Chemical EquilibriumHard
Question
The reaction A(g) + B(g) $\rightleftharpoons$C(g) + D(g) occurs in a single step. The rate constant of forward reaction is 2.0 × 10–3 mol–1 L s–1. When the reaction is started with equimolar amounts of A and B, it is found that the concentration of A is twice that of C at equilibrium. The rate constant of the backward reaction is
Options
A.5.0 × 10–4 mol–1 L s–1
B.8.0 × 10–3 mol–1 L s–1
C.1.25 × 102 mol–1 L s–1
D.2.0 × 103 mol–1 L s–1
Solution
A(g) + B(g) $\rightleftharpoons$C(g) + D(g)
Initial moles a a 0 0
Equilibrium moles a – x a – x x x
From question, [A] = 2[C] ⇒ a – x = 2x ⇒ a = 3x
Now, $K_{eq} = \frac{K_{f}}{K_{b}} = \frac{x.x}{(a - x)(a - x)} \Rightarrow \frac{2 \times 10^{- 3}}{K_{b}} = \frac{x.x}{2x.2x}$
$\therefore K_{b} = 8 \times 10^{- 3}\text{ mo}\text{l}^{- 1}\text{ L }\text{S}^{- 1}$
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