The reaction rate when the rate-determining step has a back reaction

Question

For some reason, I am reviewing general chemistry and I find that the elementary reaction for the rate-determining step (RDS) is assumed to be irreversible, that is, single arrow, in all questions that I see.

For example, think a reaction consisting of two elementary reactions:

$$\ce{A + B <=>[k_1][k_1^\prime] C} \tag{fast, reversible}$$

$$\ce{C ->[k_2] D} \tag{slow}$$

Then the overall reaction is $\ce{A + B -> D}$, and the rate is then

$$\text{rate} = k_{2}[\mathrm{C}].$$

Now, I'm curious about the rate when the equation becomes like

$$ \ce{A + B <=>[k_{1}][k_{1}^\prime] C \:(\mathrm{fast, reversible)} \\ C \color{red}{<=>[k_{2}][k_{2}^\prime]} D\:(\mathrm{slow)} } $$

In this case, I think the rate expression should be changed. From the overall chemical equation ($\ce{A + B -> D}$), the rate itself is defined as $$ \mathrm{rate} = \frac{\mathrm{d}[\mathrm{D}]}{\mathrm{d}t}, $$ and checking the reactions related to $[\mathrm{D}]$ in the RDS, $$ \frac{\mathrm{d}[\mathrm{D}]}{\mathrm{d}t} = k_{2}[\mathrm{C}] - k_{2}'[\mathrm{D}]. $$ Thus, the rate is $$ \mathrm{rate} = k_{2}[\mathrm{C}] - k_{2}'[\mathrm{D}]. $$

Though, I'm not sure if the reasoning is right. Is it correct?

Answer 1

Thus, the rate is $$ \mathrm{rate} = k_{2}[\mathrm{C}] - k{_2}'[\mathrm{D}]. $$ Though, I'm not sure if the reasoning is right. Is it correct?

This is correct but does not give you much insight into how [D] changes over time. This system will reach equilibrium eventually. If you call the equilibrium concentrations $[\ce{C}]_{eq}$ and $[\ce{D}]_{eq}$, and $\epsilon$ the difference between current and equilibrium concentration for C, you get the following:

$$[\ce{C}] = [\ce{C}]_{eq} + \epsilon\tag{1}$$ $$[\ce{D}] = [\ce{D}]_{eq} - \epsilon\tag{2}$$

Also, the rates are equal at equilibrium:

$$ k_2 [\ce{C}]_{eq} - k{_2}'[\ce{D}]_{eq} = 0\tag{3}$$

Plugging all of this into the rate equation gives:

$$\frac{d[\ce{D}]}{dt} = -\frac{d\epsilon}{dt} = k_{2}[\mathrm{C}] - k{_2}'[\mathrm{D}]\tag{4}$$

$$= k_{2}([\ce{C}]_{eq} + \epsilon) - k{_2}'([\ce{D}]_{eq} - \epsilon)$$

$$= k_{2} \epsilon - k{_2}'(- \epsilon)\tag{*}$$

$$ = (k_{2} + k{_2}') \cdot \epsilon$$

The simplification yielding the expression labelled '*' is because the net rate at equilibrium is zero (substituting (3) into the previous expression). As a result, you get a first order behavior of approching equilibrium with respect to the "excess concentration" $\epsilon$. When you integrate this, $\epsilon$ decays exponentially, with the sum of the forward and reverse rate constant in the exponent. This is somewhat surprising because we started out with these rate constants appearing with opposite signs in the (correct) rate expression given by the OP. However, this result is correct and well-known from temperature jump kinetics.

What about A and B?

The derivation above is for a two-species system $\ce{C<=>D}$. If A and B are in fast equilibrium with C, equation (1) or (2) has to be modified to account for A and B supplying more C as it reacts to D. One way is to define $[\mathcal{C}] = [\ce{A}] + [\ce{C}]$ and write (instead of (1)):

$$[\mathcal{C}] = [\mathcal{C}]_{eq} + \epsilon\tag{1'}$$

Because the first equilibrium is fast, you can calculate $\ce{A, B and C}$ from $[\mathcal{C}]$ at any point in time, and you would have to adjust equation (4) accordingly. The details depend on whether $\ce{A and B}$ are present at stoichiometric amounts or one of them is in excess. If one of them is in large excess, the result would just be different by a constant $\beta$, i.e.

$$ \mathrm{rate} = (\beta k_{2} + {k_2}') \cdot \epsilon$$

where $\beta$ depends on the equilibrium constant of the first (fast) reaction and the concentration of the reactant in excess.