Chapter 16, Problem 20QAP

Use Table 16.1 to calculate ΔS° for each of the following reactions.

(a) $\text{5NO}\left(g\right)+3{\text{MnO}}_4{}^{-}\left(aq\right)+4{\text{H}}^{+}\left(aq\right)\to 5{\text{NO}}_3{}^{-}\left(aq\right)+3{\text{Mn}}^{2+}\left(aq\right)+2{\text{H}}_2\text{O}\left(l\right)$(b) ${\text{6Fe}}^{2+}\left(aq\right)+{\text{CrO}}_4{}^{2-}\left(aq\right)+8{\text{H}}^{+}\left(aq\right)\to \text{Cr}\left(s\right)+6{\text{Fe}}^{3+}\left(aq\right)+4{\text{H}}_2\text{O}\left(l\right)$

(c) ${\text{C}}_2{\text{H}}_2\left(g\right)+\frac{5}{2}{\text{O}}_2\left(g\right)\to 2{\text{CO}}_2\left(g\right)+{\text{H}}_2\text{O}\left(g\right)$

Interpretation Introduction

(a)

Interpretation:

The entropy change ( $\Delta S°$) for the following reaction should be calculated.

${\text{5NO(g)+3MnO}}_4^{2-}{\text{(aq)+4H}}^{+}(\text{aq})\to {\text{5NO}}^{3-}{\text{(aq) + 3Mn}}^{2+}{\text{(aq)+2H}}_2\text{O(l)}$

Concept introduction:

Entropy is defined as the ratio of thermal energy to the temperature which is unavailable for work done. It is also defined as the measure of disorder of molecule of a system. It is an extensive property and state function.

Entropy is related with the number of microstates for a system and microstate is defined as the number of ways for the system to be arranged.

The standard entropy change at room temperature is equal to the difference between the standard entropy of reactant and standard entropy of product.

Answer to Problem 20QAP

$\Delta S°=-976.1\text{ J/Kmol}$

Explanation of Solution

Given process is: ${\text{5NO(g)+3MnO}}_4^{2-}{\text{(aq)+4H}}^{+}(\text{aq})\to {\text{5NO}}^{3-}{\text{(aq) + 3Mn}}^{2+}{\text{(aq)+2H}}_2\text{O(l)}$

The mathematical expression for the standard entropy value at room temperature is:

$\Delta S°=\Delta S{°}_{298}=\sum \text{n}\text{S°(products})-\sum \text{p}\text{S°(reactants)}$

Where, n and p represents the coefficients of reactants and products in the balanced chemical equation.

The value of standard entropy for $\text{NO(g)}$ is $210.7\text{ J/Kmol}$

The value of standard entropy for ${\text{MnO}}_4^{2-}\text{(aq)}$ is $191.2\text{ J/Kmol}$

The value of standard entropy for ${\text{H}}^{+}\text{(aq)}$ is $0.0\text{ J/Kmol}$

The value of standard entropy for ${\text{NO}}^{3-}\text{(aq)}$ is $\text{146}\text{.4 J/Kmol}$

The value of standard entropy for ${\text{Mn}}^{\text{2+}}\text{(aq)}$ is $\text{-73}\text{.6 J/Kmol}$

The value of standard entropy for ${\text{H}}_2\text{O(l)}$ is $\text{69}\text{.9 J/Kmol}$

Put the values, we get:

$\begin{array}{l}\Delta S°=(5\times {\text{S°(NO}}_3^{-}\text{(}aq\text{)})+3\times {\text{S°(Mn}}^{2+}(aq))+2\times {\text{S°(H}}_{\text{2}}\text{O(l))})-\\ (5\times \text{S°(NO}(g))+3\times {\text{S°(MnO}}_4^{2-}(aq))+4\times {\text{S°(H}}^{+}(aq)))\end{array}$

$\begin{array}{l}\Delta S°=[(5\times 146.4+3\times (-73.6)+2\times (69.9)-\\ (5\times 210.7+3\times 191.2+4\times 0.0)]\text{ J/Kmol}\end{array}$

$\Delta S°=[651-1627.1]\text{ J/Kmol}$

$\Delta S°=-976.1\text{ J/Kmol}$

Interpretation Introduction

(b)

Interpretation:

The entropy change ( $\Delta S°$) for the following reaction should be calculated.

${\text{6Fe}}^{\text{2+}}{\text{(aq)+CrO}}_4^{2-}{\text{(aq)+8H}}^{+}\text{(aq)}\to {\text{Cr (s) + 6Fe}}^{3+}{\text{(aq)+4H}}_{\text{2}}\text{O}(\text{l})$

Concept introduction:

Entropy is defined as the ratio of thermal energy to the temperature which is unavailable for work done. It is also defined as the measure of disorder of molecule of a system. It is an extensive property and state function.

Entropy is related with the number of microstates for a system and microstate is defined as the number of ways for the system to be arranged.

The standard entropy change at room temperature is equal to the difference between the standard entropy of reactant and standard entropy of product.

Answer to Problem 20QAP

$\Delta S°=-819\text{ J/Kmol}$

Explanation of Solution

Given process is: ${\text{6Fe}}^{\text{2+}}{\text{(aq)+CrO}}_4^{2-}{\text{(aq)+8H}}^{+}\text{(aq)}\to {\text{Cr (s) + 6Fe}}^{3+}{\text{(aq)+4H}}_{\text{2}}\text{O}(\text{l})$

The mathematical expression for the standard entropy value at room temperature is:

$\Delta S°=\Delta S{°}_{298}=\sum \text{n}\text{S°(products})-\sum \text{p}\text{S°(reactants)}$

Where, n and p represents the coefficients of reactants and products in the balanced chemical equation.

The value of standard entropy for ${\text{Fe}}^{\text{2+}}\text{(aq)}$ is $-137.2\text{ J/Kmol}$

The value of standard entropy for ${\text{CrO}}_4^{2-}\text{(aq)}$ is $50.2\text{ J/Kmol}$

The value of standard entropy for ${\text{H}}^{+}\text{(aq)}$ is $0.0\text{ J/Kmol}$

The value of standard entropy for $\text{Cr (s)}$ is $$

${\text{Fe}}^{3+}\text{(g)}$+23.8 J/Kmol

The value of standard entropy for is $-315.9\text{ J/Kmol}$

The value of standard entropy for ${\text{H}}_{\text{2}}\text{O(l)}$ is $69.9\text{ J/Kmol}$

Put the values, we get:

${\text{6Fe}}^{\text{2+}}{\text{(aq)+CrO}}_4^{2-}{\text{(aq)+8H}}^{+}\text{(aq)}\to {\text{Cr (s) + 6Fe}}^{3+}{\text{(aq)+4H}}_{\text{2}}\text{O}(\text{l})$

$\Delta S°=(1\times \text{S°(Cr(s)})+6\times {\text{S°(Fe}}^{3+}(\text{aq}))+4\times {\text{S°(H}}_2\text{O}(\text{l})))-(6\times {\text{S°(Fe}}^{2+}\text{(aq))+1}\times {\text{S°(CrO}}_4^{2-}\text{(aq)})+8\times {\text{S°(H}}^{+}\text{(aq)))}$

$\Delta S°=[(1\times 23.8+6\times (-315.9)+4\times 69.9)-(6\times (-137.2)\text{+1}\times 50.2+8\times \text{0}\text{.0)] J/Kmol}$

$\Delta S°=[-1592+773]\text{ J/Kmol}$

$\Delta S°=-819\text{ J/Kmol}$

Interpretation Introduction

(c)

Interpretation:

The entropy change ( $\Delta S°$) for the following reaction should be calculated.

${\text{C}}_{\text{2}}{\text{H}}_2\text{(g)+}\frac{5}{2}{\text{O}}_{\text{2}}\text{(g)}\to 2{\text{CO}}_2{\text{(g)+H}}_2\text{O}(\text{g)}$

Concept introduction:

Entropy is defined as the ratio of thermal energy to the temperature which is unavailable for work done. It is also defined as the measure of disorder of molecule of a system. It is an extensive property and state function.

Entropy is related with the number of microstates for a system and microstate is defined as the number of ways for the system to be arranged.

The standard entropy change at room temperature is equal to the difference between the standard entropy of reactant and standard entropy of product.

Answer to Problem 20QAP

$\Delta S°=-97.4\text{ J/Kmol}$

Explanation of Solution

Given process is: ${\text{C}}_{\text{2}}{\text{H}}_2\text{(g)+}\frac{5}{2}{\text{O}}_{\text{2}}\text{(g)}\to 2{\text{CO}}_2{\text{(g)+H}}_2\text{O}(\text{g)}$

The mathematical expression for the standard entropy value at room temperature is:

$\Delta S°=\Delta S{°}_{298}=\sum \text{n}\text{S°(products})-\sum \text{p}\text{S°(reactants)}$

Where, n and p represents the coefficients of reactants and products in the balanced chemical equation.

The value of standard entropy for ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}\text{(g)}$ is $200.8\text{ J/Kmol}$

The value of standard entropy for ${\text{O}}_{\text{2}}\text{(g)}$ is $205.0\text{ J/Kmol}$

The value of standard entropy for ${\text{CO}}_2\text{(g)}$ is $213.6\text{ J/Kmol}$

The value of standard entropy for ${\text{H}}_{\text{2}}\text{O(g)}$ is $188.7\text{ J/Kmol}$

Put the values, we get:

$\Delta S°=(2\times {\text{S°(CO}}_2\text{(g))+1}\times {\text{S°(H}}_{\text{2}}\text{O(g)))}-(1\times \text{S}°{\text{(C}}_{\text{2}}{\text{H}}_2\text{(g)})+\frac{5}{2}\times \text{S}°{\text{(O}}_{\text{2}}\text{(g)}))$

$\Delta S°=[(2\times \text{213}\text{.6+1}\times \text{188}\text{.7)}-(1\times 200.8+\frac{5}{2}\times 205.0)]\text{ J/Kmol}$

$\Delta S°=[615.9-713.3]\text{ J/Kmol}$

$\Delta S°=-97.4\text{ J/Kmol}$