Chapter 4, Problem 69E

(a)

Interpretation Introduction

To determine: The standard enthalpy change, and entropy change for the given reaction.

Solution: The standard enthalpy change for the given reaction is $-196.6\text{kJ}$ and the standard entropy change for the reaction is $189.6\text{J/K}$ .

Explanation:

Given

The balanced chemical reaction is,

$2{\text{SO}}_2\left(g\right)+{\text{O}}_2\left(g\right)\to 2{\text{SO}}_3\left(g\right)$

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

$\begin{array}{c}\Delta H^0={\displaystyle \sum n\Delta H^0\left(\text{products}\right)}-{\displaystyle \sum m\Delta H^0\left(\text{reactants}\right)}\\ =2\Delta H_{{\text{SO}}_3\left(g\right)}^0-\left(2\Delta H_{{\text{SO}}_2\left(g\right)}^0+\Delta H_{{\text{O}}_2\left(g\right)}^0\right)\end{array}$

Substitute the standard values of $\Delta H_{{\text{SO}}_2\left(g\right)}^0$ , $\Delta H_{{\text{O}}_2\left(g\right)}^0$ and $\Delta H_{{\text{SO}}_3\left(g\right)}^0$ from appendix C in above equation.

$\begin{array}{c}\Delta H^0=2\times \left(-395.2\text{kJ/mol}\right)-\left(2\text{mol}\times \left(-296.9\text{kJ/mol}\right)+1\text{mol}\times 0.0\text{kJ/mol}\right)\\ =-196.6\text{kJ}\end{array}$

Thus, the standard enthalpy change for the given reaction is $-196.6\text{kJ}$ .

Now, calculate the standard entropy change for the given reaction.

The balanced chemical reaction is,

$2{\text{SO}}_2\left(g\right)+{\text{O}}_2\left(g\right)\to 2{\text{SO}}_3\left(g\right)$

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

$\begin{array}{c}\Delta S^0={\displaystyle \sum n\Delta S^0\left(\text{products}\right)}-{\displaystyle \sum m\Delta S^0\left(\text{reactants}\right)}\\ =2\Delta S_{{\text{SO}}_3\left(g\right)}^0-\left(2\Delta S_{{\text{SO}}_2\left(g\right)}^0+\Delta S_{{\text{O}}_2\left(g\right)}^0\right)\end{array}$

Substitute the standard values of $\Delta S_{{\text{SO}}_2\left(g\right)}^0$ , $\Delta S_{{\text{O}}_2\left(g\right)}^0$ and $\Delta S_{{\text{SO}}_3\left(g\right)}^0$ from appendix C in the above equation.

$\begin{array}{c}\Delta S^0=2\times 256.2\text{J/mol}\cdot \text{K}-\left(2\text{mol}\times 248.5\text{J/mol}\cdot \text{K}+1\text{mol}\times 205.0\text{J/mol}\cdot \text{K}\right)\\ =-189.6\text{J/K}\end{array}$

Thus, the standard entropy change for the given reaction is $-189.6\text{J/K}$ .

Conclusion:

The standard enthalpy change for the given reaction is $-196.6\text{kJ}$ and the standard entropy change for the reaction is $189.6\text{J/K}$ .

(b)

Interpretation Introduction

To determine: The standard free energy change for the given reaction.