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 N2 in terms of k1, k−1, and the equilibrium concentrations of N2O and 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 N2 in terms of k1, k−1, and the equilibrium concentrations of N2O and O.
Solution
The equilibrium constant (K) for a reaction is given by the ratio of the rate constants for the forward and reverse reactions. In this case, K = k1/k−1.
At equilibrium, the rate of the forward reaction equals the rate of the reverse reaction. Therefore, we can write the following expressions for the rates of the forward and reverse reactions:
Rate of forward reaction = k1[N2O] Rate of reverse reaction = k−1[N2]*[O]
Setting these two rates equal to each other gives:
k1[N2O] = k−1[N2]*[O]
We can rearrange this equation to solve for [N2]:
[N2] = (k1/k−1)*([N2O]/[O])
So, the equilibrium concentration of N2 is given by the ratio of the rate constants times the ratio of the equilibrium concentrations of N2O and O.
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