The equilibrium composition for the reaction \(\mathrm{PCl}_{3}+\mathrm{Cl}_{2} \rightleftharpoons \mathrm{PCl}_{5}\) at 298 K is given below. \([\mathrm{PCl}_{3}]_{\mathrm{eq}}=0.2 \mathrm{~mol} \mathrm{~L}^{-1}\) \([\mathrm{Cl}_{2}]_{\mathrm{eq}}=0.1 \mathrm{~mol} \mathrm{~L}^{-1}\), \([\mathrm{PCl}_{5}]_{\mathrm{eq}}=0.40 \mathrm{~mol} \mathrm{~L}^{-1}\) If 0.2 mol of \(\mathrm{Cl}_{2}\) is added at the same temperature, the equilibrium concentrations of \(\mathrm{PCl}_{5}\) is _____ \(\times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1}\). Given: \(\mathrm{K}_{\mathrm{c}}\) for the reaction at 298 K is 20

JEE Main · 2023 · 06 Apr 2023 · Shift 2 · Equilibrium

The equilibrium composition for the reaction

PCl_{3} + Cl_{2} ⇌ PCl_{5}

at 298 K is given below.

[PCl_{3}]_{eq} = 0.2 mol L^{- 1}

[Cl_{2}]_{eq} = 0.1 mol L^{- 1}

[PCl_{5}]_{eq} = 0.40 mol L^{- 1}

If 0.2 mol of

Cl_{2}

is added at the same temperature, the equilibrium concentrations of

PCl_{5}

is _____

\times 1 0^{- 2} mol L^{- 1}

.
Given:

K_{c}

for the reaction at 298 K is 20

Answer: 49

Initially:

[PCl_{3}] = 0.2

[Cl_{2}] = 0.1 + 0.2 = 0.3

[PCl_{5}] = 0.4

M. Let x mol/L of

PCl_{3}

and

Cl_{2}

react. At new equilibrium:

[PCl_{3}] = 0.2 - x

[Cl_{2}] = 0.3 - x

[PCl_{5}] = 0.4 + x

. Using

K_{c} = 20

\frac{0.4 + x}{( 0.2 - x ) ( 0.3 - x )} = 20

. Solving:

0.4 + x = 20 (0.06 - 0.5 x + x^{2})

, which gives

20 x^{2} - 11 x + 0.8 = 0

. Solving this quadratic:

x \approx 0.086

. Therefore,

[PCl_{5}]_{eq} = 0.4 + 0.086 = 0.486

= 48.6 \times 1 0^{- 2}

M ≈ 49 × 10⁻² M.

JEE Main · 2023 · 06 Apr 2023 · Shift 2 · Integer answer