Chapter 3, Problem 3.12P

(a)

Interpretation Introduction

Interpretation:

The change in standard reaction entropy of the reaction given has to be calculated.

Explanation of Solution

Given reaction: $\text{2}{\text{H}}_{\text{2}}\left(\text{g}\right)\text{+}{\text{O}}_{\text{2}}\left(\text{g}\right)\to \text{2}{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$

Given: ${\text{C}}_{\text{p,m}}^{\text{linear}}\text{=}\frac{\text{7}}{\text{2}}\text{R}\text{;}{\text{C}}_{\text{p,m}}^{\text{non-linear}}\text{=4R}$

At temperature $\text{283}\text{K}$, the reaction becomes $\text{2}{\text{H}}_{\text{2}}\left(\text{g}\right)\text{+}{\text{O}}_{\text{2}}\left(\text{g}\right)\stackrel{{\text{ΔS}}_{\text{283}}^{\text{o}}}{\to }\text{2}{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$

At temperature $\text{273}\text{K}$, the reaction becomes $\text{2}{\text{H}}_{\text{2}}\left(\text{g}\right)\text{+}{\text{O}}_{\text{2}}\left(\text{g}\right)\stackrel{{\text{ΔS}}_{\text{273}}^{\text{o}}}{\to }\text{2}{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$

Entropy calculation for each reactant and products in both reactions are as follows,

$\begin{array}{l}{\text{ΔS}}_{{\text{H}}_{\text{2}}}\text{=}\text{n}{\text{C}}_{\text{p,m}}^{\text{linear}}\text{ln}\frac{\text{273}}{\text{283}}\text{=}\text{2}\left(\frac{\text{7}}{\text{2}}\text{R}\right)\text{ln}\frac{\text{273}}{\text{283}}\\ \\ {\text{ΔS}}_{{\text{O}}_{\text{2}}}\text{=}\text{n}{\text{C}}_{\text{p,m}}^{\text{linear}}\text{ln}\frac{\text{273}}{\text{283}}\text{=}\left(\frac{\text{7}}{\text{2}}\text{R}\right)\text{ln}\frac{\text{273}}{\text{283}}\\ \\ {\text{ΔS}}_{{\text{H}}_{\text{2}}\text{O}}\text{=}\text{n}{\text{C}}_{\text{p,m}}^{\text{non-linear}}\text{ln}\frac{\text{273}}{\text{283}}\text{=}\text{2}\left(\text{4R}\right)\text{ln}\frac{\text{273}}{\text{283}}\end{array}$

Change in entropy obtained as follows,

$\begin{array}{l}{\text{ΔS}}_{\text{283}}^{\text{o}}\text{-}{\text{ΔS}}_{\text{273}}^{\text{o}}\text{=}{\text{ΔS}}_{{\text{H}}_{\text{2}}}\text{+}{\text{ΔS}}_{{\text{O}}_{\text{2}}}\text{+}{\text{ΔS}}_{{\text{H}}_{\text{2}}\text{O}}\\ \\ \left(\text{or}\right)\\ \\ {\text{Δ}}_{\text{r}}{\text{C}}_{\text{p}}\text{ln}\frac{\text{283}}{\text{273}}\text{=}{\text{ΔS}}_{{\text{H}}_{\text{2}}}\text{+}{\text{ΔS}}_{{\text{O}}_{\text{2}}}\text{+}{\text{ΔS}}_{{\text{H}}_{\text{2}}\text{O}}\end{array}$

According to Kirchhoff’s law,

$\begin{array}{l}{\text{Δ}}_{\text{r}}{\text{C}}_{\text{p}}\text{=}{\text{ΔS}}_{{\text{H}}_{\text{2}}}\text{+}{\text{ΔS}}_{{\text{O}}_{\text{2}}}\text{+}{\text{ΔS}}_{{\text{H}}_{\text{2}}\text{O}}\\ \\ \text{=}\text{2}\left(\text{4R}\right)\text{-}\text{2}\left(\frac{\text{7}}{\text{2}}\text{R}\right)\text{-}\left(\frac{\text{7}}{\text{2}}\text{R}\right)\\ \\ \text{=}\text{-}\text{2}\text{.5}\text{R}\\ \\ \text{=}-20.785J{K}^{-1}mo{l}^{-1}\end{array}$

Change in entropy becomes,

$\begin{array}{l}{\text{ΔS}}_{\text{283}}^{\text{o}}\text{-}{\text{ΔS}}_{\text{273}}^{\text{o}}\text{=-20}\text{.785}\text{ln}\frac{\text{283}}{\text{273}}\\ \\ \text{=}-0.75J{K}^{-1}mo{l}^{-1}\end{array}$

Therefore, the change in standard reaction entropy of the reaction obtained is $-0.75J{K}^{-1}mo{l}^{-1}$

(b)

Interpretation Introduction

Interpretation:

The change in standard reaction entropy of the reaction given has to be calculated.

Explanation of Solution

Given reaction: ${\text{CH}}_4\left(\text{g}\right)\text{+}2{\text{O}}_{\text{2}}\left(\text{g}\right)\to {\text{CO}}_2\left(\text{g}\right)\text{+}\text{2}{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$

Given: ${\text{C}}_{\text{p,m}}^{\text{linear}}\text{=}\frac{\text{7}}{\text{2}}\text{R}\text{;}{\text{C}}_{\text{p,m}}^{\text{non-linear}}\text{=4R}$

At temperature $\text{283}\text{K}$, the reaction becomes ${\text{CH}}_4\left(\text{g}\right)\text{+}2{\text{O}}_{\text{2}}\left(\text{g}\right)\stackrel{{\text{ΔS}}_{\text{283}}^{\text{o}}}{\to }{\text{CO}}_2\left(\text{g}\right)\text{+}\text{2}{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$

At temperature $\text{273}\text{K}$, the reaction becomes ${\text{CH}}_4\left(\text{g}\right)\text{+}2{\text{O}}_{\text{2}}\left(\text{g}\right)\stackrel{{\text{ΔS}}_{\text{273}}^{\text{o}}}{\to }{\text{CO}}_2\left(\text{g}\right)\text{+}\text{2}{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$

Entropy calculation for each reactant and products in both reactions are as follows,

$\begin{array}{l}{\text{ΔS}}_{{\text{CH}}_{\text{4}}}\text{=}\text{n}{\text{C}}_{\text{p,m}}^{\text{non-linear}}\text{ln}\frac{\text{273}}{\text{283}}\text{=1}\left(\text{4R}\right)\text{ln}\frac{\text{273}}{\text{283}}\\ \\ {\text{ΔS}}_{{\text{O}}_{\text{2}}}\text{=}\text{n}{\text{C}}_{\text{p,m}}^{\text{linear}}\text{ln}\frac{\text{273}}{\text{283}}\text{=}\text{2}\left(\frac{\text{7}}{\text{2}}\text{R}\right)\text{ln}\frac{\text{273}}{\text{283}}\\ \\ {\text{ΔS}}_{{\text{CO}}_{\text{2}}}\text{=}\text{n}{\text{C}}_{\text{p,m}}^{\text{linear}}\text{ln}\frac{\text{273}}{\text{283}}\text{=}\left(\frac{\text{7}}{\text{2}}\text{R}\right)\text{ln}\frac{\text{273}}{\text{283}}\\ \\ {\text{ΔS}}_{{\text{H}}_{\text{2}}\text{O}}\text{=}\text{n}{\text{C}}_{\text{p,m}}^{\text{non-linear}}\text{ln}\frac{\text{273}}{\text{283}}\text{=}\text{2}\left(\text{4R}\right)\text{ln}\frac{\text{273}}{\text{283}}\end{array}$

Change in entropy obtained as follows,

$\begin{array}{l}{\text{ΔS}}_{\text{283}}^{\text{o}}\text{-}{\text{ΔS}}_{\text{273}}^{\text{o}}\text{=}{\text{ΔS}}_{{\text{CH}}_{\text{4}}}\text{+}{\text{ΔS}}_{{\text{O}}_{\text{2}}}\text{+}{\text{ΔS}}_{{\text{CO}}_{\text{2}}}\text{+}{\text{ΔS}}_{{\text{H}}_{\text{2}}\text{O}}\\ \\ \left(\text{or}\right)\\ \\ {\text{Δ}}_{\text{r}}{\text{C}}_{\text{p}}\text{ln}\frac{\text{283}}{\text{273}}\text{=}{\text{ΔS}}_{{\text{CH}}_{\text{4}}}\text{+}{\text{ΔS}}_{{\text{O}}_{\text{2}}}\text{+}{\text{ΔS}}_{{\text{CO}}_{\text{2}}}\text{+}{\text{ΔS}}_{{\text{H}}_{\text{2}}\text{O}}\end{array}$

According to Kirchhoff’s law,

$\begin{array}{l}{\text{Δ}}_{\text{r}}{\text{C}}_{\text{p}}\text{=}{\text{ΔS}}_{{\text{CH}}_{\text{4}}}\text{+}{\text{ΔS}}_{{\text{O}}_{\text{2}}}\text{+}{\text{ΔS}}_{{\text{CO}}_{\text{2}}}\text{+}{\text{ΔS}}_{{\text{H}}_{\text{2}}\text{O}}\\ \\ \text{=}\text{1}\left(\text{4R}\right)\text{-}\text{2}\left(\frac{\text{7}}{\text{2}}\text{R}\right)\text{-}\left(\frac{\text{7}}{\text{2}}\text{R}\right)\text{+}\text{2}\left(\text{4R}\right)\\ \\ \text{=}\left(\text{12R}\right)\text{-}\text{3}\left(\frac{\text{7}}{\text{2}}\text{R}\right)\text{=}\text{R}\left[\text{12-}\frac{\text{21}}{\text{2}}\right]\\ \\ \text{=}\text{1}\text{.5}\text{R}\\ \\ \text{=12}\text{.47}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

Change in entropy becomes,

$\begin{array}{l}{\text{ΔS}}_{\text{283}}^{\text{o}}\text{-}{\text{ΔS}}_{\text{273}}^{\text{o}}\text{=12}\text{.47}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{ln}\frac{\text{283}}{\text{273}}\\ \\ \text{=}0.4486J{K}^{-1}mo{l}^{-1}\end{array}$

Therefore, the change in standard reaction entropy of the reaction obtained is $0.4486J{K}^{-1}mo{l}^{-1}$