Chemical EquilibriumHard

Question

For a reversible reaction, the rate constants for the forward and backward reactions are 0.16 and 4 × 104, respectively. What is the value of equilibrium constant of the reaction?

Options

A.0.25 × 106
B.2.5 × 105
C.4 × 10–6
D.4 × 10–4

Solution

$K_{eq} = \frac{K_{f}}{K_{b}} = \frac{0.16}{4 \times 10^{4}} = 4 \times 10^{- 6}$

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