Chapter 20, Problem 20.44E

Interpretation Introduction

Interpretation:

The equilibrium constant for the reaction is to be calculated.

Concept introduction:

When any reaction is at equilibrium then a constant expresses a relationship between the reactant side and the product side. This constant is known as equilibrium constant. It is denoted by $K$ and it is independent of the initial amount of the reactant and product.

Answer to Problem 20.44E

The equilibrium constant value of a reaction is $20.9$.

Explanation of Solution

The given reaction is shown below.

$\text{A}+\text{B}\rightleftharpoons \text{C}+\text{D}$

The given concentrations and rate are shown below.

$\begin{array}{ccc}\text{Rate}\left(\text{M}/\text{s}\right)& \left[\text{A}\right],\text{M}& \left[\text{B}\right],\text{M}\\ 1.081\times {10}^{-5}& 0.660& 1.23\\ 6.577\times {10}^{-5}& 4.01& 1.23\\ 6.568\times {10}^{-5}& 4.01& 2.25\\ \text{Rate}\left(\text{M}/\text{s}\right)& \left[\text{C}\right],\text{M}& \left[\text{D}\right],\text{M}\\ 7.805\times {10}^{-7}& 2.88& 0.995\\ 1.290\times {10}^{-6}& 2.88& 1.65\\ 1.300\times {10}^{-6}& 1.01& 1.65\end{array}$

The ratio of rate of the reaction of reactant is expressed as,

$\frac{K_1}{K_2}=\frac{k{\left[{\text{A}}_1\right]}^x{\left[{\text{B}}_1\right]}^y}{k{\left[{\text{A}}_2\right]}^x{\left[{\text{B}}_2\right]}^y}$ … (1)

Where,

• $K_1$ is the rate of the reaction at initial stage.

• $K_2$ is the rate of the reaction at final stage.

• $\left[{\text{A}}_1\right]$ is the concentration of reactant $\text{A}$ at initial stage.

• $\left[{\text{A}}_2\right]$ is the concentration of reactant $\text{A}$ at final stage.

• $\left[{\text{B}}_1\right]$ is the concentration of reactant $\text{B}$ at initial stage.

• $\left[{\text{B}}_2\right]$ is the concentration of reactant $\text{B}$ at final stage.

• $x$ is the order of the reaction with respect to $\text{A}$.

• $y$ is the order of the reaction with respect to $\text{B}$.

Substitute first and second columns values for A and B in equation (1).

$\begin{array}{rll}\frac{1.081\times {10}^{-5}}{6.577\times {10}^{-5}}& =& \frac{k{\left[0.660\right]}^x{\left[1.23\right]}^y}{k{\left[4.01\right]}^x{\left[1.23\right]}^y}\\ 0.16& =& {\left(0.16\right)}^x\\ x& =& \frac{\log 0.16}{\log 0.16}\\ & =& 1\end{array}$

Hence, the order of reaction with respect to A is one.

Substitute second and third columns values for A and B in equation (1).

$\begin{array}{rll}\frac{6.577\times {10}^{-5}}{6.568\times {10}^{-5}}& =& \frac{k{\left[4.01\right]}^x{\left[1.23\right]}^y}{k{\left[4.01\right]}^x{\left[2.25\right]}^y}\\ 1.00& =& {\left(0.55\right)}^y\\ y& =& \frac{\log 1.00}{\log 0.55}\\ & =& 0\end{array}$

Hence, the order of reaction with respect to B is zero.

Hence, the expression for rate law is,

$\text{rate}=k_{\text{f}}{\left[\text{A}\right]}^1{\left[\text{B}\right]}^0$

Substitute the first column values for A and B in the above expression.

$\begin{array}{rll}1.081\times {10}^{-5}\text{M}/\text{s}& =& k_{\text{f}}{\left[0.660\text{M}\right]}^1{\left[\text{1}\text{.23}\text{M}\right]}^0\\ k_{\text{f}}& =& \frac{1.081\times {10}^{-5}\text{M}/\text{s}}{0.660\text{M}}\\ k_{\text{f}}& =& 1.638\times {10}^{-5}{\text{s}}^{-1}\end{array}$

Thus, the value of $k_{\text{f}}$ is $1.638\times {10}^{-5}{\text{s}}^{-1}$.

The ratio of rate of the reaction of products is expressed as,

$\frac{K_1}{K_2}=\frac{k{\left[{\text{C}}_1\right]}^x{\left[{\text{D}}_1\right]}^y}{k{\left[{\text{C}}_2\right]}^x{\left[{\text{D}}_2\right]}^y}$ … (2)

Where,

• $K_1$ is the rate of the reaction at initial stage.

• $K_2$ is the rate of the reaction at final stage.

• $\left[{\text{C}}_1\right]$ is the concentration of reactant $\text{C}$ at initial stage.

• $\left[{\text{C}}_2\right]$ is the concentration of reactant $\text{C}$ at final stage.

• $\left[{\text{D}}_{\text{1}}\right]$ is the concentration of reactant $\text{D}$ at initial stage.

• $\left[{\text{D}}_2\right]$ is the concentration of reactant $\text{D}$ at final stage.

• $x$ is the order of the reaction with respect to $\text{C}$.

• $y$ is the order of the reaction with respect to $\text{D}$.

Substitute first and second column values for C and D in equation (2).

$\begin{array}{rll}\frac{7.805\times {10}^{-7}}{1.290\times {10}^{-6}}& =& \frac{k{\left[2.88\right]}^x{\left[0.995\right]}^y}{k{\left[2.88\right]}^x{\left[1.65\right]}^y}\\ 0.60& =& {\left(0.60\right)}^y\\ y& =& \frac{\log 0.60}{\log 0.60}\\ & =& 1\end{array}$

Hence, the order of reaction with respect to D is one.

Substitute first and second column values for C and D in equation (2).

$\begin{array}{rll}\frac{1.290\times {10}^{-6}}{1.300\times {10}^{-6}}& =& \frac{k{\left[2.88\right]}^x{\left[1.65\right]}^y}{k{\left[1.01\right]}^x{\left[1.65\right]}^y}\\ 0.99& =& {\left(2.85\right)}^x\\ x& =& \frac{\log 0.99}{\log 2.85}\\ & \approx & 0\end{array}$

Hence, the order of reaction with respect to C is zero.

Hence, the expression for rate law is,

$\text{rate}=k_{\text{r}}{\left[\text{C}\right]}^0{\left[\text{D}\right]}^1$

Substitute the first column values for C and D in the above expression.

$\begin{array}{rll}7.805\times {10}^{-7}\text{M}/\text{s}& =& k_{\text{r}}{\left[2.88\text{M}\right]}^0{\left[\text{0}\text{.995}\text{M}\right]}^1\\ k_{\text{r}}& =& \frac{7.805\times {10}^{-7}\text{M}/\text{s}}{0.995\text{M}}\\ k_{\text{r}}& =& 7.844\times {10}^{-7}{\text{s}}^{-1}\end{array}$

Thus, the value of $k_{\text{r}}$ is $7.844\times {10}^{-7}{\text{s}}^{-1}$.

The equilibrium constant of a reaction is calculated by the expression as shown below.

$K=\frac{k_{\text{f}}}{k_{\text{r}}}$

Where,

• $k_{\text{f}}$ is the rate constant for forward reaction.

• $k_{\text{r}}$ is the rate constant for backward reaction.

The forward and reverse rate constant is $1.638\times {10}^{-5}{\text{s}}^{-1}$ and $7.844\times {10}^{-7}{\text{s}}^{-1}$ respectively.

Substitute the value of $k_{\text{f}}$ and $k_{\text{r}}$ in above formula.

$\begin{array}{rll}K& =& \frac{1.638\times {10}^{-5}{\text{s}}^{-1}}{7.844\times {10}^{-7}{\text{s}}^{-1}}\\ K& =& 20.9\end{array}$

Hence, the equilibrium constant value of a reaction is $20.9$.

Conclusion

The equilibrium constant value of a reaction is $20.9$.