Chapter 5, Problem 68QP

Interpretation Introduction

Interpretation:

The heat of combustion of the given reactions is to be calculated with reference of standard enthalpies of formation listed in appendix $2$.

Concept introduction:

Answer to Problem 68QP

Solution:

(a) $-1411\text{ kJ}/\text{mol}$

(b) $-1124\text{ kJ}/\text{mol}$

Explanation of Solution

a) ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\left(g\right)+3{\text{O}}_2\left(g\right)\to 2{\text{CO}}_2+2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The combustion of the given reaction is as follows:

${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\left(g\right)+3{\text{O}}_2\left(g\right)\to 2{\text{CO}}_2+2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

From appendix 2,

$\begin{array}{c}\Delta H{°}_f\left({\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\right)=-52.3\text{ kJ}/\text{mol}\\ \Delta H{°}_f\left({\text{H}}_{\text{2}}\text{O}\right)=-285.8\text{ kJ}/\text{mol}\\ \Delta H{°}_f\left({\text{CO}}_{\text{2}}\right)=-393.5\text{ kJ}/\text{mol}\end{array}$

The enthalpy of the formation oxygen gas is zero because it is in its most stable form.

Calculate the standard enthalpy of the reaction:

$\Delta H{°}_{\text{rxn}}=\left\{2\Delta H{°}_f\left[{\text{H}}_{\text{2}}\text{O}\left(l\right)\right]+2\Delta H{°}_f\left[{\text{CO}}_2\left(g\right)\right]\right\}-\left\{\Delta H{°}_f\left[{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\left(\text{g}\right)\right]+3\Delta H{°}_f\left[{\text{O}}_2\left(g\right)\right]\right\}$

Substitute $-52.3\text{ kJ}/\text{mol}$ for $\Delta H{°}_f\left[{\text{C}}_{\text{2}}{\text{H}}_4\left(g\right)\right]$, $-285.8\text{ kJ}/\text{mol}$ for $\Delta H{°}_f\left[{\text{H}}_{\text{2}}\text{O}\left(l\right)\right]$, $-393.5\text{ kJ}/\text{mol}$ for $\Delta H{°}_f\left[{\text{CO}}_{\text{2}}\left(g\right)\right]$, and $0$ for $\Delta H{°}_f\left[{\text{O}}_{\text{2}}\left(g\right)\right]$ in the above equation

$\begin{array}{c}\Delta H{°}_{\text{rxn}}=\left[2\left(-285.8\text{ kJ}/\text{mol}\right)+2\left(-393.5\text{ kJ}/\text{mol}\right)\right]-\left[\left(-52.3\text{ kJ}/\text{mol}\right)+3\left(0\right)\right]\\ =\left(-571.6\text{ kJ}/\text{mol}-787\text{ kJ}/\text{mol}\right)-\left(52.3\text{ kJ}/\text{mol}\right)\\ =-1411\text{ kJ}/\text{mol}\end{array}$

Hence, the value of the standard enthalpy of the reaction is $-1411\text{ kJ}/\text{mol}$.

b) $2{\text{H}}_{\text{2}}\text{S}\left(g\right)+3{\text{O}}_2\left(g\right)\to 2{\text{H}}_{\text{2}}\text{O}\left(l\right)+2{\text{SO}}_2\left(g\right)$

The combustion of the given reaction is as follows:

$2{\text{H}}_{\text{2}}\text{S}\left(g\right)+3{\text{O}}_2\left(g\right)\to 2{\text{H}}_{\text{2}}\text{O}\left(l\right)+2{\text{SO}}_2\left(g\right)$

From appendix 2,

$\begin{array}{c}\Delta H{°}_f\left({\text{H}}_{\text{2}}\text{S}\right)=-20.15\text{ kJ}/\text{mol}\\ \Delta H{°}_f\left({\text{H}}_{\text{2}}\text{O}\right)=-285.8\text{ kJ}/\text{mol}\\ \Delta H{°}_f\left({\text{SO}}_2\right)=-296.4\text{ kJ}/\text{mol}\end{array}$

The enthalpy of the formation of oxygen gas is zero because it is in its most stable form.

Calculate the standard enthalpy of the given reaction as follows:

$\Delta H{°}_{\text{rxn}}=\left\{2\Delta H{°}_f\left[{\text{SO}}_2\left(g\right)\right]+2\Delta H{°}_f\left[{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\right]\right\}-\left\{2\Delta H{°}_f\left[{\text{H}}_{\text{2}}\text{S}\left(\text{g}\right)\right]+3\Delta H{°}_f\left[{\text{O}}_2\left(g\right)\right]\right\}$

Substitute $-285.8\text{ kJ}/\text{mol}$ for $\Delta H{°}_f\left(H_{2}O\right)$, $0$ for $\Delta H{°}_f\left[{\text{O}}_{\text{2}}\left(g\right)\right]$, $-20.15\text{ kJ}/\text{mol}$ for $\Delta H{°}_f\left[{\text{H}}_{\text{2}}\text{S}\left(\text{g}\right)\right]$, and $-296.4\text{ kJ}/\text{mol}$ for $\Delta H{°}_f\left[{\text{SO}}_2\left(g\right)\right]$ in above equation

$\begin{array}{c}\Delta H{°}_{\text{rxn}}=\left[2\left(-296.4\text{ kJ}/\text{mol}\right)+2\left(-285.8\text{ kJ}/\text{mol}\right)\right]-\left[2\left(-20.15\text{ kJ}/\text{mol}\right)+3\left(0\right)\right]\\ =\left[-592.8\text{ kJ}/\text{mol}+\left(-571.6\text{ kJ}/\text{mol}\right)\right]-\left(-40.3\text{ kJ}/\text{mol}\right)\\ =-1124\text{ kJ}/\text{mol}\end{array}$

Hence, the value of the standard enthalpy of the reaction is $-1124\text{ kJ}/\text{mol}$.