Chapter 16, Problem 18QAP

Interpretation Introduction

(a)

Interpretation:

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

${\text{4NH}}_3\text{ (g)}+5{\text{O}}_{\text{2}}\text{(g)}\to {\text{4NO(g)+6H}}_{\text{2}}\text{O(g)}$

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 18QAP

$\Delta S°=+195.5\text{ J/Kmol}$

Explanation of Solution

Given process is: ${\text{4NH}}_3\text{ (g)}+5{\text{O}}_{\text{2}}\text{(g)}\to {\text{4NO(g)+6H}}_{\text{2}}\text{O(g)}$

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 ${\text{NH}}_3\text{(g)}$ is $197.6\text{ J/Kmol}$

The value of standard entropy for ${\text{O}}_{\text{2}}\text{(g)}$ is $130.7\text{ J/Kmol}$

The value of standard entropy for $\text{NO(g)}$ is $126.8\text{ J/Kmol}$

The value of standard entropy for ${\text{H}}_2\text{O(g)}$ is $188.7\text{ J/Kmol}$

Put the values, we get:

$\Delta S°=(4\times \text{S°(NO}(g))+6\times {\text{S°(H}}_2\text{O}(g)))-(4\times {\text{S°(NH}}_3(g))+5\times {\text{S°(O}}_{\text{2}}(g)))$

$\Delta S°=[(4\times 126.8+6\times 188.7)-(4\times 197.6+5\times 130.7)]\text{ J/Kmol}$

$\Delta S°=[1639.4-1443.9]\text{J/Kmol}$

$\Delta S°=+195.5\text{ J/Kmol}$

Interpretation Introduction

(b)

Interpretation:

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

${\text{2H}}_{\text{2}}{\text{O}}_2\text{(l})+{\text{N}}_{\text{2}}{\text{H}}_4\text{(l)}\to 2{\text{N}}_2\text{(g})+4{\text{H}}_{\text{2}}\text{O}(\text{g)}$

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 18QAP

$\Delta S°=+605.9\text{ J/Kmol}$

Explanation of Solution

Given process is: ${\text{2H}}_{\text{2}}{\text{O}}_2\text{(l})+{\text{N}}_{\text{2}}{\text{H}}_4\text{(l)}\to {\text{N}}_2\text{(g})+4{\text{H}}_{\text{2}}\text{O}(\text{g)}$

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 ${\text{H}}_{\text{2}}{\text{O}}_2(\text{l})$ is $109.6\text{ J/Kmol}$

The value of standard entropy for ${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}\text{(l)}$ is $121.2\text{ J/Kmol}$

The value of standard entropy for ${\text{N}}_2\text{(g)}$ is $191.5\text{ J/Kmol}$

The value of standard entropy for ${\text{H}}_2\text{O(g)}$ is $188.7\text{ J/Kmol}$

Put the values, we get:

$\Delta S°=(4\times {\text{S°(H}}_2\text{O(}g\text{)})+1\times {\text{S°(N}}_2(g)))-(2\times {\text{S°(H}}_2{\text{O}}_2(l))+1\times {\text{S°(N}}_2{\text{H}}_{\text{4}}(l)))$

$\Delta S°=[(4\times 188.7+1\times 191.5)-(2\times 109.6+1\times 121.2)]\text{ J/Kmol}$

$\Delta S°=[946.3-340.4]\text{ J/Kmol}$

$\Delta S°=+605.9\text{ J/Kmol}$

Interpretation Introduction

(c)

Interpretation:

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

${\text{C(s)+O}}_2\text{(g)}\to {\text{CO}}_2(\text{g)}$

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 18QAP

$\Delta S°=+2.9\text{ J/Kmol}$

Explanation of Solution

Given process is: ${\text{C(s)+O}}_2\text{(g)}\to {\text{CO}}_2(\text{g)}$

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 $\text{C(s)}$ is $5.7\text{ J/Kmol}$

The value of standard entropy for ${\text{O}}_2\text{(g)}$ is $205.0\text{ J/Kmol}$

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

Put the values, we get:

$\Delta S°=(1\times {\text{S°(CO}}_2(g)))-(1\times \text{S°(C}(s))+1\times {\text{S°(O}}_2(g)))$

$\Delta S°=[(1\times 213.6)-(1\times 5.7+1\times 205.0)]\text{ J/Kmol}$

$\Delta S°=[213.6-210.7]\text{ J/Kmol}$

$\Delta S°=+2.9\text{ J/Kmol}$

Interpretation Introduction

(d)

Interpretation:

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

${\text{CH}}_{\text{4}}{\text{(g)+3Cl}}_2(\text{g})\to {\text{CHCl}}_{\text{3}}\text{(l)+3HCl}(\text{g)}$

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 18QAP

$\Delta S°=-93.1\text{ J/Kmol}$

Explanation of Solution

Given process is: ${\text{CH}}_{\text{4}}{\text{(g)+3Cl}}_2(\text{g})\to {\text{CHCl}}_{\text{3}}\text{(l)+3HCl}(\text{g)}$

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 ${\text{CH}}_4\text{(g)}$ is $186.2\text{ J/Kmol}$

The value of standard entropy for ${\text{Cl}}_{\text{2}}\text{(g)}$ is $223.0\text{ J/Kmol}$

The value of standard entropy for ${\text{CHCl}}_3\text{(l)}$ is $201.7\text{ J/Kmol}$

The value of standard entropy for $\text{HCl(g)}$ is $186.8\text{ J/Kmol}$

Put the values, we get:

$\Delta S°=(1\times {\text{S°(CHCl}}_{\text{3}}(l))+3\times \text{S°(HCl}(g)))-(1\times {\text{S°(CH}}_4\text{(g)})+3\times {\text{S°(Cl}}_2\text{(g)}))$

$\Delta S°=[(1\times 201.7+3\times 186.8)-(1\times 186.2+3\times 223.0)]\text{ J/Kmol}$

$\Delta S°=[762.1-855.2]\text{ J/Kmol}$

$\Delta S°=-93.1\text{ J/Kmol}$