Chapter 20, Problem 20.45E

Interpretation Introduction

Interpretation:

The equilibrium constant for the reaction is to be calculated.

Concept introduction:

The reaction rate is defined as the speed by which the reaction is proceeding. The reaction rate depends on several factors such as the concentration of reactant and temperature. 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 $K$ it is independent of the initial amount of the reactant and product.

Answer to Problem 20.45E

The equilibrium constant value of a reaction is $1.7$.

Explanation of Solution

The given reaction is shown below.

$\text{ethyl}\text{acetate}+\text{water}\to \text{ethyl}\text{alcohol}+\text{acetic}\text{acid}$

The given concentrations and rate are shown below.

$\begin{array}{ccc}\left[\text{EtAc}\right],\text{M}& \left[{\text{H}}_2\text{O}\right],\text{M}& \text{Rate}\left(\text{M}/\text{s}\right)\\ 0.678& 0.500& 3.68\times {10}^{-3}\\ 0.987& 0.500& 5.37\times {10}^{-3}\\ 0.987& 0.309& 3.28\times {10}^{-3}\\ \left[\text{EtOH}\right],\text{M}& \left[\text{HOAc}\right],\text{M}& \text{Rate}\left(\text{M}/\text{s}\right)\\ 0.241& 0.115& 1.77\times {10}^{-4}\\ 0.241& 0.395& 6.10\times {10}^{-4}\\ 0.300& 0.115& 2.29\times {10}^{-4}\end{array}$

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

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

Where,

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

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

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

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

• ${\left[{\text{H}}_2\text{O}\right]}_1$ is the concentration of reactant ${\text{H}}_2\text{O}$ at initial stage.

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

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

• $y$ is the order of the reaction with respect to ${\text{H}}_2\text{O}$.

Substitute first and second columns values for $\text{EtAc}$ and ${\text{H}}_2\text{O}$ in equation (1).

$\begin{array}{rll}\frac{3.68\times {10}^{-3}}{5.37\times {10}^{-3}}& =& \frac{k{\left[0.678\right]}^x{\left[0.500\right]}^y}{k{\left[0.987\right]}^x{\left[0.500\right]}^y}\\ 0.68& =& {\left(0.69\right)}^x\\ x& =& \frac{\log 0.68}{\log 0.69}\\ & \approx & 1\end{array}$

Hence, the order of reaction with respect to $\text{EtAc}$ is one.

Substitute second and third columns values for $\text{EtAc}$ and ${\text{H}}_2\text{O}$ in equation (1).

$\begin{array}{rll}\frac{5.37\times {10}^{-3}}{3.28\times {10}^{-3}}& =& \frac{k{\left[0.987\right]}^x{\left[0.500\right]}^y}{k{\left[0.987\right]}^x{\left[0.309\right]}^y}\\ 1.64& =& {\left(1.62\right)}^y\\ x& =& \frac{\log 1.64}{\log 1.62}\\ & \approx & 1\end{array}$

Hence, the order of reaction with respect to ${\text{H}}_2\text{O}$ is one.

Hence, the expression for rate law is,

$\text{rate}=k_{\text{f}}{\left[\text{EtAc}\right]}^1{\left[{\text{H}}_{\text{2}}\text{O}\right]}^1$

Substitute the first column values for $\text{EtAc}$ and ${\text{H}}_2\text{O}$ in the above expression.

$\begin{array}{rll}3.68\times {10}^{-3}\text{M}/\text{s}& =& k_{\text{f}}{\left[0.678\text{M}\right]}^1{\left[\text{0}\text{.500}\text{M}\right]}^1\\ 3.68\times {10}^{-3}\text{M}/\text{s}& =& k_{\text{f}}\times 0.339{\text{M}}^2\\ k_{\text{f}}& =& \frac{3.68\times {10}^{-3}\text{M}/\text{s}}{0.339{\text{M}}^2}\\ k_{\text{f}}& =& 10.85\times {10}^{-3}{\text{M}}^{-1}{\text{s}}^{-1}\end{array}$

Thus, the value of $k_{\text{f}}$ is $10.85\times {10}^{-3}{\text{M}}^{-1}{\text{s}}^{-1}$.

The ratio of rate of the reaction at initial concentration and final concentration of reactant is expressed as,

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

Where,

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

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

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

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

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

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

Substitute first and second columns values for $\text{EtOH}$ and $\text{HOAc}$ in equation (2).

$\begin{array}{rll}\frac{1.77\times {10}^{-4}}{6.10\times {10}^{-4}}& =& \frac{k{\left[0.241\right]}^x{\left[0.115\right]}^y}{k{\left[0.241\right]}^x{\left[0.395\right]}^y}\\ 0.29& =& {\left(0.29\right)}^y\\ y& =& \frac{\log 0.29}{\log 0.29}\\ & =& 1\end{array}$

Hence, the order of the reaction with respect to $\text{HOAc}$ is one.

Substitute first and third column values for $\text{EtOH}$ and $\text{HOAc}$ in equation (2).

$\begin{array}{rll}\frac{1.77\times {10}^{-4}}{2.29\times {10}^{-4}}& =& \frac{k{\left[0.241\right]}^x{\left[0.115\right]}^y}{k{\left[0.300\right]}^x{\left[0.115\right]}^y}\\ 0.77& =& {\left(0.80\right)}^x\\ x& =& \frac{\log 0.77}{\log 0.80}\\ & \approx & 1\end{array}$

Hence, the order of the reaction with respect to $\text{EtOH}$ is one.

Hence, the expression for rate law is,

$\text{rate}=k_{\text{r}}{\left[\text{EtOH}\right]}^1{\left[\text{HOAc}\right]}^1$

Substitute the first column values for $\text{EtOH}$ and $\text{HOAc}$ in the above expression.

$\begin{array}{rll}1.77\times {10}^{-4}\text{M}/\text{s}& =& k_{\text{r}}{\left[0.241\right]}^1{\left[\text{0}\text{.115}\text{M}\right]}^1\\ 1.77\times {10}^{-4}\text{M}/\text{s}& =& k_{\text{r}}\times 0.028{\text{M}}^2\\ k_{\text{r}}& =& \frac{1.77\times {10}^{-4}\text{M}/\text{s}}{0.028{\text{M}}^2}\\ k_{\text{r}}& =& 63.21\times {10}^{-4}{\text{M}}^{-1}{\text{s}}^{-1}\end{array}$

Thus, the value of $k_{\text{r}}$ is $63.21\times {10}^{-4}{\text{M}}^{-1}{\text{s}}^{-1}$.

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 $10.85\times {10}^{-3}{\text{M}}^{-1}{\text{s}}^{-1}$ and $63.21\times {10}^{-4}{\text{M}}^{-1}{\text{s}}^{-1}$ respectively.

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

$\begin{array}{rll}K& =& \frac{10.85\times {10}^{-3}{\text{M}}^{-1}{\text{s}}^{-1}}{63.21\times {10}^{-4}{\text{M}}^{-1}{\text{s}}^{-1}}\\ K& =& 1.7\end{array}$

Hence, the equilibrium constant value of a reaction is $1.7$.

Conclusion

The equilibrium constant value of a reaction is $1.7$.