Question

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

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

Short answer

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

Step-by-step solution

Over all reaction:

If the reaction contains several steps overall reaction can be obtained by adding reactants and products on either side and canceling the common terms.

$\text{2A+B+D+F→D+E+F+G.}.....\left(1\right)$

Canceling common terms on both sides

localid="1661172446369" $\text{2A+B→E+G.}.....\left(2\right)$

Rate determining step;

If the expression contains several steps slowest step is the rate determining step

$$\begin{gathered} D+B\stackrel{k_2}{\to }\hspace{0.17em}\stackrel{}{E+F\left(Slow\right)} \\ Rate=k_2\left[D\right]\left[B\right]...........\left(3\right) \end{gathered}$$

Determination of rate law:

D is found to be equilbrium with E,A,F,B.the equilbrium constant for first reaction is

$$\begin{gathered} {\text{K}}_1\text{=}\frac{\left[D\right]}{\left[B\right]{\left[A\right]}^{\text{2}}}\text{.........}\left(4\right) \\ {\text{K}}_1=\frac{k_1}{k_{-1}}...........\left(5\right) \end{gathered}$$

From eqaution 4 and 5 following equation is derived

localid="1661173184761" $$\begin{gathered} \frac{k_1}{k_{-1}}\text{=}\frac{\left[D\right]}{\left[B\right]{\left[A\right]}^{\text{2}}}\text{.........}\left(\text{6}\right) \\ or\left[D\right]\hspace{0.17em}=\frac{k_1\left[B\right]{\left[A\right]}^{\text{2}}}{k_{-1}} \end{gathered}$$

Substituting D value in equation 3

$Rate=\frac{k_2{k}_1{\left[B\right]}^2{\left[A\right]}^2}{k_{-1}}........\left(7\right)$

Therefore rate law corresponds to above mechanism is$Rate=\frac{k_2{k}_1{\left[B\right]}^2{\left[A\right]}^2}{k_{-1}}$

Overall reaction:

b)If the reaction contains several steps overall reaction can be obtained by adding reactants and products on either side and canceling the common terms.

$\text{A+B+D+C}+\text{F→C+F+G.}.....\left(1\right)$

Canceling common terms on both sides

$\text{A+B+D→G.}.....\left(2\right)$

Rate determining step:

If the expression contains several steps s lowest step is the rate determining step

$$\begin{gathered} F\stackrel{k_3}{\to G} \\ Rate=k_3\left[F\right]........\left(3\right) \end{gathered}$$

Determination  of Rate law:

But F is in equilbrium with C,D and G then equilbrium constant K₂

is defined as

$K_2=\frac{\left[F\right]}{\left[C\right]\left[D\right]}.........\left(4\right)$

According to principle of detailed balance

$k_2=\frac{k_2}{k_{-2}}......\left(5\right)$

From eqaution 4 and 5

$$\begin{gathered} \frac{k_2}{k_{-2}}=\frac{\left[F\right]}{\left[C\right]\left[D\right]}.........\left(6\right) \\ \left[F\hspace{0.17em}\right]=\frac{\left[C\right]\left[D\right]k_2}{k_{-2}}........\left(7\right) \end{gathered}$$

Substituing the value in equation-3

$$\begin{gathered} Rate=k_3\left[F\right]..............\left(3\right) \\ Rat=\frac{k_3{k}_2\left[C\right]\left[D\right]}{k_{-2}}..........\left(8\right) \end{gathered}$$

Again C is in equilbrium with A and B in the first step,if the equilbrium constant of first step is K₁ then

$$\begin{gathered} K_1=\frac{\left[C\right]}{\left[A\right]\left[B\right]}...........\left(9\right) \\ Accordingtoprincipleofdetailedbalance \\ {K}_1=\frac{k_1}{k_{-1}}..............\left(10\right) \\ \frac{k_1}{k_{-1}}\hspace{0.17em}=\frac{\left[C\right]}{\left[A\right]\left[B\right]}.............\left(11\right) \\ or \\ \left[C\right]=\frac{k_1\left[A\right]\left[B\right]}{k_{-1}} \\ Fromequation5and6followingequationisderived \\ SubstitutingthevalueofCinequation-8 \\ Rate=\frac{k_1{k}_2{k}_3\left[D\right]\left[A\right]\left[B\right]}{k_{-1}{k}_{-2}}...............\left(13\right) \end{gathered}$$