For the reaction, $$ 2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{SO}_{3}(\mathrm{~g}) $$ $\Delta_{\mathrm{H}}=-57.2 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $\mathrm{K}_{\mathrm{C}}=1.7 \times 10^{16}$ Which of the following statement is INCORRECT?

(A) The equilibrium constant is large suggestive of reaction going to completion and so no catalyst is required.

(B) The equilibrium will shift in forward direction as the pressure increase.

(C) The equilibrium constant decreases as the temperature increases.

(D) The addition of inert gas at constant volume will not affect the equilibrium constant.

JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Equilibrium

For the reaction,

2 SO_{2} (g) + O_{2} (g) ⟶ 2 SO_{3} (g)

Δ_{H} = - 57.2 kJ mol^{- 1}

and

K_{C} = 1.7 \times 1 0^{16}

Which of the following statement is INCORRECT?

(A)The equilibrium constant is large suggestive of reaction going to completion and so no catalyst is required.✓ Correct
(B)The equilibrium will shift in forward direction as the pressure increase.
(C)The equilibrium constant decreases as the temperature increases.
(D)The addition of inert gas at constant volume will not affect the equilibrium constant.

Answer: (A)

Solution

This reaction is exothermic and runs with a fall in gas moles,

Δ n_{g} = - 1

. That genuinely means raising the pressure pushes it forward, raising the temperature lowers

K_{c}

, and adding an inert gas at fixed volume leaves

K_{c}

untouched, so those three statements stand. The faulty claim is the one arguing that a huge equilibrium constant removes the need for a catalyst: a large

K_{c}

only tells us the reaction is thermodynamically favourable, not fast, and in practice the sluggish rate still has to be helped along by a catalyst such as

V X_{2} O X_{5}

JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Single correct option

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JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Equilibrium

Solution

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