Consider the following elementary reaction:NO2(g) +O3(g) →NO3(g) +O2(g)Suppose we let k1 stand for the rate constant of this reaction, and k−1 stand for the rate constant of the reverse reaction.Write an expression that gives the equilibrium concentration of O3 in terms of k1, k−1, and the equilibrium concentrations of NO2, NO3, and O2.
Question
Consider the following elementary reaction:NO2(g) +O3(g) →NO3(g) +O2(g)Suppose we let k1 stand for the rate constant of this reaction, and k−1 stand for the rate constant of the reverse reaction.Write an expression that gives the equilibrium concentration of O3 in terms of k1, k−1, and the equilibrium concentrations of NO2, NO3, and O2.
Solution
At equilibrium, the rate of the forward reaction equals the rate of the reverse reaction.
The rate of the forward reaction can be written as: Rate1 = k1[NO2][O3]
The rate of the reverse reaction can be written as: Rate-1 = k-1[NO3][O2]
Setting these equal to each other gives: k1[NO2][O3] = k-1[NO3][O2]
We can solve this equation for [O3] to get the equilibrium concentration of O3 in terms of k1, k−1, and the equilibrium concentrations of NO2, NO3, and O2:
[O3] = (k-1/k1) * ([NO3][O2]/[NO2])
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