Chapter 10, Problem 10.35PAE

Use tabulated thermodynamic data to calculate the standard entropy change of each of the reactions listed below.

(a) $\text{Fe}\left(\text{s}\right)+\text{2HCl}\left(\text{g}\right)\to {\text{FeCl}}_{\text{2}}\left(\text{s}\right)+{\text{H}}_{\text{2}}\left(\text{g}\right)$

(b) ${\text{3NO}}_{\text{2}}\left(\text{g}\right)+H_{2}O(\mathcal{l})\to {\text{2HNO}}_{\text{3}}(\mathcal{l})+\text{NO}\left(\text{g}\right)$

(c) $\text{2K}\left(\text{s}\right)+{\text{Cl}}_{\text{2}}\left(\text{g}\right)\to \text{2KCl}\left(\text{s}\right)$

(d) $C{l}_2\left(g\right)+\text{2NO}\left(\text{g}\right)\to \text{2NOCl}\left(\text{g}\right)$

(e) ${\text{SiCl}}_{\text{4}}\left(\text{g}\right)\to \text{Si}\left(\text{s}\right)+{\text{2Cl}}_{\text{2}}\left(\text{g}\right)$

Interpretation Introduction

To determine: Standard entropy change for $\text{Fe}\left(\text{s}\right)\text{+2HCl}\left(\text{g}\right)\to {\text{FeCl}}_2\left(\text{s}\right){\text{+H}}_{\text{2}}\left(g\right)$.

Explanation of Solution

Steps to solve the problem:

When we are asked to calculate the standard entropy change for a reaction, first we should look up values for standard molar entropy on the reactants and products involve and use them in the given equation.

We should be careful about the following

1. Watch the state of the substances (in the given case all are gases)
2. Ensure that the stoichiometric coefficients are included in our calculations.
3. The following table lists the standard molar entropy $S^o$ of the reactants and products

Species	Standard molar entropy $S^o$ in $\text{J}{\text{mol}}^{\text{-1}}{\text{K}}^{\text{-1}}$
$\text{Fe}\left(\text{s}\right)$	27.3
$\text{HCl}\left(\text{g}\right)$	186.8
${\text{FeCl}}_2\left(\text{s}\right)$	117.9
${\text{H}}_{\text{2}}\left(g\right)$	130.6
${\text{NO}}_2\left(g\right)$	240.0
${\text{H}}_2\text{O}\left(l\right)$	69.91
${\text{HNO}}_{\text{3}}\left(l\right)$	155.6
$\text{NO}\left(g\right)$	210.7
$\text{K}\left(\text{s}\right)$	63.6
${\text{Cl}}_2\left(g\right)$	223.0
$\text{KCl}\left(\text{s}\right)$	82.6
$\text{NOCl}\left(g\right)$	264
${\text{SiCl}}_4\left(g\right)$	330.6
$\text{Si}\left(\text{s}\right)$	18.8

(a)$\text{Fe}\left(\text{s}\right)\text{+2HCl}\left(\text{g}\right)\to {\text{FeCl}}_2\left(\text{s}\right){\text{+H}}_{\text{2}}\left(g\right)$

Use the formula $\Delta S^o={\displaystyle \sum _i{\nu }_i{S}^o}{\left(product\right)}_i-{\displaystyle \sum _j{\nu }_j{S}^o}{\left(reactant\right)}_j$ to calculate the entropy change of the reaction. $\begin{array}{c}\Delta S^o={\displaystyle \sum _i{\nu }_i{S}^o}{\left(product\right)}_i-{\displaystyle \sum _j{\nu }_j{S}^o}{\left(reactant\right)}_j\\ =\left({\text{S}}^o\left[{\text{FeCl}}_2\left(\text{s}\right)\right]+{\text{S}}^o\left[{\text{H}}_{\text{2}}\left(g\right)\right]\right)\\ -\left(S^o\left[\text{Fe}\left(\text{s}\right)\right]{\text{+2S}}^o\left[\text{HCl}\left(\text{g}\right)\right]\right)\end{array}$

$\begin{array}{c}\Delta S^o=\left(117.9+130.6\right)-\left(27.3\text{+2}\left(186.8\right)\right)\\ =248.5-400.9\\ =-152.4\text{J}{\text{mol}}^{-1}{\text{K}}^{-1}\end{array}$

Interpretation Introduction

To determine: Standard entropy change for ${\text{3NO}}_2\left(g\right){\text{+H}}_2\text{O}\left(l\right)\to 2{\text{HNO}}_{\text{3}}\left(l\right)\text{+NO}\left(g\right)$.

Explanation of Solution

$\begin{array}{c}\Delta S^o={\displaystyle \sum _i{\nu }_i{S}^o}{\left(product\right)}_i-{\displaystyle \sum _j{\nu }_j{S}^o}{\left(reactant\right)}_j\\ =\left({\text{2S}}^o\left[{\text{HNO}}_{\text{3}}\left(l\right)\right]{\text{+S}}^o\left[\text{NO}\left(g\right)\right]\right)\\ -\left(3S^o\left[{\text{NO}}_2\left(g\right)\right]{\text{+S}}^o\left[{\text{H}}_2\text{O}\left(l\right)\right]\right)\end{array}$

$\begin{array}{c}\Delta S^o=\left(\text{2}\times \text{155}\text{.6+210}\text{.7}\right)-\left(3\times 240.0\text{+69}\text{.91}\right)\\ =\left(311.2+210.7\right)-\left(720+69.91\right)\\ =521.9-789.91\\ =-268.01\text{J}{\text{mol}}^{-1}{\text{K}}^{-1}\end{array}$

Interpretation Introduction

To determine: Standard entropy change for $\text{2K}\left(\text{s}\right){\text{+Cl}}_2\left(g\right)\to 2\text{KCl}\left(\text{s}\right)$.

Explanation of Solution

$\text{2K}\left(\text{s}\right){\text{+Cl}}_2\left(g\right)\to 2\text{KCl}\left(\text{s}\right)$

$\begin{array}{c}\Delta S^o={\displaystyle \sum _i{\nu }_i{S}^o}{\left(product\right)}_i-{\displaystyle \sum _j{\nu }_j{S}^o}{\left(reactant\right)}_j\\ ={\text{2S}}^o\left[\text{KCl}\left(\text{s}\right)\right]-\left(2S^o\left[\text{K}\left(\text{s}\right)\right]{\text{+S}}^o\left[{\text{Cl}}_2\left(g\right)\right]\right)\end{array}$

$\begin{array}{c}\Delta S^o=\text{2}\times 82.6-\left(2\times 63.6+\text{223}\text{.0}\right)\\ =165.2-\left(127.2+223.0\right)\\ =165.2-350.2\\ =-185.0\text{J}{\text{mol}}^{-1}{\text{K}}^{-1}\end{array}$

Interpretation Introduction

To determine: Standard entropy change for ${\text{Cl}}_{\text{2}}\left(g\right)\text{+2NO}\left(g\right)\to 2\text{NOCl}\left(g\right)$.

Explanation of Solution

${\text{Cl}}_{\text{2}}\left(g\right)\text{+2NO}\left(g\right)\to 2\text{NOCl}\left(g\right)$

$\begin{array}{c}\Delta S^o={\displaystyle \sum _i{\nu }_i{S}^o}{\left(product\right)}_i-{\displaystyle \sum _j{\nu }_j{S}^o}{\left(reactant\right)}_j\\ ={\text{2S}}^o\left[\text{NOCl}\left(g\right)\right]-\left(S^o\left[{\text{Cl}}_{\text{2}}\left(g\right)\right]{\text{+2S}}^o\left[\text{NO}\left(g\right)\right]\right)\end{array}$

$\begin{array}{c}\Delta S^o=\text{2}\times 264.0-\left(\text{223}\text{.0}+2\times 210.7\right)\\ =528.0-\left(223.0+421.4\right)\\ =528.0-644.4\\ =-116.4\text{J}{\text{mol}}^{-1}{\text{K}}^{-1}\end{array}$

Interpretation Introduction

To determine: Standard entropy change for ${\text{SiCl}}_4\left(g\right)\to \text{Si}\left(\text{s}\right)+{\text{2Cl}}_2\left(g\right)$.

Explanation of Solution

${\text{SiCl}}_4\left(g\right)\to \text{Si}\left(\text{s}\right)+{\text{2Cl}}_2\left(g\right)$

$\begin{array}{c}\Delta S^o={\displaystyle \sum _i{\nu }_i{S}^o}{\left(product\right)}_i-{\displaystyle \sum _j{\nu }_j{S}^o}{\left(reactant\right)}_j\\ =\left({\text{S}}^o\left[\text{Si}\left(\text{s}\right)\right]{\text{+2S}}^o\left[{\text{Cl}}_2\left(g\right)\right]\right)-S^o\left[{\text{SiCl}}_4\left(g\right)\right]\end{array}$

$\begin{array}{c}\Delta S^o=\left(18.8+\text{2}\times 223.0\right)-330.6\\ =\left(18.8+446.0\right)-330.6\\ =464.8-330.6\\ =134.2\text{J}{\text{mol}}^{-1}{\text{K}}^{-1}\end{array}$

Conclusion

Hence, all the entropy has been determined