The solids dissociate as follows: $A\left( s \right)\,\leftrightharpoons\,B\left( g \right) + C\left( g \right);\,\,{K_{{P_1}}} = x\,\,\,{\rm{at}}{{\rm{m}}^2}$ $D\left( s \right)\,\leftrightharpoons\,C\left( g \right) + E\left( g \right);\,\,{K_{{P_2}}} = y\,\,\,{\rm{at}}{{\rm{m}}^2}$ The total pressure when both the solids dissociate simultaneously is

$2\left( {\sqrt {x + y} } \right){\rm{ \,atm}}$

$\left( {x + y} \right){\rm{ \,atm}}$

The degree of dissociation of ${\rm{C}}{{\rm{H}}_3}{\rm{COOH}}$ is influenced by:

${\rm{C}}{{\rm{H}}_3}{\rm{COONa}}$

${\rm{N}}{{\rm{H}}_4}{\rm{OH}}$

Pure ammonia is placed in a vessel at temperature where its dissociation constant (${\rm{\alpha }}$) is appreciable. At equilibrium,

Concentration of ${{\rm{H}}_2}$ is less than that of ${{\rm{N}}_2}$

${K_p}$ does not change significantly with pressure

${\rm{\alpha }}$ does not change with pressure

Concentration of ${\rm{N}}{{\rm{H}}_3}$ does not change with pressure

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JEE • CHEMISTRY • Equilibrium

Equilibrium practice · Mixed difficulty

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Practice format and chapter quiz

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 1-30 of 680 questions

1 Easy

1 of 680

Remembering +4 | -1

The indicator used in titrating oxalic acid with caustic soda solution is:

View answer and explanation

Correct option: D

Phenolphthalein

Option D (Phenolphthalein) is correct because phenolphthalein is a suitable indicator for titrating a weak acid (oxalic acid) with a strong base (caustic soda).

2 Medium

2 of 680

Analyzing +4 | -1

\mathrm{K}_1

and

\mathrm{K}_2

are equilibrium constant for reactions (i) and (ii)

\mathrm{N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})\quad\ldots

(i)

\mathrm{NO}(\mathrm{g}) \rightleftharpoons \frac{1}{2} \mathrm{~N}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})\quad\ldots

(ii)

Then,

View answer and explanation

Correct option: A

\mathrm{K}_1=\left(\frac{1}{\mathrm{~K}_2}\right)^2

Option A is correct because the equilibrium constant K1 is the inverse square of K2, derived from reversing and halving the reaction.

3 Medium

3 of 680

Understanding +4 | -1

K{a_1},\,\,\,K{a_2}\,\,\,\mathrm{and}\,\,\,K{a_3}

are the respective ionisation constants for the following reactions

\begin{array}{l} {H_2}S\leftrightharpoons{H^ + } + H{S^ - }\\ H{S^ - }\leftrightharpoons{H^ + } + {S^{2 - }}\\ {H_2}S\leftrightharpoons2{H^ + } + {S^{2 - }} \end{array}

The correct relationship between

K{a_1},\,\,K{a_2}\,,K{a_3}

View answer and explanation

Correct option: B

K{a_3} = \,\,K{a_1}\, \times K{a_2}

Option B (

K{a_3} = \,K{a_1}\, \times K{a_2}

) is correct because the overall ionization constant is the product of the stepwise ionization constants.

4 Medium

4 of 680

Application +4 | -1

The

\mathrm{pH}

0.1 \mathrm{~M}-\mathrm{N}_2 \mathrm{H}_4

solution is (For

\mathrm{N}_2 \mathrm{H}_4,

K_{{b}_1}=3.6 \times 10^{-6}, K_{{b}_2}=6.4 \times 10^{-12}, \log 2=0.3

\log 3=0.48

)

View answer and explanation

Correct option: C

10.78

Option C (10.78) is correct because it accurately calculates the pH of the N2H4 solution considering the first base dissociation constant.

5 Medium

5 of 680

Understanding +4 | -1

Reaction

{\text{Ba}}{{\text{O}}_2}(s) \rightleftharpoons {\text{BaO}}(s) + {{\text{O}}_2}(g);\Delta H = - {\text{ve}}.

In equilibrium condition pressure of

\text{O}_2

depends on:

View answer and explanation

Correct option: C

Increase temperature on equilibrium

Option C (Increase temperature on equilibrium) is correct because the reaction is exothermic, and increasing the temperature will shift the equilibrium to the left, favoring the reactants and increasing the pressure of O2.

6 Medium

6 of 680

Understanding +4 | -1

A : At equilibrium concentration of the reactant and product does not change with time for a chemical reaction.
R: The rate of reaction is zero at equilibrium.

View answer and explanation

Correct option: C

If A is correct but R is incorrect

Option C is correct because at equilibrium, the concentrations of reactants and products do not change with time, but the rate of reaction is *not* zero; the forward and reverse rates are equal.

7 Easy

7 of 680

Understanding +4 | -1

\mathrm{M}{\mathrm{g}^{2 + }}

is … than

\mathrm{A}{\mathrm{l}^{3 + }}

View answer and explanation

Correct option: C

Weak Lewis acid

Option C (Weak Lewis acid) is correct because Mg²⁺ has a smaller charge density and weaker polarizing power compared to Al³⁺, making it a weaker Lewis acid.

8 Hard

8 of 680

Application +4 | -1

For the equilibrium,

\mathrm{C}(\mathrm{s})+\mathrm{CO}_2(g) \rightleftharpoons 2 \mathrm{CO}(g) ; K_p=63 \mathrm{~atm}

1000 \mathrm{~K}

if at equilibrium

p_{\mathrm{CO}}=10 p_{\mathrm{CO}_2}

then total pressure at equilibrium is

View answer and explanation

Correct option: C

6.93 atm

Option C (6.93 atm) is correct because it is the sum of the partial pressures of CO and CO2 at equilibrium, calculated using the given Kp and the relationship between the partial pressures.

9 Hard

9 of 680

Application +4 | -1

The solids dissociate as follows:

A\left( s \right)\,\leftrightharpoons\,B\left( g \right) + C\left( g \right);\,\,{K_{{P_1}}} = x\,\,\,\mathrm{at}{\mathrm{m}^2}

D\left( s \right)\,\leftrightharpoons\,C\left( g \right) + E\left( g \right);\,\,{K_{{P_2}}} = y\,\,\,\mathrm{at}{\mathrm{m}^2}

The total pressure when both the solids dissociate simultaneously is

View answer and explanation

Correct option: D

2\left( {\sqrt {x + y} } \right)\mathrm{ \,atm}

Option D (

2(\sqrt {x + y} )\mathrm{ atm}

) is correct because it represents the total pressure derived from the equilibrium constants of the two dissociation reactions.

10 Medium

10 of 680

Understanding +4 | -1

Identify the reaction in which both products concentration and reactants concentration is appreciable

View answer and explanation

Correct option: B

X(g)+Y(g) \rightleftharpoons X Y(g), {K}_{e q}=54

Option B (

X(g)+Y(g) \rightleftharpoons X Y(g), {K}_{e q}=54

) is correct because a

K_{eq}

value close to 1 (typically between 0.001 and 1000) indicates that both reactants and products are present in appreciable concentrations at equilibrium.

11 Easy

11 of 680

Application +4 | -1

The solubility of

\mathrm{S}{\mathrm{b}_2}{\mathrm{S}_3}\;\mathrm{in }\;\mathrm{water}\;\mathrm{ is}\;1.0 \times {10^{ - 5}}\;mol/L

298\;K

. What will be its solubility product?

View answer and explanation

Correct option: A

108 \times {10^{ - 25}}

Option D (120 kJ) is correct because it is calculated using the Gibbs free energy equation: ∆G = ∆H - T∆S.

12 Hard

12 of 680

Analyzing +4 | -1

\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g}) ; \mathrm{K}_1

\mathrm{~N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g}) ; \mathrm{K}_2

\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightleftharpoons \mathrm{H}_2 \mathrm{O}(\mathrm{g}) ; \mathrm{K}_3

The equilibrium constant for

2 \mathrm{NH}_3(\mathrm{~g})+\frac{5}{2} \mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})+3 \mathrm{H}_3 \mathrm{O}(\mathrm{g})

will be

View answer and explanation

Correct option: D

\frac{\mathrm{K}_2 \mathrm{~K}_3^3}{\mathrm{~K}_1}

Option D (

\frac{\mathrm{K}_2 \mathrm{~K}_3^3}{\mathrm{~K}_1}

) is correct because by manipulating the given equilibrium reactions, the equilibrium constant for the desired reaction can be obtained.

13 Medium

13 of 680

Analyzing +4 | -1

Solubility of

\mathrm{Ni}(\mathrm{OH})_2

will be maximum in

View answer and explanation

Correct option: D

0.1 \mathrm{M}

aq. solution of

\mathrm{H}_2 \mathrm{SO}_4

Option D is correct because the solubility of

\mathrm{Ni}(\mathrm{OH})_2

will be maximum in

0.1 \mathrm{M}

aq. solution of

\mathrm{H}_2 \mathrm{SO}_4

due to the acid-base reaction consuming hydroxide ions and shifting the equilibrium towards dissolution.

14 Medium

14 of 680

Application +4 | -1

Calculate the hydrolysis constant of the salt containing

\mathrm{NO}_2^{-}

ions.

\mathrm{K}_{\mathrm{a}}

for

\mathrm{HNO}_2=4.5 \times 10^{-10}

and

\mathrm{K}_{\mathrm{w}}=1 \times 10^{-14}

View answer and explanation

Correct option: A

2.2 \times 10^{-5}

Option A (

2.2 \times 10^{-5}

) is correct because it is the hydrolysis constant calculated using the formula

K_h = \frac{{K_w}}{{K_a}}

15 Easy

15 of 680

Analyzing +4 | -1

The degree of dissociation of

\mathrm{C}{\mathrm{H}_3}\mathrm{COOH}

is influenced by:

View answer and explanation

Correct option: D

Either of these

Option D (Either of these) is correct because the degree of dissociation of CH3COOH is influenced by the common ion effect from CH3COONa and HCl, and by acid-base reaction with NH4OH.

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16 Medium

16 of 680

Application +4 | -1

3000 \mathrm{~K}

, the equilibrium partial pressure of

{CO}_2, {CO}

and

{O}_2

are 0.6, 0.4 and 0.2 atm respectively.

{K}_p

for the reaction,

2 {CO}_2 \rightleftarrows 2 {CO}+{O}_2

View answer and explanation

Correct option: A

0.088

Option A (0.088) is correct because it is the result of correctly applying the equilibrium constant expression for the given reaction and partial pressures.

17 Easy

17 of 680

Understanding +4 | -1

Which of the following can act as a self-indicator?

View answer and explanation

Correct option: A

KMn{O_4}

Option A (

KMn{O_4}

) is correct because

KMn{O_4}

acts as a self-indicator due to its intense purple color, which changes upon reduction at the endpoint of the titration.

18 Medium

18 of 680

Understanding +4 | -1

Assertion :-

H_2O

is amphoteric in nature.

Reason :-

H_2O

can accept a proton to form

H_3O^+

ion and can donate a proton to form

{OH}^–

ion.

View answer and explanation

Correct option: A

If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion

Option A (If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion) is correct because water can act as both an acid and a base.

19 Medium

19 of 680

Application +4 | -1

\Delta {G^0}

for the reaction given below is

1.7\;\mathrm{ }kJ

; the equilibrium constant of the reaction,

2\mathrm{HI}\left( \mathrm{g} \right)\;\leftrightharpoons\;{\mathrm{H}_2}\left( \mathrm{g} \right) + {\mathrm{I}_2}\left( \mathrm{g} \right)

25^\circ C

is :

View answer and explanation

Correct option: D

0.5

Option D (

0.5

) is correct because it is calculated using the relationship between Gibbs free energy change and the equilibrium constant.

20 Medium

20 of 680

Understanding +4 | -1

A chemical system is in equilibrium. Addition of a catalyst would result in:

View answer and explanation

Correct option: C

A new reaction path way to reaction

Option C (A new reaction path way to reaction) is correct because a catalyst provides an alternative reaction pathway with a lower activation energy, thus speeding up the reaction without changing the equilibrium.

21 Medium

21 of 680

Understanding +4 | -1

The addition of a catalyst to the reaction system

View answer and explanation

Correct option: D

Increases the rate of forward as well as backward reaction equally

Option D (Increases the rate of forward as well as backward reaction equally) is correct because a catalyst lowers the activation energy for both forward and reverse reactions equally, thus increasing both rates.

22 Easy

22 of 680

Analyzing +4 | -1

Assertion : The equilibrium constant for the reaction

\begin{aligned} & {CaSO}_4 \cdot 5 {H}_2 {O}({s})={CaSO}_4 \cdot 3 {H}_2 {O}({s})+2 {H}_2 {O}({g}) \\ & \text { is } {K}_{{C}}=\frac{\left[{CaSO}_4 \cdot 3 {H}_2 {O}\right]\left[{H}_2 {O}\right]^2}{\left[{CaSO}_4 \cdot 5 {H}_2 {O}\right]} \end{aligned}

Reason : Equilibrium constant is the ratio of the product of molar concentration of the substances produced to the product of the molar concentrations of reactants with each concentrations term raised to the power equal to the respective stoichiometric constant.

View answer and explanation

Correct option: D

If assertion is false but reason is true

Option D is correct because the equilibrium constant for a heterogeneous reaction only includes gaseous or aqueous species, and the concentration of solids is considered unity.

23 Medium

23 of 680

Application +4 | -1

A buffer solution is prepared in which the concentration of

\mathrm{N}{\mathrm{H}_3}

0.30\;M

and the concentration of

\mathrm{NH}_4^ +

0.20\;M

. If the equilibrium constant,

{K_b}

for

\mathrm{N}{\mathrm{H}_3}

equals

1.8 \times {10^{ - 5}}

what is the pH of this solution?

(\log 2.7 = 0.43)

View answer and explanation

Correct option: C

9.43

Option C (

9.43

) is correct because it is calculated using the Henderson-Hasselbalch equation for a basic buffer solution.

24 Medium

24 of 680

Application +4 | -1

If equilibrium constant is

2.6 \times 10^8

then find the value of

\Delta {G}^{\circ}

View answer and explanation

Correct option: A

-43.36 \mathrm{~kJ}

Option A (

-43.36 \mathrm{~kJ}

) is correct because it is calculated using the Gibbs free energy equation with the given equilibrium constant.

25 Medium

25 of 680

Analyzing +4 | -1

Which of the following when mixed, will give a solution with
pH greater than 7?

View answer and explanation

Correct option: B

100 \mathrm{~mL}

0.1 \mathrm{~M} \mathrm{~CH}_3 \mathrm{COOH}+100 \mathrm{~mL}

0.1 \mathrm{~M} \mathrm{~KOH}

The correct answer is B (

100 \mathrm{~mL}

0.1 \mathrm{~M} \mathrm{~CH}_3 \mathrm{COOH}+100 \mathrm{~mL}

0.1 \mathrm{~M} \mathrm{~KOH}

) because it forms a basic buffer solution after the reaction.

26 Medium

26 of 680

Analyzing +4 | -1

A vessel contains

\mathrm{A}(\mathrm{g})

and

\mathrm{B}(\mathrm{g})

2 \mathrm{~atm}

and

4 \mathrm{~atm}

respectively at

\mathrm{TK}

, the mixture is allowed to attain equilibrium at

\mathrm{TK}

, according to the reaction

8 \mathrm{B}(\mathrm{g}) \rightleftharpoons 8 \mathrm{A}(\mathrm{g})+\mathrm{C}(\mathrm{s})

At equilibrium,

\left(\frac{\mathrm{n_A}}{\mathrm{n_B}}\right)_{\text {eq }}=\left(\frac{\mathrm{n_B}}{\mathrm{n_A}}\right)_{\text {initial }}

Find the value of

\mathrm{y}

\mathrm{K_c}=2^\mathrm{y}

View answer and explanation

Correct option: A

Option A (8) is correct because the equilibrium constant Kc is calculated to be 2⁸, thus y=8.

27 Medium

27 of 680

Application +4 | -1

In which of the following cases, pH is greater than 7?

View answer and explanation

Correct option: C

50 mL of 0.1 M

C{H_3}COOH

+ 50 mL of 0.1 M KOH

Option C (50 mL of 0.1 M

C{H_3}COOH

+ 50 mL of 0.1 M KOH) is correct because it involves the reaction of a weak acid (

CH_3COOH

) with a strong base (KOH), resulting in a salt of a weak acid and strong base, which hydrolyzes to give a pH greater than 7.

28 Medium

28 of 680

Analyzing +4 | -1

{HCl}, {HNO}_3

and

{H}_2 {SO}_4

are equally strong acids in water.
R: Water is a stronger acid than alcohols.

View answer and explanation

Correct option: B

If both A and R are correct, but R is not the correct explanation of A

Option B is correct because both the assertion and the reason are correct statements, but the reason does not explain the assertion. The leveling effect explains the assertion.

29 Medium

29 of 680

Analyzing +4 | -1

The degree of hydrolysis of which of the
following salt is independent of the
concentration of salt solution?

View answer and explanation

Correct option: D

{NaCl}

The correct answer is D (

{NaCl}

) because it is a salt of a strong acid and a strong base, and therefore does not undergo hydrolysis. Its degree of hydrolysis is zero and independent of concentration.

30 Easy

30 of 680

Analyzing +4 | -1

Pure ammonia is placed in a vessel at temperature where its dissociation constant (

\mathrm{\alpha }

) is appreciable. At equilibrium,

View answer and explanation

Correct option: D

Concentration of

{\mathrm{H}_2}

is less than that of

{\mathrm{N}_2}

Option D is correct because at equilibrium, the concentration of H2 will be less than N2 due to the stoichiometry of the ammonia dissociation reaction.

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Key Takeaways

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