Chapter 12, Problem 12.118PAE

(a)

Interpretation Introduction

To determine: Use tabulated data to calculate $\Delta H^0$ for each reaction.

Explanation of Solution

For reaction $S\left(s\right)+O_2\left(g\right)\stackrel{}{\rightleftharpoons }S{O}_2\left(g\right)$

Enthalpy change $\left(\Delta H^0\right)=\Delta H_f^0\left[\text{Product}\right]-\Delta H_f^0\left[\text{Reactants}\right]$

$\begin{array}{l}=\Delta H^0\left[S{O}_2\left(g\right)\right]-\left\{\Delta H^0\left[O_2\left(g\right)\right]+\Delta H^0\left[S\left(s\right)\right]\right\}\\ =-296.8\text{kJ/mol}-\left\{0+0\right\}\\ =-296.8\text{kJ/mol}\end{array}$

For reaction

${\text{2SO}}_2\left(g\right)+O_2\left(g\right)\stackrel{}{\rightleftharpoons }2S{O}_3\left(g\right)$

$\begin{array}{c}\Delta H^0=2\Delta H^0\left[S{O}_3\left(g\right)\right]-2\Delta H^0\left[S{O}_2\left(g\right)\right]-\Delta H^0\left[O_2\left(g\right)\right]\\ =2\times \left(-395.6\text{kJ/mol}\right)-2\times \left[0^{-296.8}\right]-0\\ =\left(-791.2+593.6\right)\text{kJ/mol}\\ =-197.6\text{kJ/mol}\end{array}$

For reaction, $S{O}_3\left(g\right)+H_{2}O\left(l\right)\stackrel{}{\rightleftharpoons }{H}_{2}S{O}_4\left(l\right)$

$\begin{array}{c}\Delta H^0=\Delta H^0\left[H_{2}S{O}_4\left(l\right)\right]-\Delta H^0\left[S{O}_3\left(g\right)\right]-\Delta H^0\left[H_{2}O\left(l\right)\right]\\ =-814\text{kJ/mol}-\left(-285.8\text{kJ/mol}\right)+\left(-395.6\text{kJ/mol}\right)\\ =-923.8\text{kJ/mol}\end{array}$

(b)

Interpretation Introduction

To determine:

Which reactions are exothermic which are endothermic?

Explanation of Solution

$\Delta H=\Delta U+P\Delta V$

$\Delta H$ For the given reaction a $aA\left(g\right)+bB\left(g\right)\stackrel{}{\rightleftharpoons }cC\left(s\right)$

$\Delta H^0=\Delta H^0\left[\text{Product}\right]-\Delta H^0\left[\text{Reactant}\right]$

If $\Delta H^0<0$, there reaction is exothermic$\Delta H^0>0$, the reaction is endothermic.

Here,

For reaction $S+O_2\stackrel{}{\rightleftharpoons }S{O}_2$

Enthalpy change $\left(\Delta H^0\right)=-296.8kj/mol$

For reaction $2S{O}_2+O_2\stackrel{}{\rightleftharpoons }2S{O}_3$

Enthalpy change $\left(\Delta H^0\right)=-197.6kj/mol$

For reaction $S{O}_3+H_{2}O\stackrel{}{\rightleftharpoons }{H}_{2}S{O}_4$

Enthalpy change $\left(\Delta H^0\right)=-923.8kj/mol$

In all reaction, $\Delta H^0<0$

So, all reactions are exothermic.

(c)

Interpretation Introduction

To determine:

In which of the reaction does entropy increases. In which does it decrease? In which does it lay about the same?

Explanation of Solution

Entropy change of a substance is defined as the measure of randomness.

If number of gaseous atoms in reactant is more than in products, the entropy of the system increases.

If number of gaseous atoms in reactant is less than in products, the entropy of the system decreases.

For reaction.

$S\left(s\right)+O_2\left(g\right)\stackrel{}{\rightleftharpoons }S{O}_2\left(g\right)$

Number of gaseous atoms in reactant $=1$

Number of gaseous atoms in product $=1$

Here, number for gaseous atoms remain the same in reactant and product so, entropy will not change but remains same.

For reaction.

$2S{O}_2\left(g\right)+O_2\left(g\right)\stackrel{}{\rightleftharpoons }2S{O}_3\left(g\right)$

Number of gaseous atom in reactant $=3$

Number of gaseous atom in product $=2$

So, entropy of the reaction decreases.

For reaction,

$S{O}_3\left(g\right)+H_{2}O\left(l\right)\stackrel{}{\rightleftharpoons }{H}_{2}S{O}_4\left(l\right)$

Number of gaseous atom in reactant $=1$

Number of gaseous atom in product $=0$

So, entropy of the reaction decreases.

(d)

Interpretation Introduction

To determine:

For which reaction do low temperatures favor formation of products?

Explanation of Solution

For first reaction.

$\Delta G=\Delta H-T\Delta S$

$\Delta G<0$ For all value of temperature so, the reaction is always spontaneous in all case.

For $2^{nd}{\text{ and 3}}^{rd}$ reaction.

$\Delta S<0$

$\Delta G=\Delta H-T\Delta S$

For low temperature,

$\Delta G<0$

For high temperature,

$\Delta G>0$

So, here low temperature favors the formation of products for last two reactions.

$\begin{array}{l}2S{O}_2\left(g\right)+O_2\left(r\right)\stackrel{}{\rightleftharpoons }2S{O}_3\left(g\right)\\ S{O}_3\left(g\right)+H_{2}O\left(l\right)\stackrel{}{\rightleftharpoons }{H}_{2}S{O}_4\left(l\right)\end{array}$

Conclusion

$\Delta H^0=\Delta H^0\left[\text{Product}\right]-\Delta H^0\left[\text{Reactant}\right]$

$\Delta G=\Delta H-T\Delta S$