For the reaction, A₂(g) + B₂(g) leftharpoons 2AB(g) ; Δ H is -ve. the following molecular scenes represent dif

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A_2(g) + B_2(g) \rightleftharpoons 2AB(g) \ ; \ \Delta H \text{ is } -\text{ve}.

(x) contains 2 molecules of

A_2

, 1 molecule of

B_2

and 6 molecules of

AB

. Scene (y) contains 3 molecules of

A_2

, 3 molecules of

B_2

and 2 molecules of

AB

i) Calculate the equilibrium constant

K_P

and

K_C

ii) For the reaction mixture represented by scene (x), (y) the reaction proceed in which directions?

iii) What is the effect of increase in pressure for the mixture at equilibrium.

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Step 1 (part i). At the equilibrium scene, the box contains 2 molecules of $A_2$ , 2 of $B_2$ , and 4 of $AB$ . Treating molecule counts as directly proportional to concentration (same fixed volume for every scene), $K_C = \dfrac{[AB]^2}{[A_2][B_2]} = \dfrac{4^2}{2\times2} = \dfrac{16}{4} = 4$ . Since $A_2(g)+B_2(g)\rightleftharpoons2AB(g)$ has $\Delta n_g = 2-2=0$ , $K_P=K_C=4$ .

Step 2 (part ii, scene x). Scene (x) contains 2 $A_2$ , 1 $B_2$ , 6 $AB$ . $Q_C = \dfrac{[AB]^2}{[A_2][B_2]} = \dfrac{6^2}{2\times1} = \dfrac{36}{2}=18$ . Since $Q_C(18) > K_C(4)$ , scene (x) must react in the REVERSE direction, converting some AB back into $A_2$ and $B_2$ .

Step 3 (part ii, scene y). Scene (y) contains 3 $A_2$ , 3 $B_2$ , 2 $AB$ . $Q_C = \dfrac{2^2}{3\times3} = \dfrac{4}{9} \approx 0.44$ . Since $Q_C(0.44) < K_C(4)$ , scene (y) must react in the FORWARD direction, forming more AB. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2022Set ANNUAL1 markMCQ
$Q. Which one of the following is incorrect statement?$
(a) presence of catalyst affects both the forward reaction and reverse reaction to the same extent.
(b) for a system at equilibrium Q is always less than the equilibrium constant.
(c) equilibrium constant varies with temperature.
(d) equilibrium can be attained from either side of the reaction.
›Reveal solutionSolution
By definition, at equilibrium the reaction quotient Q equals the equilibrium constant K (Q = K), not less than it — so the statement claiming Q < K at equilibrium is false, making it the incorrect statement among the choices.

Check each statement:

A catalyst lowers the activation energy for both the forward and reverse reactions by the same amount, so it speeds up both equally without shifting the equilibrium position — this statement is CORRECT.

The reaction quotient Q is calculated the same way as K, using the concentrations present at any given moment. When the system reaches equilibrium, by definition Q becomes equal to K (Q = K). Saying Q is 'always less than' K at equilibrium is FALSE — this is the incorrect statement. …
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CBSE 2020Set ANNUAL1 markMCQ
$Q. Which of the following is incorrect statement?$
(a) Equilibrium constant varies with temperature.
(b) For a system at equilibrium, Q is always less than the equilibrium constant.
(c) Equilibrium can be attained from either side of the reaction.
(d) Presence of catalyst affects both the forward reaction and reverse reaction to the same extent.
›Reveal solutionSolution

Statement (b) is incorrect because at equilibrium Q always equals K, never remains less than it.

The reaction quotient Q has the same mathematical form as the equilibrium constant K, but Q can be evaluated at ANY point during a reaction using the concentrations present at that instant, whereas K uses the equilibrium concentrations only. Comparing Q to K tells us the direction a reaction will shift: if Q < K the forward reaction proceeds (to increase Q), if Q > K the reverse reaction proceeds, and the reaction reaches equilibrium exactly when Q = K. So the claim that Q is 'always less than' K at equilibrium is wrong — it should be Q = K at equilibrium.

…

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