Chapter 12, Problem 42E

Ethyl acetate is synthesized in a nonreacting solvent (not water) according to the following reaction:

$\underset{\text{Acetic acid}}{{\text{CH}}_3{\text{CO}}_2\text{H}}\text{+ }\underset{\text{Ethanol}}{{\text{C}}_2{\text{H}}_5\text{OH}}\rightleftharpoons \underset{\text{Ethyl acetate}}{{\text{CH}}_3{\text{CO}}_2{\text{C}}_2{\text{H}}_5}+{\text{H}}_2\text{O}K=2.2$

For the following mixtures (a-d), will the concentration of H₂O increase, decrease, or remain the same as equilibrium is established?

a. [CH₃CO₂C₂H₅] = 0.22M, [H₂O] = 0.10M, [CH₃CO₂H] = 0.010 M, [C₂H₅OH] = 0.010 M

b. [CH₃CO₂C₂H₅] = 0.22 M, [H₂O] = 0.0020 M, [CH₃CO₂H] = 0.0020 M, [C₂H₅OH] = 0.10 M

c. [CH₃CO₂C₂H₅] = 0.88 M, [H₂O] = 0.12 M, [CH₃CO₂H) = 0.044 M, [C₂H₅OH] = 6.0 M

d. [CH₃CO₂C₂H₅] = 4.4 M, [H₂O] = 4.4 M, [CH₃CO₂H] = 0.88 M, [C₂H₅OH] = 10.0 M

e. What must the concentration of water be for a mixture with [CH₃CO₂C₂H₅] = 2.0M, [CH₃CO₂H] = 0.l0 M, and [C₂H₅OH] = 5.0 M to be at equilibrium?

f. Why is water included in the equilibrium expression for this reaction?

Interpretation Introduction

Interpretation: The reaction between Acetic acid and Ethanol is given. The answers are to be given for each option.

Concept introduction: Reaction quotient, $Q$, is the measure of concentration or partial pressure of reactants and products of a system before equilibrium point is reached.

The expression of $Q$ is,

$Q=\frac{\text{Concentration}\text{of}\text{product}}{\text{Concentration}\text{of}\text{reactant}}$

The direction of reaction depends on $Q$ and $K$,

• If $Q>K$, the reaction shift to the left.
• If $Q<K$, the reaction shift to the right.
• If $Q=K$, the reaction will reach the equilibrium.

Answer to Problem 42E

Answer

Solutions are as follows.

Explanation of Solution

Explanation

(a)

To determine: If the concentration of water $\left({\text{H}}_{\text{2}}\text{O}\right)$ increase, decrease, or remain the same for the given values.

The concentration of water $\left({\text{H}}_{\text{2}}\text{O}\right)$ decreases as the value of $Q$ is greater than $K$.

Given

Concentration of $\left[{\text{CH}}_{\text{3}}{\text{CO}}_{\text{2}}{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\right]$ is $\text{0}\text{.22}\text{M}$.

Concentration of $\left[{\text{H}}_{\text{2}}\text{O}\right]$ is $\text{0}\text{.10}\text{M}$.

Concentration of $\left[{\text{CH}}_{\text{3}}{\text{CO}}_{\text{2}}\text{H}\right]$ is $\text{0}\text{.010}\text{M}$.

Concentration of $\left[{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\right]$ is $\text{0}\text{.010}\text{M}$.

The value of $K$ is $\text{2}\text{.2}$.

The reaction that takes place is,

${\text{CH}}_{\text{3}}{\text{CO}}_{\text{2}}\text{H}+{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\stackrel{}{\rightleftharpoons }{\text{CH}}_{\text{3}}{\text{CO}}_{\text{2}}{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}+{\text{H}}_{\text{2}}\text{O}$

The reaction coefficient for the above reaction is,

$Q=\frac{{\left[{\text{H}}_{\text{2}}\text{O}\right]}_{\text{0}}{\left[{\text{CH}}_{\text{3}}{\text{CO}}_{\text{2}}{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\right]}_{\text{0}}}{{\left[{\text{CH}}_{\text{3}}{\text{CO}}_{\text{2}}\text{H}\right]}_{\text{0}}{\left[{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\right]}_{\text{0}}}$

The number zero shows the initial concentration.

Substitute the values of $\left[{\text{CH}}_{\text{3}}{\text{CO}}_{\text{2}}{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\right],\left[{\text{H}}_{\text{2}}\text{O}\right],\left[{\text{CH}}_{\text{3}}{\text{CO}}_{\text{2}}\text{H}\right]$ and $\left[{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\right]$ in the above equation.

$\begin{array}{c}Q=\frac{{\left[{\text{H}}_{\text{2}}\text{O}\right]}_{\text{0}}{\left[{\text{CH}}_{\text{3}}{\text{CO}}_{\text{2}}{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\right]}_{\text{0}}}{{\left[{\text{CH}}_{\text{3}}{\text{CO}}_{\text{2}}\text{H}\right]}_{\text{0}}{\left[{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\right]}_{\text{0}}}\\ =\frac{\left(\text{0}\text{.22}\right)\left(\text{0}\text{.10}\right)}{\left(\text{0}\text{.010}\right)\left(\text{0}\text{.010}\right)}\\ =\text{220}\end{array}$

Since, the value of $Q_{\text{p}}$ is greater than $K$. Therefore, the concentration of water $\left({\text{H}}_{\text{2}}\text{O}\right)$ decreases as the reaction shift to the left.

(b)

To determine: If the concentration of water $\left({\text{H}}_{\text{2}}\text{O}\right)$ increase, decrease, or remain the same for the given values.

The concentration of water $\left({\text{H}}_{\text{2}}\text{O}\right)$ remains same as the value of $Q$ is equal to $K$.

Given

Concentration of $\left[{\text{CH}}_{\text{3}}{\text{CO}}_{\text{2}}{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\right]$ is $\text{0}\text{.22}\text{M}$.

Concentration of $\left[{\text{H}}_{\text{2}}\text{O}\right]$ is $\text{0}\text{.0020}\text{M}$.

Concentration of $\left[{\text{CH}}_{\text{3}}{\text{CO}}_{\text{2}}\text{H}\right]$ is $\text{0}\text{.0020}\text{M}$.

Concentration of $\left[{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\right]$ is $\text{0}\text{.10}\text{M}$.

The value of $K$ is $\text{2}\text{.2}$.

The reaction that takes place is,

${\text{CH}}_{\text{3}}{\text{CO}}_{\text{2}}\text{H}+{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\stackrel{}{\rightleftharpoons }{\text{CH}}_{\text{3}}{\text{CO}}_{\text{2}}{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}+{\text{H}}_{\text{2}}\text{O}$

The reaction coefficient for the above reaction is,

$Q=\frac{{\left[{\text{H}}_{\text{2}}\text{O}\right]}_{\text{0}}{\left[{\text{CH}}_{\text{3}}{\text{CO}}_{\text{2}}{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\right]}_{\text{0}}}{{\left[{\text{CH}}_{\text{3}}{\text{CO}}_{\text{2}}\text{H}\right]}_{\text{0}}{\left[{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\right]}_{\text{0}}}$

The number zero shows the initial concentration.

Substitute the values of $\left[{\text{CH}}_{\text{3}}{\text{CO}}_{\text{2}}{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\right],\left[{\text{H}}_{\text{2}}\text{O}\right],\left[{\text{CH}}_{\text{3}}{\text{CO}}_{\text{2}}\text{H}\right]$ and $\left[{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\right]$ in the above equation.

$\begin{array}{c}Q=\frac{{\left[{\text{H}}_{\text{2}}\text{O}\right]}_{\text{0}}{\left[{\text{CH}}_{\text{3}}{\text{CO}}_{\text{2}}{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\right]}_{\text{0}}}{{\left[{\text{CH}}_{\text{3}}{\text{CO}}_{\text{2}}\text{H}\right]}_{\text{0}}{\left[{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\right]}_{\text{0}}}\\ =\frac{\left(\text{0}\text{.22}\right)\left(\text{0}\text{.0020}\right)}{\left(\text{0}\text{.0020}\right)\left(\text{0}\text{.10}\right)}\\ =\text{2}\text{.2}\end{array}$

Since, the value of $Q$ is equal to $K$. Therefore, the concentration of water $\left({\text{H}}_{\text{2}}\text{O}\right)$ remains same as the reaction reaches the equilibrium.

(c)

To determine: If the concentration of water $\left({\text{H}}_{\text{2}}\text{O}\right)$ increase, decrease, or remain the same for the given values.

The concentration of water $\left({\text{H}}_{\text{2}}\text{O}\right)$ decreases as the value of $Q$ is less than $K$.

Given

Concentration of $\left[{\text{CH}}_{\text{3}}{\text{CO}}_{\text{2}}{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\right]$ is $\text{0}\text{.88}\text{M}$.

Concentration of $\left[{\text{H}}_{\text{2}}\text{O}\right]$ is $\text{0}\text{.12}\text{M}$.

Concentration of $\left[{\text{CH}}_{\text{3}}{\text{CO}}_{\text{2}}\text{H}\right]$ is $\text{0}\text{.044}\text{M}$.

Concentration of $\left[{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\right]$ is $\text{6}\text{.0}\text{M}$.

The value of $K$ is $\text{2}\text{.2}$.

The reaction that takes place is,

${\text{CH}}_{\text{3}}{\text{CO}}_{\text{2}}\text{H}+{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\stackrel{}{\rightleftharpoons }{\text{CH}}_{\text{3}}{\text{CO}}_{\text{2}}{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}+{\text{H}}_{\text{2}}\text{O}$

The reaction coefficient for the above reaction is,

$Q=\frac{{\left[{\text{H}}_{\text{2}}\text{O}\right]}_{\text{0}}{\left[{\text{CH}}_{\text{3}}{\text{CO}}_{\text{2}}{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\right]}_{\text{0}}}{{\left[{\text{CH}}_{\text{3}}{\text{CO}}_{\text{2}}\text{H}\right]}_{\text{0}}{\left[{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\right]}_{\text{0}}}$

The number zero shows the initial concentration.

Substitute the values of $\left[{\text{CH}}_{\text{3}}{\text{CO}}_{\text{2}}{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\right],\left[{\text{H}}_{\text{2}}\text{O}\right],\left[{\text{CH}}_{\text{3}}{\text{CO}}_{\text{2}}\text{H}\right]$ and $\left[{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\right]$ in the above equation.

$\begin{array}{c}Q=\frac{{\left[{\text{H}}_{\text{2}}\text{O}\right]}_{\text{0}}{\left[{\text{CH}}_{\text{3}}{\text{CO}}_{\text{2}}{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\right]}_{\text{0}}}{{\left[{\text{CH}}_{\text{3}}{\text{CO}}_{\text{2}}\text{H}\right]}_{\text{0}}{\left[{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\right]}_{\text{0}}}\\ =\frac{\left(\text{0}\text{.88}\right)\left(\text{0}\text{.12}\right)}{\left(\text{0}\text{.044}\right)\left(\text{6}\text{.0}\right)}\\ =\text{0}\text{.4}\end{array}$

Since, the value of $Q$ is less than $K$. Therefore, the concentration of water $\left({\text{H}}_{\text{2}}\text{O}\right)$ remains increases as the reaction move to right.

(d)

To determine: If the concentration of water $\left({\text{H}}_{\text{2}}\text{O}\right)$ increase, decrease, or remain the same for the given values.

The concentration of water $\left({\text{H}}_{\text{2}}\text{O}\right)$ remains same as the value of $Q$ is equal to $K$.

Given

Concentration of $\left[{\text{CH}}_{\text{3}}{\text{CO}}_{\text{2}}{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\right]$ is $\text{4}\text{.4}\text{M}$.

Concentration of $\left[{\text{H}}_{\text{2}}\text{O}\right]$ is $\text{4}\text{.4}\text{M}$.

Concentration of $\left[{\text{CH}}_{\text{3}}{\text{CO}}_{\text{2}}\text{H}\right]$ is $\text{0}\text{.88}\text{M}$.

Concentration of $\left[{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\right]$ is $\text{10}\text{.0}\text{M}$.

The value of $K$ is $\text{2}\text{.2}$.

The reaction that takes place is,

${\text{CH}}_{\text{3}}{\text{CO}}_{\text{2}}\text{H}+{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\stackrel{}{\rightleftharpoons }{\text{CH}}_{\text{3}}{\text{CO}}_{\text{2}}{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}+{\text{H}}_{\text{2}}\text{O}$

The reaction coefficient for the above reaction is,

$Q=\frac{{\left[{\text{H}}_{\text{2}}\text{O}\right]}_{\text{0}}{\left[{\text{CH}}_{\text{3}}{\text{CO}}_{\text{2}}{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\right]}_{\text{0}}}{{\left[{\text{CH}}_{\text{3}}{\text{CO}}_{\text{2}}\text{H}\right]}_{\text{0}}{\left[{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\right]}_{\text{0}}}$

The number zero shows the initial concentration.

Substitute the values of $\left[{\text{CH}}_{\text{3}}{\text{CO}}_{\text{2}}{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\right],\left[{\text{H}}_{\text{2}}\text{O}\right],\left[{\text{CH}}_{\text{3}}{\text{CO}}_{\text{2}}\text{H}\right]$ and $\left[{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\right]$ in the above equation.

$\begin{array}{c}Q=\frac{{\left[{\text{H}}_{\text{2}}\text{O}\right]}_{\text{0}}{\left[{\text{CH}}_{\text{3}}{\text{CO}}_{\text{2}}{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\right]}_{\text{0}}}{{\left[{\text{CH}}_{\text{3}}{\text{CO}}_{\text{2}}\text{H}\right]}_{\text{0}}{\left[{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\right]}_{\text{0}}}\\ =\frac{\left(\text{4}\text{.4}\right)\left(\text{4}\text{.4}\right)}{\left(\text{0}\text{.88}\right)\left(\text{10}\text{.0}\right)}\\ =\text{2}\text{.2}\end{array}$

Since, the value of $Q$ is equal to $K$. Therefore, the concentration of water $\left({\text{H}}_{\text{2}}\text{O}\right)$ remains same as the reaction reaches the equilibrium.

(e)

To determine: The concentration of water $\left({\text{H}}_{\text{2}}\text{O}\right)$ from the given values.

The concentration of water $\left({\text{H}}_{\text{2}}\text{O}\right)$ is $\underset{\_}{0.55M}$.

Given

Concentration of $\left[{\text{CH}}_{\text{3}}{\text{CO}}_{\text{2}}{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\right]$ is $\text{2}\text{.0}\text{M}$.

Concentration of $\left[{\text{CH}}_{\text{3}}{\text{CO}}_{\text{2}}\text{H}\right]$ is $\text{0}\text{.10}\text{M}$.

Concentration of $\left[{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\right]$ is $\text{5}\text{.0}\text{M}$.

The value of equilibrium constant $\left(K\right)$ is $\text{2}\text{.2}$.

The reaction that takes place is,

${\text{CH}}_{\text{3}}{\text{CO}}_{\text{2}}\text{H}+{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\stackrel{}{\rightleftharpoons }{\text{CH}}_{\text{3}}{\text{CO}}_{\text{2}}{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}+{\text{H}}_{\text{2}}\text{O}$

Formula

The expression for equilibrium constant is,

$K=\frac{\text{Concentration}\text{of}\text{product}}{\text{Concentration}\text{of}\text{reactant}}$

The equilibrium constant for the above reaction is,

$K=\frac{\left[{\text{H}}_{\text{2}}\text{O}\right]\left[{\text{CH}}_{\text{3}}{\text{CO}}_{\text{2}}{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\right]}{\left[{\text{CH}}_{\text{3}}{\text{CO}}_{\text{2}}\text{H}\right]\left[{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\right]}$

Substitute the values of $\left[{\text{CH}}_{\text{3}}{\text{CO}}_{\text{2}}{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\right],K,\left[{\text{CH}}_{\text{3}}{\text{CO}}_{\text{2}}\text{H}\right]$ and $\left[{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\right]$ in the above equation.

$\begin{array}{c}K=\frac{\left[{\text{H}}_{\text{2}}\text{O}\right]\left[{\text{CH}}_{\text{3}}{\text{CO}}_{\text{2}}{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\right]}{\left[{\text{CH}}_{\text{3}}{\text{CO}}_{\text{2}}\text{H}\right]\left[{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\right]}\\ \text{2}\text{.2}=\frac{\left[{\text{H}}_{\text{2}}\text{O}\right]\left(\text{2}\text{.0}\right)}{\left(\text{5}\text{.0}\right)\left(\text{0}\text{.10}\right)}\\ \left[{\text{H}}_{\text{2}}\text{O}\right]=\underset{\_}{0.55M}\end{array}$

(f)

To determine: The reason for the inclusion of water $\left({\text{H}}_{\text{2}}\text{O}\right)$ in the equilibrium expression for the given reaction.

Water is included in the equilibrium expression because it does not acts as solvent in the given reaction.

The molecule that acts as solvent are not included in equilibrium expression. In this case, water does not acts as solvent. In the given reaction ethyl acetate acts as solvent. Hence, water is included in the equilibrium expression for the given reaction.

Conclusion

Conclusion

The initial value of calcium oxide depends on the value of $Q$ and $K$