0 x 10-2 Reading from a correctly plotted graph; 4.0 x 10-2 correspond to 17.0 cm3 (ii)30 seconds 1/t => 1/30 = 3.33 x 10-2 Reading from a correctly plotted graph; 3.33 x 10-2 correspond to 14.7 cm3 (iii)40 seconds 1/t => 1/40 = 2.5 x 10-2 Reading from a correctly plotted graph; 2.5 x 10-2 correspond to 12.3 cm3 4. Write the equation for the reaction taking place Na2S2O3 (aq) + 2HCl(aq) -> 2NaCl (aq)+ SO2 (g) + S(s) + H2O(l) Ionically: S2O32- (aq) + 2H+ (aq) -> SO2 (g) + S(s) + H2O(l) 5.Name the yellow precipitate Colloidal sulphur Method 1(b)
Reaction of sodium thisulphate with dilute hydrochloric acid You are provided with 2.0M Hydrochloric acid 0.4M sodium thiosulphate solution Procedure: Measure 10cm3 of sodium thisulphate into a 50cm3 glass beaker. Place the beaker on a white piece of filter paper with ink mark ‘X’ on it. Add 5.0cm3 of hydrochloric acid solution using a 10cm3 measuring cylinder into the beaker containing sodium thisulphate.
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Write the equation for the reaction taking place Na2S2O3 (aq) + 2HCl(aq) -> 2NaCl (aq)+ SO2 (g) + S(s) + H2O(l) Ionically: S2O32- (aq) + 2H+ (aq) -> SO2 (g) + S(s) + H2O(l) 5.Name the yellow precipitate Colloidal sulphur Method 1(b)
Reaction of sodium thisulphate with dilute hydrochloric acid You are provided with 2.0M Hydrochloric acid 0.4M sodium thiosulphate solution Procedure: Measure 10cm3 of sodium thisulphate into a 50cm3 glass beaker. Place the beaker on a white piece of filter paper with ink mark ‘X’ on it. Add 5.0cm3 of hydrochloric acid solution using a 10cm3 measuring cylinder into the beaker containing sodium thisulphate. Immediately start off the stop watch/clock. Determine the time taken for the ink mark ‘X’ to become invisible /obscured when viewed from above. Repeat the procedure by measuring different volumes of the thiosulphate and adding the volumes of the distilled water to complete table 1. Sample results:Table 1.
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Determine the time taken for the ink mark ‘X’ to become invisible /obscured when viewed from above. Repeat the procedure by measuring different volumes of the thiosulphate and adding the volumes of the distilled water to complete table 1. Sample results:Table 1. Volume of acid(cm3) Volume of water (cm3) Volume of sodium thiosulphate (cm3) Concentation of sodium thisulphate in molesdm-3 Time(T) taken for mark ‘X’ to be invisible/ obscured(seconds) T-1 5.0 0.0 25.0 0.4 20.0 5.0 x 10-2 5.0 5.0 20.0 0.32 23.0 4.35 x 10-2 5.0 10.0 15.0 0.24 27.0 3.7 x 10-2 5.0 15.0 10.0 0.16 32.0 3.13 x 10-2 Note concentration of diluted solution is got: C1V1=C2V2 => 0.4 x 25 = C2x 25 =0.4M C1V1=C2V2 => 0.4 x 20 = C2x 25 =0.32M C1V1=C2V2 => 0.4 x 15 = C2x 25 =0.24M C1V1=C2V2 => 0.4 x 10 = C2x 25 =0.16M Sample questions 1. On separate graph papers plot a graph of: (i)Concentration of sodium thiosulphate against time. Label this graph I
(ii)Concentration of sodium thiosulphate against against T-1.Label this graph II 2. Explain the shape of graph I Diluting/adding water causes a decrease in concentration. Decrease in concentration reduces the rate of reaction by increasing the time taken for reacting particle to collide to form products. From graph II Determine the time taken if (i)12cm3 of sodium thisulphate is diluted with 13cm3 of water.
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Explain the shape of graph I Diluting/adding water causes a decrease in concentration. Decrease in concentration reduces the rate of reaction by increasing the time taken for reacting particle to collide to form products. From graph II Determine the time taken if (i)12cm3 of sodium thisulphate is diluted with 13cm3 of water. At 12cm3 concentration of sodium thisulphate = C1V1=C2V2 => 0.4 x 1 2 = C2x 25 =0.192M From correct graph at concentration 0.192M => 2.4 x10-2 I/t = 2.4 x10-2 t = 41.6667seconds (ii)22cm3 of sodium thisulphate is diluted with 3cm3 of water. At 22cm3 concentration of sodium thisulphate = C1V1=C2V2 => 0.4 x 22 = C2x 25 =0.352M From correct graph at concentration 0.352M => 3.6 x10-2 I/t = 3.6 x10-2 t = 27.7778seconds Determine the volume of water and sodium thiosulphate if T-1 is 3.0 x10-1 From correct graph at T-1 = 3.0 x10-1 => concentration = 0.65 M = C1V1=C2V2 => 0.4 x 25 = 0.65 M x V2 = 15.3846cm3 Volume of water = 25 - 15.3846cm3 = 9.6154cm3 Determine the concentration of hydrochloric acid if 12cm3 of sodium thiosulphate and 13cm3 of water was used.
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At 22cm3 concentration of sodium thisulphate = C1V1=C2V2 => 0.4 x 22 = C2x 25 =0.352M From correct graph at concentration 0.352M => 3.6 x10-2 I/t = 3.6 x10-2 t = 27.7778seconds Determine the volume of water and sodium thiosulphate if T-1 is 3.0 x10-1 From correct graph at T-1 = 3.0 x10-1 => concentration = 0.65 M = C1V1=C2V2 => 0.4 x 25 = 0.65 M x V2 = 15.3846cm3 Volume of water = 25 - 15.3846cm3 = 9.6154cm3 Determine the concentration of hydrochloric acid if 12cm3 of sodium thiosulphate and 13cm3 of water was used. At 12cm3 concentration of sodium thisulphate = C1V1=C2V2 => 0.4 x 1 2 = C2x 25 =0.192M Mole ratio Na2S2 O3 :HCl =1:2 Moles of Na2S2 O3 = 0.192M x 12 => 2.304 x 10-3 moles 1000 Mole ratio HCl =2.304 x 10-1 moles = 1.152 x 10-3 moles 2 Molarity o f HCl = 1.152 x 10-3 moles x 1000 = 0.2304M 5.0
Method 2 Reaction of Magnesium with dilute hydrochloric acid Procedure Scub 10centimeter length of magnesium ribbon with sand paper/steel wool. Measure 40cm3 of 0.5M dilute hydrochloric acid into a flask .Fill a graduated gas jar with water and invert it into a trough. Stopper the flask and set up the apparatus to collect the gas produced as in the set up below: Carefully remove the stopper, carefully put the magnesium ribbon into the flask . cork tightly. Add the acid into the flask. Connect the delivery tube into the gas jar.
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cork tightly. Add the acid into the flask. Connect the delivery tube into the gas jar. Immediately start off the stop watch and determine the volume of the gas produced after every 30 seconds to complete table II below. Sample results: Table II Time(seconds) 0 30 60 90 120 150 180 210 240 Volume of gas produced(cm3) 0.0 20.0 40.0 60.0 80.0 90.0 95.0 96.0 96.0 Sample practice questions 1.Plot a graph of volume of gas produced (y-axis) against time Magnesium ribbon Hydrochloric acid Graduated gas jar Hydrogen gas
2.Explain the shape of the graph. The rate of reaction is faster when the concentration of the acid is high . As time goes on, the concentration of the acid decreases and therefore less gas is produced. When all the acid has reacted, no more gas is produced after 210 seconds and the graph flattens. 3.Calculate the rate of reaction at 120 seconds From a tangent at 120 seconds rate of reaction = Change in volume of gas Change in time => From the tangent at 120seconds V2 - V1 = 96-84 = 12 = 0.2cm3sec-1 T2 - T1 150-90 60 4. Write an ionic equation for the reaction taking place. Mg2+(s) + 2H+(aq) -> Mg2+(aq) + H2 (g) 5. On the same axis sketch then explain the curve that would be obtained if: (i) 0.1 M hydrochloric acid is used –Label this curve I (ii)1.0 M hydrochloric acid is used –Label this curve II
Observation: Curve I is to the right Curve II is to the left Explanation A decrease in concentration shift the rate of reaction graph to the right as more time is taken for completion of the reaction. An increase in concentration shift the rate of reaction graph to the left as less time is taken for completion of the reaction. Both graphs flatten after some time indicating the completion of the reaction. b)Influence of pressure on rate of reaction Pressure affects only gaseous reactants. An increase in pressure reduces the volume(Boyles law) in which the particles are contained.
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Both graphs flatten after some time indicating the completion of the reaction. b)Influence of pressure on rate of reaction Pressure affects only gaseous reactants. An increase in pressure reduces the volume(Boyles law) in which the particles are contained. Decrease in volume of the container bring the reacting particles closer to each other which increases their chances of effective/successful/fruitful collision to form products. An increase in pressure therefore increases the rate of reaction by reducing the time for reacting particles of gases to react. At industrial level, the following are some reactions that are affected by pressure: (a)Haber process for manufacture of ammonia N2(g) + 3H2(g) -> 2NH3(g) (b)Contact process for manufacture of sulphuric(VI)acid 2SO2(g) + O2(g) -> 2SO3(g) (c)Ostwalds process for the manufacture of nitric(V)acid 4NH3(g) + 5O2(g) -> 4NO (g) + 6H2O (l) The influence of pressure on reaction rate is not felt in solids and liquids. This is because the solid and liquid particles have fixed positions in their strong bonds and therefore no degree of freedom (Kinetic Theory of matter) c)Influence of temperature on rate of reaction
An increase in temperature increases the kinetic energy of the reacting particles by increasing their collision frequency. Increase in temperature increases the particles which can overcome the activation energy (Ea). A 10oC rise in temperature doubles the rate of reaction by reducing the time taken for the reaction to complete by a half. Practical determination of effect of Temperature on reaction rate Method 1 Reaction of sodium thisulphate with dilute hydrochloric acid Procedure: Measure 20cm3 of 0.05M sodium thisulphate into a 50cm3 glass beaker. Place the beaker on a white piece of filter paper with ink mark ‘X’ on it. Determine and record its temperature as room temperature in table 2 below. Measure 20cm3 of 0.1M hydrochloric acid solution using a 50cm3 measuring cylinder. Put the acid into the beaker containing sodium thisulphate. Immediately start off the stop watch/clock. Determine the time taken for the ink mark ‘X’ to become invisible /obscured when viewed from above.
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Put the acid into the beaker containing sodium thisulphate. Immediately start off the stop watch/clock. Determine the time taken for the ink mark ‘X’ to become invisible /obscured when viewed from above. Measure another 20cm3 separate portion of the thisulphate into a beaker, heat the solution to 30oC. Add the acid into the beaker and repeat the procedure above. Complete table 2 below using different temperatures of the thiosulphate. Sample results:Table 2. Temperature of Na2S2O3 Room temperature 30 40 50 60 Time taken for mark X to be obscured /invisible (seconds) 50.0 40.0 20.0 15.0 10.0 Reciprocal of time(1/t) 0.02 0.025 0.05 0.0667 0.1 Sample practice questions 1. Plot a graph of temperature(x-axis) against 1/t
2(a)From your graph determine the temperature at which: (i)1/t is ; I. 0.03 Reading directly from a correctly plotted graph = 32.25 oC II. 0.07 Reading directly from a correctly plotted graph = 48.0 oC (ii) t is; I. 30 seconds 30 seconds => 1/t =1/30 =0.033 Reading directly from a correctly plotted graph 0.033 => 33.5 oC II. 45 seconds 45 seconds => 1/t =1/45 =0.022 Reading directly from a correctly plotted graph 0.022 => 29.0 oC III.
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0.07 Reading directly from a correctly plotted graph = 48.0 oC (ii) t is; I. 30 seconds 30 seconds => 1/t =1/30 =0.033 Reading directly from a correctly plotted graph 0.033 => 33.5 oC II. 45 seconds 45 seconds => 1/t =1/45 =0.022 Reading directly from a correctly plotted graph 0.022 => 29.0 oC III. 25 seconds 25 seconds => 1/t =1/25 =0.04 Reading directly from a correctly plotted graph 0.04 => 36.0 oC (b) From your graph determine the time taken for the cross to become invisible at: (i) 57.5 oC Reading directly from a correctly plotted graph at 57.5 oC= 0.094 =>1/t = 0.094 t= 1/0.094 => 10.6383 seconds
(ii) 45 oC Reading directly from a correctly plotted graph at 45 oC = 0.062 =>1/t = 0.062 t= 1/0.094 => 16.1290 seconds (iii) 35 oC Reading directly from a correctly plotted graph at 35 oC = 0.047 =>1/t = 0.047 t= 1/0.047 => 21.2766 seconds Method 2 Reaction of Magnesium with dilute hydrochloric acid Procedure Scub 5centimeter length of magnesium ribbon with sand paper/steel wool. Cut the piece into five equal one centimeter smaller pieces. Measure 20cm3 of 1.0M dilute hydrochloric acid into a glass beaker . Put one piece of the magnesium ribbon into the acid, swirl. Immediately start off the stop watch/clock. Determine the time taken for the effervescence/fizzing/bubbling to stop when viewed from above. Record the time in table 2 at room temperature. Measure another 20cm3 portions of 1.0M dilute hydrochloric acid into a clean beaker. Heat separately one portion to 30oC, 40oC , 50oC and 60oC and adding 1cm length of the ribbon and determine the time taken for effervescence /fizzing /bubbling to stop when viewed from above .
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Record the time in table 2 at room temperature. Measure another 20cm3 portions of 1.0M dilute hydrochloric acid into a clean beaker. Heat separately one portion to 30oC, 40oC , 50oC and 60oC and adding 1cm length of the ribbon and determine the time taken for effervescence /fizzing /bubbling to stop when viewed from above . Record each time to complete table 2 below using different temperatures of the acid. Sample results:Table 1. Temperature of acid(oC) Room temperature 30 40 50 60 Time taken effervescence to stop (seconds) 80.0 50.0 21.0 13.5 10.0 Reciprocal of time(1/t) 0.0125 0.02 0.0476 0.0741 0.1 Sample practice questions
1. Plot a graph of temperature(x-axis) against 1/t 2.(a)Calculate the number of moles of magnesium used given that 1cm of magnesium has a mass of 1g.(Mg= 24.0) Moles = Mass of magnesium => 1.0 = 4.167 x 10 -2 moles Molar mass of Mg 24 (b)Calculate the number of moles of hydrochloric acid used Moles of acid = molarity x volume of acid 1000 => 1.0 x 20 = 2.0 x 10 -2 moles 1000 (c)Calculate the mass of magnesium that remain unreacted Mole ratio Mg: HCl = 1:2 Moles Mg = ½ moles HCl => ½ x 2.0 x 10 -2 moles = 1.0 x 10 -2 moles Mass of reacted Mg = moles x molar mass => 1.0 x 10 -2 moles x 24 = 0.24 g Mass of unreacted Mg = Original total mass - Mass of reacted Mg => 1.0 g – 0.24 = 0.76 g (b)Calculate the total volume of hydrogen gas produced during the above reactions.
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Sample results:Table 1. Temperature of acid(oC) Room temperature 30 40 50 60 Time taken effervescence to stop (seconds) 80.0 50.0 21.0 13.5 10.0 Reciprocal of time(1/t) 0.0125 0.02 0.0476 0.0741 0.1 Sample practice questions
1. Plot a graph of temperature(x-axis) against 1/t 2.(a)Calculate the number of moles of magnesium used given that 1cm of magnesium has a mass of 1g.(Mg= 24.0) Moles = Mass of magnesium => 1.0 = 4.167 x 10 -2 moles Molar mass of Mg 24 (b)Calculate the number of moles of hydrochloric acid used Moles of acid = molarity x volume of acid 1000 => 1.0 x 20 = 2.0 x 10 -2 moles 1000 (c)Calculate the mass of magnesium that remain unreacted Mole ratio Mg: HCl = 1:2 Moles Mg = ½ moles HCl => ½ x 2.0 x 10 -2 moles = 1.0 x 10 -2 moles Mass of reacted Mg = moles x molar mass => 1.0 x 10 -2 moles x 24 = 0.24 g Mass of unreacted Mg = Original total mass - Mass of reacted Mg => 1.0 g – 0.24 = 0.76 g (b)Calculate the total volume of hydrogen gas produced during the above reactions. 1/t Temperature(oC)
Mole ratio Mg : H2 = 1:1 Moles of Mg that reacted per experiment = moles H2 =1.0 x 10 -2 moles Volume of Hydrogen at s.t.p produced per experiment = moles x 24 dm3 => 1.0 x 10 -2 moles x 24 dm3 = 0.24dm3 Volume of Hydrogen at s.t.p produced in 5 experiments =0.24 dm3 x 5 = 1.2 dm3 3.(a)At what temperature was the time taken for magnesium to react equal to: (i)70seconds 70 seconds => 1/t =1/70 =0.01429 Reading directly from a correctly plotted graph 0.01429 => 28.0 oC (ii)40seconds 40 seconds => 1/t =1/40 =0.025 Reading directly from a correctly plotted graph 0.025 => 32.0 oC (b)What is the time taken for magnesium to react if the reaction was done at: (i) 55.0 oC Reading directly from a correctly plotted graph at 55.0 oC=> 1/t = 8.0 x 10-2 => t = 1/8.0 x 10-2 = 12.5 seconds (ii) 47.0 oC Reading directly from a correctly plotted graph at 47.0 oC=> 1/t = 6.0 x 10-2 => t = 1/6.0 x 10-2 = 16.6667 seconds (iii) 33.0 oC Reading directly from a correctly plotted graph at 33.0 oC=> 1/t = 2.7 x 10-2 => t = 1/2.7 x 10-2 = 37.037 seconds 4.
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Temperature of acid(oC) Room temperature 30 40 50 60 Time taken effervescence to stop (seconds) 80.0 50.0 21.0 13.5 10.0 Reciprocal of time(1/t) 0.0125 0.02 0.0476 0.0741 0.1 Sample practice questions
1. Plot a graph of temperature(x-axis) against 1/t 2.(a)Calculate the number of moles of magnesium used given that 1cm of magnesium has a mass of 1g.(Mg= 24.0) Moles = Mass of magnesium => 1.0 = 4.167 x 10 -2 moles Molar mass of Mg 24 (b)Calculate the number of moles of hydrochloric acid used Moles of acid = molarity x volume of acid 1000 => 1.0 x 20 = 2.0 x 10 -2 moles 1000 (c)Calculate the mass of magnesium that remain unreacted Mole ratio Mg: HCl = 1:2 Moles Mg = ½ moles HCl => ½ x 2.0 x 10 -2 moles = 1.0 x 10 -2 moles Mass of reacted Mg = moles x molar mass => 1.0 x 10 -2 moles x 24 = 0.24 g Mass of unreacted Mg = Original total mass - Mass of reacted Mg => 1.0 g – 0.24 = 0.76 g (b)Calculate the total volume of hydrogen gas produced during the above reactions. 1/t Temperature(oC)
Mole ratio Mg : H2 = 1:1 Moles of Mg that reacted per experiment = moles H2 =1.0 x 10 -2 moles Volume of Hydrogen at s.t.p produced per experiment = moles x 24 dm3 => 1.0 x 10 -2 moles x 24 dm3 = 0.24dm3 Volume of Hydrogen at s.t.p produced in 5 experiments =0.24 dm3 x 5 = 1.2 dm3 3.(a)At what temperature was the time taken for magnesium to react equal to: (i)70seconds 70 seconds => 1/t =1/70 =0.01429 Reading directly from a correctly plotted graph 0.01429 => 28.0 oC (ii)40seconds 40 seconds => 1/t =1/40 =0.025 Reading directly from a correctly plotted graph 0.025 => 32.0 oC (b)What is the time taken for magnesium to react if the reaction was done at: (i) 55.0 oC Reading directly from a correctly plotted graph at 55.0 oC=> 1/t = 8.0 x 10-2 => t = 1/8.0 x 10-2 = 12.5 seconds (ii) 47.0 oC Reading directly from a correctly plotted graph at 47.0 oC=> 1/t = 6.0 x 10-2 => t = 1/6.0 x 10-2 = 16.6667 seconds (iii) 33.0 oC Reading directly from a correctly plotted graph at 33.0 oC=> 1/t = 2.7 x 10-2 => t = 1/2.7 x 10-2 = 37.037 seconds 4. Explain the shape of the graph.
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Plot a graph of temperature(x-axis) against 1/t 2.(a)Calculate the number of moles of magnesium used given that 1cm of magnesium has a mass of 1g.(Mg= 24.0) Moles = Mass of magnesium => 1.0 = 4.167 x 10 -2 moles Molar mass of Mg 24 (b)Calculate the number of moles of hydrochloric acid used Moles of acid = molarity x volume of acid 1000 => 1.0 x 20 = 2.0 x 10 -2 moles 1000 (c)Calculate the mass of magnesium that remain unreacted Mole ratio Mg: HCl = 1:2 Moles Mg = ½ moles HCl => ½ x 2.0 x 10 -2 moles = 1.0 x 10 -2 moles Mass of reacted Mg = moles x molar mass => 1.0 x 10 -2 moles x 24 = 0.24 g Mass of unreacted Mg = Original total mass - Mass of reacted Mg => 1.0 g – 0.24 = 0.76 g (b)Calculate the total volume of hydrogen gas produced during the above reactions. 1/t Temperature(oC)
Mole ratio Mg : H2 = 1:1 Moles of Mg that reacted per experiment = moles H2 =1.0 x 10 -2 moles Volume of Hydrogen at s.t.p produced per experiment = moles x 24 dm3 => 1.0 x 10 -2 moles x 24 dm3 = 0.24dm3 Volume of Hydrogen at s.t.p produced in 5 experiments =0.24 dm3 x 5 = 1.2 dm3 3.(a)At what temperature was the time taken for magnesium to react equal to: (i)70seconds 70 seconds => 1/t =1/70 =0.01429 Reading directly from a correctly plotted graph 0.01429 => 28.0 oC (ii)40seconds 40 seconds => 1/t =1/40 =0.025 Reading directly from a correctly plotted graph 0.025 => 32.0 oC (b)What is the time taken for magnesium to react if the reaction was done at: (i) 55.0 oC Reading directly from a correctly plotted graph at 55.0 oC=> 1/t = 8.0 x 10-2 => t = 1/8.0 x 10-2 = 12.5 seconds (ii) 47.0 oC Reading directly from a correctly plotted graph at 47.0 oC=> 1/t = 6.0 x 10-2 => t = 1/6.0 x 10-2 = 16.6667 seconds (iii) 33.0 oC Reading directly from a correctly plotted graph at 33.0 oC=> 1/t = 2.7 x 10-2 => t = 1/2.7 x 10-2 = 37.037 seconds 4. Explain the shape of the graph. Increase in temperature increases the rate of reaction as particles gain kinetic energy increasing their frequency and intensity of collision to form products.
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1/t Temperature(oC)
Mole ratio Mg : H2 = 1:1 Moles of Mg that reacted per experiment = moles H2 =1.0 x 10 -2 moles Volume of Hydrogen at s.t.p produced per experiment = moles x 24 dm3 => 1.0 x 10 -2 moles x 24 dm3 = 0.24dm3 Volume of Hydrogen at s.t.p produced in 5 experiments =0.24 dm3 x 5 = 1.2 dm3 3.(a)At what temperature was the time taken for magnesium to react equal to: (i)70seconds 70 seconds => 1/t =1/70 =0.01429 Reading directly from a correctly plotted graph 0.01429 => 28.0 oC (ii)40seconds 40 seconds => 1/t =1/40 =0.025 Reading directly from a correctly plotted graph 0.025 => 32.0 oC (b)What is the time taken for magnesium to react if the reaction was done at: (i) 55.0 oC Reading directly from a correctly plotted graph at 55.0 oC=> 1/t = 8.0 x 10-2 => t = 1/8.0 x 10-2 = 12.5 seconds (ii) 47.0 oC Reading directly from a correctly plotted graph at 47.0 oC=> 1/t = 6.0 x 10-2 => t = 1/6.0 x 10-2 = 16.6667 seconds (iii) 33.0 oC Reading directly from a correctly plotted graph at 33.0 oC=> 1/t = 2.7 x 10-2 => t = 1/2.7 x 10-2 = 37.037 seconds 4. Explain the shape of the graph. Increase in temperature increases the rate of reaction as particles gain kinetic energy increasing their frequency and intensity of collision to form products. d)Influence of surface area on rate of reaction Surface area is the area of contact. An increase in surface area is a decrease in particle size. Practically an increase in surface area involves chopping /cutting
solid lumps into smaller pieces/chips then crushing the chips into powder.
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