Chapter 5, Problem 1DE

(a)

Interpretation Introduction

To determine: The chemical equations corresponds to the $\Delta G_f^0$ for ${\text{NH}}_3\left(g\right)$ and for $\text{CO}\left(g\right)$ .

Answer to Problem 1DE

Solution: The chemical equation corresponds to $\Delta G_f^0$ for ${\text{NH}}_3\left(g\right)$ is,

$\frac{1}{2}{\text{N}}_2\left(g\right)+\frac{3}{2}{\text{H}}_2\left(g\right)\to {\text{NH}}_3\left(g\right)$

The chemical equation corresponds to $\Delta G_f^0$ for $\text{CO}\left(g\right)$ is,

$\text{C}\left(s\right)+\frac{1}{2}{\text{O}}_2\left(g\right)\to \text{CO}\left(g\right)$

Explanation of Solution

The chemical equation corresponds to the $\Delta G_f^0$ means the equation in which only one mole of product is formed.
The chemical equation corresponds to the formation of ${\text{NH}}_3\left(g\right)$ is,

${\text{N}}_2\left(g\right)+{\text{H}}_2\left(g\right)\to {\text{NH}}_3\left(g\right)$

Since, one nitrogen atom and three hydrogen atoms are required to form one mole of ${\text{NH}}_3\left(g\right)$ . Therefore, nitrogen in the above equation is multiplied by $\frac{1}{2}$ and the hydrogen in the above equation is multiplied by $\frac{3}{2}$ .

$\frac{1}{2}{\text{N}}_2\left(g\right)+\frac{3}{2}{\text{H}}_2\left(g\right)\to {\text{NH}}_3\left(g\right)$

The chemical equation corresponds to the formation of $\text{CO}\left(g\right)$ is,

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

Since, one carbon atom and two oxygen atoms are required to form one mole of $\text{CO}\left(g\right)$ . Therefore, oxygen in the above equation is multiplied by $\frac{1}{2}$ .

$\text{C}\left(s\right)+\frac{1}{2}{\text{O}}_2\left(g\right)\to \text{CO}\left(g\right)$

Conclusion

The chemical equation corresponds to $\Delta G_f^0$ for ${\text{NH}}_3\left(g\right)$ is,

$\frac{1}{2}{\text{N}}_2\left(g\right)+\frac{3}{2}{\text{H}}_2\left(g\right)\to {\text{NH}}_3\left(g\right)$

The chemical equation corresponds to $\Delta G_f^0$ for $\text{CO}\left(g\right)$ is,

$\text{C}\left(s\right)+\frac{1}{2}{\text{O}}_2\left(g\right)\to \text{CO}\left(g\right)$

(b)

Interpretation Introduction

To determine: The formation reaction for which the value of $\Delta G_f^0$ is more positive than $\Delta H_f^0$ out of the given reactions.

Answer to Problem 1DE

Solution: The value $\Delta G_f^0$ is more positive than $\Delta H_f^0$ for the formation of ${\text{NH}}_3\left(g\right)$ .

Explanation of Solution

Given

The chemical equation corresponds to $\Delta G_f^0$ for ${\text{NH}}_3\left(g\right)$ is,

$\frac{1}{2}{\text{N}}_2\left(g\right)+\frac{3}{2}{\text{H}}_2\left(g\right)\to {\text{NH}}_3\left(g\right)$

The chemical equation corresponds to $\Delta G_f^0$ for $\text{CO}\left(g\right)$ is,

$\text{C}\left(s\right)+\frac{1}{2}{\text{O}}_2\left(g\right)\to \text{CO}\left(g\right)$

The reaction of formation of ammonia involves the formation of one gaseous product by two gaseous reactants which results in decrease in entropy. This results in the more positive value of $\Delta G_f^0$ in comparison to $\Delta H_f^0$ .
The reaction of formation of carbon monoxide involves the formation of one gaseous product by one gaseous and one solid reactant which results in increase in entropy. This results in the more negative value of $\Delta G_f^0$ in comparison to $\Delta H_f^0$ .

Conclusion

The value $\Delta G_f^0$ is more positive than $\Delta H_f^0$ for the formation of ${\text{NH}}_3\left(g\right)$ .

(c)

Interpretation Introduction

To determine: The conditions in which the value of $\Delta G_f^0$ is more positive than $\Delta H_f^0$ out of the given conditions.

Answer to Problem 1DE

Solution: The conditions when $\Delta S_f^0$ is negative, the value of $\Delta G_f^0$ is more positive than $\Delta H_f^0$ that corresponds to option (iii).

Explanation of Solution

The relation between $\Delta G_f^0$ and $\Delta H_f^0$ is defined by the formula,

$\Delta G^0=\Delta H^0-T\Delta S^0$

Where,

• $\Delta S^0$ is the standard entropy change of the process.
• $\Delta H^0$ is the standard enthalpy change of the process.
• $\Delta G^0$ is the standard free energy change of the process.
• $T$ is the temperature of the process.

The given conditions can be explained as follows;

1. At high temperature conditions the value of standard free energy change of the given reaction will be more negative as compared to the standard enthalpy change as per the above formula.
2. The reversible reaction contains negative value of standard free energy change for the spontaneous reaction and the positive value of $\Delta G^0$ for the reverse reaction.
3. At the conditions of negative $\Delta S_f^0$ value, the value of $\Delta G_f^0$ will be more positive as compared to $\Delta H_f^0$ .

Conclusion

The conditions when $\Delta S_f^0$ is negative, the value of $\Delta G_f^0$ is more positive than $\Delta H_f^0$ that corresponds to option (iii).