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JEE Main

Explore popular questions from Equilibrium for JEE Main. This collection covers Equilibrium previous year JEE Main questions hand picked by experienced teachers.

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Equilibrium

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Q 1. At a given temperature, the equilibrium constant for the reactions
{tex} \mathrm { NO } ( \mathrm { g } ) + \frac { 1 } { 2 } \mathrm { O } _ { 2 } ( \mathrm { g } ) \rightleftharpoons \mathrm { NO } _ { 2 } ( \mathrm { g } ) {/tex} and {tex}2 \mathrm { NO } _ { 2 } ( \mathrm { g } ) \rightleftharpoons {/tex} {tex} 2 \mathrm { NO } ( \mathrm { g } ) + \mathrm { O } _ { 2 } ( \mathrm { g } ) {/tex} are {tex}\mathrm K _ { 1 } {/tex} and {tex}\mathrm K _ { 2 } {/tex} respectively.
If {tex}\mathrm K _ { 1 } {/tex} is {tex} 4\times 10 ^ { - 3 } , {/tex} then {tex}\mathrm K _ { 2 } {/tex} will be

A

{tex} 8 \times 10 ^ { - 3 } {/tex}

B

{tex} 16 \times 10 ^ { - 3 } {/tex}

{tex} 6.25 \times 10 ^ { 4 } {/tex}

D

{tex} 6.25 \times 10 ^ { 6 } {/tex}

Explanation


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Q 2. In which of the following reactions {tex}\mathrm{ K _ { P } > K _ { C }} ? {/tex}

A

{tex} \mathrm { N } _ { 2 } ( \mathrm { g } ) + 3 \mathrm { H } _ { 2 } ( \mathrm { g } ) \rightleftharpoons 2 \mathrm { NH } _ { 3 } ( \mathrm { g } ) {/tex}

B

{tex} \mathrm { H } _ { 2 } ( \mathrm { g } ) + \mathrm { I } _ { 2 } ( \mathrm { g } ) \rightleftharpoons 2 \mathrm { HI } ( \mathrm { g } ) {/tex}

C

{tex} \mathrm { PCl } _ { 3 } ( \mathrm { g } ) + \mathrm { Cl } _ { 2 } ( \mathrm { g } ) \rightleftharpoons \mathrm { PCl } _ { 5 } ( \mathrm { g } ) {/tex}

{tex} 2 \mathrm { SO } _ { 3 } ( \mathrm { g } ) \rightleftharpoons 2 \mathrm { SO } _ { 2 } ( \mathrm { g } ) + \mathrm { O } _ { 2 } ( \mathrm { g } ) {/tex}

Explanation

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Q 3. At {tex} 1000 \mathrm { K } , {/tex} the value of {tex} \mathrm K _ {\mathrm p } {/tex} for the reaction {tex} \mathrm { PCl } _ { 5 } \rightleftharpoons {/tex} {tex} \mathrm { PCl } _ { 3 } + \mathrm { Cl } _ { 2 } {/tex} is {tex} 20 \mathrm { R } , {/tex} then {tex} \mathrm K _ {\mathrm C } {/tex} will be

A

0.04

0.02

C

0.01

D

0.03

Explanation

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Q 4. For the reaction {tex} \mathrm { N } _ { 2 } \mathrm { O } _ { 4 } \rightleftharpoons 2 \mathrm { NO } , {/tex} degree of dissociation is {tex} \alpha {/tex} . The number of moles at equilibirum will be

A

{tex} ( 1 - \alpha ) ^ { 2 } {/tex}

{tex} ( 1 + \alpha ) {/tex}

C

3

D

1

Explanation

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Q 5. {tex}4{/tex} mole of {tex}\mathrm A {/tex} are mixed with {tex}4{/tex} mole of {tex}\mathrm B , {/tex} when {tex}2{/tex} mole of {tex}\mathrm C {/tex} are formed at equilibrium according to the reaction, {tex}\mathrm A +\mathrm B \rightleftharpoons \mathrm C +\mathrm D . {/tex} The equilibrium constant is

A

4

1

C

{tex} \sqrt { 2 } {/tex}

D

{tex} \sqrt { 4 } {/tex}

Explanation


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Q 6. {tex}20 \% {/tex} part of {tex} \mathrm { N } _ { 2 } \mathrm { O } _ { 4 } {/tex} is dissociated in {tex} \mathrm { NO } _ { 2 } {/tex} at {tex}1{/tex} atmospheric pressure. What is the value of {tex} \mathrm K _ { \mathrm p } ? {/tex}

A

0.29

B

16

C

1.6

None of these

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Q 7. Vaporisation of one mole of {tex} \mathrm { PCl } _ { 5 } {/tex} is done in a {tex} \mathrm {1 L } {/tex} container. If {tex}0.2{/tex} mole {tex} \mathrm { Cl } _ { 2 } {/tex} is formed at equilibrium, then what will be the value of equilibrium constant?

A

0.15

0.05

C

1.78

D

0.2

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Q 8. The equilibrium constant {tex}\mathrm K _ {\mathrm c } {/tex} for {tex} \mathrm A _ { ( \mathrm g ) } \rightleftharpoons \mathrm B _ { (\mathrm g ) } {/tex} is {tex}1.1{/tex} means

A

{tex} [ \mathrm { A } ] \mathrm { eq } > [ \mathrm { B } ] \mathrm { eq } {/tex}

{tex} [ \mathrm { Aeq } ] < [ \mathrm { B } ] \mathrm { eq } {/tex}

C

{tex} [ \mathrm { Aeq } ] = 1.1 \mathrm { M } {/tex}

D

{tex} [ \mathrm { Beq } ] = 1.1 \mathrm { M } {/tex}

Explanation

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Q 9. Which of the following is not Lewis acid?

A

B

C

Explanation

All jother can accept electrons. in not electron deficient.

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Q 10. Which of the following is conjugate acid of ?

B

C

D

Explanation

Conjugate acid of is

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Q 11. Which of the following species can act both as acid as well as a base?

A

C

D

Explanation

can accept or give a proton.

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Q 12. Which of the following salts will give basic solution on hydrolysis?

A

B

KCl

D

Explanation


KOH⟶ (Basic solution),

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Q 13. What will be the ionisation constant of formic acid if its 0.001 M solution is 14.5% ionised?

B

C

D

Explanation



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Q 14. What will be the pH of a soft drink if hydrogen ion concentration in sample is M?

A

B

D

Explanation

pH = - log

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Q 15. What is pOH of an aqueous solution with hydrogen ion concentration equal to ?

B

C

D

Explanation


pOH = 14 - pH, 14 -

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Q 16. Which of the following salts does not show is correct nature mentioned against it?

A

KBr solution - Neutral

NaCN solution - Acidic

C

solution- acidic

D

KF solution - Basic

Explanation

NaCN solution is basic in nature since HCN formed is a weak acid and does not hydrolyse.

NaOH ⟶ (Basic solution)

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Q 17. The degree of ionisation of an acid HA is at M concentratin. Its. Dissociation constant will be

A

C

D

Explanation

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Q 18. Given below are the dissociation constant values of few acids. Arrange them in order of increasing acidic strength.

HCN < < <

B

< < HCN <

C

< HCN < <

D

< < < HCN

Explanation

Acidic strength,

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Q 19. Solution of a monobasic acid has a pH = 5. If one mL of it is diluted to 1 litre, what will be the pH of the resulting solution?

A

C

D

Explanation


After dilution
Total

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Q 20. is a salt of weak acid HCN and a weak base . A one molar solution of will be:

A

Neutral

B

Strongly acidic

C

Strongly basic

Weakly basic

Explanation

Since the solution will be slightly basic.

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Q 21. Which of the following is an example of homogeneous equilibrium?

B

C

D

Explanation

All the reactants and products are in same physical state.

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Q 22. For the reaction : , the standard free energy is The equilibrium constant (K) would be.

A

K = 0

B

K > 1

C

K = 1

K < 1

Explanation


then only (or /> 0).
Thus, K < 1

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Q 23. If the equilibrium constant for the reaction.
is 81, What is the value of equilibrium constant for the reaction. XY⟶

A

81

9

C

6561

D

Explanation

2XY ⟶
XY ⟶

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Q 24. At 473, K, for the reaction is . What will be the value of for the formation of at the same temperature?

A

C

D

Explanation


For

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Q 25. The value of for the following equilibrium is . Given bar at 1073 K.

B

C

D

Explanation


bar,