Chapter 17, Problem 17.12QP

Using the data in Appendix 2, calculate the standard entropy changes for the following reactions at 25°C:

1. (a) ${\text{H}}_2(g)+\text{CuO}(s)\stackrel{}{\to }\text{Cu}(s)+{\text{H}}_2\text{O}(g)$
2. (b) $\text{2Al(}s)+3\text{ZnO}(s)\stackrel{}{\to }{\text{Al}}_2{\text{O}}_3(s)+3\text{Zn}(s)$
3. (c) ${\text{CH}}_4\text{(}g)+2{\text{O}}_2(g)\stackrel{}{\to }{\text{CO}}_2(g)+2{\text{H}}_2\text{O}(l)$

Interpretation Introduction

Interpretation: The standard entropy change ${\text{(ΔS}}^{\text{°}}{}_{\text{reaction}}\text{)}$ in the given set of reactions has to be calculated.

Concept introduction: Entropy is the measure of randomness in the system.   Standard entropy change in a reaction is the difference in entropy of the products and reactants.  ${\text{(ΔS}}^{\text{°}}{}_{\text{reaction}}\text{)}$  can be calculated by the following equation.

$\begin{array}{l}{\text{(ΔS}}^{\text{°}}{}_{\text{reaction}}{\text{) = S}}^{\text{°}}{}_{\text{Products}}-{\text{ S}}^{\text{°}}{}_{\text{reactants}}\\ \end{array}$

Where

  ${\text{S}}^{\text{°}}{}_{\text{reactants}}$ is the standard entropy of the reactants

  ${\text{S}}^{\text{°}}{}_{\text{Products}}$ is the standard entropy of the products

Answer to Problem 17.12QP

The ${\text{ΔS}}^{\text{°}}{}_{\text{reaction}}$ for the reaction (a)$\text{= 47}{\text{.5 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}$

Explanation of Solution

To record the given data

Given,

$\begin{array}{l}{\text{S}}^{\text{°}}\text{(Cu)}=\text{33}{\text{.3 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ \text{S}°\left[H_{2}O(g)\right]=\text{188}{\text{.7 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ {\text{S}}^{\text{°}}\text{(}{H}_2\text{) =131}{\text{.0J K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ {\text{S}}^{\text{°}}\text{(CuO)}=\text{43}{\text{.5 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

To calculate ${\text{ΔS}}^{\text{°}}{}_{\text{reaction}}$

The ${\text{ΔS}}^{\text{°}}{}_{\text{reaction}}$ is the difference in entropy of the reactants and products.  Which is calculated by plugging in the values of standard entropy of the products and reactants  the given equation.

${\text{(ΔS}}^{\text{°}}{}_{\text{reaction}}{\text{) = S}}^{\text{°}}{}_{\text{produdcts}}{\text{- S}}^{\text{°}}{}_{\text{reacttants}}$

                 $\begin{array}{l}{\text{= S}}^{\text{°}}\text{(Cu)+S}°\left[H_{2}O(g)\right]\text{- }\left[{\text{S}}^{\text{°}}\text{(}{H}_2{\text{)+S}}^{\text{°}}\text{(CuO)}\right]\\ =\text{(1)(33}{\text{.3 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{)+(1)(188}{\text{.7 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{)-}\left[\text{(1)(131}{\text{.0J K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{)+(1)(43}{\text{.5 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{)}\right]\\ =47.5{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

Interpretation Introduction

Interpretation: The standard entropy change ${\text{(ΔS}}^{\text{°}}{}_{\text{reaction}}\text{)}$ in the given set of reactions has to be calculated.

Concept introduction: Entropy is the measure of randomness in the system.   Standard entropy change in a reaction is the difference in entropy of the products and reactants.  ${\text{(ΔS}}^{\text{°}}{}_{\text{reaction}}\text{)}$  can be calculated by the following equation.

$\begin{array}{l}{\text{(ΔS}}^{\text{°}}{}_{\text{reaction}}{\text{) = S}}^{\text{°}}{}_{\text{Products}}-{\text{ S}}^{\text{°}}{}_{\text{reactants}}\\ \end{array}$

Where

  ${\text{S}}^{\text{°}}{}_{\text{reactants}}$ is the standard entropy of the reactants

  ${\text{S}}^{\text{°}}{}_{\text{Products}}$ is the standard entropy of the products

Answer to Problem 17.12QP

The ${\text{ΔS}}^{\text{°}}{}_{\text{reaction}}$ for the reaction (b)$\text{= -12}{\text{.5 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}$

Explanation of Solution

To record the given data

Given,

$\begin{array}{l}{\text{S}}^{\text{°}}{\text{(Al}}_2{\text{O}}_3\text{)}=\text{50}{\text{.99 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ {\text{S}}^{\text{°}}\text{(Zn)}=\text{41}{\text{.6JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ {\text{S}}^{\text{°}}\text{(Al) = 28}{\text{.3 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ {\text{S}}^{\text{°}}\text{(ZnO)}=\text{43}{\text{.9 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

To calculate ${\text{ΔS}}^{\text{°}}{}_{\text{reaction}}$

The ${\text{ΔS}}^{\text{°}}{}_{\text{reaction}}$ is the difference in entropy of the reactants and products.  Which is calculated by plugging in the values of standard entropy of the reactants and products  the given equation.

 ${\text{(ΔS}}^{\text{°}}{}_{\text{reaction}}{\text{) = S}}^{\text{°}}{}_{\text{produdcts}}{\text{- S}}^{\text{°}}{}_{\text{reacttants}}$

                   $\begin{array}{l}\text{= }\left[{\text{S}}^{\text{°}}{\text{(Al}}_2{\text{O}}_3{\text{)+3S}}^{\text{°}}\text{(Zn)}\right]\text{-}\left[{\text{2S}}^{\text{°}}{\text{(Al)+3S}}^{\text{°}}\text{(ZnO)}\right]\\ \text{= (1)(50}{\text{.99 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{)+(3)(41}{\text{.6JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{)-}\left[\text{(2)(28}{\text{.3 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{)+(3)(43}{\text{.9 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{)}\right]\\ \text{= -12}{\text{.5 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

Interpretation Introduction

Interpretation: The standard entropy change ${\text{(ΔS}}^{\text{°}}{}_{\text{reaction}}\text{)}$ in the given set of reactions has to be calculated.

Concept introduction: Entropy is the measure of randomness in the system.   Standard entropy change in a reaction is the difference in entropy of the products and reactants.  ${\text{(ΔS}}^{\text{°}}{}_{\text{reaction}}\text{)}$  can be calculated by the following equation.

$\begin{array}{l}{\text{(ΔS}}^{\text{°}}{}_{\text{reaction}}{\text{) = S}}^{\text{°}}{}_{\text{Products}}-{\text{ S}}^{\text{°}}{}_{\text{reactants}}\\ \end{array}$

Where

  ${\text{S}}^{\text{°}}{}_{\text{reactants}}$ is the standard entropy of the reactants

  ${\text{S}}^{\text{°}}{}_{\text{Products}}$ is the standard entropy of the products

Answer to Problem 17.12QP

The ${\text{ΔS}}^{\text{°}}{}_{\text{reaction}}$ for the reaction (c)$\text{= - 242}{\text{.8 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}$

Explanation of Solution

To record the given data

Given,

$\begin{array}{l}{\text{S}}^{\text{°}}{\text{(CO}}_{\text{2}}\text{)}=\text{213}{\text{.6 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ \text{S}°{\text{(H}}_2\text{O(}l\text{)) = 69}{\text{.9 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ {\text{S}}^{\text{°}}{\text{(CH}}_4\text{)}=\text{186}{\text{.2 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ {\text{S}}^{\text{°}}{\text{(O}}_2\text{)}=\text{205}{\text{.0 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

To calculate ${\text{ΔS}}^{\text{°}}{}_{\text{reaction}}$

The ${\text{ΔS}}^{\text{°}}{}_{\text{reaction}}$ is the difference in entropy of the reactants and products.  Which is calculated by plugging in the values of standard entropy of the reactants and products  the given equation.

${\text{(ΔS}}^{\text{°}}{}_{\text{reaction}}{\text{) = S}}^{\text{°}}{}_{\text{produdcts}}{\text{- S}}^{\text{°}}{}_{\text{reactants}}$

                  $\begin{array}{l}{\text{= S}}^{\text{°}}{\text{(CO}}_{\text{2}}\text{)+2S}°{\text{(H}}_2\text{O(}l\text{)) - }\left[{\text{S}}^{\text{°}}{\text{(CH}}_4{\text{)+2S}}^{\text{°}}{\text{(O}}_2\text{)}\right]\\ =\text{(1)(213}{\text{.6 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{)+(2)(69}{\text{.9 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{)-}\left[\text{(1)(186}{\text{.2 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{)+(2)(205}{\text{.0 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{)}\right]\\ =-242.8{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$