In a chemical equilibrium, the rate constant for the backward reaction () is and the equilibrium con — Chemical Equilibrium Chemistry Question
Question
In a chemical equilibrium, the rate constant for the backward reaction ($k_b$) is $7.5 \times 10^{-4}$ and the equilibrium constant ($K_c$) is $1.5$. The exact rate constant for the forward reaction ($k_f$) is:
Answer: C
💡 Solution & Explanation
By definition, the equilibrium constant $K_c = k_f / k_b$. We are given $K_c = 1.5$ and $k_b = 7.5 \times 10^{-4}$. Thus, $k_f = K_c \times k_b = 1.5 \times 7.5 \times 10^{-4} = 11.25 \times 10^{-4} = 1.125 \times 10^{-3}$.
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