Chapter 4, Problem 4.16E

Calculate $\Delta G$ in two different ways for the combustion of benzene:

$2{\text{C}}_{\text{6}}{\text{H}}_{\text{6}}\left(l\right)+15{\text{O}}_2\left(\text{g}\right)\to 12{\text{CO}}_2\left(\text{g}\right)+6{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Are the two values equal?

Interpretation Introduction

Interpretation:

The value of $\Delta G$ for the combustion reaction of benzene is to be calculated in two different ways and whether the two values are equal or not is to be predicted.

Concept introduction:

The standard Gibbs-energy gives the non-expansion work achieved from the system at a constant temperature and pressure for the reaction. The standard Gibbs-energy for the reaction is represented as $\Delta G°$.

Answer to Problem 4.16E

The value of $\Delta G$ calculated by two different methods is $-6403.84\text{kJ}$ and $-6404.26\text{kJ}$. The value of $\Delta G$ calculated by two different methods is almost equal.

Explanation of Solution

The combustion reaction of benzene is shown below.

$2{\text{C}}_{\text{6}}{\text{H}}_6\left(l\right)+1{\text{5O}}_{\text{2}}\left(\text{g}\right)\to 1{\text{2CO}}_{\text{2}}\left(\text{g}\right)+6{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The value of $\Delta G$ for the combustion reaction of benzene can be calculated by the formula as shown below.

$\Delta G=\left[12{\Delta }_{\text{f}}G°\left({\text{CO}}_2\right)+6{\Delta }_{\text{f}}G°\left({\text{H}}_{\text{2}}\text{O}\right)\right]-\left[2{\Delta }_{\text{f}}G°\left({\text{C}}_{\text{6}}{\text{H}}_{\text{6}}\right)+15{\Delta }_{\text{f}}G°\left({\text{O}}_2\right)\right]$ …(1)

Where,

• ${\Delta }_{\text{f}}G°\left({\text{CO}}_{\text{2}}\right)$ is the change in Gibbs free energy for the formation of ${\text{CO}}_{\text{2}}$.

• ${\Delta }_{\text{f}}G°\left({\text{H}}_{\text{2}}\text{O}\right)$ is the change in Gibbs free energy for the formation of ${\text{H}}_{\text{2}}\text{O}$.

• ${\Delta }_{\text{f}}G°\left({\text{C}}_{\text{6}}{\text{H}}_{\text{6}}\right)$ is the change in Gibbs free energy for the formation of ${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}$.

• ${\Delta }_{\text{f}}G°\left({\text{O}}_2\right)$ is the change in Gibbs free energy for the formation of ${\text{O}}_2$.

The value of change in the Gibbs free energy for the formation of ${\text{CO}}_{\text{2}}$, ${\text{H}}_{\text{2}}\text{O}$, ${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}$, and ${\text{O}}_2$ is $-394.35\text{kJ}/\text{mol}$, $-237.14\text{kJ}/\text{mol}$, $124.4\text{kJ}/\text{mol}$ and $0\text{kJ}/\text{mol}$ respectively.

Substitute the value of change in the Gibbs free energy for the formation of ${\text{CO}}_{\text{2}}$, ${\text{H}}_{\text{2}}\text{O}$, ${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}$, and ${\text{O}}_2$ in equation (1).

$\begin{array}{rll}\Delta G& =& \left[12\times \left(-394.35\text{kJ}/\text{mol}\right)+6\times \left(-237.14\text{kJ}/\text{mol}\right)\right]-\left[2\times \left(124.4\text{kJ}/\text{mol}\right)+15\times 0\right]\\ & =& -4732.2\text{kJ}-1422.84\text{kJ}-248.8\text{kJ}\\ & =& -6403.84\text{kJ}\end{array}$

The value of ${\Delta }_{\text{f}}H°\text{and}\Delta S°$ for ${\text{CO}}_{\text{2}}$ is $-393.51\text{kJ}/\text{mol}$ and $213.785\text{J}/\text{mol}\cdot \text{K}$.

The value of ${\Delta }_{\text{f}}H°\text{and}\Delta S°$ for ${\text{H}}_{\text{2}}\text{O}$ is $-285.83\text{kJ}/\text{mol}$ and $69.91\text{J}/\text{mol}\cdot \text{K}$.

The value of ${\Delta }_{\text{f}}H°\text{and}\Delta S°$ for ${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}$ is $48.95\text{kJ}/\text{mol}$ and $173.26\text{J}/\text{mol}\cdot \text{K}$.

The value of ${\Delta }_{\text{f}}H°\text{and}\Delta S°$ for ${\text{O}}_2$ is $0\text{kJ}/\text{mol}$ and $205.14\text{J}/\text{mol}\cdot \text{K}$.

The value of $\Delta G$ for the given reaction can be calculated by the formula as shown below.

$\Delta G=\Delta H-T\Delta S$ …(2)

The value of $\Delta H$ for the given reaction is calculated by the formula as shown below.

$\Delta H=\left[12{\Delta }_{\text{f}}H°\left({\text{CO}}_2\right)+6{\Delta }_{\text{f}}H°\left({\text{H}}_{\text{2}}\text{O}\right)\right]-\left[2{\Delta }_{\text{f}}H°\left({\text{C}}_{\text{6}}{\text{H}}_{\text{6}}\right)+15{\Delta }_{\text{f}}H°\left({\text{O}}_2\right)\right]$ …(3)

Substitute the value of ${\Delta }_{\text{f}}H°$ of ${\text{CO}}_{\text{2}}$, ${\text{H}}_{\text{2}}\text{O}$, ${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}$, and ${\text{O}}_2$ in equation (3).

$\begin{array}{rll}\Delta H& =& \left[12\times \left(-393.51\text{kJ}/\text{mol}\right)+6\times \left(-285.83\text{kJ}/\text{mol}\right)\right]-\left[2\times \left(48.95\text{kJ}/\text{mol}\right)+15\times 0\right]\\ & =& -4722.12\text{kJ}-1714.98\text{kJ}-97.9\text{kJ}\\ & =& -6535\text{kJ}\end{array}$

The value of $\Delta S$ for the given reaction is calculated by the formula as shown below.

$\Delta S=\left[12\Delta S°\left({\text{CO}}_2\right)+6\Delta S°\left({\text{H}}_{\text{2}}\text{O}\right)\right]-\left[2\Delta S°\left({\text{C}}_{\text{6}}{\text{H}}_{\text{6}}\right)+15\Delta S°\left({\text{O}}_2\right)\right]$ …(4)

Substitute the value of $\Delta S°$ of ${\text{CO}}_{\text{2}}$, ${\text{H}}_{\text{2}}\text{O}$, ${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}$ and ${\text{O}}_2$ in equation (4).

$\begin{array}{rll}\Delta S& =& \left[\begin{array}{l}\left(12\times \left(213.785\text{J}/\text{mol}\cdot \text{K}\right)+6\times \left(69.91\text{J}/\text{mol}\cdot \text{K}\right)\right)-\\ \left(2\times \left(173.26\text{J}/\text{mol}\cdot \text{K}\right)+15\times 205.14\text{J}/\text{mol}\cdot \text{K}\right)\end{array}\right]\\ & =& 2565.42\text{J}/\text{K}+419.46\text{J}/\text{K}-\left(346.52\text{J}/\text{K}+3077.1\text{J}/\text{K}\right)\\ & =& -438.74\text{J}/\text{K}\end{array}$

Substitute the value of $\Delta S$ and $\Delta H$ in equation (2).

$\begin{array}{rll}\Delta G& =& -6535\text{kJ}-\left(298\text{K}\times \left(-438.74\text{J}/\text{K}\right)\right)\\ & =& -6535\text{kJ}+130.74\text{kJ}\\ & =& -6404.26\text{kJ}\end{array}$

The value of $\Delta G$ calculated by two different methods is $-6403.84\text{kJ}$ and $-6404.26\text{kJ}$. The value of $\Delta G$ calculated by two different methods is almost equal.

Conclusion

The value of $\Delta G$ calculated by two different methods is $-6403.84\text{kJ}$ and $-6404.26\text{kJ}$. The value of $\Delta G$ calculated by two different methods is almost equal.