Consider the following elementary reaction:CH33CBr(aq) →CH33C+(aq) +Br−(aq)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 CH33C+ in terms of k1, k−1, and the equilibrium concentrations of CH33CBr and Br−.
Question
Consider the following elementary reaction:CH33CBr(aq) →CH33C+(aq) +Br−(aq)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 CH33C+ in terms of k1, k−1, and the equilibrium concentrations of CH33CBr and Br−.
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 expression:
k1[CH33CBr] = k−1[CH33C+][Br−]
We can rearrange this equation to solve for the equilibrium concentration of CH33C+:
[CH33C+] = (k1/k−1)[CH33CBr]/[Br−]
This equation gives the equilibrium concentration of CH33C+ in terms of k1, k−1, and the equilibrium concentrations of CH33CBr and Br−.
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