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JEE • CHEMISTRY • Equilibrium

Equilibrium practice · Medium

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 61-90 of 401 questions
61 Medium
61 of 401
Application +4 | -1
The solubility product of Zn(OH)2\mathrm{Zn}(\mathrm{OH})_2 is 10−1410^{-14} at 25∘C.25^{\circ} \mathrm{C}. What would be the concentration Zn+2\mathrm{Zn}^{+2} ion in 0.1 M−NH4OH0.1 \, \mathrm{M}-\mathrm{NH}_4 \mathrm{OH} solution which is 50%50 \% ionized?
View answer and explanation
Correct option: B
4×10−124 \times 10^{-12}
Option B (4×10−124 \times 10^{-12}) is correct because it is calculated by dividing the solubility product by the square of the hydroxide ion concentration.
62 Medium
62 of 401
Analyzing +4 | -1
Statement (S): The pH\mathrm{pH} of human blood at body temperature is found to be 6.9.

Explanation (E): Blood is alkaline in nature.
View answer and explanation
Correct option: D
Both S and E are correct but E is not correct explanation for S
The correct answer is D (Both S and E are correct but E is not correct explanation for S) because human blood is slightly alkaline, and its pH at body temperature is around 7.4, not 6.9. The alkaline nature is due to buffer systems, not just temperature.
63 Medium
63 of 401
Application +4 | -1
The solubility of AgCNAgCN in a buffer solution of pH=3pH = 3 is xx. The value of XX is…..
[Assume: No cyano complex is formed; Ksp(AgCN)=2.2×10−16{K_{sp}}\left( {AgCN} \right) = 2.2 \times {10^{ - 16}} and Ka(HCN)=6.2×10−10{K_a}\left( {HCN} \right) = 6.2 \times {10^{ - 10}}]
View answer and explanation
Correct option: D
1.9×10−51.9 \times {10^{ - 5}}
Option D (1.9×10−51.9 \times {10^{ - 5}}) is correct because it is calculated by setting up an equilibrium expression involving the solubility product (Ksp) of AgCN, the acid dissociation constant (Ka) of HCN, and the hydrogen ion concentration at the given pH.
64 Medium
64 of 401
Application +4 | -1
3.2 moles of hydrogen iodide were heated in a sealed bulb at 444∘C444^\circ C till the equilibrium state was reached. Its degree of dissociation at this temperature was found to be 22%.22\% . The number of moles of hydrogen iodide present at equilibrium are
View answer and explanation
Correct option: B
2.496
Option B (2.496) is correct because the equilibrium moles of HI are calculated by subtracting the dissociated moles (22% of initial) from the initial moles.
65 Medium
65 of 401
Understanding +4 | -1
When acidified K2Cr2O7{K_2}C{r_2}{O_7} solution is added to Sn2+S{n^{2 + }} salt then Sn2+S{n^{2 + }} changes to
View answer and explanation
Correct option: C
Sn4+S{n^{4 + }}
The correct answer is C (Sn4+S{n^{4 + }}) because in the redox reaction, Sn2+S{n^{2 + }} is oxidized to Sn4+S{n^{4 + }} by acidified K2Cr2O7{K_2}C{r_2}{O_7}.
66 Medium
66 of 401
Application +4 | -1
A quantity of PCl5\mathrm{PC}{\mathrm{l}_5} was heated in a 10 litre vessel at 250∘C250^\circ C to show PCl5(g)⇋PCl3(g)+Cl2(g)\mathrm{PC}{\mathrm{l}_5}\left( \mathrm{g} \right)\leftrightharpoons\mathrm{PC}{\mathrm{l}_3}\left( \mathrm{g} \right) + \mathrm{C}{\mathrm{l}_2}\left( \mathrm{g} \right). At equilibrium the vessel contains 0.1 mole of PCl5\mathrm{PC}{\mathrm{l}_5}, 0.20 mole of PCl3\mathrm{PC}{\mathrm{l}_3} and 0.20 mole of Cl2\mathrm{C}{\mathrm{l}_2}. The equilibrium constant of the reaction is :
View answer and explanation
Correct option: C
0.040.04
Option C (0.04) is correct because the equilibrium constant is calculated by dividing the product of the concentrations of the products by the concentration of the reactant.
67 Medium
67 of 401
Analyzing +4 | -1
Among the following which pair of solution can form buffer solution?
View answer and explanation
Correct option: D
CH3COOH{CH}_3 {COOH} and Ca(OH)2{Ca}({OH})_2
Option D (CH3COOH{CH}_3 {COOH} and Ca(OH)2{Ca}({OH})_2) is correct because a weak acid (CH3COOH{CH}_3 {COOH}) and a base can form a buffer if the base is not in excessive amount to completely neutralize the acid.
68 Medium
68 of 401
Analyzing +4 | -1
Two equilibria, AB⇌A++B−{AB} \rightleftharpoons {A}^{+}+{B}^{-}and AB+B−⇌AB2−{AB}+{B}^{-} \rightleftharpoons {AB}_2^{-}are simultaneously maintained in a solution with equilibrium constants, K1{K}_1 and K2{K}_2 respectively. The ratio of [A+]\left[{A}^{+}\right]to [AB2−]\left[{AB}_2^{-}\right]in the solution is
View answer and explanation
Correct option: D
Inversely proportional to the square of [B−]\left[{B}^{-}\right]
Option D is correct because the ratio of [A+] to [AB2-] is inversely proportional to the square of [B-], as derived from the equilibrium constants K1 and K2.
69 Medium
69 of 401
Analyzing +4 | -1
In the decomposition of phosgene gas
COCl2( g)⇌CO(g)+Cl2( g);I2O5\mathrm{COCl}_2(\mathrm{~g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+\mathrm{Cl}_2(\mathrm{~g}) ; \mathrm{I}_2 \mathrm{O}_5 is added at constant volume, then
View answer and explanation
Correct option: A
Equilibrium will shift to the forward side
Option A is correct because decreasing the concentration of CO shifts the equilibrium towards the forward side to counteract the change, according to Le Chatelier's principle.
70 Medium
70 of 401
Analyzing +4 | -1
Statement I : A+B⇌C+DA + B \rightleftharpoons C + D,
KC=4{K_C} = 4 for the above equilibrium at 300 K when [B]=[C][B] = [C] and [D][A]=4\frac{{[D]}}{{[A]}} = 4
Statement II : At dynamic equilibrium, rate of forward and backward direction are equal.
View answer and explanation
Correct option: C
Both Statements are true
Option C is correct because both statements are true. Statement I correctly describes the relationship between concentrations at equilibrium, and Statement II accurately defines dynamic equilibrium.
71 Medium
71 of 401
Understanding +4 | -1
Statement I: If QcQ_c (Reactionquotient) <Kc< K_c ( equilabrium constant) reaction moves in direction of reactans.

Statement II : Equilitrium constant is defined in the same way as reaction quotient at end stage of the reaction.
View answer and explanation
Correct option: B
Statement I, II are incorrect
Option B (Statement I, II are incorrect) is correct because if Qc<KcQ_c < K_c, the reaction moves towards the products, and the equilibrium constant is defined at equilibrium, not at the \"end stage\" in the way described.
72 Medium
72 of 401
Analyzing +4 | -1
Assertion (A) : pH\mathrm{pH} of x M HCl\mathrm{x\,M} \mathrm{~HCl} is less than pH\mathrm{pH} of x M CHH3COOH\mathrm{x} \mathrm{~M} \mathrm{~CH} \mathrm{H}_3 \mathrm{COOH}.

Reason (R) : The degree of ionization of HCl\mathrm{HCl} and CH3COOH\mathrm{CH}_3 \mathrm{COOH} are equal.
View answer and explanation
Correct option: C
Assertion (A) is true and Reason (R) is false.
The correct answer is C because the pH of x M HCl is indeed less than that of x M CH₃COOH, but the degree of ionization of HCl and CH₃COOH are not equal, as HCl is a strong acid and CH₃COOH is a weak acid.
73 Medium
73 of 401
Application +4 | -1
The molarity of H2SO4{H}_2 {SO}_4 solution having pH=2{pH}=2 is
View answer and explanation
Correct option: B
5×10−3 M5 \times 10^{-3}\, {M}
The correct answer is B (5×10−3 M5 \times 10^{-3}\, {M}) because the concentration of H2SO4 is half the concentration of H+ ions in a solution with pH=2.
74 Medium
74 of 401
Analyzing +4 | -1
The equilibrium constants for the reaction A2⇌2A\mathrm{A}_2 \rightleftharpoons2 \mathrm{A} at 500 K500 \mathrm{~K} and 1000 K1000 \mathrm{~K} are 1×10−101 \times 10^{-10} and 1×10−5,1 \times 10^{-5}, respectively. The reaction is
View answer and explanation
Correct option: D
Endothermic
Option D (Endothermic) is correct because the equilibrium constant increases with increasing temperature, indicating an endothermic reaction.
75 Medium
75 of 401
Application +4 | -1
Number of H+{\mathrm{H}^ + } ions present in 10  mL10\;mL of a solution of pH  =  3pH\; = \;3 is:
View answer and explanation
Correct option: B
6.02×10186.02 \times {10^{18}}
Option B (6.02×10186.02 \times {10^{18}}) is correct because it accurately calculates the number of H+ ions based on the pH and volume of the solution.
76 Medium
76 of 401
Analyzing +4 | -1
Solid BaF2\mathrm{BaF}_2 is added to a solution containing 0.1 mole of Na2C2O4\mathrm{Na}_2 \mathrm{C}_2 \mathrm{O}_4 solution (1 L)(1 \mathrm{~L}) until equilibrium is reached. If the KspK_{\mathrm{sp}} of BaF2\mathrm{BaF}_2 and BaC2O4\mathrm{BaC}_2 \mathrm{O}_4 is 10−610^{-6} mol3 L−3\mathrm{mol}^3 \mathrm{~L}^{-3} and 10−10 mol2 L−2,10^{-10} \mathrm{~mol}^2 \mathrm{~L}^{-2}, respectively, then find the equilibrium concentration of Ba2+\mathrm{Ba}^{2+} in the solution. Assume that addition of BaF2\mathrm{BaF}_2 does not cause any change in volume.
View answer and explanation
Correct option: C
2.5×10−5M2.5 \times 10^{-5} \mathrm{M}
Option C is correct because the equilibrium concentration of Ba2+\mathrm{Ba}^{2+} is calculated using the solubility product of BaF2\mathrm{BaF}_2 and the fluoride ion concentration derived from the reaction with oxalate.
77 Medium
77 of 401
Application +4 | -1
3.2 moles of hydrogen iodide were heated in a sealed bulb at 444∘C444^\circ C till the equilibrium state was reached. Its degree of dissociation at this temperature was found to be 22%.22\% . The number of moles of hydrogen iodide present at equilibrium are
View answer and explanation
Correct option: B
2.496
Option B (2.496) is correct because it is the number of moles of HI remaining at equilibrium after dissociation.
78 Medium
78 of 401
Analyzing +4 | -1
Which oxide of nitrogen is the most stable?
View answer and explanation
Correct option: A
2NO2(g) ⇌ N2(g)+2O2(g);K=6.7×1016mol   litre−12N{O_2}(g)\, \rightleftharpoons \,{N_2}(g) + 2{O_2}(g);K = 6.7 \times {10^{16}}{\text{mol}}\,\,\,{\text{litr}}{{\text{e}}^{ - 1}}
Option A (2NO2(g) ⇌ N2(g)+2O2(g);K=6.7×1016mol   litre−12N{O_2}(g)\, \rightleftharpoons \,{N_2}(g) + 2{O_2}(g);K = 6.7 \times {10^{16}}{\text{mol}}\,\,\,{\text{litr}}{{\text{e}}^{ - 1}}) is correct because the lower the value of K, the more stable the oxide of nitrogen.
79 Medium
79 of 401
Application +4 | -1
For a weak electrolyte α1\alpha_1 and α2\alpha_2 are in ratio of 1:21: 2, for a given concentration ka1=2×10−4k_{\mathrm{a}_1}=2 \times 10^{-4}. What will be value of ka2\mathrm{k}_{\mathrm{a}_2} ?
View answer and explanation
Correct option: A
8×10−48 \times 10^{-4}
Option A (8×10−48 \times 10^{-4}) is correct because the degree of dissociation is proportional to the square root of the acid dissociation constant.
80 Medium
80 of 401
Application +4 | -1
Buffer capacity of a buffer solution is x, the volume of 1M added to 100 ml of this solution if the change of PH{P^H} by 1 is
View answer and explanation
Correct option: C
100x ml100x\,ml
Option C (100x ml100x\,ml) is correct because buffer capacity is defined as the number of moles of acid or base required to change the pH of 1 liter of solution by 1, and the calculation uses this definition to find the volume of 1M NaOH needed.
81 Medium
81 of 401
Application +4 | -1
Which may be added to one litre of water to act as a buffer?
View answer and explanation
Correct option: D
One mole of HC2H3O2\mathrm{H}{\mathrm{C}_2}{\mathrm{H}_3}{\mathrm{O}_2} and 0.50.5 mole of NaOH\mathrm{NaOH}
Option D (One mole of HC2H3O2\mathrm{H}{\mathrm{C}_2}{\mathrm{H}_3}{\mathrm{O}_2} and 0.50.5 mole of NaOH\mathrm{NaOH}) is correct because it results in a mixture of a weak acid and its conjugate base, which is a buffer solution.
82 Medium
82 of 401
Application +4 | -1
The equivalent conductance of M32\frac{M}{{32}} solution of a weak monobasic acid is 8.0  mhos  cm28.0\;\mathrm{mhos}\;\mathrm{c}{\mathrm{m}^2} and at infinite dilution is 400  mhos  cm2.400\;\mathrm{mhos}\;\mathrm{c}{\mathrm{m}^2}. The dissociation constant of this acid is:
View answer and explanation
Correct option: B
1.25×10−51.25 \times {10^{ - 5}}
Option B (1.25×10−51.25 \times {10^{ - 5}}) is correct because it is calculated using the formula for the dissociation constant of a weak acid, Ka=cα2K_a = c\alpha^2, where α\alpha is the degree of dissociation.
83 Medium
83 of 401
Application +4 | -1
For the reaction C(s)+CO2(g)2  CO(g)C(s) + C{O_2}(g)2\,\,CO(g) the partial pressure of CO2C{O_2} and are 2.0 and 4.0 atm resp, at equilibrium. The Kp{K_p} for the reaction is
View answer and explanation
Correct option: C
3.03.0
Option D (32.0) is incorrect. The correct answer is 8.0. The equilibrium constant Kp is calculated by taking the ratio of the square of the partial pressure of CO to the partial pressure of CO2.
84 Medium
84 of 401
Understanding +4 | -1
STATEMENT-1 : The equilibrium constant of the exothermic reaction at high temperature decreases.

STATEMENT-2 : Since ln⁡K2K1=ΔH∘R[1T1−1T2]\ln \frac{K_2}{K_1}=\frac{\Delta H^{\circ}}{R}\left[\frac{1}{T_1}-\frac{1}{T_2}\right] and for exothermic reaction, ΔH∘=−\Delta H^{\circ}=- ve and thereby; K2K1<1\frac{K_2}{K_1} < 1
View answer and explanation
Correct option: A
If both the statements are TRUE and STATEMENT-2 is the correct explanation of STATEMENT-1
Option A (If both the statements are TRUE and STATEMENT-2 is the correct explanation of STATEMENT-1) is correct because the equilibrium constant of an exothermic reaction decreases at high temperatures, and the provided equation correctly explains this relationship.
85 Medium
85 of 401
Application +4 | -1
The value of the equilibrium constant for the reaction: 2NO+O2⇌2NO22NO + {O_2} \rightleftharpoons 2N{O_2} is 0.25, then the value of equilibrium constant for 2NO2⇌2NO+O22N{O_2} \rightleftharpoons 2NO + {O_2} is
View answer and explanation
Correct option: C
4
Option C (4) is correct because the equilibrium constant for the reverse reaction is the inverse of the equilibrium constant for the forward reaction.
86 Medium
86 of 401
Understanding +4 | -1
Statement-I: Ethyl chloride reacts with sodium ethoxide to from diethyl etber. Here ethoxide ion behaves as strong base.
Statement-II: C2H5OH\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH} is weak acid.
View answer and explanation
Correct option: A
Statement I, II are correct
The correct answer is Statement I, II are correct because ethyl chloride reacts with sodium ethoxide to form diethyl ether (Williamson ether synthesis), and ethanol is a weak acid.
87 Medium
87 of 401
Application +4 | -1
Molar solubility of Zirconium phosphate is ‘S’ moles/litre. Solubility product of Zirconium phosphate [Zr3(PO4)4]\left[ {Z{r_3}{{\left( {P{O_4}} \right)}_4}} \right] is
View answer and explanation
Correct option: C
6912 S76912\,{\mathrm{S}^7}
Option C (6912 S76912\,{\mathrm{S}^7}) is correct because it is the solubility product of Zirconium phosphate calculated from its molar solubility.
88 Medium
88 of 401
Understanding +4 | -1
Which will undergo cationic hydrolysis?
View answer and explanation
Correct option: C
(NH4)2SO4\left({NH}_4\right)_2 {SO}_4
Option C ((NH4)2SO4\left({NH}_4\right)_2 {SO}_4) is correct because ammonium sulfate is a salt of a weak base (NH4OH) and a strong acid (H2SO4), and therefore undergoes cationic hydrolysis.
89 Medium
89 of 401
Analyzing +4 | -1
Consider the following reaction:

CuO(s)+H2(g)⇌Cu(s)+H2O(g);K1=2×1015H2(g)+12O2(g)⇌H2O(g);K2=5×1022\begin{aligned} & \mathrm{CuO}(\mathrm{s})+\mathrm{H}_2(\mathrm{g}) \rightleftharpoons \mathrm{Cu}(\mathrm{s})+\mathrm{H}_2 \mathrm{O}(\mathrm{g}) ; \mathrm{K}_1=2 \times 10^{15} \\ & \mathrm{H}_2(\mathrm{g})+\frac{1}{2} \mathrm{O}_2(\mathrm{g}) \rightleftharpoons \mathrm{H}_2 \mathrm{O}(\mathrm{g}) ; \mathrm{K}_2=5 \times 10^{22} \end{aligned}

The equilibrium constant for the reaction : CuO(s)⇌Cu(s)+12O2(g)\mathrm{CuO}(\mathrm{s}) \rightleftharpoons \mathrm{Cu}(\mathrm{s})+\frac{1}{2} \mathrm{O}_2(\mathrm{g}) would be:
View answer and explanation
Correct option: B
4×10−84 \times 10^{-8}
Option B (4×10−84 \times 10^{-8}) is correct because the equilibrium constant for the desired reaction is obtained by dividing K1 by K2.
90 Medium
90 of 401
Application +4 | -1
A sparingly soluble Iron (III) salt dissociates in solution as follows:

Fe2X3(s)⇌2Fe+3(aq)+3X2−(aq){Fe}_2 {X}_3(\mathrm{s}) \rightleftharpoons 2 {Fe}^{+3}(\mathrm{aq})+3 {X}^{2-}(\mathrm{aq})

If the solubility product of the iron (III) salt is Ksp{K}_{{sp}}, then [Fe+3](aq)\left[{Fe}^{+3}\right](\mathrm{aq}) at equilibrium is:
View answer and explanation
Correct option: C
(827Ksp)1/5\left(\frac{8}{27}{K}_{{sp}}\right)^{1 / 5}
Option C ((8/27)Ksp)(1/5) is correct because it is the correct expression for the concentration of Fe+3 at equilibrium, derived from the solubility product expression.
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