Chapter 16, Problem 26QAP

Calculate ΔG° at 415 K for each of the reactions in Question 18. State whether the reactions are spontaneous.

Interpretation Introduction

(a)

Interpretation:

$\text{ΔG}°$ should be calculated for the following reaction and also state whether the reaction is spontaneous or not.

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

Concept introduction:

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 mathematical expression for the standard enthalpy change value at room temperature is:

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

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

At standard states, the free energy charge which is responsible for the production of 1 mole of a molecule from its elements is known as standard free energy of formation and it is represented by ${\text{ΔG}}_{}^{\text{o}}$.

The mathematical expression for ${\text{ΔG}}_{\text{f}}^{\text{o}}$ at standard state is given by:

$\text{ΔG}°\text{=ΔH°-TΔS°}$

If both $\text{ΔH°}$ and $\text{ΔS°}$ is positive, the reaction is endothermic results in increase in entropy.

Answer to Problem 26QAP

. $\text{ΔG}°\text{= }-\text{ 986}\text{.732 kJ}$, Spontaneous

Explanation of Solution

The given value is:

$\text{ΔS°}=+195.5\text{ J/Kmol}$

Temperature = 415 K

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

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

The value of standard enthalpy for ${\text{NH}}_3\text{(g)}$ is $-46.1\text{ kJ/mol}$

The value of standard enthalpy for ${\text{O}}_{\text{2}}\text{(g)}$ is $\text{0}\text{.0 kJ/mol}$

The value of standard enthalpy for $\text{NO(g)}$ is $\text{+90}\text{.2 kJ/mol}$

The value of standard enthalpy for ${\text{H}}_2\text{O(g)}$ is $\text{-241}\text{.8 kJ/mol}$

Put the values, we get:

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

$\Delta H°=[(4\times (+90.2)+6\times (-241.8)-(4\times (-46.1)+5\times 0.0)]\text{ kJ/mol}$

$\Delta H°=[-1090+184.4]\text{J/Kmol}$

$\Delta H°=-905.6\text{ J/Kmol}$

The mathematical expression for ${\text{ΔG}}_{\text{f}}^{\text{o}}$ at standard state is given by:

$\text{ΔG}°\text{=ΔH°-TΔS°}$

Put the values,

$\text{ΔG}°\text{=}-905.6\text{ kJ-(415 K}\times (+195.5\text{ J/K))}$

Since, 1 Kilojoule = 1000 Joule

$\text{ΔG}°\text{=}-905.6\text{ kJ-(415 K}\times (+195.5\text{ J/K}\times \frac{1\text{ kJ}}{1000\text{ J}}\text{))}$

$\text{ΔG}°\text{=}-905.6\text{ kJ-(415 K}\times (0.1955\text{ kJ/K))}$

$\text{ΔG}°\text{=}-905.6\text{ kJ}-\text{81}\text{.1325 kJ}$

$\text{ΔG}°\text{= }-\text{ 986}\text{.732 kJ}$

The negative value of change in free energy denotes that the process is spontaneous.

Interpretation Introduction

(b)

Interpretation:

$\text{ΔG}°$ should be calculated for the following reaction and also state whether the reaction is spontaneous or not.

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

Concept introduction:

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 mathematical expression for the standard enthalpy change value at room temperature is:

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

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

At standard states, the free energy charge which is responsible for the production of 1 mole of a molecule from its elements is known as standard free energy of formation and it is represented by ${\text{ΔG}}_{}^{\text{o}}$.

The mathematical expression for ${\text{ΔG}}_{\text{f}}^{\text{o}}$ at standard state is given by:

$\text{ΔG}°\text{=ΔH°-TΔS°}$

If both $\text{ΔH°}$ and $\text{ΔS°}$ is positive, the reaction is endothermic results in increase in entropy.

Answer to Problem 26QAP

. $\text{ΔG}°\text{=}-893.64\text{ kJ}$, Spontaneous

Explanation of Solution

The given value is:

$\text{ΔS°}=+605.9\text{ J/Kmol}$

Temperature = 415 K

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

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

The value of standard enthalpy for ${\text{H}}_{\text{2}}{\text{O}}_2(\text{l})$ is $-187.8\text{ kJ/mol}$

The value of standard enthalpy for ${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}\text{(l)}$ is $+50.6\text{ kJ/mol}$

The value of standard enthalpy for ${\text{N}}_2\text{(g)}$ is $\text{0}\text{.0 kJ/mol}$

The value of standard enthalpy for ${\text{H}}_2\text{O(g)}$ is $\text{-241}\text{.8 kJ/mol}$

Put the values, we get:

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

$\Delta H°=[(4\times (-241.8)+1\times 0.0)-(2\times (-187.8)+1\times 50.6)]\text{ J/Kmol}$

$\Delta H°=[-967.2+325]\text{ J/Kmol}$

$\Delta H°=-642.2\text{ J/Kmol}$

The mathematical expression for ${\text{ΔG}}_{\text{f}}^{\text{o}}$ at standard state is given by:

$\text{ΔG}°\text{=ΔH°-TΔS°}$

Put the values,

$\text{ΔG}°\text{=}-642.2\text{ kJ-(415 K}\times (605.9\text{ J/K))}$

Since, 1 Kilojoule = 1000 Joule

$\text{ΔG}°\text{=}-642.2\text{ kJ-(415 K}\times (605.9\text{ J/K}\times \frac{1\text{ kJ}}{1000\text{ J}}\text{))}$

$\text{ΔG}°\text{=}-642.2\text{ kJ-(415 K}\times (0.6059\text{ kJ/K))}$

$\text{ΔG}°\text{=}-642.2\text{ kJ-(}251.45\text{ kJ)}$

$\text{ΔG}°\text{=}-893.64\text{ kJ}$

The negative value of change in free energy denotes that the process is spontaneous.

Interpretation Introduction

(c)

Interpretation:

$\text{ΔG}°$ should be calculated for the following reaction and also state whether the reaction is spontaneous or not.

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

Concept introduction:

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 mathematical expression for the standard enthalpy change value at room temperature is:

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

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

At standard states, the free energy charge which is responsible for the production of 1 mole of a molecule from its elements is known as standard free energy of formation and it is represented by ${\text{ΔG}}_{}^{\text{o}}$.

The mathematical expression for ${\text{ΔG}}_{\text{f}}^{\text{o}}$ at standard state is given by:

$\text{ΔG}°\text{=ΔH°-TΔS°}$

If both $\text{ΔH°}$ and $\text{ΔS°}$ is positive, the reaction is endothermic results in increase in entropy.

Answer to Problem 26QAP

. $\text{ΔG}°\text{= -394}\text{.7 kJ}$, Spontaneous

Explanation of Solution

The given value is:

$\text{ΔS°}=+2.9\text{ J/Kmol}$

Temperature = 415 K

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

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

The value of standard enthalpy for $\text{C(s)}$ is $0.0\text{ kJ/mol}$

The value of standard enthalpy for ${\text{O}}_2\text{(g)}$ is $\text{0}\text{.0 kJ/mol}$

The value of standard enthalpy for ${\text{CO}}_2\text{(g)}$ is $-393.5\text{ kJ/mol}$

Put the values, we get:

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

$\Delta H°=[(1\times (-393.5)-(1\times 0.0+1\times 0.0)]\text{ kJ/mol}$

$\Delta H°=[-393.5-0]\text{ kJ/mol}$

$\Delta H°=-393.5\text{ kJ/mol}$

The mathematical expression for ${\text{ΔG}}_{\text{f}}^{\text{o}}$ at standard state is given by:

$\text{ΔG}°\text{=ΔH°-TΔS°}$

Put the values,

$\text{ΔG}°\text{= -393}\text{.5 kJ}-\text{(415 K}\times (+2.9\text{ J/K)}$

Since, 1 Kilojoule = 1000 Joule

$\text{ΔG}°\text{= -393}\text{.5 kJ}-\text{(415 K}\times (+2.9\text{ J/K}\times \frac{1\text{ kJ}}{1000\text{ J}}\text{))}$

$\text{ΔG}°\text{= -393}\text{.5 kJ}-\text{(415 K}\times (+0.0029\text{ kJ/K))}$

$\text{ΔG}°\text{= -393}\text{.5 kJ}-1.20\text{ kJ}$

$\text{ΔG}°\text{= -394}\text{.7 kJ}$

The negative value of change in free energy denotes that the process is spontaneous.

Interpretation Introduction

(d)

Interpretation:

$\text{ΔG}°$ should be calculated for the following reaction and also state whether the reaction is spontaneous or not.

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

Concept introduction:

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 mathematical expression for the standard enthalpy change value at room temperature is:

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

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

At standard states, the free energy charge which is responsible for the production of 1 mole of a molecule from its elements is known as standard free energy of formation and it is represented by ${\text{ΔG}}_{}^{\text{o}}$.

The mathematical expression for ${\text{ΔG}}_{\text{f}}^{\text{o}}$ at standard state is given by:

$\text{ΔG}°\text{=ΔH°-TΔS°}$

If both $\text{ΔH°}$ and $\text{ΔS°}$ is positive, the reaction is endothermic results in increase in entropy.

Answer to Problem 26QAP

. $\text{ΔG}°\text{=}-297.964\text{ kJ}$, Spontaneous

Explanation of Solution

The given value is:

$\text{ΔS°}=-93.1\text{ J/Kmol}$

Temperature = 415K

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

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

The value of standard enthalpy for ${\text{CH}}_4\text{(g)}$ is $-74.8\text{ kJ/mol}$

The value of standard enthalpy for ${\text{Cl}}_{\text{2}}\text{(g)}$ is $\text{0 kJ/mol}$

The value of standard enthalpy for ${\text{CHCl}}_3\text{(l)}$ is $-134.5\text{ kJ/mol}$

The value of standard enthalpy for $\text{HCl(g)}$ is $-\text{92}\text{.3 kJ/mol}$

Put the values, we get:

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

$\Delta H°=[(1\times (-134.5)+3\times (-92.3))-(1\times (-74.8)+3\times 0.0)]\text{ kJ/mol}$

$\Delta H°=[-411.4+74.8]\text{ kJ/mol}$

$\Delta H°=-336.6\text{ kJ/mol}$

The mathematical expression for ${\text{ΔG}}_{\text{f}}^{\text{o}}$ at standard state is given by:

$\text{ΔG}°\text{=ΔH°-TΔS°}$

Put the values,

$\text{ΔG}°\text{=}-336.6\text{ kJ}-\text{(415 K}\times (-93.1\text{ J/K)}$

Since, 1 Kilojoule = 1000 Joule

$\text{ΔG}°\text{=}-336.6\text{ kJ}-\text{(415 K}\times (-93.1\text{ J/K}\times \frac{1\text{ kJ}}{1000\text{ J}}\text{))}$

$\text{ΔG}°\text{=}-336.6\text{ kJ}-\text{(415 K}\times (-0.0931\text{ kJ/K))}$

$\text{ΔG}°\text{=}-336.6\text{ kJ+38}\text{.636 kJ}$

$\text{ΔG}°\text{=}-297.964\text{ kJ}$

The negative value of change in free energy denotes that the process is spontaneous.