Chapter 5, Problem 5.118AP

(a)

Interpretation Introduction

Interpretation: An explanation corresponding to the fact that ${\text{C}}_2{\text{H}}_2$ and ${\text{C}}_6{\text{H}}_6$ are known as endothermic compounds is to be stated.

The standard molar enthalpy of combustion of acetylene and benzene is to be calculated.

Concept introduction: The compounds that have positive $\Delta H_{\text{f}}^{\text{ο}}$ value are termed as endothermic compounds.

The standard molar enthalpy of combustion for a reaction is the change in enthalpy when one mole of a compound goes under combustion process.

To determine: The reason for calling ${\text{C}}_2{\text{H}}_2$ and ${\text{C}}_6{\text{H}}_6$ as endothermic compounds.

Answer to Problem 5.118AP

Solution

The reason for calling ${\text{C}}_2{\text{H}}_2$ and ${\text{C}}_6{\text{H}}_6$ as endothermic compounds is stated below.

Explanation of Solution

Explanation

Given

The $\Delta H_{\text{f}}^{\text{ο}}$ value of ${\text{C}}_2{\text{H}}_2$ is $226.7\text{kJ}/\text{mol}$.

The $\Delta H_{\text{f}}^{\text{ο}}$ value of ${\text{C}}_6{\text{H}}_6$ is $49.0\text{kJ}/\text{mol}$.

The compounds that have positive $\Delta H_{\text{f}}^{\text{ο}}$ value are termed as endothermic compounds. Here, in the given the $\Delta H_{\text{f}}^{\text{ο}}$ value of both ${\text{C}}_2{\text{H}}_2$ and ${\text{C}}_6{\text{H}}_6$ are positive. Hence, ${\text{C}}_2{\text{H}}_2$ and ${\text{C}}_6{\text{H}}_6$ are referred as endothermic compounds.

(b)

Interpretation Introduction

To determine: The standard molar enthalpy of combustion of acetylene and benzene.

Answer to Problem 5.118AP

Solution

The standard molar enthalpies of combustion of acetylene and benzene are $\underset{\_}{-1,299.6kJ/mol}$ and $\underset{\_}{-3,267.7kJ/mol}$.

Explanation of Solution

Explanation

Given

The $\Delta H_{\text{f}}^{\text{ο}}$ value of ${\text{C}}_2{\text{H}}_2$ is $226.7\text{kJ}/\text{mol}$.

The $\Delta H_{\text{f}}^{\text{ο}}$ value of ${\text{C}}_6{\text{H}}_6$ is $49.0\text{kJ}/\text{mol}$.

The standard molar enthalpy of combustion for a reaction is the change in enthalpy when one mole of a compound goes under combustion process. The combustion processes of acetylene and benzene are written as,

$\begin{array}{c}{\text{C}}_2{\text{H}}_2\left(g\right)+\frac{5}{2}{\text{O}}_2\left(g\right)\to 2{\text{CO}}_2\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\\ {\text{C}}_6{\text{H}}_6\left(g\right)+\frac{15}{2}{\text{O}}_2\left(g\right)\to 6{\text{CO}}_2\left(g\right)+3{\text{H}}_{\text{2}}\text{O}\left(l\right)\end{array}$

The $\Delta H_{\text{f}}^{\text{ο}}$ value of ${\text{CO}}_2\left(g\right)$ is $-393.5\text{kJ}/\text{mol}$.

The $\Delta H_{\text{f}}^{\text{ο}}$ value of ${\text{H}}_2\text{O}\left(l\right)$ is $-285.9\text{kJ}/\text{mol}$.

The $\Delta H_{\text{f}}^{\text{ο}}$ value of ${\text{O}}_2\left(g\right)$ is $0\text{kJ}/\text{mol}$

The standard molar enthalpy change of a combustion process is calculated by the formula,

$\Delta H_{\text{comb}.}^{\text{ο}}={\displaystyle \sum \Delta H_{\text{f}}^{\text{ο}}\left(\text{Products}\right)}-{\displaystyle \sum \Delta H_{\text{f}}^{\text{ο}}\left(\text{Reactants}\right)}$

Where,

• $\Delta H_{\text{comb}.}^{\text{ο}}$ is the standard molar enthalpy change of a combustion process.
• $\Delta H_{\text{f}}^{\text{ο}}\left(\text{Products}\right)$ is the standard molar enthalpy change of products.
• $\Delta H_{\text{f}}^{\text{ο}}\left(\text{Reactants}\right)$ is the standard molar enthalpy change of reactants.

The standard molar enthalpy of combustion of acetylene and benzene are calculated in (I) and (II) part.

(I)

The standard molar enthalpy of combustion of acetylene $\left(\Delta H_{\text{comb}.,\text{Acetylene}}^{\text{ο}}\right)$ is calculated from the above expression and rewritten as,

$\Delta H_{\text{comb}.,\text{Acetylene}}^{\text{ο}}=\left(2\times \Delta H_{{\text{f,CO}}_{\text{2}}\left(g\right)}^{\text{ο}}+1\times \Delta H_{{\text{f,H}}_2\text{O}\left(l\right)}^{\text{ο}}\right)-\left(1\times \Delta H_{{\text{f,C}}_2{\text{H}}_2\left(g\right)}^{\text{ο}}+\frac{5}{2}\times \Delta H_{{\text{f,O}}_{\text{2}}\left(g\right)}^{\text{ο}}\right)$ (1)

Where,

• $\Delta H_{{\text{f,CO}}_{\text{2}}\left(g\right)}^{\text{ο}}$ is the $\Delta H_{\text{f}}^{\text{ο}}$ value of ${\text{CO}}_2\left(g\right)$.
• $\Delta H_{{\text{f,H}}_2\text{O}\left(l\right)}^{\text{ο}}$ is the $\Delta H_{\text{f}}^{\text{ο}}$ value of ${\text{H}}_2\text{O}\left(l\right)$.
• $\Delta H_{{\text{f,C}}_2{\text{H}}_2\left(g\right)}^{\text{ο}}$ is the $\Delta H_{\text{f}}^{\text{ο}}$ value of ${\text{C}}_2{\text{H}}_2\left(g\right)$.
• $\Delta H_{{\text{f,O}}_{\text{2}}\left(g\right)}^{\text{ο}}$ is the $\Delta H_{\text{f}}^{\text{ο}}$ value of ${\text{O}}_2\left(g\right)$.

Substitute the values of $\Delta H_{\text{f}}^{\text{ο}}$ for reactants and products in equation (1).

$\begin{array}{c}\Delta H_{\text{comb}.,\text{Acetylene}}^{\text{ο}}=\left(2\times \Delta H_{{\text{f,CO}}_{\text{2}}\left(g\right)}^{\text{ο}}+1\times \Delta H_{{\text{f,H}}_2\text{O}\left(l\right)}^{\text{ο}}\right)-\left(1\times \Delta H_{{\text{f,C}}_2{\text{H}}_2\left(g\right)}^{\text{ο}}+\frac{5}{2}\times \Delta H_{{\text{f,O}}_{\text{2}}\left(g\right)}^{\text{ο}}\right)\\ \Delta H_{\text{comb}.,\text{Acetylene}}^{\text{ο}}=\left(\left(2\times -393.5+1\times -285.9\right)-\left(1\times -226.7+\frac{5}{2}\times 0\right)\right)\text{kJ}/\text{mol}\\ =\left(-1,072.9-226.7\right)\text{kJ}/\text{mol}\\ =-1,299.6\text{kJ}/\text{mol}\end{array}$

Hence, the standard molar enthalpy of combustion of acetylene is $\underset{\_}{-1,299.6kJ/mol}$.

(II)

The standard molar enthalpy of combustion of acetylene $\left(\Delta H_{\text{comb}.,\text{Benzene}}^{\text{ο}}\right)$ is calculated from the above expression and rewritten as,

$\Delta H_{\text{comb}.,\text{Benzene}}^{\text{ο}}=\left(6\times \Delta H_{{\text{f,CO}}_{\text{2}}\left(g\right)}^{\text{ο}}+3\times \Delta H_{{\text{f,H}}_2\text{O}\left(l\right)}^{\text{ο}}\right)-\left(1\times \Delta H_{{\text{f,C}}_6{\text{H}}_6\left(g\right)}^{\text{ο}}+\frac{15}{2}\times \Delta H_{{\text{f,O}}_{\text{2}}\left(g\right)}^{\text{ο}}\right)$ (2)

Where,

• $\Delta H_{{\text{f,CO}}_{\text{2}}\left(g\right)}^{\text{ο}}$ is the $\Delta H_{\text{f}}^{\text{ο}}$ value of ${\text{CO}}_2\left(g\right)$.
• $\Delta H_{{\text{f,H}}_2\text{O}\left(l\right)}^{\text{ο}}$ is the $\Delta H_{\text{f}}^{\text{ο}}$ value of ${\text{H}}_2\text{O}\left(l\right)$.
• $\Delta H_{{\text{f,C}}_6{\text{H}}_6\left(g\right)}^{\text{ο}}$ is the $\Delta H_{\text{f}}^{\text{ο}}$ value of. ${\text{C}}_6{\text{H}}_6\left(g\right)$.
• $\Delta H_{{\text{f,O}}_{\text{2}}\left(g\right)}^{\text{ο}}$ is the $\Delta H_{\text{f}}^{\text{ο}}$ value of ${\text{O}}_2\left(g\right)$.

Substitute the values of $\Delta H_{\text{f}}^{\text{ο}}$ for reactants and products in equation (2).

$\begin{array}{c}\Delta H_{\text{comb}.,\text{Benzene}}^{\text{ο}}=\left(6\times \Delta H_{{\text{f,CO}}_{\text{2}}\left(g\right)}^{\text{ο}}+3\times \Delta H_{{\text{f,H}}_2\text{O}\left(l\right)}^{\text{ο}}\right)-\left(1\times \Delta H_{{\text{f,C}}_6{\text{H}}_6\left(g\right)}^{\text{ο}}+\frac{15}{2}\times \Delta H_{{\text{f,O}}_{\text{2}}\left(g\right)}^{\text{ο}}\right)\\ \Delta H_{\text{comb}.,\text{Benzene}}^{\text{ο}}=\left(\left(6\times -393.5+3\times -285.9\right)-\left(1\times 49.0+\frac{15}{2}\times 0\right)\right)\text{kJ}/\text{mol}\\ =\left(-3,218.7-49.0\right)\text{kJ}/\text{mol}\\ =-3,267.7\text{kJ}/\text{mol}\end{array}$

Hence, the standard molar enthalpy of combustion of benzene is $\underset{\_}{-3,267.7kJ/mol}$.

Conclusion

1. a. The compounds that have positive $\Delta H_{\text{f}}^{\text{ο}}$ value are termed as endothermic compounds.
2. b. The standard molar enthalpies of combustion of acetylene and benzene are $\underset{\_}{-1,299.6kJ/mol}$ and $\underset{\_}{-3,267.7kJ/mol}$