Question

25. Write the overall reaction and rate laws that correspond to the following reaction mechanisms. Be sure to eliminate intermediates from the answers.

$$\begin{gathered} a\left)\text{A+B}\underset{k_{-1}}{\overset{K_1}{\rightleftarrows }}\text{C+D}\left(Fasteqilbrium\right) \\ \text{C+\hspace{0.17em}E}\stackrel{k_2}{\to F\left(Slow\right)} \\ b\right)\text{A\hspace{0.17em}}\underset{k_{-1}}{\overset{k_1}{\rightleftarrows }}B+C\left(Fastequilbrium\right) \\ \text{C\hspace{0.17em}+D}\underset{k_{-2}}{\overset{k_2}{\rightleftarrows }}E\left(Fastequilbrium\right) \\ \text{E}\stackrel{k_3}{\to }F\left(Slow\right) \end{gathered}$$

Short answer

The rate of the reaction is $\text{Rate}=\frac{{\text{k}}_1{k}_2{k}_3\left[A\right]\left[D\right]}{{\text{k}}_{-2}{k}_{-1}\left[B\right]}$

Step-by-step solution

Overall equation:

a) If the reaction contains many steps the overall reaction can be obtained summing up all the reactants and products on either side.

$$\begin{gathered} a)\text{A+B}\underset{k_{-1}}{\overset{K_1}{\rightleftarrows }}\text{C+D}\left(Fasteqilbrium\right) \\ \text{C+\hspace{0.17em}E}\stackrel{k_2}{\to F\left(Slow\right)} \end{gathered}$$

Add all the steps to obtain the entire reaction

localid="1661145948697" $\text{A+B+C}+\text{E→D+C+F.}.......\left(1\right)$

As C is common both sides ,one would cancel this term,so the overallreactionobtained is

$\text{A+B+E→C+F.}......\left(2\right)$

Rate determining step :

The rate-determining step is the slowest step in the reaction, even though the reaction contains many steps.

$rate=k\left[C\right]\left[E\right]........\left(3\right)$

Determining the equilbrium constant:

The equilibrium constant for a reaction can be obtained by dividing the product concentration by

reactent concendration.One would consider The fast equilibrium step to calculate K as it is a reversible reaction.

$$\begin{gathered} \text{A+B}\underset{k_{-1}}{\overset{K_1}{\rightleftarrows }}\text{C+D}\left(Fasteqilbrium\right) \\ \text{K=}\frac{\left[C\right]\left[D\right]}{\left[A\right]\left[B\right]}\text{.................}\left(4\right) \end{gathered}$$

According to principle of detailed balance

$\text{K=}\frac{k_1}{k_{-1}}\text{.................}\left(5\right)$

From the equation 4 and 5 the following equation is obtained

localid="1661146817018" $$\begin{gathered} \frac{k_1}{k_{-1}}\hspace{0.17em}=\frac{\left[C\right]\left[D\right]}{\left[A\right]\left[B\right]}..........\left(6\right) \\ or \\ \left[C\right]=\frac{k_1\left[A\right]\left[B\right]}{k_{-1}\left[D\right]}...........\left(7\right) \end{gathered}$$

Substituting the value of C in the equation-6

$$\begin{gathered} Rate=\frac{k_1{k}_2\left[A\right]\left[B\right]\left[E\right]}{k_{-1}\left[D\right]}.........\left(8\right) \\ Thisistheratelawforabovereaction \end{gathered}$$

overall equation:

b) If the reaction contains many steps the overall reaction can be obtained by summing up all the reactents and products on the either side.

$$\begin{gathered} \text{A\hspace{0.17em}}\underset{k_{-1}}{\overset{k_1}{\rightleftarrows }}B+C\left(Fastequilbrium\right) \\ \text{C\hspace{0.17em}+D}\underset{k_{-2}}{\overset{k_2}{\rightleftarrows }}E\left(Fastequilbrium\right) \\ \text{E}\stackrel{k_3}{\to }F\left(Slow\right) \end{gathered}$$

Add all the steps to obtain the reaction

$\text{A+D}+\text{C}+\text{E→B+C+F+E.}.......\left(9\right)$

Cancel common terms

$\text{A+D}+\text{E→B+F+E.}.......\left(10\right)$

Rate determining step:

Rate determining step is the slowest step of several steps in the reaction path way

$$\begin{gathered} \text{E}\stackrel{k_3}{\to }\text{F}\left(Slow\right).........\left(11\right) \\ Rate\hspace{0.17em}=k_3\left[E\right]..........\left(12\right) \end{gathered}$$

Determining equilbrium constant:

The equilibrium constant for a reaction can be obtained by dividing the product concendration by reactent concendration.One would consider The fast equilbrium step to calculate K as it is reversable reaction. E is said to be equilbrium with C,D and F so the rate constant is calculated in the following way

$K_2=\frac{\left[E\right]}{\left[C\right]\left[D\right]}.......\left(13\right)$

The rate constant K₂is given as

$$\begin{gathered} {\text{K}}_2=\frac{k_2}{k_{-2}}..............\left(14\right) \\ \text{Fromtheeqaution13and14} \\ \frac{k_2}{k_{-2}}=\frac{\left[E\right]}{\left[C\right]\left[D\right]}............\left(15\right) \\ \left[E\right]=\frac{k_2\left[C\right]\left[D\right]}{k_{-2}}.............\left(16\right) \\ \text{Substituting}\left[E\right]valuein12thequationthefollowingrateisobtained \\ Rate=\frac{k_2{k}_3\left[C\right]\left[D\right]}{k_{-2}}...........\left(17\right) \\ \text{AgainCisinequilbroumwithfirstandseacondstepofthereaction}. \\ {\text{k}}_1=\frac{\left[B\right]\left[C\right]}{\left[\text{A}\right]}.......................\left(18\right) \\ Theprincipleofdetailedbalanceisgivenas \\ \text{K}=\frac{k_1}{k_{{}_{-1}}}.............\left(20\right) \\ \frac{k_1}{k_{{}_{-1}}}=\frac{\left[B\right]\left[C\right]}{\left[A\right]} \\ \left[\text{C}\right]=\frac{\left[\text{A}\right]{\text{k}}_1}{\left[B\right]k_{-1}}...............\left(21\right) \\ \text{SubstitutingCvalueinequation17toobtaintherateoftheeqaution} \\ \text{Rate}=\frac{k_1{k}_2{k}_3\left[\text{A}\right]\left[D\right]}{k_{-2}{k}_{-1}}............\left(22\right) \\ \text{Thereforetherateofthereactionismentionedasabove.} \end{gathered}$$