Chapter 17, Problem 84E

Consider the following reaction at 298 K:

$2{\text{SO}}_{\text{2}}\left(g\right)+{\text{O}}_{\text{2}}\left(g\right)\to 2{\text{SO}}_{\text{3}}\left(g\right)$

An equilibrium mixture contains O₂(g) and SO₃(g) at partial pressures of 0.50 atm and 2.0 atm, respectively. Using data from Appendix 4, determine the equilibrium partial pressure of SO₂ in the mixture. Will this reaction be most favored at a high or a low temperature, assuming standard conditions?

Interpretation Introduction

Interpretation: The equilibrium partial pressure of ${\text{SO}}_{\text{2}}$ in the given mixture is to be determined. Standard conditions are to be assumed and the most favored dependency on the high or low temperature is to be decided.

Concept introduction: The equilibrium partial pressure is defined as the total pressure experienced by the individual gases present in the equilibrium mixture. The equilibrium constant in terms of partial pressure is denoted by $K_{\text{p}}$.

Answer to Problem 84E

Answer

The equilibrium partial pressure of ${\text{SO}}_{\text{2}}$ in the given mixture is $\underset{\_}{2.98atm}$.

At the low temperature the reaction is most favored.

Explanation of Solution

Explanation

To determine: The equilibrium partial pressure of ${\text{SO}}_{\text{2}}$ in the given mixture; whether this reaction be most favored at a high or low temperature by assuming standard conditions.

Given

The reaction is given as,

$2{\text{SO}}_{\text{2}}\left(g\right)+{\text{O}}_{\text{2}}\left(g\right)\to 2{\text{SO}}_3\left(g\right)$

The standard values of $\Delta H_{\text{f}}^{\text{ο}},\Delta S_{\text{f}}^{\text{ο}}$ and $\Delta G_{\text{f}}^{\text{ο}}$ given in appendix 4 are as follows,

$\begin{array}{cccc}& \Delta H_{\text{f}}^{\text{ο}}\left(\text{kJ/mol}\right)& \Delta S_{\text{f}}^{\text{ο}}\left(\text{J/K}\cdot \text{mol}\right)& \Delta G_{\text{f}}^{\text{ο}}\left(\text{kJ/mol}\right)\\ {\text{SO}}_3& -396& 257& -371\\ {\text{O}}_{\text{2}}& 0& 205& -300\\ {\text{SO}}_2& -297& 248& 0\end{array}$

The value of standard enthalpy change $\Delta H^{\text{ο}}$ of the given reaction is calculated by the formula,

$\Delta H^{\text{ο}}={{\displaystyle \sum n_{\text{p}}\Delta H_{\text{f}}^{\text{ο}}}}_{\left(\text{products}\right)}-{{\displaystyle \sum n_{\text{r}}\Delta H_{\text{f}}^{\text{ο}}}}_{\left(\text{reactants}\right)}$

Where,

• $\Delta H_{\text{f}}^{\text{ο}}{}_{\left(\text{reactants}\right)}$ are the standard enthalpy of formation for the reactants.
• $\Delta H_{\text{f}}^{\text{ο}}{}_{\left(\text{products}\right)}$ are the standard enthalpy of formation for the products.
• $n_{\text{p}}$ is the number of products molecule.
• $n_{\text{r}}$ is the number of reactants molecule.
• ${\displaystyle \sum }$ is the symbol of summation.

For the given reaction the representation in the above form is written as,

$\begin{array}{c}\Delta H^{\text{ο}}={{\displaystyle \sum n_{\text{p}}\Delta H_{\text{f}}^{\text{ο}}}}_{\left(\text{products}\right)}-{{\displaystyle \sum n_{\text{r}}\Delta H_{\text{f}}^{\text{ο}}}}_{\left(\text{reactants}\right)}\\ \Delta H^{\text{ο}}=\left[2\Delta H_{\text{f}}^{\text{ο}}{}_{{\text{SO}}_3\left(g\right)}\right]-\left[2\Delta H_{\text{f}}^{\text{ο}}{}_{{\text{SO}}_2\left(g\right)}+\Delta H_{\text{f}}^{\text{ο}}{}_{{\text{O}}_2\left(g\right)}\right]\end{array}$

Substitute the value of standard enthalpy of formations in the above equation.

$\begin{array}{l}\Delta H^{\text{ο}}=\left[2\Delta H_{\text{f}}^{\text{ο}}{}_{{\text{SO}}_3\left(g\right)}\right]-\left[2\Delta H_{\text{f}}^{\text{ο}}{}_{{\text{SO}}_2\left(g\right)}+\Delta H_{\text{f}}^{\text{ο}}{}_{{\text{O}}_2\left(g\right)}\right]\\ \Delta H^{\text{ο}}=\left[\begin{array}{l}\left[2\text{mol}\left(-396\text{kJ/mol}\right)\right]-\\ \left[2\text{mol}\times \left(-297\text{kJ/mol}\right)+1\text{mol}\times \left(0\text{kJ/mol}\right)\right]\end{array}\right]\\ \Delta H^{\text{ο}}=\left(-792\text{kJ}\right)-\left(-594\text{kJ}\right)\\ \Delta H^{\text{ο}}=-\text{198}\text{kJ}\end{array}$

The value of standard entropy change $\Delta S^{\text{ο}}$ of the given reaction is calculated by the formula,

$\Delta S^{\text{ο}}={{\displaystyle \sum n_{\text{p}}\Delta S_{\text{f}}^{\text{ο}}}}_{\left(\text{products}\right)}-{{\displaystyle \sum n_{\text{r}}\Delta S_{\text{f}}^{\text{ο}}}}_{\left(\text{reactants}\right)}$

Where,

• $\Delta S_{\text{f}}^{\text{ο}}{}_{\left(\text{reactants}\right)}$ are the standard entropy of formation for the reactants.
• $\Delta S_{\text{f}}^{\text{ο}}{}_{\left(\text{products}\right)}$ are the standard entropy of formation for the products.
• $n_{\text{p}}$ is the number of products molecule.
• $n_{\text{r}}$ is the number of reactants molecule.
• ${\displaystyle \sum }$ is the symbol of summation.

For the given reaction the representation in the above form is written as,

$\begin{array}{c}\Delta S^{\text{ο}}={{\displaystyle \sum n_{\text{p}}\Delta S_{\text{f}}^{\text{ο}}}}_{\left(\text{products}\right)}-{{\displaystyle \sum n_{\text{r}}\Delta S_{\text{f}}^{\text{ο}}}}_{\left(\text{reactants}\right)}\\ \Delta S^{\text{ο}}=\left[2\Delta S_{\text{f}}^{\text{ο}}{}_{{\text{SO}}_3\left(g\right)}\right]-\left[2\Delta S_{\text{f}}^{\text{ο}}{}_{{\text{SO}}_2\left(g\right)}+\Delta S_{\text{f}}^{\text{ο}}{}_{{\text{O}}_2\left(g\right)}\right]\end{array}$

Substitute the values of standard entropy of formations in the above equation.

$\begin{array}{c}\Delta S^{\text{ο}}=\left[2\Delta S_{\text{f}}^{\text{ο}}{}_{{\text{SO}}_3\left(g\right)}\right]-\left[2\Delta S_{\text{f}}^{\text{ο}}{}_{{\text{SO}}_2\left(g\right)}+\Delta S_{\text{f}}^{\text{ο}}{}_{{\text{O}}_2\left(g\right)}\right]\\ \Delta S^{\text{ο}}=\left[\begin{array}{l}\left[2\text{mol}\left(257\text{J/K}\cdot \text{mol}\right)\right]-\\ \left[2\text{mol}\times \left(248\text{J/K}\cdot \text{mol}\right)+1\text{mol}\times \left(205\text{J/K}\cdot \text{mol}\right)\right]\end{array}\right]\\ \Delta S^{\text{ο}}=\left(514\text{J/K}\cdot \text{mol}\right)-\left(701\text{J/K}\cdot \text{mol}\right)\\ \Delta S^{\text{ο}}=-\text{187}\text{J/K}\end{array}$

The value of standard Gibb’s free energy $\Delta G^{\text{ο}}$ of the given reaction is calculated by the formula,

$\Delta G^{\text{ο}}={{\displaystyle \sum n_{\text{p}}\Delta G_{\text{f}}^{\text{ο}}}}_{\left(\text{products}\right)}-{{\displaystyle \sum n_{\text{r}}\Delta G_{\text{f}}^{\text{ο}}}}_{\left(\text{reactants}\right)}$

Where,

• $\Delta G_{\text{f}}^{\text{ο}}{}_{\left(\text{reactants}\right)}$ are the standard free energy of formation for the reactants.
• $\Delta G_{\text{f}}^{\text{ο}}{}_{\left(\text{products}\right)}$ are the standard free energy of formation for the products.
• $n_{\text{p}}$ is the number of products molecule.
• $n_{\text{r}}$ is the number of reactants molecule.
• ${\displaystyle \sum }$ is the symbol of summation.

For the given reaction the representation in the above form is written as,

$\begin{array}{c}\Delta G^{\text{ο}}={{\displaystyle \sum n_{\text{p}}\Delta G_{\text{f}}^{\text{ο}}}}_{\left(\text{products}\right)}-{{\displaystyle \sum n_{\text{r}}\Delta G_{\text{f}}^{\text{ο}}}}_{\left(\text{reactants}\right)}\\ \Delta G^{\text{ο}}=\left[2\Delta G_{\text{f}}^{\text{ο}}{}_{{\text{SO}}_3\left(g\right)}\right]-\left[2\Delta G_{\text{f}}^{\text{ο}}{}_{{\text{SO}}_2\left(g\right)}+\Delta G_{\text{f}}^{\text{ο}}{}_{{\text{O}}_2\left(g\right)}\right]\end{array}$

Substitute the values of Gibb’s energy of formations in the above equation.

$\begin{array}{l}\Delta G^{\text{ο}}=\left[2\Delta G_{\text{f}}^{\text{ο}}{}_{{\text{SO}}_3\left(g\right)}\right]-\left[2\Delta G_{\text{f}}^{\text{ο}}{}_{{\text{SO}}_2\left(g\right)}+\Delta G_{\text{f}}^{\text{ο}}{}_{{\text{O}}_2\left(g\right)}\right]\\ \Delta G^{\text{ο}}=\left[\begin{array}{c}\left[2\text{mol}\left(-371\text{kJ/mol}\right)\right]-\\ \left[2\text{mol}\times \left(-300\text{kJ/mol}\right)+1\text{mol}\times \left(0\text{kJ/mol}\right)\right]\end{array}\right]\\ \Delta G^{\text{ο}}=\left(-742\text{kJ}\right)+\left(600\text{kJ}\right)\\ \Delta G^{\text{ο}}=-\text{142}\text{kJ}\end{array}$

The free energy change of the given reaction in terms of equilibrium constant is given as,

$\Delta G^{\text{ο}}=\Delta G+\text{R}T\ln K$ (1)

The Gibb’s energy change in terms of entropy and enthalpy is given as,

$\Delta G^{\text{ο}}=\Delta H^{\text{ο}}-T\Delta S^{\text{ο}}$ (2)

Compare the equation (1) and (2) We get,

$\Delta G^{\text{ο}}+\text{R}T\ln K=\Delta H^{\text{ο}}-T\Delta S^{\text{ο}}$

Substitute the value is the above formula.

$\begin{array}{c}\Delta H^{\text{ο}}-T\Delta S^{\text{ο}}=\Delta G^{\text{ο}}+\text{R}T\ln K\\ \left[\begin{array}{l}\left(-198\times {10}^3\text{J}\right)-\\ \left(298\text{K}\right)\left(-187\text{J/K}\right)\end{array}\right]=\left[\begin{array}{l}\left(-142\times {10}^3\text{J}\right)+\left(8.314\text{J/K}\cdot \text{mol}\right)\\ \left(298\text{K}\right)\ln K\end{array}\right]\\ \frac{\left[\left(-142274\right)+\left(142000\right)\right]}{\left[\left(8.314\right)\left(298\right)\right]}=\left[\ln K\right]\\ \ln K=\left(-0.1105\right)\end{array}$

The equilibrium constant in terms of partial pressure for the given reaction is written as,

$K=\frac{\left(P_{{\text{SO}}_{\text{3}}\left(g\right)}^2\right)}{\left(P_{{\text{SO}}_2\left(g\right)}^2\cdot P_{{\text{O}}_{\text{2}}\left(g\right)}\right)}$

Where,

• $P_{{\text{SO}}_{\text{3}}\left(g\right)}^2$ is the partial pressure of ${\text{SO}}_{\text{3}}$ gas.
• $P_{{\text{SO}}_2\left(g\right)}^2$ is the partial pressure of ${\text{SO}}_2$ gas.
• $P_{{\text{O}}_{\text{2}\left(g\right)}}$ is the partial pressure of ${\text{O}}_{\text{2}}\left(g\right)$ gas.

Substitute the value of $K$ in the above formula.

$\begin{array}{c}\ln K=\left(-0.1105\right)\\ \ln \frac{\left(P_{{\text{SO}}_{\text{3}}\left(g\right)}^2\right)}{\left(P_{{\text{SO}}_2\left(g\right)}^2\cdot P_{{\text{O}}_{\text{2}}\left(g\right)}\right)}=\left(-0.1105\right)\\ \frac{{\left(2.0\right)}^2}{\left(P_{{\text{SO}}_2\left(g\right)}^2\right)\left(0.50\right)}=e^{\left(-0.1105\right)}\\ P_{{\text{SO}}_2\left(g\right)}^2=\frac{\left(4.0\right)}{\left(0.8953\right)\left(0.50\right)}\end{array}$

Simplify the above equation.

$\begin{array}{l}{P}_{{\text{SO}}_{\text{2}}\left(g\right)}={\left(8.9365\right)}^{1/2}\\ P_{{\text{SO}}_{\text{2}}\left(g\right)}=\underset{\_}{2.98atm}\end{array}$

This reaction is most favored at low temperature. The negative value of enthalpy and entropy clearly indicates that only at low temperature this reaction is spontaneous. Therefore, by assuming standard conditions only at low temperature reaction is most favored.

Conclusion

Conclusion

The equilibrium partial pressure of ${\text{SO}}_{\text{2}}$ in the given mixture is $\underset{\_}{2.98atm}$.

The reaction is most favored at a low temperature.