Practice Problem 7

The rate constants for the forward and reverse reactions in the following equilibrium have been measured.   At 25°C, k_f is 7.3 x 10³ liters per mole-second and k_r is 0.55 liters per mole-second.  Calculate the equilibrium constant for this reaction:

ClNO₂(g) + NO(g) NO₂(g) + ClNO(g)

Solution

We start by assuming that the rates of the forward and reverse reactions at equilibrium are the same:

At equilibrium:     rate_forward = rate_reverse

We then substitute the rate laws for these reactions into this equality:

At equilibrium:     k_f[ClNO₂][NO] = k_r[NO₂][ClNO]

We then rearrange this equation to get the equilibrium constant expression for the reaction :

The equilibrium constant for the reaction is therefore equal to the rate constant for the forward reaction divided by the rate constant for the reverse reaction: