Chapter 23, Problem 23.35P

(a)

Interpretation Introduction

Interpretation:

Value of the equilibrium constant and $\Delta G_{rxn}^o$ has to be determined for the reaction $CO\left(g\right)\text{ + 2}{H}_{\text{2}}\left(g\right)\underset{}{\to }{\text{CH}}_{\text{3}}\text{OH}\left(l\right)$.

Concept Introduction:

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by $\text{ΔG}$.

Value of $\Delta G_{rxn}^o$ can be calculated by the equation:

  $\Delta G_{rxn}^o\text{= }G°\left[\text{products}\right]-G°\left[\text{reactants}\right]$

Equilibrium constant of a reaction can be determined from the value of $\Delta G_{rxn}^o$ using the given formula,

  ${\text{ΔG}}_{rxn}^o=-\text{RT}\text{ln K}$

Answer to Problem 23.35P

Value of $\Delta G_{rxn}^o$ is $-\text{29}\text{.4 kJ/mol}$.

Value of the equilibrium constant is $\text{142}\text{.3}\times {\text{10}}^3$.

Explanation of Solution

The given reaction is shown below,

  $CO\left(g\right)\text{ + 2}{H}_{\text{2}}\left(g\right)\underset{}{\to }{\text{CH}}_{\text{3}}\text{OH}\left(l\right)$

• Calculate the value of $\Delta G_{rxn}^o$:

Standard free energy of formation values is given below,

  $\begin{array}{c}\text{Δ}G°\left[CO\left(g\right)\right]\text{=}-\text{137}\text{.2 kJ/mol}\\ \\ \text{Δ}G°\left[H_{\text{2}}\left(g\right)\right]\text{=}0\text{ kJ/mol}\\ \\ \text{Δ}G°\left[{\text{CH}}_{\text{3}}\text{OH}\left(l\right)\right]\text{=}-\text{166}\text{.6 kJ/mol}\end{array}$

Value of $\Delta G_{rxn}^o$ can be calculated by the equation:

  $\Delta G_{rxn}^o\text{= }G°\left[\text{products}\right]-G°\left[\text{reactants}\right]$

Substitute the values as follows,

  $\begin{array}{c}\Delta G_{rxn}^o\text{= }\left[\left(1\right)\left(-\text{166}\text{.6 kJ/mol}\right)\right]-\left[\left(1\right)\left(-\text{137}\text{.2 kJ/mol}\right)+\left(2\right)\left(\text{0 kJ/mol}\right)\right]\\ \\ =-\text{29}\text{.4 kJ/mol}\end{array}$

Therefore, the value of $\Delta G_{rxn}^o$ is $-\text{29}\text{.4 kJ/mol}$.

• Calculate the value of the equilibrium constant:

Value of the equilibrium constant is determined as follows,

  $\begin{array}{c}T\text{ = 25}°\text{C}\text{= }\left(25+273\right)\text{ = 298 K}\\ \\ {\text{ΔG}}_{rxn}^o=-\text{RT}\text{ln K}\\ \\ -\text{29}\text{.4 kJ/mol =}-\left(\text{8}\text{.314}\times {\text{10}}^{-3}\text{ kJ}{\text{.K}}^{-1}.mo{l}^{-1}\right)\left(\text{298 K}\right)\text{ln }K\\ \\ \text{ln }K=\frac{\text{29}\text{.4 kJ/mol}}{\left(\text{8}\text{.314}\times {\text{10}}^{-3}\text{ kJ}{\text{.K}}^{-1}.mo{l}^{-1}\right)\left(\text{298 K}\right)}\text{ = 11}\text{.866}\\ \\ \text{K = 142}\text{.3}\times {\text{10}}^3\end{array}$

Therefore, the value of the equilibrium constant is $\text{142}\text{.3}\times {\text{10}}^3$.

(b)

Interpretation Introduction

Interpretation:

Value of the equilibrium constant and $\Delta G_{rxn}^o$ has to be determined for the reaction $C\left(s,\text{ graphite}\right)\text{ + }{H}_{\text{2}}O\left(g\right)\underset{}{\to }\text{CO}\left(g\right)+{\text{ H}}_2\left(g\right)$.

Concept Introduction:

Refer to (a).

Answer to Problem 23.35P

Value of $\Delta G_{rxn}^o$ is $91.4\text{ kJ/mol}$.

Value of the equilibrium constant is $\text{9}\text{.51}\times {\text{10}}^{-17}$.

Explanation of Solution

The given reaction is shown below,

  $C\left(s,\text{ graphite}\right)\text{ + }{H}_{\text{2}}O\left(g\right)\underset{}{\to }\text{CO}\left(g\right)+{\text{ H}}_2\left(g\right)$

• Calculate the value of $\Delta G_{rxn}^o$:

Standard free energy of formation values is given below,

  $\begin{array}{c}\text{Δ}G°\left[C\left(s,\text{ graphite}\right)\right]\text{=}0\text{ kJ/mol}\\ \\ \text{Δ}G°\left[H_{\text{2}}O\left(g\right)\right]\text{=}-\text{228}\text{.6 kJ/mol}\\ \\ \text{Δ}G°\left[CO\left(g\right)\right]\text{=}-\text{137}\text{.2 kJ/mol}\\ \\ \text{Δ}G°\left[H_{\text{2}}\left(g\right)\right]\text{=}0\text{ kJ/mol}\end{array}$

Value of $\Delta G_{rxn}^o$ can be calculated by the equation:

  $\Delta G_{rxn}^o\text{= }G°\left[\text{products}\right]-G°\left[\text{reactants}\right]$

Substitute the values as follows,

  $\begin{array}{c}\Delta G_{rxn}^o\text{= }\left[\left(1\right)\left(-\text{137}\text{.2 kJ/mol}\right)+\left(1\right)\left(\text{0 kJ/mol}\right)\right]\\ -\left[\left(1\right)\left(0\text{ kJ/mol}\right)+\left(1\right)\left(-\text{228}\text{.6 kJ/mol}\right)\right]\\ \\ =91.4\text{ kJ/mol}\end{array}$

Therefore, the value of $\Delta G_{rxn}^o$ is $91.4\text{ kJ/mol}$.

• Calculate the value of the equilibrium constant:

Value of the equilibrium constant is determined as follows,

  $\begin{array}{c}T\text{ = 25}°\text{C}\text{= }\left(25+273\right)\text{ = 298 K}\\ \\ {\text{ΔG}}_{rxn}^o=-\text{RT}\text{ln K}\\ \\ 91.4\text{ kJ/mol =}-\left(\text{8}\text{.314}\times {\text{10}}^{-3}\text{ kJ}{\text{.K}}^{-1}.mo{l}^{-1}\right)\left(\text{298 K}\right)\text{ln }K\\ \\ \text{ln }K=-\frac{91.4\text{ kJ/mol}}{\left(\text{8}\text{.314}\times {\text{10}}^{-3}\text{ kJ}{\text{.K}}^{-1}.mo{l}^{-1}\right)\left(\text{298 K}\right)}\text{ = }-\text{36}\text{.891}\\ \\ \text{K = 9}\text{.51}\times {\text{10}}^{-17}\end{array}$

Therefore, the value of the equilibrium constant is $\text{9}\text{.51}\times {\text{10}}^{-17}$.

(c)

Interpretation Introduction

Interpretation:

Value of the equilibrium constant and $\Delta G_{rxn}^o$ has to be determined for the reaction $CO\left(g\right)\text{ + 2}{H}_{\text{2}}\left(g\right)\underset{}{\to }{\text{CH}}_4\left(g\right)\text{ + }{H}_{2}O\left(g\right)$.

Concept Introduction:

Refer to (a).

Answer to Problem 23.35P

Value of $\Delta G_{rxn}^o$ is $-\text{141}\text{.9 kJ/mol}$.

Value of the equilibrium constant is $\text{7}\text{.48}\times {\text{10}}^{28}$.

Explanation of Solution

The given reaction is shown below,

  $CO\left(g\right)\text{ + 2}{H}_{\text{2}}\left(g\right)\underset{}{\to }{\text{CH}}_4\left(g\right)\text{ + }{H}_{2}O\left(g\right)$

• Calculate the value of $\Delta G_{rxn}^o$:

Standard free energy of formation values is given below,

  $\begin{array}{c}\text{Δ}G°\left[CO\left(g\right)\right]\text{=}-\text{137}\text{.2 kJ/mol}\\ \\ \text{Δ}G°\left[H_{\text{2}}\left(g\right)\right]\text{=}0\text{ kJ/mol}\\ \\ \text{Δ}G°\left[{\text{CH}}_4\left(g\right)\right]\text{=}-\text{50}\text{.5 kJ/mol}\\ \\ \text{Δ}G°\left[H_{2}O\left(g\right)\right]\text{=}-\text{228}\text{.6 kJ/mol}\end{array}$

Value of $\Delta G_{rxn}^o$ can be calculated by the equation:

  $\Delta G_{rxn}^o\text{= }G°\left[\text{products}\right]-G°\left[\text{reactants}\right]$

Substitute the values as follows,

  $\begin{array}{c}\Delta G_{rxn}^o\text{= }\left[\left(1\right)\left(-\text{50}\text{.5 kJ/mol}\right)+\left(1\right)\left(-\text{228}\text{.6 kJ/mol}\right)\right]\\ -\left[\left(1\right)\left(-\text{137}\text{.2 kJ/mol}\right)+\left(3\right)\left(\text{0 kJ/mol}\right)\right]\\ \\ =-\text{141}\text{.9 kJ/mol}\end{array}$

Therefore, the value of $\Delta G_{rxn}^o$ is $-\text{141}\text{.9 kJ/mol}$.

• Calculate the value of the equilibrium constant:

Value of the equilibrium constant is determined as follows,

  $\begin{array}{c}T\text{ = 25}°\text{C}\text{= }\left(25+273\right)\text{ = 298 K}\\ \\ {\text{ΔG}}_{rxn}^o=-\text{RT}\text{ln K}\\ \\ -\text{141}\text{.9 kJ/mol =}-\left(\text{8}\text{.314}\times {\text{10}}^{-3}\text{ kJ}{\text{.K}}^{-1}.mo{l}^{-1}\right)\left(\text{298 K}\right)\text{ln }K\\ \\ \text{ln }K=\frac{\text{141}\text{.9 kJ/mol }}{\left(\text{8}\text{.314}\times {\text{10}}^{-3}\text{ kJ}{\text{.K}}^{-1}.mo{l}^{-1}\right)\left(\text{298 K}\right)}\text{ = 57}\text{.2738}\\ \\ \text{K = 7}\text{.48}\times {\text{10}}^{28}\end{array}$

Therefore, the value of the equilibrium constant is $\text{7}\text{.48}\times {\text{10}}^{28}$.