Consider the following elementary reaction:N2O(g) →N2(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 O in terms of k1, k−1, and the equilibrium concentrations of N2O and N2.=O
Question
Consider the following elementary reaction:N2O(g) →N2(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 O in terms of k1, k−1, and the equilibrium concentrations of N2O and N2.=O
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[N2O] The rate of the reverse reaction can be written as: Rate-1 = k-1[N2][O]
Setting these equal to each other gives: k1[N2O] = k-1[N2][O]
We can solve this equation for [O] to get the equilibrium concentration of O in terms of k1, k-1, and the equilibrium concentrations of N2O and N2:
[O] = (k1/k-1) * ([N2O]/[N2])
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