Chapter 6, Problem 6.91QP

Interpretation Introduction

Interpretation:

• The suggestions for preferring Methane over Water gas as fuel has to be explained.
• The heat produced by combustion of 1 mole of Methane with 1 mole of Water has to be compared.

Concept Introduction:

The change in enthalpy that is associated with the formation of one mole of a substance from its related elements being in standard state is called standard enthalpy of formation (${\text{ΔH}}_{\text{f}}\text{°}$).  The standard enthalpy of formation is used to determine the standard enthalpies of compound and element.

The standard enthalpy of reaction is the enthalpy of reaction that takes place under standard conditions.

 The equation for determining the standard enthalpies of compound and element can be given by,

${\text{ΔH°}}_{\text{reaction}}\text{=}{\displaystyle \sum {\text{nΔH°}}_{\text{f}}\text{(products)}}\text{-}{\displaystyle \sum {\text{mΔH°}}_{\text{f}}\text{(reactants)}}$

To explain: Why Methane is better fuel than Water gas

Answer to Problem 6.91QP

The heat produced by combustion of 1 mole of Water gas is less than the heat produced by combustion of 1 mole of Methane gas.

Methane can be easily obtained on compared to production of Water gas that can be obtained by high temperature process.

Explanation of Solution

To calculate the heat produced

The equation can be given as,

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

Standard enthalpy of formation of ${\text{CO}}_{\text{2}}$= $\text{-393}\text{.5}\text{kJ}{\text{mol}}^{\text{-1}}$

Standard enthalpy of formation of Water = $\text{-285}\text{.8}\text{kJ}{\text{mol}}^{\text{-1}}$

Standard enthalpy of formation of Oxygen = $\text{0}\text{kJ}{\text{mol}}^{\text{-1}}$

Standard enthalpy of formation of Methane = $\text{-74}\text{.85}\text{kJ}{\text{mol}}^{\text{-1}}$

$\begin{array}{l}{\text{ΔH°}}_{\text{reaction}}{\text{=ΔH°}}_{\text{f}}{\text{(CO}}_{\text{2}}{\text{)+2ΔH°}}_{\text{f}}{\text{(H}}_{\text{2}}{\text{O)-ΔH°}}_{\text{f}}{\text{(CH}}_{\text{4}}\text{)}\\ \\ {\text{ΔH°}}_{\text{reaction}}\text{=(1)(-393}\text{.5}\text{kJ}{\text{mol}}^{\text{-1}}\text{)+(2)(-285}\text{.8}\text{kJ}{\text{mol}}^{\text{-1}}\text{)-(1)(-74}\text{.85}\text{kJ}{\text{mol}}^{\text{-1}}\text{)}\\ \\ {\text{ΔH°}}_{\text{reaction}}\text{=-890}\text{.3}\text{kJ}{\text{mol}}^{\text{-1}}\end{array}$

Heat produced = $\text{-890}\text{.3}\text{kJ}{\text{mol}}^{\text{-1}}$

To calculate the enthalpy of Water

${\text{H}}_{\text{2(g)}}\text{+}\frac{\text{1}}{\text{2}}{\text{O}}_{\text{2(g)}}\to {\text{H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}{\text{ΔH°}}_{\text{f}}\text{=-285}\text{.8}\text{kJ}{\text{mol}}^{\text{-1}}$

To calculate the ${\text{ΔH°}}_{\text{reaction}}$ of reaction

From combustion of ${\text{CO}}_{\left(\text{g}\right)}$ and ${\text{ΔH°}}_{\text{f}}$ , ${\text{ΔH°}}_{\text{reaction}}$ can be calculated,

${\text{CO}}_{\left(\text{g}\right)}\text{+}\frac{\text{1}}{\text{2}}{\text{O}}_{\text{2(g)}}\to {\text{CO}}_{\text{2(g)}}{\text{ΔH°}}_{\text{reaction}}\text{=?}$

$\begin{array}{l}{\text{ΔH°}}_{\text{reaction}}{\text{=ΔH°}}_{\text{f}}{\text{(CO}}_{\text{2}}{\text{)-ΔH°}}_{\text{f}}\text{(CO)-}\frac{\text{1}}{\text{2}}{\text{ΔH°}}_{\text{f}}{\text{(O}}_{\text{2}}\text{)}\\ \\ {\text{ΔH°}}_{\text{reaction}}\text{=(1)(-393}\text{.5}\text{kJ}{\text{mol}}^{\text{-1}}\text{)-(1)(-110}\text{.5}\text{kJ}{\text{mol}}^{\text{-1}}\text{)}\\ \\ {\text{ΔH°}}_{\text{reaction}}\text{=-283}\text{.0}\text{kJ}{\text{mol}}^{\text{-1}}\end{array}$

Enthalpy of reaction = $\text{-283}\text{.0}\text{kJ}{\text{mol}}^{\text{-1}}$

To calculate the total heat produced and compare

Enthalpy of reaction = $\text{-283}\text{.0}\text{kJ}{\text{mol}}^{\text{-1}}$

Enthalpy of Water = $\text{-285}\text{.8 kJ}{\text{mol}}^{\text{-1}}$

Heat produced during the combustion of 1 mole of Water gas = $\frac{\text{-(285}\text{.8+283}\text{.0)kJ}{\text{mol}}^{\text{-1}}}{\text{2}}$

     = $\text{-284}\text{.4}\text{kJ}{\text{mol}}^{\text{-1}}$

The heat produced by combustion of 1 mole of Water gas is less than the heat produced by combustion of 1 mole of Methane gas.

To explain why Methane is better fuel than Water gas

Methane can be easily obtained on compared to production of Water gas that can be obtained by high temperature process.

The heat produced by combustion of 1 mole of Water gas is less than the heat produced by combustion of 1 mole of Methane gas.

Methane that is natural gas can be easily obtained when compared to production of Water gas that requires a very high temperature process.

Conclusion

• The suggestions for preferring Methane over Water gas as fuel was explained.
• The heat produced by combustion of 1 mole of Methane with 1 mole of Water was compared.