Chapter 18, Problem 3E

(a)

Interpretation Introduction

Interpretation:

The value of $\text{ΔH°}$ , $\Delta S^o$ and $\Delta G^o$ should be calculated for the following reaction.

  ${\text{3H}}_2\text{(g)}+{\text{N}}_2\text{(g)}\to 2{\text{NH}}_{\text{3}}(\text{g)}$

Concept Introduction:

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

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

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:

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

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

Spontaneity depends upon the temperature and also depends upon the sign of free energy change.

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

  ${\text{ΔG}}_{\text{298}}^{\text{o}}\text{=ΔH°-TΔS°}$

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

When the magnitude of $\text{TΔS}$ is more than the $\text{ΔH}$ , then $\text{ΔG}$ will be negative, and when the magnitude of $\text{TΔS}$ is less than the $\text{ΔH}$ , then $\text{ΔG}$ will be positive. If the value of ${\text{ΔG}}_{298}^{\text{o}}$ is more than zero, then non-spontaneous will be the reaction whereas if the value of ${\text{ΔG}}_{298}^{\text{o}}$ is less than zero, then spontaneous will be the reaction.

Therefore, reaction is non- spontaneous at low temperature and spontaneous at high temperature.

Answer to Problem 3E

  $\Delta S{°}_{298}=-199\text{ J/Kmol}$

  $\Delta H{°}_{298}=-91.6\text{ kJ/mol}$

  $\Delta G{°}_{298}=-32.8\text{ kJ/mol}$

Explanation of Solution

The given reaction is:

  ${\text{3H}}_2\text{(g)}+{\text{N}}_2\text{(g)}\to 2{\text{NH}}_{\text{3}}(\text{g)}$

The value of standard entropy for ${\text{NH}}_3\text{(g)}$ is $192.8\text{ J/Kmol}$

The value of standard entropy for ${\text{N}}_2\text{(g)}$ is $191.6\text{ J/Kmol}$

The value of standard entropy for ${\text{H}}_2(g)$ is $131\text{ J/Kmol}$

Put the values, in below formula.

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

  $\Delta S{°}_{298}=(2\times {\text{S°}}_{\text{298}}{\text{(NH}}_3\text{(g)}))-(1\times {\text{S°}}_{\text{298}}{\text{(N}}_2(g))+3\times {\text{S°}}_{\text{298}}{\text{(H}}_{\text{2}}\text{(g)}))$

  $\Delta S{°}_{298}=(2\times 192.8\text{ J/Kmol})-(1\times 191.6\text{ J/Kmol}+3\times \text{131 J/Kmol})$

  $\Delta S{°}_{298}=[385.6-584.6]\text{ J/Kmol}$

  $\Delta S{°}_{298}=-199\text{ J/Kmol}$

The value of standard enthalpy for ${\text{NH}}_3\text{(g)}$ is $-45.9\text{ kJ/mol}$

The value of standard enthalpy for ${\text{N}}_2\text{(g)}$ is $0.0\text{ kJ/mol}$

The value of standard enthalpy for ${\text{H}}_2(g)$ is $0.0\text{ kJ/mol}$

Put the values, in below formula.

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

  $\Delta H{°}_{298}=(2\times H{\text{°}}_{\text{298}}{\text{(NH}}_3\text{(g)}))-(1\times H{\text{°}}_{\text{298}}{\text{(N}}_2(g))+3\times H{\text{°}}_{\text{298}}{\text{(H}}_{\text{2}}\text{(g)}))$

  $\Delta H{°}_{298}=(2\times (-45.8)\text{ kJ/mol})-(1\times 0.0\text{ kJ/mol}+3\times \text{0}\text{.0 kJ/mol})$

  $\Delta H{°}_{298}=-91.6\text{ kJ/mol}$

The value of standard Gibbs free energy for ${\text{NH}}_3\text{(g)}$ is $-16.4\text{ kJ/mol}$

The value of standard Gibbs free energy for ${\text{N}}_2\text{(g)}$ is $0.0\text{ kJ/mol}$

The value of standard Gibbs free energy for ${\text{H}}_2(g)$ is $0.0\text{ kJ/mol}$

Put the values, in below formula.

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

  $\Delta G{°}_{298}=(2\times G{\text{°}}_{\text{298}}{\text{(NH}}_3\text{(g)}))-(1\times G{\text{°}}_{\text{298}}{\text{(N}}_2(g))+3\times G{\text{°}}_{\text{298}}{\text{(H}}_{\text{2}}\text{(g)}))$

  $\Delta G{°}_{298}=(2\times (-16.4)\text{ kJ/mol})-(1\times 0.0\text{ kJ/mol}+3\times \text{0}\text{.0 kJ/mol})$

  $\Delta G{°}_{298}=-32.8\text{ kJ/mol}$

(b)

Interpretation Introduction

Interpretation:

Whether the reaction is spontaneous or not should be determined.

Concept Introduction:

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

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

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:

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

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

Spontaneity depends upon the temperature and also depends upon the sign of free energy change.

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

  ${\text{ΔG}}_{\text{298}}^{\text{o}}\text{=ΔH°-TΔS°}$

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

When the magnitude of $\text{TΔS}$ is more than the $\text{ΔH}$ , then $\text{ΔG}$ will be negative, and when the magnitude of $\text{TΔS}$ is less than the $\text{ΔH}$ , then $\text{ΔG}$ will be positive. If the value of ${\text{ΔG}}_{298}^{\text{o}}$ is more than zero, then non-spontaneous will be the reaction whereas if the value of ${\text{ΔG}}_{298}^{\text{o}}$ is less than zero, then spontaneous will be the reaction.

Therefore, reaction is non- spontaneous at low temperature and spontaneous at high temperature.

Answer to Problem 3E

For the given reaction, $\Delta G{°}_{298}<0$ thus the reaction is spontaneous.

Explanation of Solution

The given reaction is:

  ${\text{3H}}_2\text{(g)}+{\text{N}}_2\text{(g)}\to 2{\text{NH}}_{\text{3}}(\text{g)}$

From part (a):

  $\Delta G{°}_{298}=-32.8\text{ kJ/mol}$

The negative value of Gibbs free energy represents the given reaction is spontaneous.

(c)

Interpretation Introduction

Interpretation:

The temperature at which reaction is spontaneous at standard conditions should be determined by considering $\text{ΔH°}$ and $\Delta S^o$ don’t depend upon temperature.

Concept Introduction:

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

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

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:

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

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

Spontaneity depends upon the temperature and also depends upon the sign of free energy change.

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

  ${\text{ΔG}}_{\text{298}}^{\text{o}}\text{=ΔH°-TΔS°}$

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

When the magnitude of $\text{TΔS}$ is more than the $\text{ΔH}$ , then $\text{ΔG}$ will be negative, and when the magnitude of $\text{TΔS}$ is less than the $\text{ΔH}$ , then $\text{ΔG}$ will be positive. If the value of ${\text{ΔG}}_{298}^{\text{o}}$ is more than zero, then non-spontaneous will be the reaction whereas if the value of ${\text{ΔG}}_{298}^{\text{o}}$ is less than zero, then spontaneous will be the reaction.

Therefore, reaction is non- spontaneous at low temperature and spontaneous at high temperature.

Answer to Problem 3E

The temperature at which reaction is spontaneous must be greater than $460.3\text{ K}$ .

Explanation of Solution

The given reaction is:

  ${\text{3H}}_2\text{(g)}+{\text{N}}_2\text{(g)}\to 2{\text{NH}}_{\text{3}}(\text{g)}$

From part (a)

  $\Delta S{°}_{298}=-199\text{ J/Kmol}$

  $\Delta H{°}_{298}=-91.6\text{ kJ/mol}$

Here, sign of $\text{ΔH°}$ is negative and $\text{ΔS°}$ is also negative.

  ${\text{ΔG}}_{\text{298}}^{\text{o}}\text{=ΔH°-TΔS°}$

Put the values,

  $\text{ΔH°-TΔS° = (}-91.6\text{ kJ/mol)}-\text{T}(-199\text{ J/Kmol})$

Now,

Let ${\text{ΔG}}_{\text{298}}^{\text{o}}=0$

  $0=(-91.6\text{ kJ/mol)}-\text{T}(-199\text{ J/Kmol})$

Since, 1 Kilojoule = 1000 Joule

  $0=(-91.6\text{ kJ/mol)}-\text{T}(-199\text{ J/Kmol}\times \frac{1\text{ kJ}}{1000\text{ J}})$

  $0=(-91.6\text{ kJ/mol)}-\text{T}(-0.199\text{ kJ/K}\cdot \text{mol})$

  $+91.6\text{ kJ/mol}=-\text{T}(-0.199\text{ kJ/K}\cdot \text{mol})$

  $\text{T =}\frac{91.6\text{ kJ/mol}}{0.199\text{ kJ/K}\cdot \text{mol}}=460.3\text{ K}$

The process is spontaneous when ${\text{ΔG}}_{\text{298}}^{\text{o}}<0$ , thus the temperature must be greater than $\text{T >}460.3\text{ K}$