Chapter 3, Problem 3C.6P

Interpretation Introduction

Interpretation: The standard enthalpy and standard entropy for the given reaction at $298\text{K}$ and $500\text{K}$ has to be calculated.

Concept introduction: Standard entropy of the reaction increases as the numbers of gaseous constituents in the reaction increases.  Standard entropy of reaction predicts the feasibility of the reaction.  For a reaction to be feasible, the change in the standard entropy of the reaction must be positive.  The enthalpy of formation for substances in their free state is zero.

Answer to Problem 3C.6P

Standard enthalpy of formation for the given reaction at $298\text{K}$ and $500\text{K}$ are $-46.11kJmo{l}^{-1}$ and $-39.028kJmo{l}^{-1}$ respectively.

Standard entropy for the given reaction at $298\text{K}$ and $500\text{K}$ are $-151.1J{K}^{-1}mo{l}^{-1}$ and $-111.148J{K}^{-1}mo{l}^{-1}$ respectively.

Explanation of Solution

Given reaction is-

    $\frac{1}{2}{\text{N}}_{\text{2}}\left(\text{g}\right)+\frac{3}{2}{\text{H}}_{\text{2}}\left(\text{g}\right)\to {\text{NH}}_{\text{3}}\left(\text{g}\right)$

The values of standard enthalpy of formations are:

• • The standard enthalpy of formation for ${\text{N}}_{\text{2}}$ at $298\text{K}$ is $0\text{kJ}{\text{mol}}^{\text{-1}}$.
• • The standard enthalpy of formation for ${\text{H}}_{\text{2}}$ at $298\text{K}$ is $0\text{kJ}{\text{mol}}^{\text{-1}}$.
• • The standard enthalpy of formation for ${\text{NH}}_3$ at $298\text{K}$ is $-99.38\text{kJ}{\text{mol}}^{\text{-1}}$.

The values of standard entropies are:

• • The standard entropy of ${\text{N}}_{\text{2}}$ at $298\text{K}$ is $191.61\text{J}{\text{K}}^{-1}{\text{mol}}^{\text{-1}}$.
• • The standard entropy of ${\text{H}}_{\text{2}}$ at $298\text{K}$ is $130.684\text{J}{\text{K}}^{-1}{\text{mol}}^{\text{-1}}$.
• • The standard entropy of ${\text{NH}}_3$ at $298\text{K}$ is $140.6\text{J}{\text{K}}^{-1}{\text{mol}}^{\text{-1}}$.

The values of heat capacities are:

• • The heat capacity of ${\text{N}}_{\text{2}}$ at $298\text{K}$ is $29.125\text{J}{\text{K}}^{-1}{\text{mol}}^{\text{-1}}$.
• • The heat capacity of ${\text{H}}_{\text{2}}$ at $298\text{K}$ is $28.824\text{J}{\text{K}}^{-1}{\text{mol}}^{\text{-1}}$.
• • The heat capacity of ${\text{NH}}_3$ at $298\text{K}$ is $35.06\text{J}{\text{K}}^{-1}{\text{mol}}^{\text{-1}}$.

For $298K$:

Enthalpy of formation reaction at $298\text{K}$ is calculated by the following formula.

    ${\Delta }_{\text{f}}{H}^{\Theta }{}_{\text{reaction}}=\sum {\Delta }_{\text{f}}{H}^{\Theta }{}_{\text{product}}-\sum {\Delta }_{\text{f}}{H}^{\Theta }{}_{\text{reactants}}$

Substitute the reactants and products in the above equation.

    $\begin{array}{c}{\Delta }_{\text{f}}{H}^{\Theta }{}_{\text{reaction}}={\Delta }_{\text{f}}{H}^{\Theta }\left({\text{NH}}_{\text{3}}\left(\text{g}\right)\right)-\left(\frac{3}{2}\times {\Delta }_{\text{f}}{H}^{\Theta }\left({\text{H}}_{\text{2}}\left(\text{g}\right)\right)+\frac{1}{2}\times {\Delta }_{\text{f}}{H}^{\Theta }\left({\text{N}}_{\text{2}}\left(\text{g}\right)\right)\right)\\ =\left(-46.11\text{kJ}{\text{mol}}^{\text{-1}}\right)-\left(\frac{3}{2}\times 0\text{kJ}{\text{mol}}^{\text{-1}}+\frac{1}{2}0\text{kJ}{\text{mol}}^{\text{-1}}\right)\\ =\underset{\_}{-46.11kJmo{l}^{-1}}\end{array}$

Hence, the enthalpy of formation for the given reaction at $298\text{K}$ is $-46.11kJmo{l}^{-1}$

Standard entropy of reaction at $298\text{K}$ is calculated by the following equation.

    $\begin{array}{c}\Delta S^{\Theta }{}_{\text{reaction}}=\sum \Delta S^{\Theta }{}_{\text{product}}-\sum \Delta S^{\Theta }{}_{\text{reactants}}\\ =\Delta S^{\Theta }\left({\text{NH}}_{\text{3}}\left(\text{g}\right)\right)-\left(\frac{3}{2}\times \Delta S^{\Theta }\left({\text{H}}_{\text{2}}\left(\text{g}\right)\right)+\frac{1}{2}\times \Delta S^{\Theta }\left({\text{N}}_{\text{2}}\left(\text{g}\right)\right)\right)\end{array}$

Substitute the corresponding values in the above equation.

    $\begin{array}{c}\Delta S^{\Theta }{}_{\text{reaction}}=\left(140.6\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\right)-\left(\frac{3}{2}\times 130.6\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}+\frac{1}{2}\times 191.6\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\right)\\ =\left(\text{140}\text{.6}\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\right)-\left(291.7\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\right)\\ =\underset{\_}{-151.1J{K}^{-1}mo{l}^{-1}}\end{array}$

Hence, the standard entropy for the given reaction at $298\text{K}$ is $-151.1J{K}^{-1}mo{l}^{-1}$

For $500K$:

The enthalpy of formation of ${\text{NH}}_3$ reaction at $500\text{K}$ is calculated by the following formula.

    $\begin{array}{c}{\Delta }_{\text{f}}{H}^{\Theta }{}_{{\text{NH}}_{\text{3}}}={\Delta }_{\text{f}}{H}^{\Theta }{}_{{\text{NH}}_{\text{3}}\text{at}\text{298}\text{K}}+C_P\Delta T\\ =-46.11{\text{kJmol}}^{\text{-1}}+35.06{\text{kJK}}^{-1}{\text{mol}}^{\text{-1}}\times \frac{1}{1000\text{kg}}\times \left(500\text{K}-298\text{K}\right)\\ =-46.11{\text{kJmol}}^{\text{-1}}+7.082{\text{kJmol}}^{\text{-1}}\\ =-39.028{\text{kJmol}}^{\text{-1}}\end{array}$

In the free state, standard enthalpy of formation of ${\text{N}}_{\text{2}}$ and ${\text{H}}_{\text{2}}$ at $500\text{K}$ are $0{\text{kJmol}}^{\text{-1}}$.

Standard enthalpy of formation of the given reaction at $500\text{K}$ is calculated by the following formula.

    $\begin{array}{l}{\Delta }_{\text{f}}{H}^{\Theta }{}_{\text{reaction}}=\sum {\Delta }_{\text{f}}{H}^{\Theta }{}_{\text{product}}-\sum {\Delta }_{\text{f}}{H}^{\Theta }{}_{\text{reactants}}\\ {\Delta }_{\text{f}}{H}^{\Theta }{}_{\text{reaction}}=\left(\begin{array}{l}{\Delta }_{\text{f}}{H}^{\Theta }\left({\text{NH}}_{\text{3}}\left(\text{g}\right)\right)-\\ \left(\frac{3}{2}\times {\Delta }_{\text{f}}{H}^{\Theta }\left({\text{H}}_{\text{2}}\left(\text{g}\right)\right)+\frac{1}{2}{\Delta }_{\text{f}}{H}^{\Theta }\left({\text{N}}_{\text{2}}\left(\text{g}\right)\right)\right)\end{array}\right)\end{array}$

Substitute the corresponding values in the above equation.

    $\begin{array}{c}{\Delta }_{\text{f}}{H}^{\Theta }{}_{\text{reaction}}=\left(-39.028{\text{kJmol}}^{\text{-1}}\right)-\left(\frac{3}{2}\times 0{\text{kJmol}}^{\text{-1}}+\frac{1}{2}0{\text{kJmol}}^{\text{-1}}\right)\\ =\underset{\_}{-39.028kJmo{l}^{-1}}\end{array}$

Hence, the enthalpy of formation for the given reaction at $500\text{K}$ is $-39.028kJmo{l}^{-1}$

The entropy of ${\text{NH}}_3$ at $500\text{K}$ is calculated by the following formula.

    $\Delta S^{\Theta }{}_{{\text{NH}}_{\text{3}}}=\Delta S^{\Theta }{}_{{\text{NH}}_{\text{3}}\text{at}\text{298}\text{K}}+C_P\ln \left(\frac{T_2}{T_1}\right)$

Substitute the corresponding values in the above equation.

    $\begin{array}{c}\Delta S^{\Theta }{}_{{\text{NH}}_{\text{3}}}=192.45{\text{JK}}^{-1}{\text{mol}}^{\text{-1}}+35.06{\text{JK}}^{-1}{\text{mol}}^{\text{-1}}\times \ln \left(\frac{500\text{K}}{298\text{K}}\right)\\ =192.45{\text{JK}}^{-1}{\text{mol}}^{\text{-1}}+18.144{\text{JK}}^{-1}{\text{mol}}^{\text{-1}}\\ =210.594{\text{JK}}^{-1}{\text{mol}}^{\text{-1}}\end{array}$

The entropy of ${\text{N}}_2$ at $500\text{K}$ is calculated by the following formula.

    $\Delta S^{\Theta }{}_{{\text{N}}_2}=\Delta S^{\Theta }{}_{{\text{N}}_2\text{at}\text{298}\text{K}}+C_P\ln \left(\frac{T_2}{T_1}\right)$

Substitute the corresponding values in the above equation.

    $\begin{array}{c}\Delta S^{\Theta }{}_{{\text{N}}_2}=191.61\text{J}{\text{K}}^{-1}{\text{mol}}^{\text{-1}}+29.125\text{J}{\text{K}}^{-1}{\text{mol}}^{\text{-1}}\times \ln \left(\frac{500\text{K}}{298\text{K}}\right)\\ =191.61\text{J}{\text{K}}^{-1}{\text{mol}}^{\text{-1}}+15.0761\text{J}{\text{K}}^{-1}{\text{mol}}^{\text{-1}}\\ =206.68261\text{J}{\text{K}}^{-1}{\text{mol}}^{\text{-1}}\end{array}$

The entropy of ${\text{H}}_2$ at $500\text{K}$ is calculated by the following formula.

    $\Delta S^{\Theta }{}_{{\text{H}}_2}=\Delta S^{\Theta }{}_{{\text{H}}_2\text{at}\text{298}\text{K}}+C_P\ln \left(\frac{T_2}{T_1}\right)$

Substitute the corresponding values in the above equation.

    $\begin{array}{c}\Delta S^{\Theta }{}_{{\text{H}}_2}=130.684{\text{JK}}^{-1}{\text{mol}}^{\text{-1}}+28.824{\text{JK}}^{-1}{\text{mol}}^{\text{-1}}\times \ln \left(\frac{500\text{K}}{298\text{K}}\right)\\ =130.684{\text{JK}}^{-1}{\text{mol}}^{\text{-1}}+14.9168{\text{JK}}^{-1}{\text{mol}}^{\text{-1}}\\ =145.60084{\text{JK}}^{-1}{\text{mol}}^{\text{-1}}\end{array}$

The standard entropy of reaction is calculated by the following formula.

    $\begin{array}{l}\Delta S^{\Theta }{}_{\text{reaction}}=\sum \Delta S^{\Theta }{}_{\text{product}}-\sum \Delta S^{\Theta }{}_{\text{reactants}}\\ \Delta S^{\Theta }{}_{\text{reaction}}=\left(\begin{array}{l}\Delta S^{\Theta }\left({\text{NH}}_{\text{3}}\left(\text{g}\right)\right)-\\ \left(\frac{3}{2}\times \Delta S^{\Theta }\left({\text{H}}_{\text{2}}\left(\text{g}\right)\right)+\frac{1}{2}\times \Delta S^{\Theta }\left({\text{N}}_{\text{2}}\left(\text{g}\right)\right)\right)\end{array}\right)\end{array}$

Substitute the corresponding values in the above equation.

  $\begin{array}{c}\Delta S^{\Theta }{}_{\text{reaction}}=\left(\begin{array}{c}\left(210.594\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\right)\\ -\left(\frac{3}{2}\times 145.60084\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}+\frac{1}{2}\times 206.68261\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\right)\end{array}\right)\\ =\left(210.594\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\right)-\left(321.742565\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\right)\\ =\underset{\_}{-111.1486J{K}^{-1}mo{l}^{-1}}\end{array}$

Hence, the standard entropy for the given reaction at $500\text{K}$ is $-111.148J{K}^{-1}mo{l}^{-1}$.