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 CH33CBr in terms of k1, k−1, and the equilibrium concentrations of CH33C+ 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 CH33CBr in terms of k1, k−1, and the equilibrium concentrations of CH33C+ and Br−.
Solution
The equilibrium constant (K) for a reaction is the ratio of the rate constants for the forward and reverse reactions. In this case, the equilibrium constant for the reaction would be 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 CH33CBr:
[CH33CBr] = (k−1/k1)[CH33C+][Br−]
So, the equilibrium concentration of CH33CBr is given by the ratio of the rate constant of the reverse reaction to the rate constant of the forward reaction, multiplied by the equilibrium concentrations of CH33C+ and Br−.
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