The rate constant for the forward reaction: A(g)$\rightleftharpoons$ 2B(g) is 1.5 × 10−3 s−1 at 300

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The rate constant for the forward reaction: A(g)$\rightleftharpoons$ 2B(g) is 1.5 × 10−3 s−1 at 300 K. If 10−5 moles of ‘A’ and 100 moles of ‘B’ are present in a 10 litre vessel at equilibrium, then the rate constant of the backward reaction at this temperature is

Accepted Answer

$K_{eq} = \frac{K_{f}}{K_{b}} = \frac{\lbrack B\rbrack^{2}}{\lbrack A\rbrack}$

$\Rightarrow \frac{1.5 \times 10^{- 3}}{K_{b}} = \frac{(100/10)^{2}}{\left( 10^{- 5}/10 \right)} \Rightarrow K_{b} = 1.5 \times 10^{- 11}\text{ }\text{M}^{- 1}\text{ }\text{S}^{- 1}$

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