Chapter 18, Problem 98QP

Interpretation Introduction

Interpretation:

The equilibrium pressure constant and temperature for the given reaction is to be calculated.

Concept introduction:

All the energy available with the system, utilized in doing useful work, is called Gibbs free energy.

Entropy is the direct measurement of randomness or disorder. It is an extensive property and a state function.

The enthalpy of the system defined as the sum of the internal energy and the product of the pressure and the volume. It is a state function and an extensive property.

The standard Gibbs free energy change for the given reaction at temperature is calculated using the following expression:

$\Delta G^{\text{o}}=\Delta H^{\text{o}}-T\times \Delta S^{\text{o}}$

The standard enthalpy change of the reaction, $\Delta H_{\text{rxn}}^{\text{o}}$, is calculated using the following expression:

$\Delta H_{\text{rxn}}^{\text{o}}={\displaystyle \sum n}\Delta H_f^{\text{o}}\left(\text{product}\right)-{\displaystyle \sum m\Delta H_f^{\text{o}}}\left(\text{reactant}\right)$

The standard entropy change for this reaction is calculated using the following expression:

$\Delta S_{\text{rxn}}^{\text{o}}={\displaystyle \sum n}{S}^{\text{o}}\left(\text{product}\right)-{\displaystyle \sum m{S}^{\text{o}}}\left(\text{reactant}\right)$

The equilibrium pressure constant is calculated using expression as follows:

$\Delta G_{}^{\text{o}}=-RT\ln K_p$

Answer to Problem 98QP

Solution: The equilibrium pressure constant for the given reaction is $36$.

At a temperature higher than $981\text{K}$, the reaction favors the formation of ${\text{H}}_2\text{O}\left(g\right)$

and $\text{CO}\left(g\right)$. The value ${\text{K}}_{\text{p}}$ remains same.

Explanation of Solution

Given information: The reaction is as follows:

${\text{H}}_2\text{O}\left(g\right)+\text{CO}\left(g\right)\rightleftharpoons {\text{CO}}_2\left(g\right)+{\text{H}}_2\left(g\right)$.

Temperature, $\text{T}={300}^{\text{o}}\text{C}$.

The equation for the reaction of ${\text{H}}_2\text{O}\left(g\right)$ with $\text{CO}\left(g\right)$ is as follow:

${\text{H}}_2\text{O}\left(g\right)+\text{CO}\left(g\right)\rightleftharpoons {\text{CO}}_2\left(g\right)+{\text{H}}_2\left(g\right).$

The standard Gibbs free energy change for the given reaction at temperature ${300}^{\text{o}}\text{C}$ is calculated using the following expression:

$\Delta G^{\text{o}}=\Delta H^{\text{o}}-T\times \Delta S^{\text{o}}$.

Here, $\Delta H^{\text{o}}$ is the standard enthalpy change, $T$ is the temperature, $\Delta S^o$ is the standard entropy change and $\Delta G^o$ is the standard Gibbs free energy change.

The standard enthalpy change of the system, $\Delta H^{\text{o}}$, is calculated using the following expression:

$\Delta H^{\text{o}}=\Delta H_{\text{rxn}}^{\text{o}}$.

The standard enthalpy change of the reaction, $\Delta H_{\text{rxn}}^{\text{o}}$, is calculated using the following expression:

$\Delta H_{\text{rxn}}^{\text{o}}={\displaystyle \sum n}\Delta H_f^{\text{o}}\left(\text{product}\right)-{\displaystyle \sum m\Delta H_f^{\text{o}}}\left(\text{reactant}\right)$.

Here, $\Delta H_{\text{rxn}}^{\text{o}}$ is the standard enthalpy change for the reaction, $\Delta H_f^{\text{o}}$ is the standard enthalpy change for the formation of the substance, $n$ is the stoichiometric coefficient of products, and $m$ is the stoichiometric coefficient of reactants.

The enthalpy change for the reaction is as follows:

$\Delta H_{rxn}^o=\left(\Delta H_f^o\left[{\text{CO}}_2\left(g\right)\right]+\Delta H_f^o\left[{\text{H}}_2\left(g\right)\right]\right)-\left(\Delta H_f^o\left[{\text{H}}_2\text{O}\left(g\right)\right]+\Delta H_f^o\left[\text{CO}\left(g\right)\right]\right)$.

From appendix $2$, the standard enthalpy changes of formation for the substances are as follows:

$\Delta H_f^o\left[\text{CO}\left(g\right)\right]=\text{}-110.5\text{kJ}/\text{mol,}$

$\Delta H_f^o\left[{\text{H}}_2\left(g\right)\right]=0\text{kJ}/\text{mol,}$

$\Delta H_f^o\left[{\text{CO}}_2\left(g\right)\right]=-393.5\text{kJ}/\text{mol,}$

$\Delta H_f^o\left[{\text{H}}_2\text{O}\left(g\right)\right]=-241.8\text{kJ}/\text{mol}\text{.}$

Substitute the standard enthalpy change of the formation value of the substance in the above expression.

$\begin{array}{l}\Delta H_{\text{rxn}}^{\text{o}}=\left[\left(-393.5\text{kJ}/\text{mol}\right)+\left(0\right)\right]-\left[-\left(241.8\text{kJ}/\text{mol}\right)+\left(-110.5\text{kJ}/\text{mol}\right)\right]\\ =-41.2\text{kJ}/\text{mol}\\ \Delta H^{\text{o}}=-41.2\text{kJ}/\text{mol}\text{.}\end{array}$

Therefore, the standard enthalpy changes for the given reaction are $-41.2\text{kJ}/\text{mol}$.

The entropy change of the system, $\Delta S^{\text{o}}$, is calculated using the following expression:

$\Delta S^{\text{o}}=\Delta S_{\text{rxn}}^{\text{o}}$

The standard entropy change for this reaction is calculated using the following expression:

$\Delta S_{\text{rxn}}^{\text{o}}={\displaystyle \sum n}{S}^{\text{o}}\left(\text{product}\right)-{\displaystyle \sum m{S}^{\text{o}}}\left(\text{reactant}\right).$

Here, $\Delta S_{\text{rxn}}^{\text{o}}$ is the standard entropy change for the reaction, $\Delta S^{\text{o}}$ is the standard entropy change of the substance, $n$ is the stoichiometric coefficient of products, and $m$ is the stoichiometric coefficient of reactants.

The entropy change for the reaction is as follows:

$\Delta S_{\text{rxn}}^{\text{o}}=\left(S^{\text{o}}\left[{\text{CO}}_2\left(g\right)\right]+S^{\text{o}}\left[{\text{H}}_2\left(g\right)\right]\right)-\left(S^{\text{o}}\left[{\text{H}}_2\text{O}\left(g\right)\right]+S^{\text{o}}\left[\text{CO}\left(g\right)\right]\right)$.

From appendix2, the standard entropy values of the substances are as follows:

$S_{}^{\text{o}}\left[\text{CO}\left(g\right)\right]=\text{}197.9\text{J}/\text{K}\cdot \text{mol}$;

$S_{}^{\text{o}}\left[{\text{H}}_2\left(g\right)\right]=131.0\text{J}/\text{K}\cdot \text{mol}$;

$S_{}^{\text{o}}\left[{\text{H}}_2\text{O}\left(g\right)\right]=188.7\text{kJ}/\text{mol}$;

$S^{\text{o}}\left[{\text{CO}}_2\left(g\right)\right]=213.79\text{J}/\text{K}\cdot \text{mol}$.

Substitute the standard entropy values of the substances in the above expression.

$\begin{array}{l}\Delta S_{\text{rxn}}^{\text{o}}=\left[\left(213.79\text{J}/\text{K}\cdot \text{mol}\right)+\left(131.0\text{J}/\text{K}\cdot \text{mol}\right)\right]-\left[\left(188.7\text{J}/\text{K}\cdot \text{mol}\right)+\left(197.9\text{J}/\text{K}\cdot \text{mol}\right)\right].\\ =-42.0\text{J}/\text{K}\cdot \text{mol}\\ \Delta S^{\text{o}}=-42.0\text{J}/\text{K}\cdot \text{mol}\end{array}$

Therefore, the standard entropy change for the given reaction is $-42.0\text{J}/\text{mol}$.

The standard Gibbs free energy change for the given reaction at temperature ${300}^o\text{C}$ is calculated using the following expression:

$\Delta G^{\text{o}}=\Delta H^{\text{o}}-T\times \Delta S^{\text{o}}.$

Here, $\Delta H^o$ is the standard enthalpy changes, $T$ is the temperature, $\Delta S^o$ is the standard entropy change and $\Delta G^{\text{o}}$ is the standard Gibbs free energy change.

Substitutes the value of $\Delta H^o$, $T$ and $\Delta S^{\text{o}}$ in the above expression.

$\begin{array}{l}\Delta G^{\text{o}}=\left(-41.2\times {10}^3\text{J}/\text{mol}\right)-\left(573\text{K}\right)\times \left(-42.0\text{J}/\text{mol}\right)\\ =-1.71\times {10}^4\text{J}/\text{mol}\text{.}\end{array}$

Therefore, standard Gibbs free energy change for the given reaction is $-1.71\times {10}^4\text{J}/\text{mol}$.

The equilibrium pressure constant is calculated using expression as follows:

$\Delta G_{}^{\text{o}}=-RT\ln {\text{K}}_{\text{p}}$

Here, $\text{R}$

is the constant, $T$

is the temperature, ${\text{K}}_{\text{p}}$ is the equilibrium pressure constant and $\Delta G^o$ is the Gibbs free energy change.

Substitute the value of $\text{R}$, $T,$ and $\Delta G^{\text{o}}$ in the above expression.

$\begin{array}{c}\ln {\text{K}}_{\text{p}}=-\left(\frac{-1.71\times {10}^4\cancel{\text{J}}/\text{mol}}{\left(8.314\cancel{\text{J}}/\cancel{\text{K}}\cdot \text{mol}\right)\left(\text{573}\cancel{\text{K}}\right)}\right)\\ \ln {\text{K}}_{\text{p}}=3.59\\ {\text{K}}_{\text{p}}=e^{3.59}\\ {\text{K}}_{\text{p}}=36\end{array}$

Therefore, the equilibrium pressure constant for the given reaction is $36$.

Entropy change for this reaction is negative. The Gibbs free energy is positive at higher temperature.

The temperature is calculated using the following expression:

$\Delta G^{\text{o}}=\Delta H^{\text{o}}-T\times \Delta S^{\text{o}}.$

Here, $\Delta H^{\text{o}}$ is the standard enthalpy changes, $T$ is the temperature, $\Delta S^{\text{o}}$ is the standard entropy change and $\Delta G^{\text{o}}$ is the standard Gibbs free energy change.

Substitute the value of $\Delta H^{\text{o}}$, $\Delta G^{\text{o}}$ and $\Delta S^{\text{o}}$ in the above expression.

$\begin{array}{l}0=\Delta H^{\text{o}}-T\times \Delta S^{\text{o}}\\ T=\frac{\Delta H^{\text{o}}}{\Delta S^{\text{o}}}\\ =\frac{-41.2\times {10}^3\cancel{\text{J}/\text{mol}}}{-40.\times {10}^3\cancel{\text{J}/\text{mol}}}\\ =981\text{K}\text{.}\end{array}$

Therefore, $\Delta G^{\text{o}}$ is zero is $981\text{K}$. Above this temperature $\Delta G^{\text{o}}$ is positive and ${\text{K}}_{\text{p}}$

will be smaller than one. The reaction moves in the backward direction. The temperature higher than $981\text{K}$ the reaction favors the formation of ${\text{H}}_2\text{O}\left(g\right)$ and $\text{CO}\left(g\right)$.

On adding a catalyst to the reaction, the value of ${\text{K}}_{\text{p}}$ does not increase. The catalyst increases both, the rate of forward and the rate of backward reaction. The value ${\text{K}}_{\text{p}}$ remains the same.

Conclusion

The equilibrium pressure constant for the given reaction is $36$. Temperature higher than $981\text{K}$ the reaction favours the formation of ${\text{H}}_2\text{O}\left(g\right)$ and $\text{CO}\left(g\right)$. The value of ${\text{K}}_{\text{p}}$ remains the same.