Chapter 5, Problem 102AP

Interpretation Introduction

Interpretation:

The standard enthalpy of formation of benzene and acetylene, and hence, the enthalpy change for the formation of benzene from acetylene are to be calculated.

Concept introduction:

The standard enthalpy for a reaction is the amount of enthalpy that occursunderstandard conditions

The standard enthalpy of a reaction is determined by using the equation given below:

$\Delta H{°}_{\text{rxn}}={\displaystyle \sum n}\Delta H_{\text{f}}^{°}\left(\text{products}\right)-{\displaystyle \sum m}\Delta H_{\text{f}}^{°}\left(\text{reactants}\right)$

Here, the stoichiometric coefficients are represented by m for the reactants and n for the productsand the enthalpy of formation understandard conditions isrepresented by $\Delta H_{\text{f}}^{°}$.

The value of the enthalpy of formation of an element is zero at its most stable state.

Answer to Problem 102AP

Solution: $\Delta H{°}_{\text{f}}\left[{\text{C}}_{\text{6}}{\text{H}}_6\left(l\right)\right]=49.0\text{ kJ}/\text{mol}$, $\Delta H{°}_{\text{f}}\left[{\text{C}}_2{\text{H}}_2\left(g\right)\right]=226.6\text{ kJ}/\text{mol}$, and $\Delta H{°}_{\text{rxn}}=-630.8\text{ kJ}/\text{mol}$.

Explanation of Solution

Given information:

$\begin{array}{l}\Delta H°\left[{\text{C}}_2{\text{H}}_2\left(s\right)\right]=-1299.4\text{ kJ}/\text{mol}\\ \Delta H°\left[{\text{C}}_6{\text{H}}_6\left(s\right)\right]=-3267.4\text{ kJ}/\text{mol}\end{array}$

The given reaction is as follows:

${\text{3C}}_2{\text{H}}_2\left(g\right)\to {\text{C}}_6{\text{H}}_6\left(l\right)$

The reaction for the combustion of acetylene isas follows:

${\text{C}}_{\text{2}}{\text{H}}_2\left(g\right)+\frac{5}{2}{\text{O}}_2\left(g\right) \to 2{\text{CO}}_2\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right) \Delta H° = -1299.4\text{ kJ}/\text{mol}$

The reaction for the combustion of benzene isas follows:

${\text{C}}_{\text{6}}{\text{H}}_6\left(l\right)+\frac{15}{2}{\text{O}}_2\left(g\right) \to 6{\text{CO}}_2\left(g\right)+3{\text{H}}_{\text{2}}\text{O}\left(l\right) \Delta H° = -3267.4\text{ kJ/mol}$

From appendix 2,

$\begin{array}{c}\Delta H{°}_{\text{f}}\left[{\text{CO}}_2\left(g\right)\right]=-393.5\text{ kJ}/\text{mol}\\ \Delta H{°}_{\text{f}}\left[{\text{H}}_2\text{O}\left(l\right)\right]=-285.8\text{ kJ}/\text{mol}\\ \Delta H{°}_{\text{f}}\left[{\text{O}}_2\left(g\right)\right]=0\text{ kJ}/\text{mol}\end{array}$

The oxygen gas is in its most stable form, so its enthalpy of formation is zero.

Calculate the standard enthalpy for the combustion of acetylene as follows:

$\Delta H{°}_{\text{rxn}}=\left\{2\Delta H{°}_{\text{f}}\left[{\text{CO}}_2\left(g\right)\right]+\Delta H{°}_{\text{f}}\left[{\text{H}}_{\text{2}}\text{O}\left(l\right)\right]\right\}-\left\{\Delta H{°}_{\text{f}}\left[{\text{C}}_{\text{2}}{\text{H}}_2\left(g\right)\right]+\frac{5}{2}\Delta H{°}_{\text{f}}\left[{\text{O}}_2\left(g\right)\right]\right\}$

Substitute $-393.5\text{ kJ}/\text{mol}$ for $\Delta H{°}_{\text{f}}\left[{\text{CO}}_2\left(g\right)\right]$, $-285.8\text{ kJ}/\text{mol}$ for $\Delta H{°}_{\text{f}}\left[{\text{H}}_2\text{O}\left(l\right)\right]$, $-1299.4\text{ kJ}/\text{mol}$ for $\Delta H{°}_{\text{rxn}}$, and $0\text{ kJ}/\text{mol}$ for $\Delta H{°}_{\text{f}}\left[{\text{O}}_2\left(g\right)\right]$ in the above equation

$\begin{array}{c}-1299.4\text{ kJ}/\text{mol}=\left\{2\left(-393.5\text{ kJ}/\text{mol}\right)+\left(-285.8\text{ kJ}/\text{mol}\right)\right\}-\left\{\Delta H{°}_{\text{f}}\left[{\text{C}}_{\text{2}}{\text{H}}_2\left(g\right)\right]+\frac{5}{2}\left(0\right)\right\}\\ =\left[-787\text{ kJ}/\text{mol}-285.8\text{ kJ}/\text{mol}\right]-\Delta H{°}_{\text{f}}\left[{\text{C}}_{\text{2}}{\text{H}}_2\left(g\right)\right]\\ =\left(-1072.8\text{ kJ}/\text{mol}\right)-\Delta H{°}_{\text{f}}\left[{\text{C}}_{\text{2}}{\text{H}}_2\left(g\right)\right]\end{array}$

Rearrange the above equation for $\Delta H{°}_{\text{f}}\left[{\text{C}}_{\text{2}}{\text{H}}_2\left(g\right)\right]$ as follows:

$\begin{array}{c}\Delta H{°}_{\text{f}}\left[{\text{C}}_{\text{2}}{\text{H}}_2\left(g\right)\right]=-1072.8\text{ kJ}/\text{mol}+1299.4\text{ kJ}/\text{mol}\\ =226.6\text{ kJ}/\text{mol}\end{array}$

Calculate the standard enthalpy for thecombustion of benzene as follows:

$\Delta H{°}_{\text{rxn}}=\left\{6\Delta H{°}_{\text{f}}\left[{\text{CO}}_2\left(g\right)\right]+3\Delta H{°}_{\text{f}}\left[{\text{H}}_{\text{2}}\text{O}\left(l\right)\right]\right\}-\left\{\Delta H{°}_{\text{f}}\left[{\text{C}}_{\text{6}}{\text{H}}_6\left(l\right)\right]+\frac{15}{2}\Delta H{°}_{\text{f}}\left[{\text{O}}_2\left(g\right)\right]\right\}$

Substitute $-393.5\text{ kJ}/\text{mol}$ for $\Delta H{°}_{\text{f}}\left[{\text{CO}}_2\left(g\right)\right]$, $-285.8\text{ kJ}/\text{mol}$ for $\Delta H{°}_{\text{f}}\left[{\text{H}}_2\text{O}\left(l\right)\right]$, $-3267.4\text{ kJ}/\text{mol}$ for $\Delta H{°}_{\text{rxn}}$, and $0\text{ kJ}/\text{mol}$ for $\Delta H{°}_{\text{f}}\left[{\text{O}}_2\left(g\right)\right]$ in the above equation

$\begin{array}{c}-3267.4\text{ kJ}/\text{mol}=\left\{6\left(-393.5\text{ kJ}/\text{mol}\right)+3\left(-285.8\text{ kJ}/\text{mol}\right)\right\}-\left\{\Delta H{°}_{\text{f}}\left[{\text{C}}_{\text{6}}{\text{H}}_6\left(l\right)\right]+\frac{15}{2}\left(0\right)\right\}\\ =\left[-2361\text{ kJ}/\text{mol}-857.4\text{ kJ}/\text{mol}\right]-\Delta H{°}_{\text{f}}\left[{\text{C}}_{\text{6}}{\text{H}}_6\left(l\right)\right]\\ =\left(-3218.4\text{ kJ}/\text{mol}\right)-\Delta H{°}_{\text{f}}\left[{\text{C}}_{\text{6}}{\text{H}}_6\left(l\right)\right]\end{array}$

Rearrange the above equation for $\Delta H{°}_{\text{f}}\left[{\text{C}}_{\text{6}}{\text{H}}_6\left(l\right)\right]$ as follows:

$\begin{array}{c}\Delta H{°}_{\text{f}}\left[{\text{C}}_{\text{6}}{\text{H}}_6\left(l\right)\right]=-3218.4\text{ kJ}/\text{mol}+3267.4\text{ kJ}/\text{mol}\\ =49\text{ kJ}/\text{mol}\end{array}$

Calculate the enthalpy of the reaction for the formation of benzene from acetylene as follows:

$\Delta H{°}_{\text{rxn}}=\left\{\Delta H{°}_{\text{f}}\left[{\text{C}}_6{\text{H}}_6\left(l\right)\right]\right\}-\left\{3\Delta H{°}_{\text{f}}\left[{\text{C}}_2{\text{H}}_2\left(g\right)\right]\right\}$

Substitute $\text{49 kJ}/\text{mol}$ for $\Delta H{°}_{\text{f}}\left[{\text{C}}_6{\text{H}}_6\left(l\right)\right]$ and $226.6\text{ kJ}/\text{mol}$ for $\Delta H{°}_{\text{f}}\left[{\text{C}}_2{\text{H}}_2\left(g\right)\right]$ in the above equation

$\begin{array}{c}\Delta H{°}_{\text{rxn}}=\left\{\left(49\text{ kJ}/\text{mol}\right)\right\}-\left\{3\left(226.6\text{ kJ}/\text{mol}\right)\right\}\\ =49\text{ kJ}/\text{mol}-679.8\text{ kJ}/\text{mol}\\ =-630.8\text{ kJ}/\text{mol}\end{array}$

Conclusion

Theenthalpy change for theformation of benzene from acetylene is $-630.8\text{ kJ}/\text{mol}$.