Chapter 18, Problem 18.87QP

(a)

Interpretation Introduction

Interpretation:

For the given reaction the values of ${\text{ΔH}}^{\text{o}}\text{,}{\text{ΔS}}^{\text{o}}\text{and}{\text{ΔG}}^{\text{o}}{}_{\text{f}}$ has to be calculated.

Concept introduction:

Standard free energy change:

Standard free energy change is measured by subtracting the product of temperature and standard entropy change from the standard enthalpy change of a system.

$\begin{array}{l}{\text{ΔG}}^{\text{o}}\text{=}{\text{ΔH}}^{\text{o}}\text{-}{\text{TΔS}}^{\text{o}}\\ \text{where,}\\ {\text{ΔG}}^{\text{o}}\text{-}\text{standard}\text{free}\text{energy change}\\ {\text{ΔH}}^{\text{o}}\text{-}\text{standard}\text{enthalpy}\text{change}\\ {\text{ΔS}}^{\text{o}}\text{-}\text{standard}\text{entropy change}\text{and T}\text{-}\text{temperature}\text{.}\end{array}$

Answer to Problem 18.87QP

The value of enthalpy change $\left({\text{ΔH}}^{\text{o}}\right)$ is $\text{-75}\text{kJ}\text{.}$

Explanation of Solution

To calculate: the value of ${\text{ΔH}}^{\text{o}}$

Given reaction and information

$\begin{array}{l}{\text{ΔH}}^{\text{o}}\left(\text{kJ}\right)\\ \underset{\_}{\begin{array}{l}{\text{CO}}_{\text{2}}{}_{\text{(g)}}\text{+}{\text{2H}}_{\text{2}}{}_{\left(\text{g}\right)}\stackrel{}{\to }{\text{ HCHO}}_{\left(\text{g}\right)}\text{+}{\text{H}}_{\text{2}}{\text{O}}_{\left(\text{g}\right)}\text{35}\\ {\text{HCHO}}_{\left(\text{g}\right)}\text{+}{\text{2H}}_{\text{2}}{}_{\left(\text{g}\right)}\stackrel{}{\to }{\text{ CH}}_{\text{4}}{}_{\left(\text{g}\right)}\text{+}{\text{H}}_{\text{2}}{\text{O}}_{\left(\text{g}\right)}\text{-201}\\ {\text{C}}_{\left(\text{s}\right)}\text{+}{\text{O}}_{\text{2}}{}_{\left(\text{g}\right)}\stackrel{}{\to }{\text{ CO}}_{\text{2}}{}_{\left(\text{g}\right)}\text{-393}\\ {\text{2H}}_{\text{2}}{\text{O}}_{\left(\text{g}\right)}\stackrel{}{\to }{\text{ 2H}}_{\text{2}}{}_{\left(\text{g}\right)}\text{+}{\text{O}}_{\text{2}}{}_{\left(\text{g}\right)}\text{484}\end{array}}\\ {\text{C}}_{\left(\text{s}\right)}\text{+}{\text{2H}}_{\text{2}}{}_{\left(\text{g}\right)}\stackrel{}{\to }{\text{ CH}}_{\text{4}}{}_{\left(\text{g}\right)}\text{-75}\text{kJ}\text{.}\end{array}$

The value of enthalpy change $\left({\text{ΔH}}^{\text{o}}\right)$ is $\text{-75}\text{kJ}\text{.}$

(b)

Interpretation Introduction

Interpretation:

For the given reaction the values of ${\text{ΔH}}^{\text{o}}\text{,}{\text{ΔS}}^{\text{o}}\text{and}{\text{ΔG}}^{\text{o}}{}_{\text{f}}$ has to be calculated.

Concept introduction:

Standard free energy change:

Standard free energy change is measured by subtracting the product of temperature and standard entropy change from the standard enthalpy change of a system.

$\begin{array}{l}{\text{ΔG}}^{\text{o}}\text{=}{\text{ΔH}}^{\text{o}}\text{-}{\text{TΔS}}^{\text{o}}\\ \text{where,}\\ {\text{ΔG}}^{\text{o}}\text{-}\text{standard}\text{free}\text{energy change}\\ {\text{ΔH}}^{\text{o}}\text{-}\text{standard}\text{enthalpy}\text{change}\\ {\text{ΔS}}^{\text{o}}\text{-}\text{standard}\text{entropy change}\text{and T}\text{-}\text{temperature}\text{.}\end{array}$

Answer to Problem 18.87QP

The value of entropy change $\left({\text{ΔS}}^{\text{o}}\right)$ is $\text{-0}\text{.08084}\text{kJ/K}\text{.}$

Explanation of Solution

To calculate: the value of ${\text{ΔS}}^{\text{o}}$

Given reaction and information

$\begin{array}{l}{\text{C}}_{\left(\text{s}\right)}\text{+}{\text{2H}}_{\text{2}}{}_{\left(\text{g}\right)}\stackrel{}{\to }{\text{ CH}}_{\text{4}}{}_{\left(\text{g}\right)}\\ {\text{S}}^{\text{o}}\text{:}\text{5}\text{.740}\text{2}\text{×}\text{130}\text{.6}\text{186}\text{.1}\text{J/K}\end{array}$

Calculate the value of ${\text{ΔS}}^{\text{o}}$

$\begin{array}{l}{\text{ΔS}}^{\text{o}}\text{=}{\displaystyle \sum {\text{nS}}^{\text{o}}}\left(\text{products}\right)\text{-}{\displaystyle \sum {\text{mS}}^{\text{o}}}\left(\text{reactants}\right)\\ \text{=}\left[\text{186}\text{.1}\text{-}\text{5}\text{.740}\text{-}\left(\text{2}\text{×}\text{130}\text{.6}\right)\right]\text{J/K}\\ \text{=}\left[\text{186}\text{.1}\text{-}\text{266}\text{.94}\right]\text{J/K}\\ \text{=}\text{-80}\text{.74}\text{J/K}\text{.}\\ \text{=}\text{0}\text{.08084}\text{kJ/K}\text{.}\end{array}$

The value of entropy change $\left({\text{ΔS}}^{\text{o}}\right)$ is $\text{-0}\text{.08084}\text{kJ/K}\text{.}$

(c)

Interpretation Introduction

Interpretation:

For the given reaction the values of ${\text{ΔH}}^{\text{o}}\text{,}{\text{ΔS}}^{\text{o}}\text{and}{\text{ΔG}}^{\text{o}}{}_{\text{f}}$ has to be calculated.

Concept introduction:

Standard free energy change:

Standard free energy change is measured by subtracting the product of temperature and standard entropy change from the standard enthalpy change of a system.

$\begin{array}{l}{\text{ΔG}}^{\text{o}}\text{=}{\text{ΔH}}^{\text{o}}\text{-}{\text{TΔS}}^{\text{o}}\\ \text{where,}\\ {\text{ΔG}}^{\text{o}}\text{-}\text{standard}\text{free}\text{energy change}\\ {\text{ΔH}}^{\text{o}}\text{-}\text{standard}\text{enthalpy}\text{change}\\ {\text{ΔS}}^{\text{o}}\text{-}\text{standard}\text{entropy change}\text{and T}\text{-}\text{temperature}\text{.}\end{array}$

To calculate: the value of ${\text{ΔG}}^{\text{o}}{}_{\text{f}}$

Answer to Problem 18.87QP

The value of free energy change $\left({\text{ΔG}}^{\text{o}}{}_{\text{f}}\right)$ at $\text{25}{}^{\text{o}}\text{C}$ is $\text{-51}\text{kJ}\text{.}$

Explanation of Solution

Calculate the value of ${\text{ΔG}}^{\text{o}}{}_{\text{f}}$ using the values of ${\text{ΔH}}^{\text{o}}\text{,}{\text{ΔS}}^{\text{o}}\text{and}\text{T}\text{(298}\text{K)}$

$\begin{array}{l}{\text{ΔG}}^{\text{o}}{}_{\text{f}}\text{=}{\text{ΔG}}^{\text{o}}\text{=}{\text{ΔH}}^{\text{o}}\text{-}{\text{TΔS}}^{\text{o}}\\ \text{=}\text{-75}\text{kJ}\text{-}\left(\text{298}\text{K}\right)\left(\text{-0}\text{.08084}\text{kJ/K}\right)\\ \text{=}\left(\text{109}\text{.6}\text{-}\text{24}\text{.09032}\right)\text{kJ}\\ \text{=}\text{-51}\text{kJ}\text{.}\end{array}$

The value of free energy change $\left({\text{ΔG}}^{\text{o}}{}_{\text{f}}\right)$ at $\text{25}{}^{\text{o}}\text{C}$ is $\text{-51}\text{kJ}\text{.}$