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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 afte... | {
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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... | {
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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 dissol... | {
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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... | {
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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(... | {
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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 th... | {
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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... | {
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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... | {
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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 permuti... | {
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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 oth... | {
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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... | {
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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 potenti... | {
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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 Potass... | {
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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) 3... | {
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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: ∆... | {
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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/... | {
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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 ... | {
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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 bo... | {
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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 temperatur... | {
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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 dissociatio... | {
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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 ... | {
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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 (endoth... | {
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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 ... | {
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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) ... | {
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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) ... | {
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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, e... | {
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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 + ethan... | {
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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 = m... | {
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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 nit... | {
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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 = m... | {
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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... | {
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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 capac... | {
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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 ... | {
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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 so... | {
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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 temperatu... | {
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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... | {
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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 ... | {
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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 =... | {
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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 Tempera... | {
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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 ... | {
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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 b... | {
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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 a... | {
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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 t... | {
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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 O... | {
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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... | {
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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 Mas... | {
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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 volum... | {
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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 soluti... | {
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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 h... | {
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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.... | {
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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... | {
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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 -165... | {
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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... | {
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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 ... | {
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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 easi... | {
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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 = - ... | {
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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... | {
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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 diss... | {
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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 chan... | {
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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... | {
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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)Illust... | {
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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/coor... | {
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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 car... | {
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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 prod... | {
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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 ... | {
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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) En... | {
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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 Expe... | {
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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 Expe... | {
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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 ... | {
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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 sol... | {
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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... | {
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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 temp... | {
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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)... | {
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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... | {
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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 +... | {
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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... | {
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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.... | {
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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... | {
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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... | {
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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 reac... | {
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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 p... | {
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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 Met... | {
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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.... | {
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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... | {
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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... | {
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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 Fr... | {
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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... | {
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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... | {
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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 sodiu... | {
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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 ... | {
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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 = C1V... | {
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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.... | {
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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 particle... | {
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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 int... | {
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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 second... | {
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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 ab... | {
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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 magnes... | {
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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 ... | {
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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 vo... | {
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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 tempe... | {
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