In a reversible reaction A $\leftrightarrows$ B at a particular temperature, the rate constant for the reverse reaction is 4 times the rate constant for the forward reaction. What is the value of the equilibrium constant at that temperature?

Question

Quizplus · Accepted Answer

The equilibrium constant $ K $ is the ratio of the rate constant of the forward reaction $ k_f $ to the rate constant of the reverse reaction $ k_r $. Given $ k_r = 4k_f $, $ K = \frac{k_f}{k_r} = \frac{k_f}{4k_f} = 0.25 $.

In a Reversible Reaction A ⇆\leftrightarrows⇆ B at a Particular Temperature, the Rate Constant for the Temperature

In a Reversible Reaction A $\leftrightarrows$ B at a Particular Temperature, the Rate Constant for the Temperature