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luz2uxam | chemistry | chemical-bonding-and-molecular-structure | molecular-orbital-theory | <p>The total number of species from the following in which one unpaired electron is present, is _______.</p>
<p>$$\mathrm{N}_2, \mathrm{O}_2, \mathrm{C}_2^{-}, \mathrm{O}_2^{-}, \mathrm{O}_2^{2-}, \mathrm{H}_2^{+}, \mathrm{CN}^{-}, \mathrm{He}_2^{+}$$</p> | [] | null | 4 | <p><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial,... | integer | jee-main-2024-online-9th-april-morning-shift | 988 |
lv0vytmr | chemistry | chemical-bonding-and-molecular-structure | molecular-orbital-theory | <p>Number of molecules/species from the following having one unpaired electron is ________.</p>
<p>$$\mathrm{O}_2, \mathrm{O}_2^{-1}, \mathrm{NO}, \mathrm{CN}^{-1}, \mathrm{O}_2^{2-}$$</p> | [] | null | 2 | <p>$$\mathrm{O}_2^{-} \text {and } \mathrm{NO} \text { have } 1 \text { unpaired electron. }$$</p> | integer | jee-main-2024-online-4th-april-morning-shift | 989 |
lv3xm5q9 | chemistry | chemical-bonding-and-molecular-structure | molecular-orbital-theory | <p>When $$\psi_{\mathrm{A}}$$ and $$\psi_{\mathrm{B}}$$ are the wave functions of atomic orbitals, then $$\sigma^*$$ is represented by :</p> | [{"identifier": "A", "content": "$$\\psi_A+2 \\psi_B$$\n"}, {"identifier": "B", "content": "$$\\psi_{\\mathrm{A}}-\\psi_{\\mathrm{B}}$$\n"}, {"identifier": "C", "content": "$$\\psi_A-2 \\psi_B$$\n"}, {"identifier": "D", "content": "$$\\psi_{\\mathrm{A}}+\\psi_{\\mathrm{B}}$$"}] | ["B"] | null | <p>In Molecular Orbital Theory, molecular orbitals are formed by the linear combination of atomic orbitals (LCAO). The wave functions of atomic orbitals $\psi_A$ and $\psi_B$ can combine in two ways:</p>
<ol>
<li><strong>Constructive Interference</strong>: This leads to the formation of a bonding molecular orbital ($\... | mcq | jee-main-2024-online-8th-april-evening-shift | 990 |
lv40v9yu | chemistry | chemical-bonding-and-molecular-structure | molecular-orbital-theory | <p>Number of molecules having bond order 2 from the following molecules is _________.</p> <p>$$\mathrm{C}_2, \mathrm{O}_2, \mathrm{Be}_2, \mathrm{Li}_2, \mathrm{Ne}_2, \mathrm{~N}_2, \mathrm{He}_2$$</p> | [] | null | 2 | <p>To determine the number of molecules with a bond order of 2 from the given molecules, we need to first calculate the bond order for each molecule. The bond order can be determined using Molecular Orbital Theory (MOT). The bond order is given by the formula:</p>
<p>
<p>$$\text{Bond Order} = \frac{\text{Number of bo... | integer | jee-main-2024-online-8th-april-evening-shift | 991 |
1krx8h5dq | chemistry | chemical-bonding-and-molecular-structure | resonance | Identify the species having one $$\pi$$-bond and maximum number of canonical forms from the following : | [{"identifier": "A", "content": "SO<sub>3</sub>"}, {"identifier": "B", "content": "O<sub>2</sub>"}, {"identifier": "C", "content": "SO<sub>2</sub>"}, {"identifier": "D", "content": "CO$$_3^{2 - }$$"}] | ["D"] | null | Among SO<sub>3</sub>, O<sub>2</sub>, SO<sub>2</sub> and CO$$_3^{2 - }$$, only O<sub>2</sub> and CO$$_3^{2 - }$$ has only one $$\pi$$-bond.<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264623/exam_images/ta7fcu4vvzhmhpraq0w4.webp" style="max-width: 100%;height: auto;display: block;margin: 0 au... | mcq | jee-main-2021-online-25th-july-evening-shift | 992 |
jaoe38c1lsfjlsrq | chemistry | chemical-bonding-and-molecular-structure | resonance | <p>The difference in energy between the actual structure and the lowest energy resonance structure for the given compound is</p> | [{"identifier": "A", "content": "electromeric energy\n"}, {"identifier": "B", "content": "resonance energy\n"}, {"identifier": "C", "content": "ionization energy\n"}, {"identifier": "D", "content": "hyperconjugation energy"}] | ["B"] | null | <p>The difference in energy between the actual structure and the lowest energy resonance structure for the given compound is known as resonance energy.</p> | mcq | jee-main-2024-online-29th-january-morning-shift | 993 |
SH9c4822qbHkKs4MGODP9 | chemistry | chemical-bonding-and-molecular-structure | valence-bond-theory | The pair that contains two P – H bonds in each of the oxoacids is | [{"identifier": "A", "content": "H<sub>3</sub>PO<sub>2</sub> and H<sub>4</sub>P<sub>2</sub>O<sub>5</sub>"}, {"identifier": "B", "content": "H<sub>4</sub>P<sub>2</sub>O<sub>5</sub> and H<sub>4</sub>P<sub>2</sub>O<sub>6</sub>"}, {"identifier": "C", "content": "H<sub>4</sub>P<sub>2</sub>O<sub>5</sub> and H<sub>3</sub>PO<s... | ["A"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263401/exam_images/a7dwpsa6n7njmos7lcxb.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263999/exam_images/xrv5xps3c0fvcjfmqcev.webp" style="max-width: 100%;height: auto;display: block;margi... | mcq | jee-main-2019-online-10th-january-evening-slot | 994 |
lsbn3ilu | chemistry | chemical-bonding-and-molecular-structure | valence-bond-theory | The lowest oxidation number of an atom in a compound $\mathrm{A}_2 \mathrm{B}$ is -2 . The number of electrons in its valence shell is _______. | [] | null | 6 | $\mathrm{A}_2 \mathrm{~B} \rightarrow 2 \mathrm{~A}^{+}+\mathrm{B}^{-2}$
<p>When an atom has the lowest oxidation number of -2 in a compound, it means the atom has gained two electrons beyond its neutral state to achieve this oxidation state. This is because gaining electrons makes the oxidation number more negative. I... | integer | jee-main-2024-online-1st-february-morning-shift | 995 |
VLonTGvaaXvC2Rn2Px7k9k2k5dz61w9 | chemistry | chemical-bonding-and-molecular-structure | van-der-walls-forces | The relative strength of interionic/intermolecular forces in decreasing order is :
| [{"identifier": "A", "content": "dipole-dipole $$>$$ ion-dipole $$>$$ ion-ion"}, {"identifier": "B", "content": "ion-dipole $$>$$ dipole-dipole $$>$$ ion-ion"}, {"identifier": "C", "content": "ion-dipole $$>$$ ion-ion $$>$$ dipole-dipole"}, {"identifier": "D", "content": "ion-ion $$>$$ ion-dipole $... | ["D"] | null | Ionic interactions are stronger as compared to
van der waal interactions.
<br><br>So, correct order is
<br><br>ion-ion $$>$$ ion-dipole $$>$$ dipole-dipole | mcq | jee-main-2020-online-7th-january-morning-slot | 996 |
zIzC0tNqd3aPhBvL | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | For the reaction CO (g) + (1/2) O<sub>2</sub> (g) $$\leftrightharpoons$$ CO<sub>2</sub> (g), K<sub>p</sub>/K<sub>c</sub> is : | [{"identifier": "A", "content": "RT"}, {"identifier": "B", "content": "(RT)<sup>-1</sup>"}, {"identifier": "C", "content": "(RT)<sup>-1/2</sup>"}, {"identifier": "D", "content": "(RT)<sup>1/2</sup>"}] | ["C"] | null | $${K_p} = {K_c}{\left( {RT} \right)^{\Delta n}};$$
<br><br>$$\Delta n = 1 - \left( {1 + {1 \over 2}} \right)$$
<br><br>$$ = 1 - {3 \over 2} = - {1 \over 2}.$$
<br><br>$$\therefore$$ $$\,\,\,\,{{{K_p}} \over {{K_c}}} = {\left( {RT} \right)^{ - 1/2}}$$ | mcq | aieee-2002 | 997 |
xnMc4pIlDWh0AoYp | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | For the reaction, CO(g) + Cl<sub>2</sub>(g) $$\leftrightharpoons$$ COCl<sub>2</sub>(g) the $${{{K_p}} \over {{K_c}}}$$ is equal to : | [{"identifier": "A", "content": "$$\\sqrt {RT} $$"}, {"identifier": "B", "content": "RT"}, {"identifier": "C", "content": "1/RT"}, {"identifier": "D", "content": "1.0"}] | ["C"] | null | $${K_p} = {K_c}{\left( {RT} \right)^{\Delta n}};$$
<br><br>Here $$\Delta n = 1 - 2 = - 1$$
<br><br>$$\therefore$$ $${{{k_p}} \over {{K_c}}} = {1 \over {RT}}$$ | mcq | aieee-2004 | 999 |
FOPtW9AewX3ZEoEr | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | What is the equilibrium expression for the reaction <br/>
P<sub>4</sub> (s) + 5O<sub>2</sub> $$\leftrightharpoons$$ P<sub>4</sub>O<sub>10</sub> (s)? | [{"identifier": "A", "content": "K<sub>c</sub> = [P<sub>4</sub>O<sub>10</sub>] / 5[P<sub>4</sub>] [O<sub>2</sub>] "}, {"identifier": "B", "content": "K<sub>c</sub> = 1/[O<sub>2</sub>]<sup>5</sup>"}, {"identifier": "C", "content": "K<sub>c</sub> = [P<sub>4</sub>O<sub>10</sub>] / [P<sub>4</sub>] [O<sub>2</sub>]<sup>5</su... | ["B"] | null | For $${P_4}\left( s \right) + 5{O_2}\left( g \right)\,\rightleftharpoons\,{P_4}{O_{10}}\left( 8 \right)$$
<br><br>$${K_c} = {1 \over {{{\left( {{O_2}} \right)}^5}}}.$$ The solids have concentration unity | mcq | aieee-2004 | 1,000 |
jBYItI7R5Faztuq9 | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | The equilibrium constant for the reaction N<sub>2</sub>(g) + O<sub>2</sub>(g) $$\leftrightharpoons$$ 2NO(g) at temperature T is
4 $$\times$$ 10<sup>-4</sup>. The value of K<sub>c</sub> for the reaction NO(g) $$\leftrightharpoons$$ $$1 \over 2$$N<sub>2</sub> (g) + $$1 \over 2$$O<sub>2</sub> (g) at the same temperature i... | [{"identifier": "A", "content": "2.5 $$\\times$$ 10<sup>2</sup>"}, {"identifier": "B", "content": "4 $$\\times$$ 10<sup>-4</sup>"}, {"identifier": "C", "content": "50"}, {"identifier": "D", "content": "0.02"}] | ["C"] | null | $${K_c} = {{{{\left[ {NO} \right]}^2}} \over {\left[ {{N_2}} \right]\left[ {{O_2}} \right]}} = 4 \times {10^{ - 4}}$$
<br><br>$$K{'_c} = {{{{\left[ {{N_2}} \right]}^{1/2}}{{\left[ {{Q_2}} \right]}^{1/2}}} \over {\left[ {NO} \right]}}$$
<br><br>$$ = {1 \over {\sqrt {{K_c}} }}$$
<br><br>$$ = {1 \over {\sqrt {4 \times {{... | mcq | aieee-2004 | 1,001 |
es64H9nSDhNRSnA4 | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | For the reaction
2NO<sub>2</sub> (g) $$\leftrightharpoons$$ 2NO (g) + O<sub>2</sub> (g), (K<sub>c</sub> = 1.8 $$\times$$ 10<sup>-6</sup> at 184<sup>o</sup>C) (R = 0.0831 kJ/(mol. K))<br/>
When K<sub>p</sub> and K<sub>c</sub> are compared at 184<sup>o</sup>C , it is found that : | [{"identifier": "A", "content": "K<sub>p</sub> is greater than K<sub>c</sub>"}, {"identifier": "B", "content": "K<sub>p</sub> is less than K<sub>c</sub>"}, {"identifier": "C", "content": "K<sub>p</sub> = K<sub>c</sub>"}, {"identifier": "D", "content": "Whether K<sub>p</sub> is greater than, less than or equal to K<sub>... | ["A"] | null | For the reaction : -
<br><br>$$2N{O_2}\left( g \right)\rightleftharpoons2NO\left( g \right) + {O_2}\left( g \right)$$
<br><br>Given $${K_c} = 1.8 \times {10^{ - 6}}\,\,$$ at $$\,\,{184^ \circ }C$$
<br><br>$$R=0.0831$$ $$\,\,kJ/mol.k$$
<br><br>$${K_p} = 1.8 \times {10^{ - 6}} \times 0.0831 \times 457$$
<br><br>$$ = 6... | mcq | aieee-2005 | 1,002 |
7T1O7dmxQQdrgE6B | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | Phosphorus pentachloride dissociates as follows, in a closed reaction vessel<br/>
PCl<sub>5</sub> (g) $$\leftrightharpoons$$ PCl<sub>3</sub> (g) + Cl<sub>2</sub> (g) <br/>
If total pressure at equilibrium of the reaction mixture is P and degree of dissociation of PCl<sub>5</sub> is x, the
partial pressure of PCl<sub>3<... | [{"identifier": "A", "content": "$$\\left( {{x \\over {x + 1}}} \\right)P$$ "}, {"identifier": "B", "content": "$$\\left( {{2x \\over {1 - x}}} \\right)P$$ "}, {"identifier": "C", "content": "$$\\left( {{x \\over {x - 1}}} \\right)P$$ "}, {"identifier": "D", "content": "$$\\left( {{x \\over {1 - x}}} \\right)P$$ "}] | ["A"] | null | $$\mathop {PC{l_5}\left( g \right)}\limits_{1 - x} \mathop {\,\rightleftharpoons\,PC{l_3}\left( g \right)}\limits_x \,\, + \,\,\mathop {C{l_2}\left( g \right)}\limits_x $$
<br><br>Total moles after dissociation
<br><br>$$1 - x + x + x = 1 + x$$
<br><br>$${P_{PC{l_3}}} = $$ mole fraction of $$PCl{}_3 \times $$ Total pr... | mcq | aieee-2006 | 1,004 |
BN4EdeIy0Gczcmki | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | The equilibrium constant for the reaction <br/>
SO<sub>3</sub> (g) $$\leftrightharpoons$$ SO<sub>2</sub> (g) + $$1 \over 2$$ O<sub>2</sub> (g)<br/>
is K<sub>c</sub> = 4.9 $$\times$$ 10<sup>–2</sup>. The value of K<sub>c</sub> for the reaction<br/>
2SO<sub>2</sub> (g) + O<sub>2</sub> (g) $$\leftrightharpoons$$ 2SO<sub>... | [{"identifier": "A", "content": "416"}, {"identifier": "B", "content": "9.8 $$\\times$$ 10<sup>-2</sup>"}, {"identifier": "C", "content": "4.9 $$\\times$$ 10<sup>-2</sup>"}, {"identifier": "D", "content": "2.40 $$\\times$$ 10<sup>-3</sup>"}] | ["A"] | null | $$S{O_3}\left( g \right)\,\rightleftharpoons\,S{O_2}\left( g \right)\,\, + \,\,{1 \over 2}{O_2}\left( g \right)$$
<br><br>$${K_c} = {{\left[ {S{O_2}} \right]{{\left[ {{O_2}} \right]}^{{\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}}}} \over {\left[ {S{O_3}} \right]}}$$
... | mcq | aieee-2006 | 1,005 |
3yX4MhgoPTGVHeB5 | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | The equilibrium constants K<sub>P1</sub> and K<sub>P2</sub> for the reactions X $$\leftrightharpoons$$ 2Y and Z $$\leftrightharpoons$$ P + Q, respectively are in
the ratio of 1 : 9. If the degree of dissociation of X and Z be equal then the ratio of total pressure at
these equilibria is : | [{"identifier": "A", "content": "1 : 36"}, {"identifier": "B", "content": "1 : 1"}, {"identifier": "C", "content": "1 : 3"}, {"identifier": "D", "content": "1 : 9"}] | ["A"] | null | Let the initial moles of $$X$$ be $$'a'$$
<br><br>and that of $$Z$$ be $$'b'$$ the for the given reactions,
<br><br>we have $$X\,\,\,\,\,\,\,\,\,\rightleftharpoons\,\,\,\,\,\,\,\,\,2Y$$
<br><br>$$\eqalign{
& Initial\,\,\,\,\,\,\,\,\,\,\,\,a\,\,moles\,\,\,\,\,\,\,0 \cr
& At\,\,equi.\,\,\,\,\,\,\,a\lef... | mcq | aieee-2008 | 1,006 |
07XXSkPNgWGKLYw1 | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | A vessel at 1000 K contains CO<sub>2</sub> with a pressure of 0.5 atm. Some of the CO<sub>2</sub> is converted into CO on
the addition of graphite. If the total pressure at equilibrium is 0.8 atm, the value of K is : | [{"identifier": "A", "content": "3 atm "}, {"identifier": "B", "content": "0.3 atm "}, {"identifier": "C", "content": "0.18 atm"}, {"identifier": "D", "content": "1.8 atm "}] | ["D"] | null | $$C{O_2} + {C_{\left( {grapnite} \right)}}\,\rightleftharpoons\,2CO$$
<br><br>$${P_{initial}}\,\,0.5atm\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0$$
<br><br>$${P_{final}}\,\,\,\left( {0.5 - x} \right)atm\,\,\,\,\,\,\,2x\,atm$$
<br><br>Total $$P$$ at equilibrium
<br><br>$$ = 0.5 - x + 2x = 0.5 + x\,atm$$
<br><br>$$... | mcq | aieee-2011 | 1,009 |
OZsFoYVKT9unygaB | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | The equilibrium constant (K<sub>C</sub>) for the reaction N<sub>2</sub>(g) + O<sub>2</sub>(g) $$\to$$ 2NO(g) at temperature T is 4 $$\times$$ 10<sup>–4</sup>. The value of K<sub>C</sub> for the reaction, NO(g) $$\to$$ 1/2N<sub>2</sub>(g) + 1/2O<sub>2</sub>(g) at the same temperature is : | [{"identifier": "A", "content": "0.02"}, {"identifier": "B", "content": "2.5 $$\\times$$ 10<sup>2</sup>"}, {"identifier": "C", "content": "4 $$\\times$$ 10<sup>-4</sup>"}, {"identifier": "D", "content": "50.0"}] | ["D"] | null | For the reaction
<br><br>$${N_2} + {O_2} \to 2NO$$
<br><br>$$\,K = 4 \times {10^{ - 4}}$$
<br><br>Hence for the reaction
<br><br>$$NO \to {1 \over 2}{N_2} + {1 \over 2}{O_2}$$
<br><br>$$K' = {1 \over {\sqrt K }}$$
<br><br>$$ = {1 \over {\sqrt {4 \times {{10}^{ - 4}}} }}$$
<br><br>$$ = 50$$ | mcq | aieee-2012 | 1,010 |
I2FfcEHEOHxjLXgs | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | For the reaction SO<sub>2</sub> (g) + $${1 \over 2} O_2(g) \leftrightharpoons$$ SO<sub>3</sub>(g). <br/>
if K<sub>P</sub> = K<sub>C</sub>(RT)<sup>x</sup> where the symbols have usual meaning then
the value of x is: (assuming ideality) | [{"identifier": "A", "content": "-1"}, {"identifier": "B", "content": "-1/2"}, {"identifier": "C", "content": "1/2"}, {"identifier": "D", "content": "1"}] | ["B"] | null | $$S{O_2}\left( g \right) + {1 \over 2}{O_2}\left( g \right)\,\rightleftharpoons\,S{O_3}\left( g \right)$$
<br><br>$${K_p} = {K_C}{\left( {RT} \right)^x}$$
<br><br>where $$x = \Delta {n_g} = $$ number of gaseous moles in product
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$$$-$$ number of gaseou... | mcq | jee-main-2014-offline | 1,011 |
slcAxWaeZqwRmZga | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | The standard Gibbs energy change at 300 K for the reaction 2A $$\leftrightharpoons$$ B + C is 2494.2 J. At a given time,
the composition of the reaction mixture is [A] = 1/2, [B] = 2 and [C] = 1/2. The reaction proceeds in the: [R = 8.314 J/K/mol, e = 2.718] | [{"identifier": "A", "content": "reverse direction because Q > K<sub>c</sub>"}, {"identifier": "B", "content": "forward direction because Q < K<sub>c</sub>"}, {"identifier": "C", "content": "reverse direction because Q < K<sub>c</sub>"}, {"identifier": "D", "content": "forward direction because Q > K<sub>c<... | ["A"] | null | $$\Delta {G^ \circ } = 2494.2J$$
<br><br>$$2A\,\rightleftharpoons\,B + C$$
<br><br>$$R = 8.314\,J/K/mol.$$
<br><br>$$e = 2.718$$
<br><br>$$\left[ A \right] = {1 \over 2},\left[ B \right] = 2,$$
<br><br>$$\left[ C \right] = {1 \over 2};Q = {{\left[ B \right]\left[ C \right]} \over {{{\left[ A \right]}^2}}}$$
<br><br>... | mcq | jee-main-2015-offline | 1,012 |
iBWkn6XhSK3d12VG | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | The equilibrium constant at 298 K for a reaction A + B $$\leftrightharpoons$$ C + D is 100. If the initial concentration of
all the four species were 1M each, then equilibrium concentration of D (in mol L<sup>–1</sup>) will be: | [{"identifier": "A", "content": "0.818"}, {"identifier": "B", "content": "1.818"}, {"identifier": "C", "content": "1.182"}, {"identifier": "D", "content": "0.182"}] | ["B"] | null | <b>Given, </b>
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267730/exam_images/gpdkhpydwpbnbn6omhdy.webp" loading="lazy" alt="JEE Main 2016 (Offline) Chemistry - Chemical Equilibrium Question 78 English Explanation">
<br><br>$$\therefore$$ $$\,\,\,{K_c} = {\left( {{{1... | mcq | jee-main-2016-offline | 1,013 |
ax46xkqWgYsItJK5jVP6o | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | A solid XY kept in an evacuated sealed container undergoes decomposition to form a mixture of gases X and Y at temperature T. The equilibrium pressure is 10 bar in this vessel. K<sub>p</sub> for this reaction is :
| [{"identifier": "A", "content": "5"}, {"identifier": "B", "content": "10"}, {"identifier": "C", "content": "25"}, {"identifier": "D", "content": "100"}] | ["C"] | null | <p>To determine the equilibrium constant, $K_p$, for the decomposition reaction of the solid compound XY into its gaseous components X and Y, we need to consider the following reaction:</p>
<p>
<p>$$ \text{XY (s)} \rightleftharpoons \text{X (g)} + \text{Y (g)} $$</p>
</p>
<p>For a reaction where a solid decomposes ... | mcq | jee-main-2016-online-10th-april-morning-slot | 1,014 |
CprqaIhBf2B6XthNgHb7s | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | At a certain temperature in a $$5$$ $$L$$ vessel, 2 moles of carbon monoxide and 3 moles of chlorine were allowed to reach equilibrium according to the reaction,
<br/> CO + Cl<sub>2</sub> $$\rightleftharpoons$$ COCl<sub>2</sub>
<br/>At equilibrium, if one mole of CO is present then equilibrium constant (K<sub... | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "2.5"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "4"}] | ["B"] | null | <table class="tg">
<tbody><tr>
<th class="tg-p7ly"></th>
<th class="tg-p7ly">CO</th>
<th class="tg-p7ly">+</th>
<th class="tg-fbrz">Cl<sub>2</sub></th>
<th class="tg-fbrz">$$\rightleftharpoons$$</th>
<th class="tg-fbrz">COCl<sub>2</sub></th>
</tr>
<tr>
<td class="tg-13k7">Initially num... | mcq | jee-main-2018-online-15th-april-evening-slot | 1,015 |
5DyBjOOhCKpSEcMoeo3rsa0w2w9jx97n6gj | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | In which one of the following equilibria, K<sub>p</sub> $$ \ne $$ K<sub>C</sub> ? | [{"identifier": "A", "content": "2NO(g) \u21cb N<sub>2</sub>(g) + O<sub>2</sub>(g)"}, {"identifier": "B", "content": "2C(s) + O<sub>2</sub>(g) \u21cb 2CO(g)"}, {"identifier": "C", "content": "2HI(g) \u21cb H<sub>2</sub>(g) + I<sub>2</sub>(g)"}, {"identifier": "D", "content": "NO<sub>2</sub>(g) + SO<sub>2</sub>(g) \u21c... | ["B"] | null | We know,
<br>K<sub>p</sub> = K<sub>C</sub>(RT)<sup>$$\Delta $$n<sub>g</sub></sup>
<br><br>K<sub>p</sub> = K<sub>C</sub> when $$\Delta $$n<sub>g</sub> = 0
<br>K<sub>p</sub> $$ \ne $$ K<sub>C</sub> when $$\Delta $$n<sub>g</sub> $$ \ne $$ 0 and T $$ \ne $$ 12 K
<br><br>In this reaction, 2C(s) + O<sub>2</sub>(g) ⇋ 2CO(g)
<... | mcq | jee-main-2019-online-12th-april-evening-slot | 1,016 |
WTrxdDGwVVa5QCJNsLbQe | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | For the following reactions, equilibrium
constants are given :<br/>
<br/>S(s) + O<sub>2</sub>(g) ⇋ SO<sub>2</sub>(g); K<sub>1</sub> = 10<sup>52</sup><br/>
<br/>2S(s) + 3O<sub>2</sub>(g) ⇋ 2SO<sub>3</sub>(g); K<sub>2</sub> = 10<sup>129</sup><br/>
<br/>The equilibrium constant for the reaction,<br/>
<br/>2SO<sub>2</su... | [{"identifier": "A", "content": "10<sup>181</sup>"}, {"identifier": "B", "content": "10<sup>25</sup>"}, {"identifier": "C", "content": "10<sup>77</sup>"}, {"identifier": "D", "content": "10<sup>154</sup>"}] | ["B"] | null | S(s) + O<sub>2</sub>(g) ⇋ SO<sub>2</sub>(g); K<sub>1</sub> = 10<sup>52</sup>
<br><br>By reversing the equation, we get
<br><br>SO<sub>2</sub>(g); ⇋ S(s) + O<sub>2</sub>(g) ; $${1 \over {{K_1}}} = {1 \over {{{10}^{52}}}}$$ ..............(1)
<br><br>2S(s) + 3O<sub>2</sub>(g) ⇋ 2SO<sub>3</sub>(g); K<sub>2</sub> = 10<su... | mcq | jee-main-2019-online-8th-april-evening-slot | 1,017 |
FvoCSr01x8ku0onxV2jEr | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | Two solids dissociate as follows –
<br/><picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266268/exam_images/uqjldhdlltvid4j6irjs.webp"/><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266103/exam_images/vtezzbhj... | [{"identifier": "A", "content": "(x + y) atm "}, {"identifier": "B", "content": "$$\\left( {\\sqrt {x + y} } \\right)$$ atm"}, {"identifier": "C", "content": "2$$\\left( {\\sqrt {x + y} } \\right)$$ atm"}, {"identifier": "D", "content": "x<sup>2</sup> + y<sup>2</sup> atm"}] | ["C"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263582/exam_images/l0unbj43efmzlu6ogqyh.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266285/exam_images/g3hpfyrd6hvz7dcm9qmy.webp"><img src="https://res.c... | mcq | jee-main-2019-online-12th-january-morning-slot | 1,018 |
JX0ZZZbwalKbgS6Jp2vRQ | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | 5.1 g NH<sub>4</sub>SH is introduced in 3.0 L evacuated flask at 327ºC. 30% of the solid NH<sub>4</sub>SH decomposed to NH<sub>3</sub> and H<sub>2</sub>S as gases . The Kp of the reaction at 327<sup>o</sup>C is (R = 0.082 L atm mol<sup>–1</sup> K<sup>–1</sup>, Molar mass of S = 32 g mol<sup>–1</sup> molar mass of N = 1... | [{"identifier": "A", "content": "0.242 $$ \\times $$ 10<sup>$$-$$4</sup> atm<sup>2</sup>"}, {"identifier": "B", "content": "1 $$ \\times $$ 10<sup>\u20134</sup> atm<sup>2</sup>\n"}, {"identifier": "C", "content": "4.9 $$ \\times $$ 10<sup>$$-$$3</sup> atm<sup>2</sup>"}, {"identifier": "D", "content": "0.242 atm<sup>2</... | ["D"] | null | NH<sub>4</sub>SH(s) $$\rightleftharpoons$$ NH<sub>3</sub>(g) + H<sub>2</sub>S(g)
<br><br>$$n = {{5.1} \over {51}} = .1\,mole\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,$$
<br><br>$$.1\left( { - 1 - \alpha } \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\... | mcq | jee-main-2019-online-10th-january-evening-slot | 1,020 |
sCuCIyzQctkzU3Yg4cnVF | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | The values of K<sub>p</sub>/K<sub>c</sub> for the following reactions at 300 K are, respectively : (At 300 K, RT = 24.62 dm<sup>3</sup> atm mol<sup>–1</sup>)
<br/><br/>N<sub>2</sub>(g) + O<sub>2</sub>(g) $$\rightleftharpoons$$ 2 NO(g)
<br/><br/>N<sub>2</sub>O<sub>4</sub>(g) $$\rightleftharpoons$$ 2 NO(g)
<br/><br... | [{"identifier": "A", "content": "24.62 dm<sup>3</sup> atm mol<sup>\u20131</sup>, 606.0 dm<sup>6</sup> atm<sup>2</sup> mol<sup>\u20132</sup> 1.65 $$ \\times $$ 10<sup>\u20133</sup> dm<sup>\u20136</sup> atm<sup>\u20132</sup> mol<sup>2</sup>"}, {"identifier": "B", "content": "1,4.1 $$ \\times $$ 10<sup>\u20132</sup> dm<su... | ["C"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267661/exam_images/nflpizlvl6pwqgygmlb2.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 10th January Morning Slot Chemistry - Chemical Equilibrium Question 71 English Explanation"> | mcq | jee-main-2019-online-10th-january-morning-slot | 1,021 |
OXleJcmajnlCRgGiZwAFa | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | Consider the following reversible chemical reactions :
<br/><br/><img src="data:image/png;base64,UklGRlwSAABXRUJQVlA4IFASAAAwlACdASoAA+kAP4G+1WS2MCunIlKaYsAwCWlu6B/JCGLZbneRv8Z6gvHY/Zefd8fb37/OyPgHZG9cyAXudZuMqjlwHl/C9+8jK1NgoXoCF+VC4UOFp6mwUL0BC/KhcKHCy4m7AZyhyDQ5Bocgw5XLSoXChwtPU2ChegIX4++UOQaHINDkGhyDQkDsW7NEMmlh+u... | [{"identifier": "A", "content": "K<sub>1</sub>K<sub>2</sub> = $${1 \\over 3}$$"}, {"identifier": "B", "content": "K<sub>2</sub> = K<sub>1</sub><sup>3</sup>"}, {"identifier": "C", "content": "K<sub>2</sub> = K<sub>1</sub><sup>$$-$$3</sup>"}, {"identifier": "D", "content": "K<sub>1</sub>K<sub>2</sub> = 3"}] | ["C"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267683/exam_images/j4h1nyops5hm9wtmf5en.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 9th January Evening Slot Chemistry - Chemical Equilibrium Question 72 English Explanation"> | mcq | jee-main-2019-online-9th-january-evening-slot | 1,022 |
fvvAe6DDxfHLuSybPhg5E | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | In a chemical reaction,
<br/><br/><img src="data:image/png;base64,UklGRr4HAABXRUJQVlA4ILIHAADwOwCdASr0AZIAPm02mEkkIyKhIbMJsIANiWlu/HyZN+tQ0P0d/nH4zeDn+A6Lj0F7fZNx7j/Ov3I9PP83/QPIvgEeo/8R/Dv6j/rvyw9uf6Z91dkP/U/yj+beSr/LfwD1A+pvmb/oh/TPIA/gHoLx7/pd/OPZS/0/45/O/Vn9Af9X+Vf3P3X/+D/Hf7R7zPsyFNpSZ0X6Ii4UmdF+iIuFJnRfoiLhSZHkG... | [{"identifier": "A", "content": "16"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "1/4"}, {"identifier": "D", "content": "4"}] | ["D"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264739/exam_images/q9upzfam6hyhsoh6m2o7.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265850/exam_images/bms988tdon2t2diysvau.webp" style="max-width: 100%;height: auto;display: block;margi... | mcq | jee-main-2019-online-12th-january-morning-slot | 1,023 |
FBD5i44Aa7d4LIF4CCjgy2xukfq9iah8 | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | For a reaction X + Y ⇌ 2Z , 1.0 mol of X, 1.5 mol <br/>of Y and 0.5 mol of Z were taken in a 1 L
vessel and<br/> allowed to react. At equilibrium, the concentration<br/> of Z was 1.0 mol L<sup>–1</sup>. The equilibrium constant of reaction<br/> is $${x \over {15}}$$. The value of x is _________. | [] | null | 16 | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266846/exam_images/epachbfsh2pmgpid9cpi.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 5th September Evening Slot Chemistry - Chemical Equilibrium Question 57 English Explanation">
<br>... | integer | jee-main-2020-online-5th-september-evening-slot | 1,024 |
dfT78WspfSOKCkAFl9jgy2xukg3fgqra | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | The value of K<sub>C</sub> is 64 at 800 K for the reaction
<br/><br/>N<sub>2</sub>(g) + 3H<sub>2</sub>(g) ⇌ 2NH<sub>3</sub>(g)
<br/><br/>The value of K<sub>C</sub> for the following reaction is :
<br/><br/>NH<sub>3</sub>(g) ⇌ $${1 \over 2}$$N<sub>2</sub>(g) + $${3 \over 2}$$H<sub>2</sub>(g) | [{"identifier": "A", "content": "8"}, {"identifier": "B", "content": "$${1 \\over 8}$$"}, {"identifier": "C", "content": "$${1 \\over 4}$$"}, {"identifier": "D", "content": "$${1 \\over {64}}$$"}] | ["B"] | null | N<sub>2</sub>(g) + 3H<sub>2</sub>(g) ⇌ 2NH<sub>3</sub>(g) ; K<sub>C</sub>
<br><br>2NH<sub>3</sub>(g) ⇌ N<sub>2</sub>(g) + 3H<sub>2</sub>(g) ; $${1 \over {{K_C}}}$$
<br><br>Multiplying by $${1 \over 2}$$, reaction becomes
<br><br>NH<sub>3</sub>(g) ⇌ $${1 \over 2}$$N<sub>2</sub>(g) + $${3 \over 2}$$H<sub>2</sub>(g) ;
<... | mcq | jee-main-2020-online-6th-september-evening-slot | 1,025 |
tGqFP5AUBgiAasKCDRjgy2xukfuptcel | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | The variation of equilibrium constant with
temperature is given below :
<br/><br/><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.... | [{"identifier": "A", "content": "28.4, \u20135.71 and \u201314.29"}, {"identifier": "B", "content": "0.64, \u20137.14 and \u20135.71"}, {"identifier": "C", "content": "28.4, \u20137.14 and \u20135.71"}, {"identifier": "D", "content": "0.64, \u20135.71 and \u201314.29"}] | ["A"] | null | ln $$\left[ {{{{k_2}} \over {{k_1}}}} \right]$$ = $${{\Delta H^\circ } \over R}\left\{ {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right\}$$
<br><br>$$ \Rightarrow $$ ln(10) = $${{\Delta H^\circ } \over R}\left\{ {{1 \over {298}} - {1 \over {373}}} \right\}$$
<br><br>$$ \Rightarrow $$ $${\Delta H^\circ }$$ = 28.37 kJ/mol
... | mcq | jee-main-2020-online-6th-september-morning-slot | 1,026 |
Wymv2TIWJaheArgnEdjgy2xukfi6jvls | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | Consider the following reaction:
<br/><br/>N<sub>2</sub>O<sub>4</sub>(g) ⇌ 2NO<sub>2</sub>(g); $$\Delta $$H<sup>o</sup> = +58 kJ
<br/><br/>For each of the following cases (a, b), the direction in which the equilibrium shifts is :
<br/>(a) Temperature is decreased.
<br/>(b) Pressure is increased by adding N<sub>2</sub>
... | [{"identifier": "A", "content": "(a) towards reactant, (b) towards product"}, {"identifier": "B", "content": "(a) towards reactant, (b) no change"}, {"identifier": "C", "content": "(a) towards product, (b) towards reactant"}, {"identifier": "D", "content": "(a) towards product, (b) no change"}] | ["B"] | null | $$ \because $$ Given reaction is endothermic.
<br><br>$$ \therefore $$ On decreasing temperature backward
reaction will be favoured.
<br><br>On adding N<sub>2</sub>, pressure is increased at
constant T, and volume would also be constant
so no change is observed. | mcq | jee-main-2020-online-5th-september-morning-slot | 1,028 |
TIkZn72VO1Voziu4yojgy2xukfc91f9y | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | If the equilibrium constant for<br/> A ⇌ B + C is $$K_{eq}^{(1)}$$ and that of
<br/>B + C ⇌ P is $$K_{eq}^{(2)}$$, the equilibrium
<br/>constant for A ⇌ P is : | [{"identifier": "A", "content": "$${{K_{eq}^{(1)}} \\over {K_{eq}^{(2)}}}$$"}, {"identifier": "B", "content": "$${K_{eq}^{(1)}}$$ + $${K_{eq}^{(2)}}$$"}, {"identifier": "C", "content": "$${K_{eq}^{(2)}}$$ - $${K_{eq}^{(1)}}$$"}, {"identifier": "D", "content": "$${K_{eq}^{(1)}}$$ $${K_{eq}^{(2)}}$$"}] | ["D"] | null | A ⇌ B + C $$K_{eq}^{(1)}$$ = $${{\left[ B \right]\left[ C \right]} \over {\left[ A \right]}}$$ ....(1)
<br><br>B + C ⇌ P $$K_{eq}^{(2)}$$ = $${{\left[ P \right]} \over {\left[ B \right]\left[ C \right]}}$$ ....(2)
<br><br>For A ⇌ P K<sub>eq</sub> = $${{\left[ P \right]} \over {\left[ A \right]}}$$
<br... | mcq | jee-main-2020-online-4th-september-evening-slot | 1,029 |
be9mFEh0uaOIM9Asfgjgy2xukf92vfgc | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | For the equilibrium A ⇌ B , the variation of the
rate of the forward (a) and reverse (b) reaction
with time is given by : | [{"identifier": "A", "content": "<img src=\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263534/exam_images/cpqestv7mxua9v68f391.webp\" style=\"max-width: 100%;height: auto;display: block;margin: 0 auto;\" loading=\"lazy\" alt=\"JEE Main 2020 (Online) 4th September Morning Slot Chemistry - Chemical Equilibriu... | ["C"] | null | At equilibrium,
<br><br>rate of forward reaction = Rate of backward
reaction | mcq | jee-main-2020-online-4th-september-morning-slot | 1,030 |
yEdUCUZPSH3TjZAnJAjgy2xukevgnseq | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | An open beaker of water in equilibrium with
water vapour is in a sealed container. When a
few grams of glucose are added to the beaker
of water, the rate at which water molecules : | [{"identifier": "A", "content": "leaves the solution increases"}, {"identifier": "B", "content": "leaves the vapour increases"}, {"identifier": "C", "content": "leaves the vapour decreases"}, {"identifier": "D", "content": "leaves the solution decreases"}] | ["D"] | null | With addition of solute in solvent, surface area for vapourisation decreases causes lowering in vapour
pressure. | mcq | jee-main-2020-online-2nd-september-morning-slot | 1,031 |
UZbTpLs7LFbcmlQe4b7k9k2k5ll44t0 | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | In the figure shown below reactant A
(represented by square) is in equilibrium with
product B (represented by circle). The
equilibrium constant is :
<img src="data:image/png;base64,UklGRkgLAABXRUJQVlA4IDwLAACwTgCdASpHAb4APm0ylUgkIqIhI3E7+IANiWlu4W/RG/Ol8i/y/8W/B/+2f0vpnPJ/tz/WfYr+M+ph+D+5o+H/jP7Rf1/9zvvZ+Z/4r+Xftr+N3ur... | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "8"}, {"identifier": "D", "content": "4"}] | ["B"] | null | In the figure, Reactant A is represented by square.
<br>And there are 6 squares. So at equilibrium
<br>[A]<sub>eq</sub> = 6
<br><br>In the figure, Product B is represented by circle.
<br>And there are 11 circles. So at equilibrium
<br>[B]<sub>eq</sub> = 11
<br><br>For reaction, A ⇌ B
<br><br>K<sub>eq</sub> = $${{\left[... | mcq | jee-main-2020-online-9th-january-evening-slot | 1,032 |
aYXsHNIdb8o9MrAUiR1klrgxhot | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | At 1990 K and 1 atm pressure, there are equal number of Cl<sub>2</sub>, molecules and Cl atoms in the
reaction mixture. The value of K<sub>p</sub> for the reaction Cl<sub>2 (g)</sub> $$ \rightleftharpoons $$ 2Cl<sub>(g)</sub> under the above conditions
is x $$ \times $$ 10<sup>-1</sup>.<br/>
The value of x is _______. ... | [] | null | 5 | Cl<sub>2</sub>(g) $$\rightleftharpoons$$ 2Cl(g)<br/><br/>Let moles of both of Cl<sub>2</sub> and Cl molecule be x.<br/><br/>Partial pressure of Cl is, $${p_{Cl}} = {x \over {2x}} \times 1 = {1 \over 2}$$<br/><br/>Partial pressure of Cl<sub>2</sub> is, $${p_{C{l_2}}} = {x \over {2x}} \times 1 = {1 \over 2}$$<br/><br/>No... | integer | jee-main-2021-online-24th-february-morning-slot | 1,034 |
nfBK2z0vo8pk8hKlIF1klueivgm | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | A homogeneous ideal gaseous reaction $$A{B_{2(g)}} \rightleftharpoons {A_{(g)}} + 2{B_{(g)}}$$ is carried out in a 25 litre flask at 27$$^\circ$$C. The initial amount of AB<sub>2</sub> was 1 mole and the equilibrium pressure was 1.9 atm. The value of K<sub>p</sub> is x $$\times$$ 10<sup>$$-$$2</sup>. The value of x is ... | [] | null | 72TO75 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l3l0tjjc/ca37916d-b8bf-49b7-9aee-4573c6e9abc7/ef4a1a80-dbd9-11ec-adf6-01b3c1852b0a/file-1l3l0tjjd.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l3l0tjjc/ca37916d-b8bf-49b7-9aee-4573c6e9abc7/ef4a1a80-dbd9-11ec-adf6-01b3c1852b0a... | integer | jee-main-2021-online-26th-february-morning-slot | 1,036 |
9T536VrN9LRuXuLDCM1kmhuu6ix | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | For the reaction $$A(g) \rightleftharpoons B(g)$$ at 495 K, $$\Delta$$<sub>r</sub>G$$^\circ$$ = $$-$$9.478 kJ mol<sup>$$-$$1</sup>.<br/>If we start the reaction in a closed container at 495 K with 22 millimoles of A, the amount of B in the equilibrium mixture is ____________ millimoles.<br/> (Round off to the Nearest I... | [] | null | 20 | $$\Delta$$G<sup>o</sup> = $$-$$RT ln Keq<br><br>$$-$$9.478 $$\times$$ 10<sup>3</sup> = $$-$$495 $$\times$$ 8.314 ln Keq<br><br>ln Keq = 2.303 = ln 10<br><br>So, Keq = 10<br><br>Now, A(g) $$\rightleftharpoons$$ B(g)<br><br>$$\matrix{
{t = 0} & {22} & 0 \cr
{t = t} & {22 - x} & x \cr
} $$<br><... | integer | jee-main-2021-online-16th-march-morning-shift | 1,037 |
AALr4KCta5vwmOHeFi1kmj8rwvb | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | 0.01 moles of a weak acid HA (K<sub>a</sub> = 2.0 $$\times$$ 10<sup>$$-$$6</sup>) is dissolved in 1.0 L of 0.1 M HCl solution. The degree of dissociation of HA is __________ $$\times$$ 10<sup>$$-$$5</sup> (Round off to the Nearest Integer). <br/><br/>[Neglect volume change on adding HA. Assume degree of dissociation &l... | [] | null | 2 | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263732/exam_images/apw95glwmk7vsc8xzgye.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 17th March Morning Shift Chemistry - Chemical Equilibrium Question 48 English Explanation"><br><br... | integer | jee-main-2021-online-17th-march-morning-shift | 1,038 |
5FwMvVp0Oixhc7DsGj1kmkjmcj4 | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | Consider the reaction<br/><br/> $$N_{2}O_{4}\left( g\right) \rightleftharpoons 2NO_{2}\left( g\right) $$<br/><br/>The temperature at which K<sub>C</sub> = 20.4 and K<sub>P</sub> = 600.1, is ____________ K. (Round off to the Nearest Integer). [Assume all gases are ideal and R = 0.0831 L bar K<sup>$$-$$1</sup> mol<sup... | [] | null | 354 | N<sub>2</sub>O<sub>4</sub>(g) $$\rightleftharpoons$$ 2NO<sub>2</sub>(g)<br><br>$$\Delta$$n<sub>g</sub> = 2 $$-$$ 1 = 1<br><br>K<sub>P</sub> = K<sub>C</sub>(RT)<sup>$$\Delta$$n<sub>g</sub></sup><br><br>600.1 = 20.4 (0.0831 $$\times$$ T)<sup>1</sup><br><br>$$ \Rightarrow $$ T = $${{600.1} \over {20.4 \times 0.0831}}$$ = ... | integer | jee-main-2021-online-17th-march-evening-shift | 1,039 |
3c3jXS1kKN0TMCSAPU1kmm24oyg | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | The gas phase reaction $$2A(g) \rightleftharpoons {A_2}(g)$$ at 400 K has $$\Delta$$G<sup>o</sup> = + 25.2 kJ mol<sup>-1</sup>.<br/><br/>The equilibrium constant K<sub>C</sub> for this reaction is ________ $$\times$$ 10<sup>$$-$$2</sup>. (Round off to the Nearest Integer).<br/><br/>[Use : R = 8.3 J mol<sup>$$-$$1</sup>... | [] | null | 2 | Using formula,<br><br>$$\Delta$$G$$^\circ$$ = $$-$$ RTln(K<sub>p</sub>)<br><br>$$ \Rightarrow $$ 25.2 $$\times$$ 10<sup>3</sup> = $$-$$8.3 $$\times$$ 400 $$\times$$ 2.3 log (K<sub>p</sub>)<br><br>$$ \Rightarrow $$ K<sub>p</sub> = 10<sup>$$-$$3.3</sup><br><br>= 10<sup>$$-$$3</sup> $$\times$$ 0.501<br><br>= 5.01 $$\times... | integer | jee-main-2021-online-18th-march-evening-shift | 1,040 |
1krq6szsp | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | 2SO<sub>2</sub>(g) + O<sub>2</sub>(g) $$\rightleftharpoons$$ 2SO<sub>3</sub>(g)<br/><br/>In an equilibrium mixture, the partial pressures are <br/><br/>P<sub>SO<sub>3</sub></sub> = 43 kPa; P<sub>O<sub>2</sub></sub> = 530 Pa and P<sub>SO<sub>2</sub></sub> = 45 kPa. The equilibrium constant K<sub>P</sub> = ___________ $$... | [] | null | 172 | On reaction, 2SO<sub>2</sub>(g) + O<sub>2</sub>(g) $$\rightarrow$$ 2SO<sub>3</sub>(g)<br/><br/>Given values are : p<sub>SO<sub>3</sub></sub> = 45kPa, p<sub>SO<sub>2</sub></sub> = 530 Pa = 0.53 kPa<br/><br/>p<sub>SO<sub>2</sub></sub> = 43 kPa<br/><br/>Now, $${K_p} = {{{{[{p_{S{O_3}(g)}}]}^2}} \over {{{[{p_{S{O_2}(g)}}]}... | integer | jee-main-2021-online-20th-july-morning-shift | 1,041 |
1krt6vg2f | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | Value of K<sub>P</sub> for the equilibrium reaction<br/><br/>N<sub>2</sub>O<sub>4(g)</sub> $$\rightleftharpoons$$ 2NO<sub>2(g)</sub> at 288 K is 47.9. The K<sub>C</sub> for this reaction at same temperature is ____________. (Nearest integer)<br/><br/>(R = 0.083 L bar K<sup>$$-$$1</sup> mol<sup>$$-$$1</sup>) | [] | null | 2 | $${K_C} = {{{K_P}} \over {RT}} = {{47.9} \over {0.083 \times 288}} = 2$$ | integer | jee-main-2021-online-22th-july-evening-shift | 1,042 |
1krutyzlo | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | For the reaction<br/><br/>A + B $$\rightleftharpoons$$ 2C<br/><br/>the value of equilibrium constant is 100 at 298 K. If the initial concentration of all the three species is 1 M each, then the equilibrium concentration of C is x $$\times$$ 10<sup>$$-$$1</sup> M. The value of x is ____________. (Nearest integer) | [] | null | 25 | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265358/exam_images/wpk0kq9fjtjueiqpsv61.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 25th July Morning Shift Chemistry - Chemical Equilibrium Question 43 English Explanation"> <br><br... | integer | jee-main-2021-online-25th-july-morning-shift | 1,043 |
1krxcezww | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | Assuming that Ba(OH)<sub>2</sub> is completely ionised in aqueous solution under the given conditions the concentration of H<sub>3</sub>O<sup>+</sup> ions in 0.005 M aqueous solution of Ba(OH)<sup>2</sup> at 298 K is ______________ $$\times$$ 10<sup>$$-$$12</sup> mol L<sup>$$-$$1</sup>. (Nearest integer) | [] | null | 1 | $$Ba{(OH)_2} \to B{a^{ + 2}} + 2O{H^ - } \downarrow 2 \times 0.005 = 0.01 = {10^{ - 2}}$$<br><br>At 298 K : in aq. solution $$[{H_3}{O^ + }][O{H^ - }] = {10^{ - 14}}$$<br><br>$$[{H_3}{O^ + }] = {{{{10}^{ - 14}}} \over {{{10}^{ - 2}}}} = {10^{ - 12}}$$ | integer | jee-main-2021-online-25th-july-evening-shift | 1,044 |
1krz31m2x | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | PCl<sub>5</sub> $$\rightleftharpoons$$ PCl<sub>3</sub> + Cl<sub>2</sub><br/><br/>K<sub>c</sub> = 1.844<br/><br/>3.0 moles of PCl<sub>5</sub> is introduced in a 1 L closed reaction vessel at 380 K. The number of moles of PCl<sub>5</sub> at equilibrium is ______________ $$\times$$ 10<sup>$$-$$3</sup>. (Round off to the N... | [] | null | 1400 | PCl<sub>5(g)</sub> $$\rightleftharpoons$$ PCl<sub>3(g)</sub> + Cl<sub>2(g)</sub> K<sub>2</sub> = 1.844<br><br>t = 0 3moles<br><br>t = $$\infty$$ x x<br><br>$$ \Rightarrow {{[PC{l_3}][C{l_2}]} \over {[PC{l_5}]}} = {{{x^2}} \over {3 - x}} = 1.844$$<br><br>$$ \Rightarrow {x^2} + 1.844 - 5.532 = 0$$<br><br>$$ \Rightarrow x... | integer | jee-main-2021-online-27th-july-morning-shift | 1,045 |
1ks1ixqw8 | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | The equilibrium constant for the reaction<br/><br/>A(s) $$\rightleftharpoons$$ M(s) + $${1 \over 2}$$O<sub>2</sub>(g)<br/><br/>is K<sub>p</sub> = 4. At equilibrium, the partial pressure of O<sub>2</sub> is _________ atm. (Round off to the nearest integer) | [] | null | 16 | k<sub>p</sub> = Po$$_2^{1/2}$$ = 4<br><br>$$\therefore$$ Po<sub>2</sub> = 16 bar = 16 atm | integer | jee-main-2021-online-27th-july-evening-shift | 1,046 |
1ktb5zysh | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | The OH<sup>$$-$$</sup> concentration in a mixture of 5.0 mL of 0.0504 M NH<sub>4</sub>Cl and 2 mL of 0.0210 M NH<sub>3</sub> solution is x $$\times$$ 10<sup>$$-$$6</sup> M. The value of x is ___________. (Nearest integer)<br/><br/>[Given K<sub>w</sub> = 1 $$\times$$ 10<sup>$$-$$14</sup> and K<sub>b</sub> = 1.8 $$\times... | [] | null | 3 | $$\left[ {NH_4^ + } \right]$$ = 0.0504 & [NH<sub>3</sub>] = 0.0210<br><br>So, $${K_b} = {{[NH_4^ + ][H{O^ - }]} \over {[N{H_3}]}}$$<br><br>$$[H{O^ - }] = {{{K_b} \times [N{H_3}]} \over {[NH_4^ + ]}} = 1.8 \times {10^{ - 5}} \times {2 \over 5} \times {{210} \over {504}} = 3 \times {10^{ - 6}}$$ | integer | jee-main-2021-online-26th-august-morning-shift | 1,047 |
1ktcry9cn | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | The equilibrium constant K<sub>c</sub> at 298 K for the reaction A + B $$\rightleftharpoons$$ C + D is 100. Starting with an equimolar solution with concentrations of A, B, C and D all equal to 1M, the equilibrium concentration of D is ___________ $$\times$$ 10<sup>$$-$$2</sup> M. (Nearest integer) | [] | null | 182 | <p> <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1ky4hs1vt/7e63d559-006a-4e38-b17c-173400ecb985/d91ab790-6fc5-11ec-8887-d75613c1bf3a/file-1ky4hs1vu.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1ky4hs1vt/7e63d559-006a-4e38-b17c-173400ecb985/d91ab790-6fc5-11ec-8887-d75613c1bf3... | integer | jee-main-2021-online-26th-august-evening-shift | 1,048 |
1ktctc6nr | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | The reaction rate for the reaction<br/><br/>[PtCl<sub>4</sub>]<sup>2$$-$$</sup> + H<sub>2</sub>O $$\rightleftharpoons$$ [Pt(H<sub>2</sub>O)Cl<sub>3</sub>]<sup>$$-$$</sup> + Cl<sup>$$-$$</sup><br/><br/>was measured as a function of concentrations of different species. It was observed that $${{ - d\left[ {{{\left[ {PtC{l... | [] | null | 0 | The rate equation provided in your question describes the rates of the forward and reverse reactions for this chemical system. The equilibrium constant, $K_c$, is the ratio of the forward rate constant ($k_f$) to the reverse rate constant ($k_r$). At equilibrium, the rate of the forward reaction is equal to the rate of... | integer | jee-main-2021-online-26th-august-evening-shift | 1,049 |
1kteel4k2 | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | The number of moles of NH<sub>3</sub>, that must be added to 2L of 0.80 M AgNO<sub>3</sub> in order to reduce the concentration of Ag<sup>+</sup> ions to 5.0 $$\times$$ 10<sup>$$-$$8</sup> M (K<sub>formation</sub> for [Ag(NH<sub>3</sub>)<sub>2</sub>]<sup>+</sup> = 1.0 $$\times$$ 10<sup>8</sup>) is ____________. (Neares... | [] | null | 4 | Let moles added = a<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267050/exam_images/zpyygyyimoipd1vht27c.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 27th August Morning Shift Chemistry - Chemical Equilibrium Question 36 ... | integer | jee-main-2021-online-27th-august-morning-shift | 1,050 |
1ktfupc62 | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | When 5.1 g of solid NH<sub>4</sub>HS is introduced into a two litre evacuated flask at 27$$^\circ$$C, 20% of the solid decomposes into gaseous ammonia and hydrogen sulphide. The K<sub>p</sub> for the reaction at 27$$^\circ$$C is x $$\times$$ 10<sup>$$-$$2</sup>. The value of x is _____________. (Integer answer) [Given ... | [] | null | 6 | moles of NH<sub>4</sub>HS initially taken = $${{5.1g} \over {51g/mol}}$$ = 0.1 mol<br><br>volume of vessel = 2 litre<br><br> <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l0p3p902/e913b719-7346-4ca3-8c6e-3f5922dcc83d/afc32410-a2b3-11ec-a4c5-57354ee38743/file-1l0p3p903.png?format=png" data-orsrc="http... | integer | jee-main-2021-online-27th-august-evening-shift | 1,051 |
1l54y48r6 | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | <p>4.0 moles of argon and 5.0 moles of PCl<sub>5</sub> are introduced into an evacuated flask of 100 litre capacity at 610 K. The system is allowed to equilibrate. At equilibrium, the total pressure of mixture was found to be 6.0 atm. The K<sub>p</sub> for the reaction is :</p>
<p>[Given : R = 0.082 L atm K<sup>$$-$$1<... | [{"identifier": "A", "content": "2.25"}, {"identifier": "B", "content": "6.24"}, {"identifier": "C", "content": "12.13"}, {"identifier": "D", "content": "15.24"}] | ["A"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5mkagsh/cdd0b8a6-eb19-44ed-9d57-1b85785143ab/4ac8df10-044b-11ed-b6d5-555c254d75e0/file-1l5mkagsi.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5mkagsh/cdd0b8a6-eb19-44ed-9d57-1b85785143ab/4ac8df10-044b-11ed-b6d5-555c254d75e0... | mcq | jee-main-2022-online-29th-june-evening-shift | 1,052 |
1l54z1uy4 | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | <p>A box contains 0.90 g of liquid water in equilibrium with water vapour at 27$$^\circ$$C. The equilibrium vapour pressure of water at 27$$^\circ$$C is 32.0 Torr. When the volume of the box is increased, some of the liquid water evaporates to maintain the equilibrium pressure. If all the liquid water evaporates, then ... | [] | null | 29 | <p>We know, 760 Torr = 1 atm</p>
<p>$$\therefore$$ 32 Torr = $${{32} \over {760}}$$ atm</p>
<p>As all the liquid water evaporates so entire water is in gaseous state.</p>
<p>$$\therefore$$ Weight of water vapour = 0.9 g</p>
<p>$$\therefore$$ Moles of water vapour (n) = $${{0.9} \over {18}}$$</p>
<p>Pressure (P) = $${{3... | integer | jee-main-2022-online-29th-june-evening-shift | 1,053 |
1l57szlay | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | <p>2NOCl(g) $$\rightleftharpoons$$ 2NO(g) + Cl<sub>2</sub>(g)</p>
<p>In an experiment, 2.0 moles of NOCl was placed in a one-litre flask and the concentration of NO after equilibrium established, was found to be 0.4 mol/L. The equilibrium constant at 30$$^\circ$$C is ______________ $$\times$$ 10<sup>$$-$$4</sup>.</p> | [] | null | 125 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5mje0dm/e9d41c32-a58b-4de7-8d83-428404631f67/c436d3b0-0447-11ed-9d9a-21bdd0f00aa4/file-1l5mje0dn.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5mje0dm/e9d41c32-a58b-4de7-8d83-428404631f67/c436d3b0-0447-11ed-9d9a-21bdd0f00aa4... | integer | jee-main-2022-online-27th-june-morning-shift | 1,054 |
1l5ammdi1 | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | <p>The standard free energy change ($$\Delta$$G$$^\circ$$) for 50% dissociation of N<sub>2</sub>O<sub>4</sub> into NO<sub>2</sub> at 27$$^\circ$$C and 1 atm pressure is $$-$$ x J mol<sup>$$-$$1</sup>. The value of x is ___________. (Nearest Integer)</p>
<p>[Given : R = 8.31 J K<sup>$$-$$1</sup> mol<sup>$$-$$1</sup>, lo... | [] | null | 710 | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l8mdj5pa/8120212a-92fd-4e3f-937a-70f4e4ec93ae/82bbdbe0-3f95-11ed-8fcd-e94b6c00f3b4/file-1l8mdj5pb.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l8mdj5pa/8120212a-92fd-4e3f-937a-70f4e4ec93ae/82bbdbe0-3f95-11ed-8fcd-e94b6c00f3b4/fi... | integer | jee-main-2022-online-25th-june-morning-shift | 1,055 |
1l5bdwvkz | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | <p>PCl<sub>5</sub> dissociates as</p>
<p>PCl<sub>5</sub>(g) $$\rightleftharpoons$$ PCl<sub>3</sub>(g) + Cl<sub>2</sub>(g)</p>
<p>5 moles of PCl<sub>5</sub> are placed in a 200 litre vessel which contains 2 moles of N<sub>2</sub> and is maintained at 600 K. The equilibrium pressure is 2.46 atm. The equilibrium constant ... | [] | null | 1107 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5mkc6w5/9bcd3b0f-bd25-4002-a7df-7f91aec78977/7ac1c150-044b-11ed-b6d5-555c254d75e0/file-1l5mkc6w6.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5mkc6w5/9bcd3b0f-bd25-4002-a7df-7f91aec78977/7ac1c150-044b-11ed-b6d5-555c254d75e0... | integer | jee-main-2022-online-24th-june-evening-shift | 1,056 |
1l5c689au | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | <p>For a reaction at equilibrium</p>
<p>A(g) $$\rightleftharpoons$$ B(g) + $${1 \over 2}$$ C(g)</p>
<p>the relation between dissociation constant (K), degree of dissociation ($$\alpha$$) and equilibrium pressure (p) is given by :</p> | [{"identifier": "A", "content": "$$K = {{{\\alpha ^{{1 \\over 2}}}{p^{{3 \\over 2}}}} \\over {{{\\left( {1 + {3 \\over 2}\\alpha } \\right)}^{{1 \\over 2}}}(1 - \\alpha )}}$$"}, {"identifier": "B", "content": "$$K = {{{\\alpha ^{{3 \\over 2}}}{p^{{1 \\over 2}}}} \\over {{{\\left( {2 + \\alpha } \\right)}^{{1 \\over 2}}... | ["B"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5mjgrml/bddc4d35-1e47-45de-af09-7895683e36e4/10e198d0-0448-11ed-9d9a-21bdd0f00aa4/file-1l5mjgrmm.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5mjgrml/bddc4d35-1e47-45de-af09-7895683e36e4/10e198d0-0448-11ed-9d9a-21bdd0f00aa4... | mcq | jee-main-2022-online-24th-june-morning-shift | 1,057 |
1l5c7268y | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | <p>2O<sub>3</sub>(g) $$\rightleftharpoons$$ 3O<sub>2</sub>(g)</p>
<p>At 300 K, ozone is fifty percent dissociated. The standard free energy change at this temperature and 1 atm pressure is ($$-$$) ____________ J mol<sup>$$-$$1</sup>. (Nearest integer)</p>
<p>[Given : ln 1.35 = 0.3 and R = 8.3 J K<sup>$$-$$1</sup> mol<s... | [] | null | 747 | $\underset{1-x}{2 \mathrm{O}_3(\mathrm{~g})} \rightleftharpoons \underset{\frac{3 \mathrm{x}}{2}}{3 \mathrm{O}_2(\mathrm{~g})}$<br/><br/>
Given, $x=0.5$
<br/><br/>
$\therefore \mathrm{k}_{\mathrm{p}}=\frac{[3(0.5)]^{3} \times 1}{[2]^{3} \times(0.5)^{2} \times 1.25}$
<br/><br/>
$\therefore \mathrm{k}_{\mathrm{p}}=\frac{... | integer | jee-main-2022-online-24th-june-morning-shift | 1,058 |
1l5w5kyld | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | <p>The equilibrium constant for the reversible reaction</p>
<p>2A(g) $$\rightleftharpoons$$ 2B(g) + C(g) is K<sub>1</sub></p>
<p>$${3 \over 2}$$A(g) $$\rightleftharpoons$$ $${3 \over 2}$$B(g) + $${3 \over 4}$$C(g) is K<sub>2</sub>.</p>
<p>K<sub>1</sub> and K<sub>2</sub> are related as :</p> | [{"identifier": "A", "content": "$${K_1} = \\sqrt {{K_2}} $$"}, {"identifier": "B", "content": "$${K_2} = \\sqrt {{K_1}} $$"}, {"identifier": "C", "content": "$${K_2} = K_1^{3/4}$$"}, {"identifier": "D", "content": "$${K_1} = K_2^{3/4}$$"}] | ["C"] | null | $2 \mathrm{A}(\mathrm{g})=2 \mathrm{B}(\mathrm{~g})+\mathrm{C}(\mathrm{g}) \quad K_{1} \quad\quad ...(i)$
<br/><br/>
$$
\frac{3}{2} \mathrm{A}(\mathrm{g})=\frac{3}{2} \mathrm{B}(\mathrm{g})+\frac{3}{4} \mathrm{C}(\mathrm{g}) \quad K_{2}\quad\quad...(ii)
$$
<br/><br/>
eq. (ii) is $\frac{3}{4}$ times of eq. (i), hence,
<... | mcq | jee-main-2022-online-30th-june-morning-shift | 1,059 |
1l6nxtp6c | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | <p>At $$600 \mathrm{~K}, 2 \mathrm{~mol}$$ of $$\mathrm{NO}$$ are mixed with $$1 \mathrm{~mol}$$ of $$\mathrm{O}_{2}$$.</p>
<p>$$2 \mathrm{NO}_{(\mathrm{g})}+\mathrm{O}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{NO}_{2}(\mathrm{g})$$</p>
<p>The reaction occurring as above comes to equilibrium under a total pressure of ... | [] | null | 2 | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7vyiyym/b3d0ff82-278a-44ca-8925-29649ed4cb48/73b469e0-310e-11ed-9c98-ad39d53b642b/file-1l7vyiyyn.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7vyiyym/b3d0ff82-278a-44ca-8925-29649ed4cb48/73b469e0-310e-11ed-9c98-ad39d53b642b/fi... | integer | jee-main-2022-online-28th-july-evening-shift | 1,061 |
1ldoltihj | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | <p>(i) $$\mathrm{X}(\mathrm{g}) \rightleftharpoons \mathrm{Y}(\mathrm{g})+\mathrm{Z}(\mathrm{g}) \quad \mathrm{K}_{\mathrm{p} 1}=3$$</p>
<p>(ii) $$\mathrm{A}(\mathrm{g}) \rightleftharpoons 2 \mathrm{~B}(\mathrm{g}) \quad \mathrm{K}_{\mathrm{p} 2}=1$$</p>
<p>If the degree of dissociation and initial concentration of bot... | [] | null | 12 | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1le7i662y/833b3985-b140-440c-8deb-848011ab2b75/31afaaa0-ae31-11ed-b364-6dadf34ccd07/file-1le7i662z.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1le7i662y/833b3985-b140-440c-8deb-848011ab2b75/31afaaa0-ae31-11ed-b364-6dadf34ccd07/fi... | integer | jee-main-2023-online-1st-february-morning-shift | 1,062 |
1ldpqfe1q | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | <p>For reaction : $$\mathrm{SO}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_{3}(\mathrm{~g})$$</p>
<p>$$\mathrm{K}_{\mathrm{p}}=2 \times 10^{12}$$ at $$27^{\circ} \mathrm{C}$$ and $$1 \mathrm{~atm}$$ pressure. The $$\mathrm{K}_{\mathrm{c}}$$ for the same reaction is _________... | [] | null | 1 | $$
\begin{aligned}
& \mathrm{SO}_{2(\mathrm{~g})}+\frac{1}{2} \mathrm{O}_{2(\mathrm{~g})} \rightleftharpoons \mathrm{SO}_{3(\mathrm{~g})} \\\\
& \mathrm{K}_{\mathrm{P}}=2 \times 10^{12} \text { at } 300 \mathrm{~K} \\\\
& \mathrm{~K}_{\mathrm{P}}=\mathrm{K}_{\mathrm{C}} \times(\mathrm{RT})^{\Delta \mathrm{n}_{\mathrm{g... | integer | jee-main-2023-online-31st-january-morning-shift | 1,063 |
1lgsza2h5 | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | <p>4.5 moles each of hydrogen and iodine is heated in a sealed ten litre vessel. At equilibrium, 3 moles of $$\mathrm{HI}$$ were found. The equilibrium constant for $$\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})$$ is _________.</p> | [] | null | 1 | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1liclq95i/c8c19de4-06d3-45eb-a68d-4a6e0ce940bd/e4643760-002f-11ee-a99f-43454a56aaa3/file-1liclq95j.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1liclq95i/c8c19de4-06d3-45eb-a68d-4a6e0ce940bd/e4643760-002f-11ee-a99f-43454a56aaa3/fi... | integer | jee-main-2023-online-11th-april-evening-shift | 1,066 |
1lgv10uu5 | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | <p>A mixture of 1 mole of $$\mathrm{H}_{2} \mathrm{O}$$ and 1 mole of $$\mathrm{CO}$$ is taken in a 10 litre container and heated to $$725 \mathrm{~K}$$. At equilibrium $$40 \%$$ of water by mass reacts with carbon monoxide according to the equation :
<br/><br/>$$\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\math... | [] | null | 44 | The given reaction is :
<br/><br/>$$\mathrm{CO(g)} + \mathrm{H_{2}O(g)} \rightleftharpoons \mathrm{CO_{2}(g)} + \mathrm{H_{2}(g)} $$
<br/><br/>We are given that $40\%$ of water by mass reacts with carbon monoxide.
<br/><br/>The molecular weight of water (H<sub>2</sub>O) is approximately $18 \, \mathrm{g/mol}$, so $... | integer | jee-main-2023-online-11th-april-morning-shift | 1,067 |
1lgvveuca | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | <p>$$\mathrm{A}(g) \rightleftharpoons 2 \mathrm{~B}(g)+\mathrm{C}(g)$$</p>
<p>For the given reaction, if the initial pressure is $$450 \mathrm{~mm} ~\mathrm{Hg}$$ and the pressure at time $$\mathrm{t}$$ is $$720 \mathrm{~mm} ~\mathrm{Hg}$$ at a constant temperature $$\mathrm{T}$$ and constant volume $$\mathrm{V}$$. The... | [] | null | 3 | <p>Given the reaction:</p>
<p>$$\mathrm{A}_{(\mathrm{g})} \rightleftharpoons 2 \mathrm{~B}_{(\mathrm{g})}+\mathrm{C}_{(\mathrm{g})}$$</p>
<p>At the beginning of the reaction (at time = 0), the total pressure is solely due to A, hence it is 450 mmHg.</p>
<p>As the reaction progresses, let's denote 'x' as the... | integer | jee-main-2023-online-10th-april-evening-shift | 1,068 |
1lh27t8xf | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | <p> For a concentrated solution of a weak electrolyte ($$\mathrm{K}_{\text {eq }}=$$ equilibrium constant) $$\mathrm{A}_{2} \mathrm{B}_{3}$$ of concentration '$$c$$', the degree of dissociation '$$\alpha$$' is :</p> | [{"identifier": "A", "content": "$$\\left(\\frac{K_{e q}}{25 c^{2}}\\right)^{\\frac{1}{5}}$$"}, {"identifier": "B", "content": "$$\\left(\\frac{K_{e q}}{108 c^{4}}\\right)^{\\frac{1}{5}}$$"}, {"identifier": "C", "content": "$$\\left(\\frac{K_{e q}}{5 c^{4}}\\right)^{\\frac{1}{5}}$$"}, {"identifier": "D", "content": "$$... | ["B"] | null | $$
\begin{aligned}
& \mathrm{A}_2 \mathrm{~B}_3(\mathrm{aq} .) \rightleftharpoons 2 \mathrm{~A}_{\text {(aq.) }}^{3+}+3 \mathrm{~B}_{\text {(aq) }}^{2-} \\\\
& \mathrm{c}(1-\alpha) \quad\quad\quad 2 \mathrm{c} \alpha\quad\quad\quad 3 \mathrm{c}\alpha\\\\
& \mathrm{K}_{\mathrm{eq}}=\frac{\left[\mathrm{A}^{3+}\right]^2\l... | mcq | jee-main-2023-online-6th-april-morning-shift | 1,069 |
1lh332lfs | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | <p>The equilibrium composition for the reaction $$\mathrm{PCl}_{3}+\mathrm{Cl}_{2} \rightleftharpoons \mathrm{PCl}_{5}$$ at $$298 \mathrm{~K}$$ is given below:</p>
<p>$$\left[\mathrm{PCl}_{3}\right]_{\mathrm{eq}}=0.2 \mathrm{~mol} \mathrm{~L}^{-1},\left[\mathrm{Cl}_{2}\right]_{\mathrm{eq}}=0.1 \mathrm{~mol} \mathrm{~L}... | [] | null | 49 | <p>The initial equilibrium concentrations are given as:<br/><br/>
$[\mathrm{PCl}_{3}]_{\text{eq}} = 0.2 \, \mathrm{mol/L}$, $[\mathrm{Cl}_{2}]_{\text{eq}} = 0.1 \, \mathrm{mol/L}$, $[\mathrm{PCl}_{5}]_{\text{eq}} = 0.4 \, \mathrm{mol/L}$.</p>
<p>After adding $0.2 \, \mathrm{mol}$ of $\mathrm{Cl_2}$, the new concentrati... | integer | jee-main-2023-online-6th-april-evening-shift | 1,070 |
jaoe38c1lse7hf3x | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | <p>For the given reaction, choose the correct expression of $$\mathrm{K}_{\mathrm{C}}$$ from the following :-</p>
<p>$$\mathrm{Fe}_{(\mathrm{aq})}^{3+}+\mathrm{SCN}_{(\mathrm{aq})}^{-} \rightleftharpoons(\mathrm{FeSCN})_{(\mathrm{aq})}^{2+}$$</p> | [{"identifier": "A", "content": "$$\\mathrm{K}_{\\mathrm{C}}=\\frac{\\left[\\mathrm{Fe}^{3+}\\right]\\left[\\mathrm{SCN}^{-}\\right]}{\\left[\\mathrm{FeSCN}^{2+}\\right]}$$\n"}, {"identifier": "B", "content": "$$\\mathrm{K}_{\\mathrm{C}}=\\frac{\\left[\\mathrm{FeSCN}^{2+}\\right]}{\\left[\\mathrm{Fe}^{3+}\\right]\\left... | ["B"] | null | <p>$$\begin{aligned}
& \mathrm{K}_{\mathrm{C}}=\frac{\text { Products ion conc. }}{\text { Reactants ion conc. }} \\
& \mathrm{K}_{\mathrm{C}}=\frac{\left[\mathrm{FeSCN}^{2+}\right]}{\left[\mathrm{Fe}^{3+}\right]\left[\mathrm{SCN}^{-}\right]}
\end{aligned}$$</p> | mcq | jee-main-2024-online-31st-january-morning-shift | 1,072 |
jaoe38c1lsfk477k | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | <p>For the reaction $$\mathrm{N}_2 \mathrm{O}_{4(\mathrm{~g})} \rightleftarrows 2 \mathrm{NO}_{2(\mathrm{~g})}, \mathrm{K}_{\mathrm{p}}=0.492 \mathrm{~atm}$$ at $$300 \mathrm{~K} . \mathrm{K}_{\mathrm{c}}$$ for the reaction at same temperature is _________ $$\times 10^{-2}$$.</p>
<p>(Given : $$\mathrm{R}=0.082 \mathrm{... | [] | null | 2 | <p>$$\begin{aligned}
& \mathrm{K}_{\mathrm{P}}=\mathrm{K}_{\mathrm{C}} \cdot(\mathrm{RT})^{\Delta \mathrm{n}_{\mathrm{g}}} \\
& \Delta \mathrm{n}_{\mathrm{g}}=1 \\
& \Rightarrow \mathrm{K}_{\mathrm{c}}=\frac{\mathrm{K}_{\mathrm{P}}}{\mathrm{RT}}=\frac{0.492}{0.082 \times 300}=2 \times 10^{-2}
\end{aligned}$$</p> | integer | jee-main-2024-online-29th-january-morning-shift | 1,073 |
lv2er9pz | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | <p>The equilibrium constant for the reaction</p>
<p>$$\mathrm{SO}_3(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})$$</p>
<p>is $$\mathrm{K}_{\mathrm{c}}=4.9 \times 10^{-2}$$. The value of $$\mathrm{K}_{\mathrm{c}}$$ for the reaction given below is $$2 \mathrm{SO}_2(\mat... | [{"identifier": "A", "content": "49"}, {"identifier": "B", "content": "416"}, {"identifier": "C", "content": "41.6"}, {"identifier": "D", "content": "4.9"}] | ["B"] | null | <p>The reaction</p>
<p>$$2 \mathrm{SO}_2+\mathrm{O}_2 \rightleftharpoons 2 \mathrm{SO}_3$$</p>
<p>can be formed from the given reaction by reverting it and multiplying coefficients by 2.</p>
<p>Thus,</p>
<p>$$\begin{aligned}
\mathrm{K}_c^{\prime} & =\mathrm{K}_{\mathrm{c}}^{-2}=\frac{1}{\mathrm{~K}_{\mathrm{c}}^2}=\lef... | mcq | jee-main-2024-online-4th-april-evening-shift | 1,075 |
lv5gspnf | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | <p>For the given hypothetical reactions, the equilibrium constants are as follows :</p>
<p>$$\begin{aligned}
& \mathrm{X} \rightleftharpoons \mathrm{Y} ; \mathrm{K}_1=1.0 \\
& \mathrm{Y} \rightleftharpoons \mathrm{Z} ; \mathrm{K}_2=2.0 \\
& \mathrm{Z} \rightleftharpoons \mathrm{W} ; \mathrm{K}_3=4.0
\end{al... | [{"identifier": "A", "content": "12.0"}, {"identifier": "B", "content": "8.0"}, {"identifier": "C", "content": "6.0"}, {"identifier": "D", "content": "7.0"}] | ["B"] | null | <p>To find the equilibrium constant for the overall reaction $$\mathrm{X} \rightleftharpoons \mathrm{W}$$, we need to combine the equilibrium constants for the individual reactions given. Let's analyze this step-by-step:</p>
<p>The given reactions and their equilibrium constants are:</p>
<p>$$\begin{aligned} & \mathr... | mcq | jee-main-2024-online-8th-april-morning-shift | 1,076 |
lv9s2805 | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | <p>Given below are two statements :</p>
<p>Statement I : On passing $$\mathrm{HCl}_{(\mathrm{g})}$$ through a saturated solution of $$\mathrm{BaCl}_2$$, at room temperature white turbidity appears.</p>
<p>Statement II : When $$\mathrm{HCl}$$ gas is passed through a saturated solution of $$\mathrm{NaCl}$$, sodium chlori... | [{"identifier": "A", "content": "Both Statement I and Statement II are correct\n"}, {"identifier": "B", "content": "Statement I is incorrect but Statement II is correct\n"}, {"identifier": "C", "content": "Both Statement I and Statement II are incorrect\n"}, {"identifier": "D", "content": "Statement I is correct but St... | ["D"] | null | <p>To determine the correctness of the statements, we'll analyze each one individually by applying principles of solubility, common ion effect, and chemical equilibria.</p>
<hr />
<h3><strong>Statement I:</strong></h3>
<p><strong>On passing $\mathrm{HCl}_{(g)}$ through a saturated solution of $\mathrm{BaCl}_2$ at room ... | mcq | jee-main-2024-online-5th-april-evening-shift | 1,077 |
lvb2a7x3 | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | <p>The ratio $$\frac{K_P}{K_C}$$ for the reaction :</p>
<p>$$\mathrm{CO}_{(\mathrm{g})}+\frac{1}{2} \mathrm{O}_{2(\mathrm{~g})} \rightleftharpoons \mathrm{CO}_{2(\mathrm{~g})}$$ is :</p> | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "$$\n(\\mathrm{RT})^{1 / 2}\n$$"}, {"identifier": "C", "content": "RT"}, {"identifier": "D", "content": "$$\\mathrm{\n\\frac{1}{\\sqrt{R T}}}\n$$"}] | ["D"] | null | <p>To solve this problem, we need to understand the relationship between the equilibrium constant in terms of pressure, $$K_P$$, and the equilibrium constant in terms of concentration, $$K_C$$. The relationship between these two constants for a general reaction is given by:
<p>$$ K_P = K_C (RT)^{\Delta n} $$</p>
<p>w... | mcq | jee-main-2024-online-6th-april-evening-shift | 1,078 |
lvc587ob | chemistry | chemical-equilibrium | chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant | <p>At $$-20^{\circ} \mathrm{C}$$ and $$1 \mathrm{~atm}$$ pressure, a cylinder is filled with equal number of $$\mathrm{H}_2, \mathrm{I}_2$$ and $$\mathrm{HI}$$ molecules for the reaction
$$\mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})$$, the $$\mathrm{K}_{\mathrm{p}}$$... | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "10"}, {"identifier": "D", "content": "0.01"}] | ["C"] | null | <p>At $$-20^{\circ} \mathrm{C}$$ and a pressure of $$1 \mathrm{~atm}$$, a cylinder contains equal quantities of $$\mathrm{H}_2, \mathrm{I}_2,$$ and $$\mathrm{HI}$$ molecules. The equilibrium constant for the reaction</p>
<p>$$\mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{... | mcq | jee-main-2024-online-6th-april-morning-shift | 1,079 |
lAbL1dOH9madyBTF | chemistry | chemical-equilibrium | le-chatelier's-principle-and-factors-affecting-chemical-equilibrium | Change in volume of the system does not alter which of the following equilibria? | [{"identifier": "A", "content": "N<sub>2</sub>(g) + O<sub>2</sub>(g) $$\\leftrightharpoons$$ 2NO (g)"}, {"identifier": "B", "content": "PCl<sub>5</sub>(g) $$\\leftrightharpoons$$ PCl<sub>3</sub> (g) + Cl<sub>2</sub> (g)"}, {"identifier": "C", "content": "N<sub>2</sub>(g) + 3H<sub>2</sub>(g) $$\\leftrightharpoons$$ 2NH<... | ["A"] | null | In this reaction the ratio of number of moles of reactants to products is same $$i.e.\,\,2:2,$$ hence change in volume will not alter the number of moles. | mcq | aieee-2002 | 1,080 |
3F69yMnkhThKnTAU | chemistry | chemical-equilibrium | le-chatelier's-principle-and-factors-affecting-chemical-equilibrium | Consider the reaction equilibrium<br/>
2 SO<sub>2</sub> (g) + O<sub>2</sub> (g) $$\leftrightharpoons$$ 2 SO<sub>3</sub> (g); $$\Delta H^o$$ = -198 kJ<br/>
One the basis of Le Chatelier's principle, the condition favourable for the forward reaction is : | [{"identifier": "A", "content": "increasing temperature as well as pressure"}, {"identifier": "B", "content": "lowering the temperature and increasing the pressure"}, {"identifier": "C", "content": "any value of temperature and pressure"}, {"identifier": "D", "content": "lowering temperature as well as pressure"}] | ["B"] | null | Due to exothermicity of reaction low or optimum temperature will be required. Since $$3$$ moles are changing to $$2$$ moles.
<br><br>$$\therefore$$ High pressure will be required. | mcq | aieee-2003 | 1,081 |
iCB432yoEFUORD9G | chemistry | chemical-equilibrium | le-chatelier's-principle-and-factors-affecting-chemical-equilibrium | The exothermic formation of ClF<sub>3</sub> is represented by the equation: <br/>
Cl<sub>2</sub> (g) + 3F<sub>2</sub> (g) $$\leftrightharpoons$$ 2ClF<sub>3</sub> (g); $$\Delta H$$ = -329 kJ<br/>
Which of the following will increase the quantity of ClF<sub>3</sub> in an equilibrium mixture of
Cl<sub>2</sub>, F<sub>2</su... | [{"identifier": "A", "content": "Increasing the temperature"}, {"identifier": "B", "content": "Removing Cl<sub>2</sub>"}, {"identifier": "C", "content": "Increasing the volume of the container "}, {"identifier": "D", "content": "Adding F<sub>2</sub>"}] | ["D"] | null | The reaction given is an exothermic reaction thus accordingly to Lechatalier's principle lowering of temperature, addition of $${F_2}$$ and or $$C{l_2}$$ favour the for ward direction and hence the production of $$Cl{F_3}.$$ | mcq | aieee-2005 | 1,082 |
6vmuqmKCuf8FOH8rAjOWz | chemistry | chemical-equilibrium | le-chatelier's-principle-and-factors-affecting-chemical-equilibrium | The following reaction occurs in the Blast Furnace where iron ore is reduced to iron
metal :
<br/><br/>Fe<sub>2</sub>O<sub>3</sub>(s) + 3 CO(g) $$\rightleftharpoons$$ 2 Fe(1) + 3 CO<sub>2</sub>(g)
<br/><br/>Using the Le Chatelier’s principle, predict which one of the following will not disturb the equilibrium ? | [{"identifier": "A", "content": "Removal of CO "}, {"identifier": "B", "content": "Removal of CO<sub>2</sub> "}, {"identifier": "C", "content": "Addition of CO<sub>2</sub>"}, {"identifier": "D", "content": "Addition of Fe<sub>2</sub>O<sub>3</sub>"}] | ["D"] | null | <p>Addition of a solid component to a system at constant
pressure has no effect on the equilibrium. Therefore,
addition of Fe<sub>2</sub>O<sub>3</sub> will not disturb the equilibrium.</p> | mcq | jee-main-2017-online-9th-april-morning-slot | 1,083 |
O2mqgIpwasZAg7L4e5sLG | chemistry | chemical-equilibrium | le-chatelier's-principle-and-factors-affecting-chemical-equilibrium | In which of the following reactions, an increase in the volume of the container will favour the formation of products? | [{"identifier": "A", "content": "2NO<sub>2</sub>(g) $$\\rightleftharpoons$$ 2NO(g) + O<sub>2</sub>(g)"}, {"identifier": "B", "content": "H<sub>2</sub>(g) + I<sub>2</sub>(g) $$\\rightleftharpoons$$ 2HI(g)"}, {"identifier": "C", "content": "4NH<sub>3</sub>(g) + 5O<sub>2</sub>(g) $$\\rightleftharpoons$$ 4NO(g) + 6H<sub>2<... | ["A"] | null | From Boyle's law, we know when volume increase then pressure decreases, and to keep the pressure on original value the reaction should proceeds in the direction where the number of moles of gases increases.
<br><br>In this reaction, only satisfy this.
<br><br>2 NO<sub>2</sub>(g) $$\rightleftharpoons$$ 2 NO(g) + O<sub... | mcq | jee-main-2018-online-15th-april-morning-slot | 1,084 |
ujFoukI0ZKA9R7QeWXVur | chemistry | chemical-equilibrium | le-chatelier's-principle-and-factors-affecting-chemical-equilibrium | The gas phase reaction 2NO<sub>2</sub>(g) $$ \to $$ N<sub>2</sub>O<sub>4</sub>(g) is an exothermic reaction. The decomposition of N<sub>2</sub>O<sub>4</sub>, in equilibrium mixture of NO<sub>2</sub>(g) and N<sub>2</sub>O<sub>4</sub>(g), can be increased by : | [{"identifier": "A", "content": "lowering the temperature."}, {"identifier": "B", "content": "increasing the pressure."}, {"identifier": "C", "content": "addition of an inert gas at constant volume. "}, {"identifier": "D", "content": "addition of an inert gas at constant pressure. "}] | ["D"] | null | 2NO<sub>2</sub> (g) $$\buildrel \, \over
\longrightarrow $$ N<sub>2</sub>O<sub>4</sub> (g); $$\Delta $$H = $$-$$ ve
<br><br>At equilibrium, N<sub>2</sub>O<sub>4</sub> $$\rightleftharpoons$$ 2NO<sub>2</sub> ; $$\Delta $$H = + ve
<br><br> On adding the inert gas at constant pressure, the number of moles per unit volum... | mcq | jee-main-2018-online-16th-april-morning-slot | 1,085 |
d02h5fIw8PMD42WGT93rsa0w2w9jx0x1gkb | chemistry | chemical-equilibrium | le-chatelier's-principle-and-factors-affecting-chemical-equilibrium | For the reaction,
<br/> 2SO<sub>2</sub>(g) + O<sub>2</sub>(g) = 2SO<sub>3</sub>(g), $$\Delta $$H = –57.2 kJ mol<sup>–1</sup>
and K<sub>C</sub> = 1.7 × 10<sup>16</sup>
<br/>Which of the following statement is incorrect ? | [{"identifier": "A", "content": "The equilibrium will shift in forward direction as the pressure increase."}, {"identifier": "B", "content": "The addition of inert gas at constant volume will be not affect the equilibrium constant."}, {"identifier": "C", "content": "The equilibrium constant is large suggestive of react... | ["C"] | null | Equilibrium constant has no relation with
catalyst.
<br><br>Catalyst only affects the rate with
which a reaction proceeds.
<br><br>Here we use catalyst V<sub>2</sub>O<sub>5</sub> to
speed up the reaction. | mcq | jee-main-2019-online-10th-april-evening-slot | 1,086 |
bkjyti2pc4gu9E5DFQ3rsa0w2w9jx83y3v4 | chemistry | chemical-equilibrium | le-chatelier's-principle-and-factors-affecting-chemical-equilibrium | The INCORRECT match in the following is : | [{"identifier": "A", "content": "$$\\Delta $$G<sup>o</sup>\n = 0, K = 1"}, {"identifier": "B", "content": "$$\\Delta $$G<sup>o</sup>\n < 0, K < 1"}, {"identifier": "C", "content": "$$\\Delta $$G<sup>o</sup>\n > 0, K < 1"}, {"identifier": "D", "content": "$$\\Delta $$G<sup>o</sup>\n < 0, K > 1"}] | ["B"] | null | We know, $$\Delta {G^o} = - RT\ln K$$
<br>Case 1 :
<br>If $$\Delta $$G<sup>o</sup> < 0
<br>$$ \Rightarrow $$ $$- RT\ln K$$ < 0
<br>$$ \Rightarrow $$ $$\ln K$$ > 0
<br>$$ \Rightarrow $$ K > 1
<br><br>Case 2 :
<br>If $$\Delta $$G<sup>o</sup> > 0
<br>$$ \Rightarrow $$ $$- RT\ln K$$ > 0
<br>$$ \R... | mcq | jee-main-2019-online-12th-april-evening-slot | 1,087 |
ldqy3up7 | chemistry | chemical-equilibrium | le-chatelier's-principle-and-factors-affecting-chemical-equilibrium | Consider the following equation:
<br/><br/>
$2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g), \Delta H=-190 \mathrm{~kJ}$
<br/><br/>
The number of factors which will increase the yield of $\mathrm{SO}_{3}$ at equilibrium from the following is _______.<br/><br/>
A. Increasing temperature<... | [] | null | 3 | <p>$$\mathrm{2SO_2+O_2\rightleftharpoons 2SO_3~~\Delta H=-190~kJ}$$</p>
<p>It is an exothermic reaction</p>
<p>$$\therefore$$ factor B, C, D will increase the amount of SO$$_3$$.</p> | integer | jee-main-2023-online-30th-january-evening-shift | 1,090 |
1ldsdsiwh | chemistry | chemical-equilibrium | le-chatelier's-principle-and-factors-affecting-chemical-equilibrium | <p>At 298 K</p>
<p>$$\mathrm{N_2~(g)+3H_2~(g)\rightleftharpoons~2NH_3~(g),~K_1=4\times10^5}$$</p>
<p>$$\mathrm{N_2~(g)+O_2~(g)\rightleftharpoons~2NO~(g),~K_2=1.6\times10^{12}}$$</p>
<p>$$\mathrm{H_2~(g)+\frac{1}{2}O_2~(g)\rightleftharpoons~H_2O~(g),~K_3=1.0\times10^{-13}}$$</p>
<p>Based on above equilibria, then equili... | [] | null | 4 | <p>$$\mathrm{2NH_3(g)\frac{5}{2}O_2(g)\rightleftharpoons 2NO(g)+3H_2O(g)}$$</p>
<p>Clearly, $$\mathrm{{K_{eq}} = {1 \over {{k_1}}} \times {k_2} \times k_3^3}$$</p>
<p>$$ = {{1.6 \times {{10}^{12}} \times {{10}^{ - 39}}} \over {4 \times {{10}^5}}}$$</p>
<p>$$ = 0.4 \times {10^{ - 32}}$$</p>
<p>$$ = 4 \times {10^{ - 33}}... | integer | jee-main-2023-online-29th-january-evening-shift | 1,091 |
1lgyhl1dn | chemistry | chemical-equilibrium | le-chatelier's-principle-and-factors-affecting-chemical-equilibrium | <p>The number of correct statement/s involving equilibria in physical processes from the following is ________</p>
<p>(A) Equilibrium is possible only in a closed system at a given temperature.</p>
<p>(B) Both the opposing processes occur at the same rate.</p>
<p>(C) When equilibrium is attained at a given temperature,... | [] | null | 3 | <p><b>(A) Equilibrium is possible only in a closed system at a given temperature.</b></p>
<ul>
<li>This statement is true. In a closed system, there is no exchange of matter with the surroundings. This allows the reaction to reach equilibrium at a constant temperature.</li>
</ul>
<p><b>(B) Both the opposing processes o... | integer | jee-main-2023-online-10th-april-morning-shift | 1,092 |
lv7v4nyl | chemistry | chemical-equilibrium | le-chatelier's-principle-and-factors-affecting-chemical-equilibrium | <p>The following reaction occurs in the Blast furnance where iron ore is reduced to iron metal</p>
<p>$$\mathrm{Fe}_2 \mathrm{O}_{3(s)}+3 \mathrm{CO}_{(g)} \rightleftharpoons \mathrm{Fe}_{(j)}+3 \mathrm{CO}_{2(g)}$$</p>
<p>Using the Le-chatelier's principle, predict which one of the following will not disturb the equil... | [{"identifier": "A", "content": "Addition of $$\\mathrm{CO}_2$$\n"}, {"identifier": "B", "content": "Removal of $$\\mathrm{CO}$$\n"}, {"identifier": "C", "content": "Addition of $$\\mathrm{Fe}_2 \\mathrm{O}_3$$\n"}, {"identifier": "D", "content": "Removal of $$\\mathrm{CO}_2$$"}] | ["C"] | null | <p>For the reaction :</p>
<p>$$\mathrm{Fe}_2 \mathrm{O}_{3(\mathrm{~s})}+3 \mathrm{CO}_{(\mathrm{g})} \rightleftharpoons \mathrm{Fe}_{(\mathrm{l})}+3 \mathrm{CO}_{2(\mathrm{~g})}$$</p>
<p>Addition or removal of $$\mathrm{Fe}_2 \mathrm{O}_{3(\mathrm{s})}$$ and/or $$\mathrm{Fe}_{\text {(l) }}$$ will not affect the equili... | mcq | jee-main-2024-online-5th-april-morning-shift | 1,093 |
VcCAgyN3fplIli6M | chemistry | chemical-kinetics-and-nuclear-chemistry | arrhenius-equation | In respect of the equation k = Ae<sup>-E<sub>a</sub>/RT</sup> in chemical kinetics, which one of the following statements is correct? | [{"identifier": "A", "content": "A is adsorption factor"}, {"identifier": "B", "content": "E<sub>a</sub> is energy of activation"}, {"identifier": "C", "content": "R is Rydberg\u2019s constant"}, {"identifier": "D", "content": "k is equilibrium constant"}] | ["B"] | null | In equation $$K = A{e^{ - {E_a}/RT}};A = $$ Frequency factor
<br><br>$$K = $$ velocity constant, $$R=$$ gas constant and
<br><br>$${E_b} = $$ energy of activation | mcq | aieee-2003 | 1,094 |
kJAXWKzFjLzd8NnZ | chemistry | chemical-kinetics-and-nuclear-chemistry | arrhenius-equation | A schematic plot of $$ln$$ $${K_{eq}}$$ versus inverse of temperature for a reaction is shown below
<br/><br/><img src="data:image/png;base64,UklGRjgMAABXRUJQVlA4ICwMAAAwnACdASoAA70BP4HA2mY2MKwnIbLIqsAwCWlu4W+1OmNwvx5E/0f9e70/95tfV/Otd21eyP5QfOejWgC/PPMfnMfgbNaeV/+O/uCoCJaxwo9K3x5boaSOFHpW+PLdDSRwj0NsO0Br7yRYGzEelond9N... | [{"identifier": "A", "content": "highly spontaneous at ordinary temperature"}, {"identifier": "B", "content": "one with negligible enthalpy change "}, {"identifier": "C", "content": "endothermic "}, {"identifier": "D", "content": "exothermic "}] | ["D"] | null | The graph show that reaction is exothermic.
<br><br>$$\log \,k = {{ - \Delta H} \over {RT}} + 1$$
<br><br>For exothermic reaction $$\Delta H < 0$$
<br><br>$$\therefore$$ $$\,\,\,\,\,log\,\,k\,\,Vs{1 \over T}\,\,$$ would be negative straight line with positive slope. | mcq | aieee-2005 | 1,095 |
bFM8KL83corCGEsg | chemistry | chemical-kinetics-and-nuclear-chemistry | arrhenius-equation | Rate of a reaction can be expressed by Arrhenius equation as: <br/>
$$$k = A\,{e^{ - E/RT}}$$$
In this equation, E represents | [{"identifier": "A", "content": "the energy above which all the colliding molecules will react"}, {"identifier": "B", "content": "the energy below which colliding molecules will not react "}, {"identifier": "C", "content": "the total energy of the reacting molecules at a temperature, T "}, {"identifier": "D", "content"... | ["A"] | null | In Arrhenius equation $$K = A\,{e^{ - E/RT}},\,\,E$$ is the energy of activation, which is required by the colliding molecules to react resulting in the formation of products. | mcq | aieee-2006 | 1,096 |
HCILgcwKXnYVtRGp | chemistry | chemical-kinetics-and-nuclear-chemistry | arrhenius-equation | The energies of activation for forward and reverse reactions for A<sub>2</sub> + B<sub>2</sub> $$\leftrightharpoons$$ 2AB are 180 kJ mol<sup>−1</sup>
and 200 kJ mol<sup>−1</sup> respectively. The presence of catalyst lowers the activation energy of both (forward and reverse) reactions by 100 kJ mol<sup>−1</sup>. The en... | [{"identifier": "A", "content": "300"}, {"identifier": "B", "content": "120"}, {"identifier": "C", "content": "200"}, {"identifier": "D", "content": "20"}] | ["D"] | null | $$\Delta {H_R} = {E_f} - {E_b} = 180 - 200 = - 20kJ/mol$$
<br><br>The nearest correct answer given in choices may be obtained by neglecting sign. | mcq | aieee-2007 | 1,097 |
hfgD93hASiXKWbBb | chemistry | chemical-kinetics-and-nuclear-chemistry | arrhenius-equation | The rate of a reaction doubles when its temperature changes from 300K to 310K. Activation energy of such
a reaction will be:(R = 8.314 JK<sup>–1</sup> mol<sup>–1</sup> and log 2 = 0.301) | [{"identifier": "A", "content": "48.6 kJ mol<sup>\u20131</sup> "}, {"identifier": "B", "content": "58.5 kJ mol<sup>\u20131</sup>"}, {"identifier": "C", "content": "60.5 kJ mol<sup>\u20131</sup> "}, {"identifier": "D", "content": "53.6 kJ mol<sup>\u20131</sup>"}] | ["D"] | null | Activation energy can be calculated from the equation
<br><br>$${{\log \,{k_2}} \over {\log \,{k_1}}} = {{{E_a}} \over {2.303R}}\left( {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right)$$
<br><br>given $${{{k_2}} \over {{k_1}}} = 2\,\,{T_2} = 310\,K\,\,{T_1} = 300\,K$$
<br><br>$$ = \log 2 = {{ - {E_a}} \over {2.303 \tim... | mcq | jee-main-2013-offline | 1,098 |
dwGTF5etVJOvFtl9 | chemistry | chemical-kinetics-and-nuclear-chemistry | arrhenius-equation | Two reactions R<sub>1</sub> and R<sub>2</sub> have identical pre-exponential factors. Activation energy of R<sub>1</sub> exceeds that of R<sub>2</sub> by 10 kJ mol<sup>–1</sup>. If k<sub>1</sub> and k<sub>2</sub> are rate constants for reactions R<sub>1</sub> and R<sub>2</sub> respectively at 300 K, then ln(k<sub>2</su... | [{"identifier": "A", "content": "12"}, {"identifier": "B", "content": "6"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "8"}] | ["C"] | null | We know, from arrhenius equation,
<br><br>k = A.$${e^{{{ - {E_a}} \over {RT}}}}$$
<br><br>$$ \therefore $$ k<sub>1</sub> = A.$${e^{{{ - {E_{{a_1}}}} \over {RT}}}}$$ ......(1)
<br><br>k<sub>2</sub> = A.$${e^{{{ - {E_{{a_2}}}} \over {RT}}}}$$ ......(2)
<br><br>On dividing equation (2) by (1), we get
<br><br>$${{{k_2}} \o... | mcq | jee-main-2017-offline | 1,099 |
GXzTlAJFalSuW7k6TcPVS | chemistry | chemical-kinetics-and-nuclear-chemistry | arrhenius-equation | The rate of a reaction A doubles on increasing the temperature from 300 to 310 K. By how much, the temperature of reaction B should be Increased from 300 K so that rate doubles if activation energy of the reaction B is twice to that of reaction A. | [{"identifier": "A", "content": "9.84 K"}, {"identifier": "B", "content": "4.92 K"}, {"identifier": "C", "content": "2.45 K"}, {"identifier": "D", "content": "19.67 K"}] | ["B"] | null | For reaction A, T<sub>1</sub> = 300 K, T<sub>2</sub> = 310 K, k<sub>2</sub> = 2 k<sub>1</sub>
<br><br>$$\log {{{k_2}} \over {{k_1}}} = {{{E_{{a_1}}}} \over {2.303R}}\left[ {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right]$$
<br><br>$$ \therefore $$ $$\log {{2{k_1}} \over {{k_1}}} = \log 2 = {{{E_{{a_1}}}} \over {2.303R}}... | mcq | jee-main-2017-online-8th-april-morning-slot | 1,100 |
OU0GIizMN5uNzlPOBbG31 | chemistry | chemical-kinetics-and-nuclear-chemistry | arrhenius-equation | The rate of a reaction quadruples when the temperature changes from 300 to 310 K. The activation energy of this reaction is :
<br/><br/>(Assume activation energy and preexponential factor are independent of
temperature; ln 2 = 0.693; R = 8.314 J mol<sup>−1</sup> K<sup>−1</sup>)
| [{"identifier": "A", "content": "107.2 kJ mol<sup>$$-$$1</sup>"}, {"identifier": "B", "content": "53.6 kJ mol<sup>$$-$$1</sup>"}, {"identifier": "C", "content": "26.8 kJ mol<sup>$$-$$1</sup>"}, {"identifier": "D", "content": "214.4 kJ mol<sup>$$-$$1</sup>"}] | ["A"] | null | <p>According to Arrhenius equation,</p>
<p>$$\log {{{k_2}} \over {{k_1}}} = {{{E_a}} \over {2.303R}}\left( {{{{T_2} - {T_1}} \over {{T_1}{T_2}}}} \right)$$</p>
<p>Substituting the given values in the equation, we get</p>
<p>$$\log 4 = {{{E_a}} \over {2.303 \times 8.314\,J\,{K^{ - 1}}mo{l^{ - 1}}}}\left( {{{310 - 300} \... | mcq | jee-main-2017-online-9th-april-morning-slot | 1,101 |
BuRyLYEbdumiFeMieMLSG | chemistry | chemical-kinetics-and-nuclear-chemistry | arrhenius-equation | Consider the given plots for a reaction obeying Arrhenius equation (0<sup>o</sup>C < T < 300<sup>o</sup>C) : (K and Ea are rate constant and activation energy, respectively)
<br/><br/><img src="data:image/png;base64,UklGRkYLAABXRUJQVlA4IDoLAACQmwCdASoAA8UBP4HA2WY2MCynIVKZCsAwCWlu/Eb5M1fHZ1/fsj2mZBcbQdHdPZ4X3uzMz... | [{"identifier": "A", "content": "I is right but II is wrong "}, {"identifier": "B", "content": "Both I and II are correct "}, {"identifier": "C", "content": "Both I and II are wrong "}, {"identifier": "D", "content": "I is wrong but II is right "}] | ["B"] | null | Arrhenius Equation -
<br><br>
Arrhenius gave the quantitative dependence of rate constant on temperature by the Arrhenius equation.
<br><br>
- wherein<br><br>
$$k = A{e^{ - {E_a}/RT}}$$<br><br>
$${\mathop{\rm lnk}\nolimits} = lnA - {{{E_a}} \over {RT}}$$<br><br>
k = Rate constant
<br><br>
as we have learned in Arrheni... | mcq | jee-main-2019-online-10th-january-morning-slot | 1,102 |
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