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

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

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

Prove that the pressure necessary to obtain $50 \%$ dissociation of $\mathrm{PCl}_5$ at $250^{\circ} \mathrm{C}$ is numerically three times of $K_p$.

$\mathrm{P}=4 \mathrm{~K}_{\mathrm{P}}$

$\mathrm{P}=5 \mathrm{~K}_{\mathrm{P}}$

$\mathrm{P}=6 \mathrm{~K}_{\mathrm{P}}$

For the equilibrium $\mathrm{A}(\mathrm{g})+2 \mathrm{~B}(\mathrm{~g}) \rightleftharpoons \mathrm{AB}_2(\mathrm{~g})$ the value of equilibrium constant at $300 \mathrm{k}$ is $3 \times 10^{-3} \mathrm{~mole}^{-2} \mathrm{~L}^2$ and value of equilibrium constant at $500 \mathrm{k}$ is $5 \times 10^{-5} \mathrm{~mol}^{-2} \mathrm{~L}^2$. What is the sign of enthalpy change for this reaction

Both (a) and (b) are possible at lower temperature it is positive and at high temperature it is negative.

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JEE • CHEMISTRY • Equilibrium

Equilibrium practice · Hard

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Showing 1-30 of 118 questions

1 Hard

1 of 118

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.

2 Hard

2 of 118

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.

3 Hard

3 of 118

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.

4 Hard

4 of 118

Application +4 | -1

For the equilibrium

{SrCl}_2 \cdot 6 {H}_2{O}(\mathrm{s}) \rightleftharpoons {SrCl}_2 \cdot 2 {H}_2 {O}(\mathrm{s})+4 {H}_2 {O}(\mathrm{g}),

K_{{P}}=8.1 \times 10^{-7} \mathrm{~atm}^4

27^{\circ} \mathrm{C}.

1.642 \mathrm{~L}

of air saturated with water vapour at

27^{\circ} \mathrm{C}

is exposed to a large quantity of

{SrCl}_2 \cdot 2 {H}_2 {O} \text{(s)},

then what mass of water vapour will be absorbed? Saturated vapour pressure of water at

27^{\circ} \mathrm{C}=30.4 \text{ torr}.

View answer and explanation

Correct option: A

12 \mathrm{~mg}

Option A (12 mg) is correct because it calculates the mass of water vapor absorbed by SrCl2.2H2O(s) by considering the difference between the actual and equilibrium partial pressures of water vapor.

5 Hard

5 of 118

Remembering +4 | -1

The correct equation of equilibrium constant in terms of Gibbs energy is

View answer and explanation

Correct option: A

K=e^{-\Delta G^{\circ} / R T}

Option A (

K=e^{-\Delta G^{\circ} / R T}

) is correct because it represents the correct relationship between the equilibrium constant (K) and the standard Gibbs free energy change (Δ G°).

6 Hard

6 of 118

Application +4 | -1

298 \mathrm{~K}

for the gas phase reaction -

2{SO}_2(\mathrm{~g})+{O}_2(\mathrm{~g}) \rightleftharpoons 2{SO}_3(\mathrm{~g}) ;\, {K}_{{P}}=4 \times 10^{24} \mathrm{~atm}^{-1}

{SO}_2(\mathrm{~g}),{O}_2(\mathrm{~g})

and

{SO}_3(\mathrm{~g})

at partial pressures of 2,1 and 2 atm respectively, are mixed and allowed to attain equilibrium, the approximate partial pressure of

{SO}_2(\mathrm{~g})

at equilibrium is-

View answer and explanation

Correct option: B

2 \times 10^{-8}

Option A (

2.01 \times 10^{21}

atoms) is correct because it represents the number of magnesium atoms calculated from the mass percentage in the given amount of chlorophyll.

7 Hard

7 of 118

Application +4 | -1

Find the concentration of

\mathrm{H}^{+}, \mathrm{HCO}_3

, and

\mathrm{CO}_3^{-2}

in a

0.01 \mathrm{M}

solution of carbonic acid if the

\mathrm{pH}

of solution is 4.18 .

\left(\mathrm{K}_1=4.45 \times 10^{-7}, \mathrm{~K}_2=4.69 \times 10^{-11}\right)

View answer and explanation

Correct option: B

4.78 \times 10^{-11} \mathrm{~mol} \mathrm{~L}^{-1}

Option D (Below

20.9 \%

) is correct because for CuSO4.5H2O to effloresce, the partial pressure of water vapor in the air must be less than that at equilibrium.

8 Hard

8 of 118

Analyzing +4 | -1

Equilibrium constant

({K})

for the reaction

{Ni}({s})+4 {CO}({g}) \rightleftharpoons {Ni}({CO})_4({~g})

can be written in terms of

(1)

{Ni}({s})+2 {CO}_2({~g})+2 {C}(\mathrm{s}) \rightleftharpoons {Ni}({CO})_4({~g})

; equilibrium constant

={K}_{{1}}

.

(2)

{CO}_2({~g})+{C}({s}) \rightleftharpoons 2 {CO}({g})

; equilibrium constant

={K}_2

.

What is the relation between

{K}, {K}_2

and

{K}_2

View answer and explanation

Correct option: A

{K}=\left({K}_1\right) /\left({K}_2\right)^2

Option A is correct because the equilibrium constant K can be expressed as K1/(K2)² by manipulating the given equilibrium constants.

9 Hard

9 of 118

Application +4 | -1

K_{\mathrm{a}}

\mathrm{HCN}

and

\mathrm{CH}_3 \mathrm{COOH}

are respectively

4.5 \times 10^{-10}

and

1.5 \times 10^{-5}

298 \mathrm{~K}

.
Thus, equilibrium constant for

\mathrm{CN}^{-}+\mathrm{CH}_3 \mathrm{COOH}\rightleftharpoons \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{HCN}

would be

View answer and explanation

Correct option: D

3.3 \times 10^4

Option D (

3.3 \times 10^4

) is correct because the equilibrium constant is calculated by multiplying the acid dissociation constant of acetic acid by the reciprocal of the acid dissociation constant of HCN.

10 Hard

10 of 118

Application +4 | -1

For the reaction,

{CuSO}_4 \cdot 5 {H}_2 {O}({s}) \rightleftharpoons {CuSO}_4 \cdot 3{H}_2 {O}({s})+2 {H}_2 {O}({g})

K_p

298 \mathrm{~K}

1.086 \times 10^{-4} \mathrm{~atm}^2

and vapour pressure of water is 23.8 torr. The salt

{CuSO}_4 \cdot 5 {H}_2 {O}

will be efforescent when the relative humidity is

View answer and explanation

Correct option: D

Less than

33.1 \%

Option D (Less than 33.1 %) is correct because the salt will be efflorescent when the relative humidity is less than the equilibrium vapor pressure of water for the hydrated salt.

11 Hard

11 of 118

Analyzing +4 | -1

The solubility of silver benozate (

\mathrm{PhCOOAg}

) in

\mathrm{H}_2 \mathrm{O}

and in solution of

\mathrm{pH}=4

, 5 , and 6 are

S_1, S_2, S_3

and

S_4

, respectively. Given the decreasing order of their solubilities.

View answer and explanation

Correct option: C

\mathrm{S}_2 > \mathrm{S}_3 > \mathrm{S}_4 > \mathrm{S}_1

The correct answer is A (

\mathrm{S}_2 > \mathrm{S}_3 > \mathrm{S}_4 > \mathrm{S}_1

) because the solubility of silver benzoate, a salt of a weak acid, increases with increasing acidity (lower pH).

12 Hard

12 of 118

Application +4 | -1

The solubility product of a sparingly salt

A{X_2}\;\mathrm{is}\,3.2 \times {10^{ - 11}}

. Its solubility (in mol/L) is

View answer and explanation

Correct option: C

2 \times {10^{ - 4}}

Option C (

2 \times {10^{ - 4}}

) is correct because it is derived from the solubility product expression for

A{X_2}

, where

K_{sp} = 4S^3

13 Hard

13 of 118

Analyzing +4 | -1

The dissociation constant of

\mathrm{NH}_3

27^{\circ} \mathrm{C}

from the following data:

\begin{aligned} & \mathrm{NH}_3+\mathrm{H}^{+} \rightleftharpoons \mathrm{NH}_4^{+} ; \Delta H^{\circ}=-52.21 \mathrm{~kJ} / \mathrm{mol} ; \Delta S^{\circ}=+1.6 \mathrm{~JK}^{-1} \mathrm{~mol}^{-1} \\ & \mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H}^{+}+\mathrm{OH}^{-} ; \Delta H^{\circ}=54.70 \mathrm{~kJ} / \mathrm{mol} ; \Delta S^{\circ}=-76.3 \mathrm{~JK}^{-1} \mathrm{~mol}^{-1} \end{aligned}

Given:

\mathrm{R}=8.3 \mathrm{~J} / \mathrm{K}-\mathrm{mol}

View answer and explanation

Correct option: B

\mathrm{e}^{-10}

Option B (e^-10) is correct because it is the calculated dissociation constant of NH3 based on the given thermodynamic data.

14 Hard

14 of 118

Application +4 | -1

The percentage error in

\mathrm{H}^{+}

concentration provided by

10^{-8} \mathrm{~M} \mathrm{~HCl}

if ionisation of water is not neglected is :

View answer and explanation

Correct option: C

5 \%

Option A (

3 \%

) is correct because the percentage error in H⁺ concentration is calculated by considering the autoionization of water.

15 Hard

15 of 118

Application +4 | -1

A weak acid

\mathrm{H}X\left( {{K_a} = 1 \times {{10}^{ - 5}}} \right)

on reaction with

\mathrm{NaOH}

gives

\mathrm{Na}X

. For

0.1\;M

aqueous solution of

\mathrm{Na}X

, the

\%

hydrolysis is:

View answer and explanation

Correct option: B

0.01\%

Option B (

0.01\%

) is correct because it is calculated by finding the hydrolysis constant and then the percentage of hydrolysis.

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16 Hard

16 of 118

Application +4 | -1

Consider the following equilibrium at

25^{\circ} \mathrm{C}

\mathrm{CuSO}_4.5 \mathrm{H}_2 \mathrm{O}{(\mathrm{s})} \rightleftharpoons \mathrm{CuSO}_4.3 \mathrm{H}_2 \mathrm{O}{(\mathrm{s})}+2 \mathrm{H}_2 \mathrm{O}{(\mathrm{g})}

If this reaction takes place at two cities

\mathrm{A}

and

\mathrm{B}

having relative humidity

60 \%

and

80 \%

respectively. The vapour pressure of water at

25^{\circ} \mathrm{C}

20 \mathrm{~mm} \mathrm{~Hg}

and

\mathrm{K}_{\mathrm{P}}

for the above reaction is

4 \times 10^{-2}(\mathrm{atm})^2

25^{\circ} \mathrm{C}

then

\mathrm{CuSO}_4.5 \mathrm{H}_2 \mathrm{O}

acts as efflorescent in which city?

View answer and explanation

Correct option: A

At A only

Option A (At A only) is correct because CuSO4.5H2O acts as efflorescent when the partial pressure of water vapor in the atmosphere is less than the equilibrium vapor pressure, which is the case in city A.

17 Hard

17 of 118

Application +4 | -1

For the reaction

\mathrm{H}_2(\mathrm{~g})+\mathrm{CO}_2(\mathrm{~g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+\mathrm{H}_2 \mathrm{O}(\mathrm{g})

, If the initial concentration of

\left[\mathrm{H}_2\right]=\left[\mathrm{CO}_2\right]

and

\mathrm{~x} \mathrm{~mol} / \mathrm{L}

of hydrogen is consumed at equilibrium, the correct expression of

k_{\mathrm{p}}

View answer and explanation

Correct option: C

\frac{x^2}{1-x^2}

Option C (

\frac{x^2}{1-x^2}

) is correct because it is the correct expression for Kp based on the equilibrium concentrations of the reactants and products.

18 Hard

18 of 118

Application +4 | -1

20 \%

\mathrm{N}_2 \mathrm{O}_4

molecules are dissociated in a sample of gas at

27^{\circ} \mathrm{C}

and 760 torn. Calculate the density of the equilibrium mixture.

View answer and explanation

Correct option: A

3.11 \mathrm{~g} \mathrm{~litre}^{-1}

Option A (

3.11 \mathrm{~g} \mathrm{~litre}^{-1}

) is correct because it is calculated using the ideal gas law and the concept of degree of dissociation to find the density of the equilibrium mixture.

19 Hard

19 of 118

Application +4 | -1

An aqueous solution contains an unknown concentration of

B{a^{2 + }}.

When 50 mL of a 1 M solution of

N{a_2}S{O_4}

is added,

BaS{O_4}

just begins to precipitate. The final volume is 500 mL. The solubility product of

BaS{O_4}

1 \times {10^{ - 10}}

. What is the original concentration of

B{a^{2 + }}?

View answer and explanation

Correct option: C

1.1 \times {10^{ - 9}}M

Option C (1) is correct because after finding the roots and substituting them into the expression, the result simplifies to 1.

20 Hard

20 of 118

Application +4 | -1

The ionisation constant of an acid,

K_a

is the measure of strength of an acid. The

K_a

values of acetic acid, hypochlorous acid and formic acid are

1.74 \times 10^{-5}

3.0 \times 10^{-8}

and

1.8 \times 10^{-4}

respectively. Which of the following order of

\mathrm{pH}

0.1 \mathrm{~mol}\,\mathrm{dm}^{-3}

solutions of these acids is correct?

View answer and explanation

Correct option: B

Hypochlorous acid

>

acetic acid

>

formic acid

Option B (Hypochlorous acid

>

acetic acid

>

formic acid) is correct because pH increases with decreasing Ka value, indicating weaker acidity.

21 Hard

21 of 118

Application +4 | -1

Prove that the pressure necessary to obtain

50 \%

dissociation of

\mathrm{PCl}_5

250^{\circ} \mathrm{C}

is numerically three times of

K_p

View answer and explanation

Correct option: A

\mathrm{P}=3 \mathrm{~K}_{\mathrm{P}}

Option A (P = 3Kp) is correct because the equilibrium constant Kp is related to the total pressure P by the equation Kp = P/3 when the degree of dissociation is 50%.

22 Hard

22 of 118

Analyzing +4 | -1

An amount of 3 moles of

\mathrm{N}_2

and some

\mathrm{H}_2

is introduced into an evacuated vessel. The reaction starts at

t=0

and equilibrium is attained at

t=t_1

. The amount of ammonia at

t=2 t_1

is found to be

34 \mathrm{~g}.

It is observed that

\frac{w\left(\mathrm{N}_2\right)}{w\left(\mathrm{H}_2\right)}=\frac{14}{3}

t=\frac{t_1}{3}

and

t=\frac{t_1}{2}.

The only correct statement is

View answer and explanation

Correct option: B

w\left(\mathrm{N}_2\right)+w\left(\mathrm{H}_2\right)+w\left(\mathrm{NH}_3\right)=102

g at

t=2 t_1

Option B (

w(N_2) + w(H_2) + w(NH_3) = 102

g at

t=2t_1

) is correct because the total mass remains constant throughout the reaction, and the initial mass is calculable from the given moles of

N_2

and the stoichiometric amount of

H_2

23 Hard

23 of 118

Analyzing +4 | -1

The following equilibria are given :

\begin{aligned} & {N}_2+3 {H}_2 \rightleftharpoons 2 {NH}_3 ; {K}_1 \\ & {~N}_2+{O}_2 \rightleftharpoons 2 {NO} ; {K}_2 \\ & {H}_2+\frac{1}{2} {O}_2 \rightleftharpoons {H}_2 {O} ; {K}_3 \end{aligned}

The equilibrium constant of the reaction

2 {NH}_3+\frac{5}{2} {O}_2 \rightleftharpoons 2 {NO}+3 {H}_2 {O}

in terms of

{K}_1, {~K}_2

and

{K}_3

View answer and explanation

Correct option: C

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

The correct answer is C because the equilibrium constant for the target reaction can be obtained by manipulating the given equilibrium reactions and their corresponding equilibrium constants.

24 Hard

24 of 118

Application +4 | -1

A solution contains a mixture of

\mathrm{Ag}^{+}(0.10 \mathrm{~M})

and

\mathrm{Hg}_2^{2+}(0.10 \mathrm{~M}),

which are to be separated by selective precipitation. Calculate the maximum concentration of iodide ion at which one of them gets precipitated almost completely. What per cent of that metal ion is precipitated before the start of precipitation of second metal ion?

K_{\mathrm{sp}}(\mathrm{AgI})=8.5\times 10^{-17}

and

K_{\text {sp }}\left(\mathrm{Hg}_2 \mathrm{I}_2\right)=2.5 \times 10^{-26}.

View answer and explanation

Correct option: A

5 \times 10^{-13} \mathrm{~M}, 99.83 \%

Option A (

5 \times 10^{-13}

M, 99.83 %) is correct because it identifies the maximum iodide concentration before the second precipitation and calculates the percentage of the first metal ion precipitated.

25 Hard

25 of 118

Application +4 | -1

Calculate

\mathrm{pH}

of solution obtained by mixing equal vol. of

0.02 \mathrm{~M} \, {HOCl} \,\, \& \, \, 0.2 \mathrm{~M} \, {CH}_3 {COOH}

solution given that

\mathrm{K}_{\mathrm{a}_1}({HOCl})=2 \times 10^{-4}, \mathrm{~K}_{\mathrm{a}_2}\left({CH}_3 {COOH}\right)=2 \times 10^{-5}.

View answer and explanation

Correct option: A

2.69

Option A (2.69) is correct because the pH is calculated considering the contributions of both weak acids, HOCl and CH3COOH.

26 Hard

26 of 118

Analyzing +4 | -1

For the equilibrium

\mathrm{A}(\mathrm{g})+2 \mathrm{~B}(\mathrm{~g}) \rightleftharpoons \mathrm{AB}_2(\mathrm{~g})

the value of equilibrium constant at

300 \mathrm{k}

3 \times 10^{-3} \mathrm{~mole}^{-2} \mathrm{~L}^2

and value of equilibrium constant at

500 \mathrm{k}

5 \times 10^{-5} \mathrm{~mol}^{-2} \mathrm{~L}^2

.

What is the sign of enthalpy change for this reaction

View answer and explanation

Correct option: B

Negative

The sign of the enthalpy change is negative because the equilibrium constant decreases as temperature increases, indicating an exothermic reaction.

27 Hard

27 of 118

Analyzing +4 | -1

3 \mathrm{~mol}

ammonia is heated in presence of

1 \mathrm{~mol}

of neon. At equilibrium, the mole fraction of hydrogen is found to be 0.5. The value of

\left(K_C\right) \times(V)

for

\mathrm{NH}_3(\mathrm{g}) \rightleftharpoons \frac{1}{2} \mathrm{N}_2(\mathrm{g})+\frac{3}{2} \mathrm{H}_2(\mathrm{g})

View answer and explanation

Correct option: C

3 \sqrt{3}

Option C (

3 \sqrt{3}

) is correct because, after calculating the equilibrium constant

K_C

based on the given mole fraction of hydrogen, the value of

K_C \times V

is found to be

3 \sqrt{3}

28 Hard

28 of 118

Application +4 | -1

K_p

has the value of

10^{-6} {~atm}^3

and

10^{-4} {~atm}^3

298 {~K}

and

323 {~K}

respectively for the reaction

{CuSO}_4 \cdot 3 {H}_2 {O}({s}) \rightleftharpoons {CuSO}_4({~s})+3 {H}_2 {O}({g})

\Delta_{{r}} H^{\circ}

for the reaction is :

View answer and explanation

Correct option: C

147.41 {~kJ} / {mol}

Option C (

147.41 {~kJ} / {mol}

) is correct because it is calculated using the Van't Hoff equation relating the change in Kp with temperature to the enthalpy change.

29 Hard

29 of 118

Application +4 | -1

\Delta G^{\circ}

25^{\circ} \mathrm{C}

for the reaction

\frac{1}{2} \mathrm{N}_2(\mathrm{g})+\frac{3}{2} \mathrm{H}_2(\mathrm{g}) \rightleftharpoons \mathrm{NH}_3(\mathrm{g})

-16.5 \mathrm{~kJ} \mathrm{~mol}^{-1}.

Hence, thermodynamic equilibrium constant is

View answer and explanation

Correct option: B

7.80 \times 10^2

Option B (

7.80 \times 10^2

) is correct because it is the thermodynamic equilibrium constant calculated using the given Gibbs free energy change.

30 Hard

30 of 118

Application +4 | -1

Calculate the percentage ionization of

0.01 \mathrm{~M}

acetic acid in

0.1 \mathrm{~M} {~HCl} {~K}_{{a}}

of acetic acid is

1.8 \times 10^{-5}

View answer and explanation

Correct option: B

0.018 \%

Option B (

0.018 \%

) is correct because it is calculated by considering the common ion effect of HCl on the ionization of acetic acid.

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