Equilibrium — CBSE Class 12 · 2006 · All India Compartment Board Question with Solution

At 300°C the thermal dissociation of HI,

2 HI ⇌ H_{2} + I_{2}

, is found to be 20%. What will be the equilibrium concentrations of H

_{2}

and I

_{2}

in the system at this temperature if the equilibrium concentration of HI in it is 0.96 mol L

^{- 1}

Answer: [H₂] = [I₂] = 0.12 mol L⁻¹

Solution

Let the initial concentration of HI be

a

. With 20% dissociation, HI left:

0.8 a = 0.96 \Rightarrow a = 1.2 mol L^{- 1}

HI dissociated =

0.2 \times 1.2 = 0.24

mol L

^{- 1}

.
From

2 HI \to H_{2} + I_{2}

, each mole pair of HI gives half a mole of each product:

[H_{2}] = [I_{2}] = \frac{0.24}{2}

[H_{2}] = [I_{2}] = 0.12 mol L^{- 1}

CBSE Class 12 Board · 2006 · All India Compartment · Subjective