Chapter 12, Problem 12.40QA

Interpretation Introduction

To calculate:

The entropy change for the given reactions using the standard entropies.

Answer to Problem 12.40QA

Solution:

a) $H_{2}S\left(g\right)+\frac{3}{2} O_2\left(g\right) \to H_{2}O\left(g\right) + {SO}_2\left(g\right)$; $∆S_{rxn}^o=-76.1 J/K$

b) ${2 SO}_2\left(g\right)+O_2\left(g\right)\to {2SO}_3\left(g\right)$; $∆S_{rxn}^o=-187.8 J/K$

c) ${SO}_3\left(g\right)+ H_{2}O\left(l\right) \to H_2{SO}_4\left(aq\right)$;  $∆S_{rxn}^o=- 306.6 J/K$

d) $S\left(g\right)+ O_2\left(g\right) \to {SO}_2\left(g\right)$;  $∆S_{rxn}^o=-124.6 J/K$

Explanation of Solution

1) Concept:

$∆S_{rxn}^o$(entropy change for the reaction) can be calculated from the difference between the standard molar entropies of $n_{reactant}$ moles of reactants and the standard molar entropies $n_{product}$ moles of products.

2) Formula:

$∆S_{rxn}^o= \sum n_{product} { S}_{product}^o- \sum n_{reactant} { S}_{reactant}^o$

3) Given:

Substance	S0 [J/(mol.K)]
$H_{2}S\left(g\right)$	205.6
$O_2\left(g\right)$	205.0
$H_{2}O\left(g\right)$	188.8
${SO}_2\left(g\right)$	248.2
${SO}_3\left(g\right)$	256.8
$H_{2}O\left(l\right)$	69.9
$H_2{SO}_4\left(aq\right)$	20.1
$S\left(g\right)$	167.8

4) Calculations:

a) $H_{2}S\left(g\right)+\frac{3}{2} O_2\left(g\right) \to H_{2}O\left(g\right) + {SO}_2\left(g\right)$

The values of standard molar entropies are,

$S^o for H_{2}S\left(g\right)=205.6 J/mol.K$

$S^o for O_2\left(g\right)=205.0 J/mol.K$

$S^o for H_{2}O\left(g\right)=188.8 J/mol.K$

$S^o for {SO}_2\left(g\right)=248.2 J/mol.K$

Therefore,

$∆S_{rxn}^o= \left[\left(n_{H_{2}O\left(g\right)} { S}_{H_{2}O\left(g\right)}^o\right)+ \left(n_{{SO}_2\left(g\right)} { S}_{{SO}_2\left(g\right)}^o\right)\right]-\left[\left(n_{ H_{2}S\left(g\right)} { S}_{ H_{2}S\left(g\right)}^o\right)+\left(n_{ O_2\left(g\right)} { S}_{ O_2\left(g\right)}^o\right)\right]$

$∆S_{rxn}^o=\left[\left(1 mol\times 188.8 J/mol.K\right)+\left(1 mol\times 248.2 J/mol.K\right)\right]$

$-\left[\left(1 mol\times 205.6 J/mol.K\right)+ \left(\frac{3}{2}mol\times 205.0 J/mol.K\right)\right]$

$∆S_{rxn}^o=437 J/K-513.1 J/K$

$∆S_{rxn}^o=-76.1 J/K$

The value of $∆{\mathrm{S}}_{\mathrm{r}\mathrm{x}\mathrm{n}}^{\mathrm{o}}$ is $-76.1\mathrm{ }\mathrm{J}/\mathrm{K}.$

b) ${2 SO}_2\left(g\right)+O_2\left(g\right)\to {2SO}_3\left(g\right)$

The values of standard molar entropies are,

$S^o for {SO}_2\left(g\right)=248.2 J/mol.K$

$S^o for O_2\left(g\right)=205.0 J/mol.K$

$S^o for {SO}_3\left(g\right)=256.8 J/mol.K$

$∆S_{rxn}^o= \left[\left(n_{{SO}_3\left(g\right)} { S}_{{SO}_3\left(g\right)}^o\right)\right]- \left[\left(n_{ {SO}_2\left(g\right)} { S}_{ {SO}_2\left(g\right)}^o\right)+\left(n_{ O_2\left(g\right)} { S}_{ O_2\left(g\right)}^o\right)\right]$

$∆S_{rxn}^o=\left[\left(2 mol\times 256.8 J/mol.K\right)\right]-\left[\left(2 mol\times 248.2 J/mol.K\right)+\left(1 mol\times 205.0 J/mol.K\right)\right]$

$∆S_{rxn}^o=513.6 J/K-701.4 J/K$

$∆S_{rxn}^o=-187.8 J/K$

The value of $∆{\mathrm{S}}_{\mathrm{r}\mathrm{x}\mathrm{n}}^{\mathrm{o}}$ is $-187.8\mathrm{ }\mathrm{J}/\mathrm{K}.$

c) ${SO}_3\left(g\right)+ H_{2}O\left(l\right) \to H_2{SO}_4\left(aq\right)$

The values of standard molar entropies are,

$S^o for {SO}_3\left(g\right)=256.8 J/mol.K$

$S^o for H_{2}O\left(l\right)= 69.9 J/mol.K$

$S^o for H_2{SO}_4\left(aq\right)=20.1 J/mol.K$

$∆S_{rxn}^o= \left[\left(n_{ H_2{SO}_4\left(aq\right)} { S}_{ H_2{SO}_4\left(aq\right)}^o\right)\right]- \left[\left(n_{{SO}_3\left(g\right)} { S}_{{SO}_3\left(g\right)}^o\right)+ \left(n_{H_{2}O\left(g\right)} { S}_{H_{2}O\left(g\right)}^o\right)\right]$

$∆S_{rxn}^o=\left[\left(1 mol\times 20.1 J/mol.K\right)\right]-\left[\left(1 mol\times 256.8 J/mol.K\right)+\left(1 mol\times 69.9 J/mol.K\right)\right]$

$∆S_{rxn}^o=20.1J/K-326.7 J/K$

$∆S_{rxn}^o=- 306.6 J/K$

The value of $∆{\mathrm{S}}_{\mathrm{r}\mathrm{x}\mathrm{n}}^{\mathrm{o}}$ is $-306.6\mathrm{ }\mathrm{J}/\mathrm{K}.$

d) $S\left(g\right)+ O_2\left(g\right) \to {SO}_2\left(g\right)$

The values of standard molar entropies are,

$S^o for S\left(g\right)= 167.8 J/mol.K$

$S^o for O_2\left(g\right)=205.0 J/mol.K$

$S^o for {SO}_2\left(g\right)=248.2 J/mol.K$

$∆S_{rxn}^o= \left[\left(n_{ {SO}_2\left(g\right)} { S}_{ {SO}_2\left(g\right)}^o\right)\right]- \left[\left(n_{S\left(g\right)} { S}_{S\left(g\right)}^o\right)+ \left(n_{O_2\left(g\right)} { S}_{O_2\left(g\right)}^o\right)\right]$

$∆S_{rxn}^o=\left[\left(1 mol\times 248.2 J/mol.K\right)\right]-\left[\left(1 mol\times 167.8 J/mol.K\right)+\left(1 mol\times 205.0J/mol.K\right)\right]$

$∆S_{rxn}^o=248.2 J/K-372.8 J/K$

$∆S_{rxn}^o=-124.6 J/K$

The value of $∆{\mathrm{S}}_{\mathrm{r}\mathrm{x}\mathrm{n}}^{\mathrm{o}}$ is $-124.6\mathrm{ }\mathrm{J}/\mathrm{K}.$

Conclusion:

$∆S_{rxn}^o$(entropy change for the reaction) can be calculated from the difference between the standard molar entropies of $n_{reactant}$ moles of reactants and the standard molar entropies $n_{product}$ moles of products,

$∆S_{rxn}^o= \sum n_{product} { S}_{product}^o- \sum n_{reactant} { S}_{reactant}^o$