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Problem PageQuestionConsider the following elementary reaction:NO(g) +O2(g) →NO2(g) +O(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 NO in terms of k1, k−1, and the equilibrium concentrations of O2, NO2, and O

Question

Problem PageQuestionConsider the following elementary reaction:NO(g) +O2(g) →NO2(g) +O(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 NO in terms of k1, k−1, and the equilibrium concentrations of O2, NO2, and O

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Solution

The rate of the forward reaction can be expressed as: Rate1 = k1[NO][O2]

The rate of the reverse reaction can be expressed as: Rate-1 = k-1[NO2][O]

At equilibrium, the rate of the forward reaction equals the rate of the reverse reaction. Therefore, we can set Rate1 equal to Rate-1 and solve for [NO]:

k1[NO][O2] = k-1[NO2][O]

Rearranging to solve for [NO] gives:

[NO] = (k-1/k1) * ([NO2][O]/[O2])

This problem has been solved

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