Chapter 6, Problem 6B.7P

Interpretation Introduction

Interpretation:

The equilibrium constant for the dimerization of the acid in the vapour and the standard enthalpy of the dimerization reaction has to be calculated.

Concept introduction:

If the equilibrium constant of a reaction is measured in terms of the activities of the products and the reactants, it is called thermodynamic equilibrium constant and it is dimensionless as activities are also dimensionless.  Equilibrium constant for a reaction is unique at a certain temperature and does not change throughout the reaction at a given temperature.

Answer to Problem 6B.7P

The equilibrium constant for the dimerization of the acid in the vapour, at $437\text{K}$ is $\underset{\_}{0.450}$.

The equilibrium constant for the dimerization of the acid in the vapour, at $471\text{K}$ is $\underset{\_}{0.142}$.

The enthalpy of dimerization of acetic acid is $\underset{\_}{-9.72\times 10^{-4}Jmol}$.

Explanation of Solution

In the first experiment, the mass of the acetic acid present in the container was $0.0519\text{g}$.

The molar mass of the acetic acid is $60.052\text{g}{\text{mol}}^{\text{-1}}$.

The number of moles of the acetic acid is calculated by the following formula.

    $\text{Moles}\text{of}\text{acetic}\text{acid}=\frac{\text{Given}\text{quantity}\text{of}\text{acetic}\text{acid}}{\text{Molar}\text{mass}}$                                 (1)

Substitute the value of the quantity of acetic acid and molar mass of acetic acid in equation (1).

    $\begin{array}{c}\text{Moles}\text{of}\text{acetic}\text{acid}=\frac{0.0519\text{g}}{60.052\text{g}{\text{mol}}^{\text{-1}}}\\ =8.64\times {10}^{-4}\text{mol}\end{array}$

Therefore, the total number of moles of the monomer of acetic acid and the dimer of acetic acid is $8.64\times {10}^{-4}\text{mol}$.  Mathematically the relation between the moles of monomer and the moles of dimer is shown below.

    $n_{\text{monomer}}+n_{\text{dimer}}=8.64\times {10}^{-4}\text{mol}$                                                                    (2)

The number of moles of monomer of the acetic acid in the container is calculated by the following formula.

    $n=\frac{pV}{RT}$                                                                                                          (3)

It is given that the value of temperature in the first experiment was $437\text{K}$.

It is given that the volume of the container in the first experiment was $21.45{\text{cm}}^{\text{3}}$.

The conversion of ${\text{cm}}^{\text{3}}$ to $\text{L}$ is done as,

    $1{\text{cm}}^{\text{3}}=1\times {10}^{-3}\text{L}$

Therefore, the conversion of $\text{21}\text{.45}{\text{cm}}^{\text{3}}$ to $\text{L}$ is done as,

    $\begin{array}{c}21.45{\text{cm}}^{\text{3}}=21.45\times {10}^{-3}\text{L}\\ =\text{0}\text{.0214}\text{L}\end{array}$

Therefore, the volume of the container in the first experiment was $\text{0}\text{.0214}\text{L}$.

It is given that the value of external pressure in the first experiment was $101.9\text{kPa}$.

The conversion of $\text{kPa}$ to $\text{atm}$ is done as,

    $1\text{kPa}=0.00986\text{atm}$

Therefore, the conversion of $101.9\text{kPa}$ to $\text{atm}$ is done as,

    $\begin{array}{c}101.9\text{kPa}=101.9\times 0.00986\text{atm}\\ =1.004\text{atm}\end{array}$

Therefore, the external pressure was $1.004\text{atm}$.

The value of the gas constant is $0.08206\text{L}\text{atm}{\text{K}}^{\text{-1}}{\text{mol}}^{-1}$.

Substitute the value of pressure, temperature, volume and gas constant in equation (3), from the data of first experiment.

    $\begin{array}{c}n=\frac{1.004\text{atm}\times \text{0}\text{.0214}\text{L}}{0.08206\text{L}\text{atm}{\text{K}}^{\text{-1}}{\text{mol}}^{-1}\times 437\text{K}}\\ =\frac{0.0214}{35.86{\text{mol}}^{-1}}\\ =5.96\times {10}^{-4}\text{mol}\end{array}$

Therefore, the moles of monomer of acetic acid is $5.96\times {10}^{-4}\text{mol}$.

Substitute the moles of monomer in equation (2).

    $\begin{array}{c}5.96\times {10}^{-4}\text{mol}+n_{\text{dimer}}=8.64\times {10}^{-4}\text{mol}\\ n_{\text{dimer}}=8.64\times {10}^{-4}\text{mol}-5.96\times {10}^{-4}\text{mol}\\ =\text{2}\text{.68}\times {10}^{-4}\text{mol}\end{array}$

Therefore, the moles of the dimer of acetic acid is $2.68\times {10}^{-4}\text{mol}$.

The dimerization reaction of acetic acid is shown below.

    $\begin{array}{l}2\left({\text{CH}}_{\text{3}}\text{COOH}\right)\rightleftharpoons {\left({\text{CH}}_{\text{3}}\text{COOH}\right)}_2\\ \text{Monomer}\text{of }\text{Dimer}\text{of }\\ \text{acetic}\text{acid}\text{acetic acid }\end{array}$

The equilibrium constant for the dimerization reaction of acetic acid is calculated by the following formula.

    $K=\frac{p_{\dim er}}{{\left(p_{\text{monomer}}\right)}^2}$                                                                                             (4)

Where,

• $p_{\dim er}$ is the partial pressure of the dimer of acetic acid.
• $p_{\text{monomer}}$ is the partial pressure of the monomer of acetic acid.

The partial pressure of monomer is calculated by the following formula.

    $\begin{array}{c}{p}_{\text{monomer}}=\frac{n_{\text{monomer}}RT}{V}\\ =\frac{5.96\times {10}^{-4}\text{mol}\times 0.08206\text{L}\text{atm}{\text{K}}^{\text{-1}}{\text{mol}}^{-1}\times 437\text{K}}{\text{0}\text{.0214}\text{L}}\\ =\frac{213.72\times {10}^{-4}\text{L}\text{atm}}{\text{0}\text{.0214}}\\ =0.998\text{atm}\end{array}$

Therefore, the partial pressure of monomer is $0.998\text{atm}$.

The partial pressure of dimer is calculated by the following formula.

    $\begin{array}{c}{p}_{\text{dimer}}=\frac{n_{\text{dimer}}RT}{V}\\ =\frac{\text{2}\text{.68}\times {10}^{-4}\text{mol}\times 0.08206\text{L}\text{atm}{\text{K}}^{\text{-1}}{\text{mol}}^{-1}\times 437\text{K}}{\text{0}\text{.0214}\text{L}}\\ =\frac{96.10\times {10}^{-4}\text{L}\text{atm}}{\text{0}\text{.0214}}\\ =0.449\text{atm}\end{array}$

Therefore, the partial pressure of dimer of the acetic acid is $0.449\text{atm}$.

The equilibrium constant at temperature $437\text{K}$ is $K_1$.  Substitute the partial pressure of monomer and dimer of acetic acid in equation (4).

    $\begin{array}{c}{K}_1=\frac{0.449}{{\left(0.998\right)}^2}\\ =\frac{0.449}{0.996}\\ =\underset{\_}{0.450}\end{array}$

Therefore, the equilibrium constant of the first experiment is $\underset{\_}{0.450}$.

In the second experiment, the mass of the acetic acid present in the container was $0.0380\text{g}$.

The molar mass of the acetic acid is $60.052\text{g}{\text{mol}}^{\text{-1}}$.

Substitute the value of the quantity of acetic acid and molar mass of acetic acid in equation (1).

    $\begin{array}{c}\text{Moles}\text{of}\text{acetic}\text{acid}=\frac{0.0380\text{g}}{60.052\text{g}{\text{mol}}^{\text{-1}}}\\ =6.32\times {10}^{-4}\text{mol}\end{array}$

Therefore, the total number of moles of the monomer and dimer of acetic acid is $6.32\times {10}^{-4}\text{mol}$.  Mathematically the relation between the moles of monomer and the moles of the dimer of acetic acid is shown below.

    $n_{\text{monomer}}+n_{\text{dimer}}=6.32\times {10}^{-4}\text{mol}$                                                                    (5)

It is given that the value of temperature in the second experiment was $471\text{K}$.

The volume of the container was $\text{0}\text{.0214}\text{L}$.

The external pressure was $1.004\text{atm}$.

The value of the gas constant is $0.08206\text{L}\text{atm}{\text{K}}^{\text{-1}}{\text{mol}}^{-1}$.

Substitute the value of pressure, temperature, volume and gas constant in equation (3).

    $\begin{array}{c}n=\frac{1.004\text{atm}\times \text{0}\text{.0214}\text{L}}{0.08206\text{L}\text{atm}{\text{K}}^{\text{-1}}{\text{mol}}^{-1}\times 471\text{K}}\\ =\frac{0.0214}{38.65{\text{mol}}^{-1}}\\ =5.53\times {10}^{-4}\text{mol}\end{array}$

Therefore, the moles of monomer is $5.53\times {10}^{-4}\text{mol}$.

Substitute the moles of monomer in equation (5).

    $\begin{array}{c}5.53\times {10}^{-4}\text{mol}+n_{\text{dimer}}=6.32\times {10}^{-4}\text{mol}\\ n_{\text{dimer}}=6.32\times {10}^{-4}\text{mol}-5.53\times {10}^{-4}\text{mol}\\ =0.79\times {10}^{-4}\text{mol}\end{array}$

Therefore, the moles of the dimer of acetic acid is $0.79\times {10}^{-4}\text{mol}$.

The partial pressure of monomer is calculated by the following formula.

    $\begin{array}{c}{p}_{\text{monomer}}=\frac{n_{\text{monomer}}RT}{V}\\ =\frac{5.53\times {10}^{-4}\text{mol}\times 0.08206\text{L}\text{atm}{\text{K}}^{\text{-1}}{\text{mol}}^{-1}\times 471\text{K}}{\text{0}\text{.0214}\text{L}}\\ =\frac{213.73\times {10}^{-4}\text{L}\text{atm}}{\text{0}\text{.0214}}\\ =0.998\text{atm}\end{array}$

Therefore, the partial pressure of monomer is $0.998\text{atm}$.

The partial pressure of dimer is calculated by the following formula.

    $\begin{array}{c}{p}_{\text{dimer}}=\frac{n_{\text{dimer}}RT}{V}\\ =\frac{0.79\times {10}^{-4}\text{mol}\times 0.08206\text{L}\text{atm}{\text{K}}^{\text{-1}}{\text{mol}}^{-1}\times 471\text{K}}{\text{0}\text{.0214}\text{L}}\\ =\frac{30.53\times {10}^{-4}\text{L}\text{atm}}{\text{0}\text{.0214}}\\ =0.142\text{atm}\end{array}$

Therefore, the partial pressure of dimer of acetic acid is $0.142\text{atm}$.

The equilibrium constant at temperature $471\text{K}$ is $K_1$.  Substitute the partial pressure of monomer and dimer of acetic acid in equation (4).

    $\begin{array}{c}{K}_2=\frac{0.142}{{\left(0.998\right)}^2}\\ =\frac{0.142}{0.996}\\ =\underset{\_}{0.142}\end{array}$

Therefore, the equilibrium constant of the second experiment is $\underset{\_}{0.142}$.

The standard enthalpy of the dimerization reaction is calculated by the following formula.

    $\ln \left(\frac{K_2}{K_1}\right)=\frac{{\Delta }_{\dim }{H}^{\Theta }}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)$                                                                         (6)

Where,

• $K_1$ and $K_2$ are the equilibrium constants at temperature $T_1$ and $T_2$.
• ${\Delta }_{\dim }{H}^{\Theta }$ is the standard enthalpy of dimerization of the acetic acid.

The value of $K_1$ is $0.450$.

The value of $K_2$ is $0.142$.

The temperature $T_1$ is $437\text{K}$ ant the temperature $T_{}$ is $471\text{K}$.

The value of the gas constant is $8.314\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}$.

Substitute the value of equilibrium constants, gas constant and temperatures in equation (6).

    $\begin{array}{l}\ln \left(\frac{0.142}{0.450}\right)=\frac{{\Delta }_{\dim }{H}^{\Theta }}{8.314\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}}\left(\frac{1}{437\text{K}}-\frac{1}{471\text{K}}\right)\\ \ln \left(0.31\right)=\frac{{\Delta }_{\dim }{H}^{\Theta }}{8.314\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}}\left(0.0022{\text{K}}^{\text{-1}}-0.0021{\text{K}}^{\text{-1}}\right)\\ -1.17=\frac{{\Delta }_{\dim }{H}^{\Theta }}{8.314\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}}\times 0.0001{\text{K}}^{\text{-1}}\end{array}$

Rearrange the above equation as shown below.

    $\begin{array}{c}{\Delta }_{\dim }{H}^{\Theta }=\frac{-1.17\times 8.314\text{J}{\text{mol}}^{\text{-1}}}{0.0001}\\ =\frac{-9.72\text{J}\text{mol}}{0.0001}\\ =\underset{\_}{-9.72\times 10^{-4}Jmol}\end{array}$

Hence, the enthalpy of dimerization of acetic acid is $\underset{\_}{-9.72\times 10^{-4}Jmol}$.