Chemical EquilibriumHard

Question

Forty percent of a mixture of 0.2 mol of N2 and 0.6 mol of H2 reacts to give NH3 according to the equation

N2(g) + 3H2 (g) $\rightleftharpoons$2NH3(g) at constant temperature and pressure. Then, the ratio of the final volume to the initial volumes of gases is

Options

A.4: 5
B.5: 4
C.7: 10
D.8: 5

Solution

N2 + 3H2 $\rightleftharpoons$ 2NH3

0.2 mole 0.6 mole 0

Equilibrium $0.2 - 0.2 \times \frac{40}{100}$ 0.6 – 0.08 × 3 2 × 0.08

= 0.12 = 0.36 = 0.16

$\therefore\frac{V_{\text{final}}}{V_{\text{initial}}} = \frac{n_{\text{final}}}{n_{\text{initial}}} = \frac{0.64}{0.80} = \frac{4}{5}$

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