The exothermic formation of $\mathrm{ClF}_{3}$ is represented by the equation: $\mathrm{Cl}_{2(\mathrm{g})}+3 \mathrm{~F}_{2(\mathrm{g})} \rightleftharpoons 2 \mathrm{ClF}_{3(\mathrm{g})} ; \quad \Delta \mathrm{rH}=-329 \mathrm{~kJ}$ Which of the following will increase the quantity of $\mathrm{ClF}_{3}$ in an equilibrium mixture of $\mathrm{Cl}_{2}, \mathrm{~F}_{2}$ and $\mathrm{ClF}_{3}$ ?

(C) Increasing the volume of the container

JEE Main · 2005 · 08 May 2005 · Offline · Equilibrium

The exothermic formation of

ClF_{3}

is represented by the equation:

Cl_{2 (g)} + 3 F_{2 (g)} ⇌ 2 ClF_{3 (g)}; Δ rH = - 329 kJ

Which of the following will increase the quantity of

ClF_{3}

in an equilibrium mixture of

Cl_{2}, F_{2}

and

ClF_{3}

(A)Increasing the temperature
(B)Removing $Cl_{2}$
(C)Increasing the volume of the container
(D)Adding $F_{2}$ ✓ Correct

Answer: (D)

Solution

Each option is a Le Chatelier nudge on

Cl X_{2} + 3 F X_{2} 2 ClF X_{3}

, which is exothermic. Raising the temperature pushes an exothermic equilibrium backward, removing

Cl X_{2}

takes away a reactant and also pulls it backward, and expanding the volume favours the side with more gas moles, the four-mole reactant side, so all three lower the

ClF X_{3}

. Adding

F X_{2}

does the opposite: feeding in more reactant drives the equilibrium forward and builds up the

ClF X_{3}

JEE Main · 2005 · 08 May 2005 · Offline · Single correct option

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JEE Main · 2005 · 08 May 2005 · Offline · Equilibrium

Solution

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