Q.1 mol of $CH_4$, 1 mole of $CS_2$ and 2 mol of $H_2S$ and 2 mol of $H_2$ are mixed in a 500 ml flask. The equilibrium constant for the reaction $CH_4(g) + 2H_2S(g) \rightleftharpoons CS_2(g) + 4H_2(g)$ is $K_C = 4 \times 10^{-2}$ mol$^2$ lit$^{-2}$. In which direction will the reaction proceed to reach equilibrium?

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:

1. 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.
2. 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. …

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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