At a certain temperature, only 50% HI is dissociated into H2 and I2 at equilibrium. The equilibrium constant is :

Correct answer: (D) 0.25. 2HIg ⇌ H_2g + I_2g Moles 1-α (α)/2 (α)/2 Concentration (1-α)/V (α)/2V (α)/2V K_C=(H_2I_2HI^2=α2Vα2V1-αV^2 ∴α=0.50α2=0.25 =α^241-α^2=0.50×0.50)/(4×0.50×0.50)=1/4 K_C = 0.25

Q26 / 90 · JEE Main 2014 (09 Apr Online) · Chemistry

At a certain temperature, only 50% HI is dissociated into H₂ and I₂ at equilibrium. The equilibrium constant is :

A
0.5
B
4.0
C
1.0
Correct answer
D
0.25

Solution

\begin{matrix} 2 H I (g) & ⇌ & H_{2} (g) & + & I_{2} (g) \\ Moles & 1 - α & \frac{α}{2} & \frac{α}{2} \\ Concentration & \frac{1 - α}{V} & \frac{α}{2V} & \frac{α}{2V} \end{matrix}

K_{C} = \frac{[H_{2}] [I_{2}]}{{[H I]}^{2}} = \frac{[\frac{α}{2V}] [\frac{α}{2V}]}{{[\frac{1 - α}{V}]}^{2}}

(\begin{matrix} ∴ α & = & 0.50 \\ \frac{α}{2} & = & 0.25 \end{matrix})

= \frac{α^{2}}{4 {(1 - α)}^{2}} = \frac{0 .50 \times 0 .50}{4 \times0.50 \times 0 .50} = \frac{1}{4}

K_{C} = 0.25

Hence, the correct answer is option: 'D'.

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