Chapter 16, Problem 25QAP

Calculate ΔG° at 355 K for each of the reactions in Question 17. 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.

${\text{CO(g)+2H}}_{\text{2}}\text{(g)}\to {\text{CH}}_{\text{3}}\text{OH(l)}$

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 25QAP

. $\text{ΔG}°\text{=}-10.34\text{ kJ}$, Spontaneous

Explanation of Solution

The given value is:

$\text{ΔS°}=-332.2\text{ J/Kmol}$

Temperature = 355 K

${\text{CO(g)+2H}}_{\text{2}}\text{(g)}\to {\text{CH}}_{\text{3}}\text{OH(l)}$

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

The value of standard enthalpy for ${\text{CH}}_3\text{OH(l)}$ is $-238.7\text{ kJ/mol}$

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

The value of standard enthalpy for $\text{CO(g)}$ is $\text{-110}\text{.5 kJ/mol}$

Put the values, we get:

$\text{ΔH}°=(1\times {\text{H°(CH}}_3\text{OH}(l)))-(1\times \text{H°(CO}(g))+2\times {\text{H°(H}}_{\text{2}}(g)))$

$\Delta \text{H}°=[(1\times (-238.7))-(1(-110.5)+2\times 0)]\text{ kJ/mol}$

$\Delta \text{H}°=-128.2\text{ kJ/mol}$

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

$\text{ΔG}°\text{=ΔH°-TΔS°}$

Put the values,

$\text{ΔG}°\text{=}-128.2\text{ kJ-(355 K}\times (-332.2\text{ J/K)}$

Since, 1 Kilojoule = 1000 Joule

$\text{ΔG}°\text{=}-128.2\text{ kJ-(355 K}\times (-332.2\text{ J/K}\times \frac{1\text{ kJ}}{1000\text{ J}}\text{)}$

$\text{ΔG}°\text{=}-128.2\text{ kJ}-\text{(355 K}\times (-0.332\text{ kJ/K)}$

$\text{ΔG}°\text{=}-128.2\text{ kJ+117}\text{.86 kJ}$

$\text{ΔG}°\text{=}-10.34\text{ kJ}$

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.

${\text{N}}_2(\text{g})+{\text{O}}_{\text{2}}\text{(g)}\to 2\text{NO(g})$

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 25QAP

. $\text{ΔG}°\text{=171}\text{.56 kJ}$, Non-spontaneous

Explanation of Solution

The given value is:

$\text{ΔS°}=+24.9\text{ J/Kmol}$

Temperature = 355 K

${\text{N}}_2(\text{g})+{\text{O}}_{\text{2}}\text{(g)}\to 2\text{NO(g})$

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

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

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

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

Put the values, we get:

$\Delta H°=(2\times \text{H°(NO}(g)))-(1\times {\text{H°(N}}_2(g))+1\times {\text{H°(O}}_2(g)))$

$\Delta H°=[(2\times 90.2)-(1\times 0+1\times 0)]\text{ kJ/mol}$

$\Delta S°=[180.4-0]\text{ kJ/mol}$

$\Delta S°=180.4\text{ kJ/mol}$

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

$\text{ΔG}°\text{=ΔH°-TΔS°}$

Put the values,

$\text{ΔG}°\text{=}180.4\text{ kJ-(355 K}\times (-24.9\text{ J/K)}$

Since, 1 Kilojoule = 1000 Joule

$\text{ΔG}°\text{=}180.4\text{ kJ-(355 K}\times (+24.9\text{ J/K}\times \frac{1\text{ kJ}}{1000\text{ J}}\text{)}$

$\text{ΔG}°\text{=180}\text{.4 kJ}-\text{(355 K}\times (+0.0249\text{ kJ/K)}$

$\text{ΔG}°\text{=180}\text{.4 kJ}-\text{8}\text{.8395 kJ }$

$\text{ΔG}°\text{=171}\text{.56 kJ}$

The positive value of change in free energy denotes that the process is non-spontaneous.

Interpretation Introduction

(c)

Interpretation:

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

${\text{BaCO}}_{\text{3}}\text{(s)}\to {\text{BaO(s)+CO}}_{\text{2}}(\text{g)}$

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 25QAP

. $\text{ΔG}°\text{=}-208.2755\text{ kJ}$, Spontaneous

Explanation of Solution

The given value is:

$\text{ΔS°}=+171.9\text{ J/Kmol}$

Temperature = 355 K

${\text{BaCO}}_{\text{3}}\text{(s)}\to {\text{BaO(s)+CO}}_{\text{2}}(\text{g)}$

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

The value of standard enthalpy for ${\text{BaCO}}_3\text{(s)}$ is $-1216.3\text{ kJ/mol}$

The value of standard enthalpy for $\text{BaO(s)}$ is $\text{-553}\text{.5 kJ/mol}$

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

Put the values, we get:

$\Delta H°=(1\times \text{H°(BaO}(s))+1\times {\text{H°(CO}}_2(g)))-(1\times {\text{H°(BaCO}}_3(s)))$

$\Delta H°=[(1\times (-553.5)+1\times (-393.5)-(1\times (-1216.3)]\text{ kJ/mol}$

$\Delta H°=[-947+1216.3]\text{ kJ/mol}$

$\Delta H°=+269.3\text{ kJ/mol}$

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

$\text{ΔG}°\text{=ΔH°-TΔS°}$

Put the values,

$\text{ΔG}°\text{= 269}\text{.3 kJ}-\text{(355 K}\times (+171.9\text{ J/K)}$

Since, 1 Kilojoule = 1000 Joule

$\text{ΔG}°\text{= 269}\text{.3 kJ}-\text{(355 K}\times (+171.9\text{ J/K}\times \frac{1\text{ kJ}}{1000\text{ J}}\text{)}$

$\text{ΔG}°\text{= 269}\text{.3 kJ}-\text{(355 K}\times (0.1719\text{ kJ/K)}$

$\text{ΔG}°\text{= 269}\text{.3 kJ}-\text{61}\text{.0245 kJ }$

$\text{ΔG}°\text{=}-208.2755\text{ kJ}$

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.

$2{\text{NaCl(s)+F}}_2(\text{g})\to 2{\text{NaF(s)+Cl}}_2(\text{g)}$

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 25QAP

. $\text{ΔG}°\text{=}-\text{317}\text{.38 kJ}$, Spontaneous

Explanation of Solution

The given value is:

$\text{ΔS°}=-20.9\text{ J/Kmol}$

Temperature = 355 K

$2{\text{NaCl(s)+F}}_2(\text{g})\to 2{\text{NaF(s)+Cl}}_2(\text{g)}$

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

The value of standard enthalpy for $\text{NaCl(s)}$ is $-411.2\text{ kJ/mol}$

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

The value of standard enthalpy for $\text{NaF(s)}$ is $-573.6\text{ kJ/mol}$

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

Put the values, we get:

$\Delta \text{H}°=(2\times \text{H°(NaF}(s))+1\times {\text{H°(Cl}}_2(g))-(2\times \text{H°(NaCl(s)})+1\times {\text{H°(F}}_2\text{(g)}))$

$\text{ΔH}°=[(2\times (-573.6)+1\times 0)-(2\times (-411.2)+1\times 0)]\text{ kJ/mol}$

$\text{ΔH}°=[-1147.2+822.4]\text{ J/Kmol}$

$\text{ΔH}°=-324.8\text{ J/Kmol}$

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

$\text{ΔG}°\text{=ΔH°-TΔS°}$

Put the values,

$\text{ΔG}°\text{=}-324.8\text{ kJ}-\text{(355 K}\times (-20.9\text{ J/K)}$

Since, 1 Kilojoule = 1000 Joule

$\text{ΔG}°\text{=}-324.8\text{ kJ}-\text{(355 K}\times (-20.9\text{ J/K}\times \frac{1\text{ kJ}}{1000\text{ J}}\text{)}$

$\text{ΔG}°\text{=-324}\text{.8 kJ}-\text{(355 K}\times (-0.0209\text{ kJ/K)}$

$\text{ΔG}°\text{=}-\text{324}\text{.8 kJ}+\text{7}\text{.4195 kJ }$

$\text{ΔG}°\text{=}-\text{317}\text{.38 kJ}$

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