Chapter 19, Problem 87CWP

(a)

Interpretation Introduction

Interpretation: Standard enthalpy of the reaction and change in standard entropy of the given reaction are to be determined and the temperature at which the change in standard Gibb’s energy is zero, is to be calculated.

Concept introduction: The standard enthalpy of the reaction is calculated by the formula,

$\Delta H^{\circ }{}_{\text{reaction}}=\Delta H^{\circ }{}_{\text{formation}}\left(\text{product}\right)-\Delta H^{\circ }{}_{\text{formation}}\left(\text{reactant}\right)$

The change in standard Gipp’s free energy of the reaction is calculated as,

$\Delta G^{\circ }=\Delta H^{\circ }-T\Delta S^{\circ }$

To determine: Standard enthalpy of the reaction and change in standard entropy of the given reaction.

Answer to Problem 87CWP

Standard enthalpy of the reaction and change in standard entropy of the given reaction is $\underset{\_}{+131.5kJ/mol}$ and $\underset{\_}{134J/K\cdot mol}$, respectively.

Explanation of Solution

Hydrogen gas is produced by reacting graphite with water.

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

The standard enthalpy of formation of ${\text{H}}_{\text{2}}\text{O}\left(g\right)$ is $-242\text{kJ}/\text{mol}$.

The standard enthalpy of formation of $\text{CO}\left(g\right)$ is $-110.5\text{kJ}/\text{mol}$.

The standard enthalpy of formation of $\text{C}\left(s\right)$ and ${\text{H}}_{\text{2}}\left(g\right)$ is zero since the standard enthalpy of formation of free elements in their standard state is zero.

The standard enthalpy of the reaction is calculated by the formula,

$\Delta H^{\circ }{}_{\text{reaction}}=\Delta H^{\circ }{}_{\text{formation}}\left(\text{product}\right)-\Delta H^{\circ }{}_{\text{formation}}\left(\text{reactant}\right)$

Therefore, the above equation becomes,

$\Delta H^{\circ }{}_{\text{reaction}}=\Delta H^{\circ }{}_{\text{formation}}\left(\text{CO}\left(g\right)\right)-\Delta H^{\circ }{}_{\text{formation}}\left({\text{H}}_{\text{2}}\text{O}\left(g\right)\right)$

Substitute the value of $\Delta H^{\circ }{}_{\text{formation}}\left(\text{CO}\left(g\right)\right)$ and $\Delta H^{\circ }{}_{\text{formation}}\left({\text{H}}_{\text{2}}\text{O}\left(g\right)\right)$ in the above equation.

$\begin{array}{c}\Delta H^{\circ }{}_{\text{reaction}}=\left(-110.5\text{kJ}/\text{mol}\right)-\left(-242\text{kJ}/\text{mol}\right)\\ =-110.5\text{kJ}/\text{mol}+242\text{kJ}/\text{mol}\\ =\underset{\_}{+131.5kJ/mol}\end{array}$

Therefore, the standard enthalpy of the reaction is $+131.5\text{kJ}/\text{mol}$.

The standard entropy of ${\text{H}}_{\text{2}}\text{O}\left(g\right)$ is $189\text{J}/\text{K}\cdot \text{mol}$.

The standard entropy of $\text{CO}\left(g\right)$ is $198\text{J}/\text{K}\cdot \text{mol}$.

The standard entropy of $\text{C}\left(s\right)$ is $6\text{J}/\text{K}\cdot \text{mol}$.

The standard entropy of ${\text{H}}_{\text{2}}\left(g\right)$ is $131\text{J}/\text{K}\cdot \text{mol}$.

The standard entropy change of the reaction is calculated by the formula,

$\Delta S^{\circ }{}_{\text{reaction}}=S^{\circ }\left(\text{product}\right)-S^{\circ }\left(\text{reactant}\right)$

Therefore, the above equation becomes,

$\Delta S^{\circ }{}_{\text{reaction}}=\left[S^{\circ }\left(\text{CO}\left(g\right)\right)+S^{\circ }\left({\text{H}}_{\text{2}}\left(g\right)\right)\right]-\left[S^{\circ }\left({\text{H}}_{\text{2}}\text{O}\left(g\right)\right)+S^{\circ }\left(\text{C}\left(s\right)\right)\right]$

Substitute the value of $S^{\circ }\left(\text{CO}\left(g\right)\right)$, $S^{\circ }\left({\text{H}}_{\text{2}}\left(g\right)\right)$, $S^{\circ }\left({\text{H}}_{\text{2}}\text{O}\left(g\right)\right)$ and $S^{\circ }\left(\text{C}\left(s\right)\right)$ in the above equation.

$\begin{array}{c}\Delta S^{\circ }{}_{\text{reaction}}=\left[198\text{J}/\text{K}\cdot \text{mol}+131\text{J}/\text{K}\cdot \text{mol}\right]-\left[189\text{J}/\text{K}\cdot \text{mol}+6\text{J}/\text{K}\cdot \text{mol}\right]\\ =329\text{J}/\text{K}\cdot \text{mol}-195\text{J}/\text{K}\cdot \text{mol}\\ =\underset{\_}{134J/K\cdot mol}\end{array}$

Therefore, the standard entropy change of the reaction is $\underset{\_}{134J/K\cdot mol}$.

(b)

Interpretation Introduction

Interpretation: Standard enthalpy of the reaction and change in standard entropy of the given reaction are to be determined and the temperature at which the change in standard Gibb’s energy is zero, is to be calculated.

Concept introduction: The standard enthalpy of the reaction is calculated by the formula,

$\Delta H^{\circ }{}_{\text{reaction}}=\Delta H^{\circ }{}_{\text{formation}}\left(\text{product}\right)-\Delta H^{\circ }{}_{\text{formation}}\left(\text{reactant}\right)$

The change in standard Gipp’s free energy of the reaction is calculated as,

$\Delta G^{\circ }=\Delta H^{\circ }-T\Delta S^{\circ }$

To determine: The temperature at which the change in standard Gipp’s energy of the given reaction is zero.

Answer to Problem 87CWP

The temperature at which the change in standard Gipp’s energy of the given reaction is zero is $\underset{\_}{981.3K}$

Explanation of Solution

Given

The change in standard Gipp’s energy of the given reaction is zero.

Standard enthalpy of the reaction and change in standard entropy of the given reaction is $+131.5\text{kJ}/\text{mol}$ and $134\text{J}/\text{K}\cdot \text{mol}$, respectively.

The change in standard Gipp’s free energy of the reaction is calculated as,

$\Delta G^{\circ }=\Delta H^{\circ }-T\Delta S^{\circ }$

Substitute the value of $\Delta G^{\circ }$, $\Delta H^{\circ }$ and $\Delta S^{\circ }$ in the above equation.

$\begin{array}{c}0=+131.5\times {10}^3\text{J}/\text{mol}-T\times 134\text{J}/\text{K}\cdot \text{mol}\\ T\times 134\text{J}/\text{K}\cdot \text{mol}=131.5\times {10}^3\text{J}/\text{mol}\\ T=\frac{131.5\times {10}^3\text{J}/\text{mol}}{134\text{J}/\text{K}\cdot \text{mol}}\\ =\underset{\_}{981.3K}\end{array}$

Therefore, the temperature at which the change in standard Gipp’s energy of the given reaction is zero is $\underset{\_}{981.3K}$