Chapter 5, Problem 82AP

Interpretation Introduction

Interpretation:

The heat formedfrommethane as well as water gas combustionis to be compared. To check whether methane is preferred over water gas as a fuel ornot, the reason for the preference of methane over water gas is to be explained.

Concept introduction:

The standard enthalpy for a reaction is the change in enthalpy that occurs under standard conditions, that is, under room temperature and pressure (${\text{27}}^0\text{C}\text{and}1\text{atm}$).

The standard enthalpy of the reaction is to be determined using the equation given below:

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

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

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

Answer to Problem 82AP

Solution: Methane is preferred over water gas.

Methane is preferred because it is a natural gas and nontoxic in nature.

Explanation of Solution

Given information:

One mole of water gas, which is equal to $0.5{\text{ mol H}}_{\text{2}}$ and $0.5\text{ mol CO}$.

The combustion of methane is expressed as

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

From appendix 2,

$\begin{array}{l}\Delta H{°}_{\text{f}}\left({\text{CH}}_{\text{4}}\right)=-74.85\text{}\text{ kJ}/\text{mol}\\ \Delta H{°}_{\text{f}}\left({\text{H}}_{\text{2}}\text{O}\right)=-285.8\text{}\text{kJ}/\text{mol}\\ \Delta H{°}_{\text{f}}\left({\text{CO}}_{\text{2}}\right)=-393.5\text{kJ}/\text{mol}\end{array}$

The enthalpy of formation of oxygen gas is zero because it is in itsmost stable form.

Calculate the standard enthalpy of reactionas follows:

$\Delta H{°}_{\text{rxn}}=\left[\Delta H{°}_{\text{f}}\left({\text{CO}}_{\text{2}}\right)+2\Delta H{°}_{\text{f}}\left({\text{H}}_2\text{O}\right)\right]-\left[\Delta H{°}_{\text{f}}\left({\text{CH}}_{\text{4}}\right)+2\text{}\Delta H{°}_{\text{f}}\left({\text{O}}_{\text{2}}\right)\right]$

Substitute $-74.85\text{}\text{ kJ}/\text{mol}$ for $\Delta H{°}_{\text{f}}\left({\text{CH}}_{\text{4}}\right)$, $-285.8\text{}\text{kJ}/\text{mol}$ for $\Delta H{°}_{\text{f}}\left({\text{H}}_{\text{2}}\text{O}\right)$, and $-393.5\text{kJ}/\text{mol}$ for $\Delta H{°}_{\text{f}}\left({\text{CO}}_{\text{2}}\right)$ in the above equation

$\begin{array}{c}\Delta H{°}_{\text{rxn}}=\left[\left(1\right)\left(-393.5\text{ kJ}/mol\right)+\left(2\right)\left(-285.8\text{kJ}/mol\right)\right]-\left[\left(1\right)\left(-74.85\text{kJ}/mol\right)+2\left(0\right)\right]\\ =-890.3\text{ kJ}/\text{mol}\end{array}$

The heat produced by water gas can be calculated from combustion of ${\text{H}}_{\text{2}}\text{ and CO}$, which are calculated in distinct steps.

The combustion of ${\text{H}}_{\text{2}}$ is shown below:

${\text{H}}_{\text{2}}\left(g\right)+\frac{1}{2}{\text{O}}_2\left(g\right)\to {\text{H}}_2\text{O}\left(l\right)$

From appendix 2,

$\Delta H{°}_{\text{f}}\left({\text{H}}_{\text{2}}\text{O}\right)=-285.8\text{}\text{ kJ}/\text{mol}$

Here, the enthalpy of formation is also the enthalpy of the reaction as reactants are in their most stable state and have the enthalpy of formation equal to zero.

$\Delta H{°}_{\text{rxn}}=\Delta H{°}_{\text{f}}\left({\text{H}}_{\text{2}}\text{O}\right)$

Substitute $-285.8\text{}\text{ kJ}/\text{mol}$ for $\Delta H{°}_{\text{rxn}}$ in above equation

$\Delta H{°}_{\text{rxn}}=-285.8\text{ kJ}/\text{mol}$

The combustion of $\text{CO}$ isshown below:

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

From appendix 2,

$\begin{array}{l}\Delta H{°}_{\text{f}}\left(\text{CO}\right)=-110.5\text{ kJ}/mol\\ \Delta H{°}_{\text{f}}\left({\text{CO}}_{\text{2}}\right)=-393.5\text{ kJ}/mol\end{array}$

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

Calculate the standard enthalpy of reaction for the combustion of $\text{CO}$ as follows:

$\Delta H{\text{°}}_{\text{rxn}}=\left[\Delta H{°}_{\text{f}}\left({\text{CO}}_2\right)\right]-\left[\Delta H{°}_{\text{f}}\left(\text{CO}\right)+\frac{1}{2}\Delta H{°}_{\text{f}}\left({\text{O}}_2\right)\right]$

Substitute $-110.5\text{ kJ}/mol$ for $\Delta H{°}_{\text{f}}\left(\text{CO}\right)$, $-393.5\text{ kJ}/mol$ for $\Delta H{°}_{\text{f}}\left({\text{CO}}_{\text{2}}\right)$ in the above equation

$\begin{array}{c}\Delta H{\text{°}}_{\text{rxn}}=\left[\left(-393.5\text{ kJ}/mol\right)\right]-\left[\left(-110.5\text{ kJ}/mol\right)+2\left(0\right)\right]\\ =-283\text{ kJ}/mol\end{array}$

The total heat produced during the combustion of one mole of water gas can be calculated as

$\begin{array}{c}\Delta H{°}_{\text{rxn}}=\frac{1}{2}\left(285.8\right)+\frac{1}{2}\left(283\right)\\ =142.9+141.5\\ =284.4\text{ kJ}/\text{mol}\end{array}$

It is clear from the values that the combustion of 1mole of methane produces more heat as compared to the combustion of1 mole of water gas that has half mole of each ${\text{H}}_{\text{2}}\text{ and CO}$. Hence, methane is preferred over water gas.

Other reasons for the preference of methane over water gas are as follows:

Methane gas is easier to obtain than water gas because it can be organically obtained.

Water gas contains $\text{CO}$, which is highly toxic in nature and makes it more harmful than methane.

Therefore, methane is preferred over water gas.

Conclusion

Methane is preferred over water gasbecause methane gas organically obtained, it produces more heat in combustion and nontoxic. On the other hand, water gas contains toxic $\text{CO }$ and produce less heat in combustion.