Chapter 6, Problem 6B.2P

(a)

Interpretation Introduction

Interpretation:

The value of equilibrium constant for the dissociation of methane, ${\text{CH}}_{\text{4}}\left(\text{g}\right)$ has to be determined.

Concept introduction:

The equilibrium constant of a reaction represents the rate of the forward and backward reactions.  If the equilibrium constant is small, the rate of the backward reaction will be more than the rate of the forward reaction.  If the equilibrium constant is large, the rate of the forward reaction is more than the rate of backward reaction.

Answer to Problem 6B.2P

The value of equilibrium constant for the dissociation of methane, ${\text{CH}}_{\text{4}}\left(\text{g}\right)$ is $\underset{\_}{1.250\times 10^{-9}}$.

Explanation of Solution

The decomposition of ${\text{CH}}_{\text{4}}\left(\text{g}\right)$ is represented by the reaction given below.

    ${\text{CH}}_{\text{4}}\left(\text{g}\right)\to \text{C}\left(\text{s}\right)+2{\text{H}}_{\text{2}}\left(\text{g}\right)$

The standard enthalpy change of the reaction (${\Delta }_{\text{r}}{H}^{\Theta }$) is given by the formula below.

    $\begin{array}{l}{\Delta }_{\text{r}}{H}^{\Theta }={\Delta }_{\text{f}}{H}^{\Theta }{}_{\text{products}}-{\Delta }_{\text{f}}{H}^{\Theta }{}_{\text{reactants}}\\ {\Delta }_{\text{r}}{H}^{\Theta }=\left(\left(2\times {\Delta }_{\text{f}}{H}^{\Theta }\left({\text{H}}_{\text{2}}\left(\text{g}\right)\right)+{\Delta }_{\text{f}}{H}^{\Theta }\left(\text{C}\left(\text{s}\right)\right)\right)-\left({\Delta }_{\text{f}}{H}^{\Theta }\left({\text{CH}}_4\left(\text{g}\right)\right)\right)\right)\end{array}$        (1)

The values of standard enthalpy of formations are:

• ${\Delta }_{\text{f}}{H}^{\Theta }\left(\text{C}\left(\text{s}\right)\right)=0\text{kJ}{\text{mol}}^{-1}$
• ${\Delta }_{\text{f}}{H}^{\Theta }\left({\text{H}}_{\text{2}}\left(\text{g}\right)\right)=0{\text{mol}}^{-1}$
• ${\Delta }_{\text{f}}{H}^{\Theta }\left({\text{CH}}_4\left(\text{g}\right)\right)=-74.85\text{kJ}{\text{mol}}^{-1}$

Substitute the corresponding values in the equation (1).

    $\begin{array}{c}{\Delta }_{\text{r}}{H}^{\Theta }=\left(\left(2\times 0+0\right)-\left(-74.85\text{kJ}{\text{mol}}^{-1}\right)\right)\\ =74.85\text{kJ}{\text{mol}}^{-1}\end{array}$

The standard entropy change of the reaction (${\Delta }_{\text{r}}{S}^{\Theta }$) is given by the formula below.

    ${\Delta }_{\text{r}}{S}^{\Theta }=\Delta S_{\text{m}}{{}^{\Theta }}_{\text{products}}-\Delta S_{\text{m}}{{}^{\Theta }}_{\text{reactants}}$

    ${\Delta }_{\text{r}}{S}^{\Theta }=\left(\left(2\times \Delta S_{\text{m}}{}^{\Theta }\left({\text{H}}_{\text{2}}\left(\text{g}\right)\right)+\Delta S_{\text{m}}{}^{\Theta }\left(\text{C}\left(\text{s}\right)\right)\right)-\left(\Delta S_{\text{m}}{}^{\Theta }\left({\text{CH}}_4\left(\text{g}\right)\right)\right)\right)$        (2)

The values of standard molar entropies are:

• $\Delta S_{\text{m}}{}^{\Theta }\left(\text{C}\left(\text{s}\right)\right)=5.740\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}$
• $\Delta S_{\text{m}}{}^{\Theta }\left({\text{H}}_{\text{2}}\left(\text{g}\right)\right)=130.68\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}$
• $\Delta S_{\text{m}}{{}^{\Theta }}^{\Theta }\left({\text{CH}}_4\left(\text{g}\right)\right)=186.26\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}$

Substitute the corresponding values in the equation (2).

    $\begin{array}{c}{\Delta }_{\text{r}}{S}^{\Theta }=\left(\left(\left(2\times 130.68\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\right)+5.740\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\right)-\left(186.26\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\right)\right)\\ =\left(261.36\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}+5.740\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\right)-\left(186.26\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\right)\\ =80.84\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\end{array}$

The formula to calculate ${\Delta }_{\text{r}}{G}^{\Theta }$ is given below.

    ${\Delta }_{\text{r}}{G}^{\Theta }={\Delta }_{\text{r}}{H}^{\Theta }-T{\Delta }_{\text{r}}{S}^{\Theta }$        (3)

Where,

• $T$ is the temperature of the reaction

The temperature of the reaction ($T$) is $\text{298}\text{K}$.

The value of ${\Delta }_{\text{r}}{H}^{\Theta }$ is $74.85\text{kJ}{\text{mol}}^{-1}$.

The value of ${\Delta }_{\text{r}}{S}^{\Theta }$ is $80.84\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}$.

Substitute the value of $T$, ${\Delta }_{\text{r}}{H}^{\Theta }$ and ${\Delta }_{\text{r}}{S}^{\Theta }$ in equation (3).

    $\begin{array}{c}{\Delta }_{\text{r}}{G}^{\Theta }=74.85\text{kJ}{\text{mol}}^{-1}-\left(\text{298}\text{K}\times 80.85\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\right)\\ =\left(74.85\text{kJ}{\text{mol}}^{-1}\times \frac{1000\text{J}}{\text{1}\text{kJ}}\right)-\left(\text{298}\text{K}\times 80.84\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\right)\\ =74850\text{J}{\text{mol}}^{-1}-24093.3\text{J}{\text{mol}}^{-1}\\ =5.08\times {10}^4\text{J}{\text{mol}}^{-1}\end{array}$

The equilibrium constant ($K$) is calculated by the formula given below.

    $\ln K=-\frac{{\Delta }_{\text{r}}{G}^{\Theta }}{RT}$        (4)

Where,

• $R$ is the gas constant. $\left(8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\right)$.

Substitute the corresponding values in equation (4).

    $\begin{array}{c}\ln K=-\frac{5.08\times {10}^4\text{J}{\text{mol}}^{-1}}{8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\times 298\text{K}}\\ K=e^{-20.50}\\ =\underset{\_}{1.250\times 10^{-9}}\end{array}$

Therefore, the equilibrium constant is $\underset{\_}{1.250\times 10^{-9}}$.

(b)

Interpretation Introduction

Interpretation:

The equilibrium constant at $50\text{°C}$ has to be calculated.

Concept introduction:

The equilibrium constant of a reaction represents the rate of the forward and backward reactions.  If the equilibrium constant is small, the rate of the backward reaction will be more than the rate of the forward reaction.  If the equilibrium constant is large, the rate of the forward reaction is more than the rate of backward reaction.

Answer to Problem 6B.2P

The equilibrium constant at $50\text{°C}$ is $\underset{\_}{1.216\times 10^{-10}}$.

Explanation of Solution

The value of temperature is $50\text{°C}$.

The temperature values in degree Celsius is converted to Kelvin using the equation given below.

$T\left(\text{K}\right)=T\left(°\text{C}\right)+273.15$

Substitute the temperature value in the above equation as follows:

$\begin{array}{c}T\left(\text{K}\right)=T\left(°\text{C}\right)+273\\ =50+273\\ =323\text{K}\end{array}$

The formula to calculate equilibrium constant ($K$) at a given temperature is shown below.

    $\ln K_2-\ln K_1=-\frac{{\Delta }_{\text{r}}H}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$        (5)

Where,

• ${\Delta }_{\text{r}}{H}^{\Theta }$ is the standard enthalpy of reaction.
• $T_1,T_2$ are temperatures.
• $K_1,K_2$ are the equilibrium constant corresponding to the temperatures $T_1,T_2$.
• $R$ is the gas constant. $\left(8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\right)$.

The value of ${\Delta }_{\text{r}}{H}^{\Theta }$ is $74.85\text{kJ}{\text{mol}}^{-1}$.

The conversion of $\text{kJ}{\text{mol}}^{-1}$ to $\text{J}{\text{mol}}^{-1}$ is shown below.

    $\text{1}\text{kJ}{\text{mol}}^{-1}={10}^3\text{J}{\text{mol}}^{-1}$

Therefore, the conversion of $74.85\text{kJ}{\text{mol}}^{-1}$ to $\text{J}{\text{mol}}^{-1}$ is shown below.

    $74.85\text{kJ}{\text{mol}}^{-1}=74.85\times {10}^3\text{J}{\text{mol}}^{-1}$

The value of $\ln K_1$ is $-20.50$.

The value of $T_1$ is $\text{298}\text{K}$.

The value of $T_2$ is $\text{323}\text{K}$.

Substitute the value of ${\Delta }_{\text{r}}{H}^{\Theta }$, $\ln K_1$, $T_1$, $T_2$ and $R$ in equation (5).

    $\begin{array}{c}\ln K_2-\left(-20.50\right)=-\frac{74.85\times {10}^3\text{J}{\text{mol}}^{-1}}{8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}}\left(\frac{1}{\text{323}\text{K}}-\frac{1}{\text{298}\text{K}}\right)\\ \ln K_2+20.50=-2.3383\\ \ln K_2=-22.83\\ K_2=\underset{\_}{1.216\times 10^{-10}}\end{array}$

Therefore, the equilibrium constant at $50\text{°C}$ is $\underset{\_}{1.216\times 10^{-10}}$.

(c)

Interpretation Introduction

Interpretation:

The degree of dissociation of methane, $\alpha$ at $\text{298}\text{K}$ and a total pressure of $\text{0}\text{.010}\text{bar}$ has to be calculated.

Concept introduction:

Degree of disassociation of a molecule is the extent to which it has dissociated in its reaction.  Degree of dissociation is directly proportional to the equilibrium constant.

Answer to Problem 6B.2P

The degree of dissociation of methane, $\alpha$ at $\text{298}\text{K}$ and a total pressure of $\text{0}\text{.010}\text{bar}$ is $\underset{\_}{1.76\times 10^{-4}}$.

Explanation of Solution

The disassociation of ${\text{CH}}_{\text{4}}\left(\text{g}\right)$ is represented by the reaction given below.

    ${\text{CH}}_{\text{4}}\left(\text{g}\right)\to \text{C}\left(\text{s}\right)+2{\text{H}}_{\text{2}}\left(\text{g}\right)$

The equilibrium composition is expressed in terms of dissociation constant, $\alpha$ and total pressure, $p$ as shown below.

$\begin{array}{l}{\text{CH}}_{\text{4}}\left(\text{g}\right)\to \text{C}\left(\text{s}\right)+2{\text{H}}_{\text{2}}{\left(\text{g}\right)}_{\text{2}}\\ \text{Initial}\text{n}\text{ 0}\text{0}\\ \text{Change }\text{-nα }\text{0 }2\text{nα }\\ \text{Equilibrium:}\left(\text{1-α}\right)\text{n}02\text{nα}\end{array}$

Hence, the total number of moles of reaction at equilibrium is given below.

    $\left(1+\text{α}\right)\text{n}$

The formula for mole fraction is given below.

    $\text{Mole}\text{fraction}\text{of}\text{component}=\frac{\text{Number}\text{of}\text{moles}\text{of}\text{the}\text{component}}{\text{Total}\text{number}\text{of}\text{moles}}$

Therefore,

The mole fraction of ${\text{CH}}_{\text{4}}\left(\text{g}\right)$ is given below.

    $\text{Mole}\text{fraction}\text{of}{\text{CH}}_{\text{4}}\left(\text{g}\right)=\frac{\text{1-α}}{1+\text{α}}$

The mole fraction of ${\text{H}}_{\text{2}}\left(\text{g}\right)$ is given below.

    $\text{Mole}\text{fraction}\text{of}{\text{H}}_{\text{2}}\left(\text{g}\right)=\frac{\text{2α}}{1+\text{α}}$

The partial pressure is calculated by the formula given below.

    $\text{Partial}\text{pressure}=\text{Mole}\text{fraction}\times \text{Total}\text{pressure}$

Hence,

The partial pressure of ${\text{CH}}_{\text{4}}\left(\text{g}\right)$ is given below.

    $\text{Partial pressure}\text{of}{\text{CH}}_{\text{4}}\left(\text{g}\right)=\frac{\left(\text{1}-\text{α}\right)p}{1+\text{α}}$

The partial pressure of ${\text{H}}_{\text{2}}\left(\text{g}\right)$ is given below

    $\text{Partial pressure}\text{of}{\text{H}}_{\text{2}}\left(\text{g}\right)=\frac{\text{2α}p}{1+\text{α}}$

Where,

• $p$ is the total pressure.

The equilibrium constant for the given reaction in terms of partial pressure of each component is calculated by the formula shown below.

    $K=\frac{{\left(p_{{\text{H}}_{\text{2}}}\right)}^2}{p_{{\text{CH}}_4}}$        (6)

Where,

• $p_{{\text{H}}_{\text{2}}}$ is the partial pressure of ${\text{H}}_{\text{2}}$.
• $p_{{\text{CH}}_4}$ is the partial pressure of ${\text{CH}}_{\text{4}}$.

Substitute the value of $p_{{\text{H}}_{\text{2}}}$ and $p_{{\text{CH}}_4}$ in equation (6).

    $\begin{array}{c}K=\frac{{\left(\frac{\text{2α}p}{1+\text{α}}\right)}^2}{\frac{\left(\text{1}-\text{α}\right)p}{1+\text{α}}}\\ =\frac{{\text{4α}}^{2}p}{{\left(\text{1}-\text{α}\right)}^2}\end{array}$        (7)

As $\alpha$ is very less than $1$, it can be neglected in the denominator of equation (7).

Therefore,

    $\begin{array}{c}K={\text{4α}}^{2}p\\ \text{α}=\sqrt{\frac{K}{4p}}\end{array}$        (8)

The value of $K$ is $1.250\times {10}^{-9}$.

The value of $p$ is $\text{0}\text{.010}\text{bar}$.

Substitute the value of $K$ and $p$ in equation (8).

    $\begin{array}{c}\text{α}=\sqrt{\frac{1.250\times {10}^{-9}}{4\times \text{0}\text{.010}}}\\ =\underset{\_}{1.76\times 10^{-4}}\end{array}$

Therefore, the degree of dissociation of methane, $\alpha$ at $\text{298}\text{K}$ and a total pressure of $\text{0}\text{.010}\text{bar}$ is $\underset{\_}{1.76\times 10^{-4}}$.

(d)

Interpretation Introduction

Interpretation:

The explanation for the variation of degree of dissociation with pressure and temperature has to be stated.

Concept introduction:

Degree of disassociation of a molecule is the extent to which it has dissociated in its reaction.  Degree of dissociation is directly proportional to the equilibrium constant.

Answer to Problem 6B.2P

As the pressure increases, the degree of dissociation decreases.  As the temperature increases, the degree of dissociation increases.

Explanation of Solution

The disassociation of ${\text{CH}}_{\text{4}}\left(\text{g}\right)$ is represented by the reaction given below.

    ${\text{CH}}_{\text{4}}\left(\text{g}\right)\to \text{C}\left(\text{s}\right)+2{\text{H}}_{\text{2}}\left(\text{g}\right)$

The moles of gaseous species are higher on the product side than the reactant side.  According to Le Chatelier’s principle, as the pressure of a system is increased the reaction shifts in the direction in which the number of moles of gaseous species decreases.  Therefore, reaction goes backward and degree of dissociation decreases on increasing pressure.

The value of ${\Delta }_{\text{r}}{H}^{\Theta }$ is positive for the reaction.  Therefore, the reaction is endothermic. Increasing temperature favors the endothermic reaction according to the Le Chatelier’s principle.  The reaction goes forward and the degree of dissociation increases on increasing the temperature.