Chapter 6, Problem 6.85QP

A gaseous mixture consists of 28.4 mole percent of hydrogen and 71.6 mole percent of methane. A 15.6-L gas sample, measured at 19.4°C and 2.23 atm, is burned in air. Calculate the heat released.

Interpretation Introduction

Interpretation:

The heat released has to be calculated.

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,

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

Answer to Problem 6.85QP

The heat released is calculated as $\text{1}{\text{.04×10}}^{\text{3}}\text{kJ}\text{.}$

Explanation of Solution

Given,

Mole percent of Hydrogen is $\text{28}\text{.4}\text{mole}\text{.}$

Mole percent of Methane is $\text{71}\text{.6}\text{mole}\text{.}$

Volume of the gas sample is $\text{15}\text{.6}\text{L}\text{.}$

Temperature is $\text{19}\text{.4°C}$

Pressure is $\text{2}\text{.23}\text{atm}$

The reactions are written as,

$\begin{array}{l}{\text{2H}}_{\text{2}\left(\text{g}\right)}{\text{+O}}_{\text{2(g)}}\stackrel{}{\to }{\text{2H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\text{(1)}\\ {\text{CH}}_{\text{4(g)}}{\text{+2O}}_{\text{2(g)}}\stackrel{}{\to }{\text{CO}}_{\text{2(g)}}{\text{+2H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\text{(2)}\end{array}$

The standard enthalpies of the reactions are calculated for both the equation,

For equation (1), the standard enthalpy of the reaction is,

$\begin{array}{l}\text{ΔH°}\text{=}\left(\text{2}\right){\text{ΔH°}}_{\text{f}}\left[{\text{H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\right]\\ \text{ΔH°}\text{=}\left(\text{2}\right)\left(\text{-285}\text{.8}\text{kJ/mole}\right)\\ \text{ΔH°}\text{=}\text{-571}\text{.6}\text{kJ/mole}\end{array}$

For reaction (2), the standard enthalpy of the reaction is,

$\begin{array}{l}\text{ΔH°}\text{=}{\text{ΔH°}}_{\text{f}}\left({\text{CO}}_{\text{2}}\right){\text{+2ΔH°}}_{\text{f}}\left[{\text{H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\right]\text{-ΔH°}\left({\text{CH}}_{\text{4}}\right)\\ \text{ΔH°}\text{=}\left(\text{-393}\text{.5}\text{kJ/mole}\right)\text{+2}\left(\text{-285}\text{.8}\text{kJ/mole}\right)\text{-}\left(\text{-74}\text{.85}\text{kJ/mole}\right)\\ \text{ΔH°}\text{=}\text{-890}\text{.3}\text{kJ/mole}\end{array}$

The total moles of gas in the mixture is calculated using the ideal gas equation

$\begin{array}{l}\text{n}\text{=}\frac{\text{PV}}{\text{RT}}\\ \\ \text{n}\text{=}\frac{\left(\text{2}\text{.23}\text{atm}\right)\left(\text{15}\text{.6}\text{L}\right)}{\left(\text{0}\text{.0821L}\text{.atm/K}\text{.mole}\right)\left(\text{273+19}\text{.4}\text{K}\right)}\\ \\ \text{n}\text{=}\text{1}\text{.45}\text{mole}\\ \\ \text{mole}{\text{H}}_{\text{2}}\text{=}\left(\text{0}\text{.284}\right)\left(\text{1}\text{.45}\text{mole}\right)\text{=}\text{0}\text{.412}\text{mole}\\ \text{mole}{\text{CH}}_{\text{4}}\text{=}\left(\text{0}\text{.716}\right)\left(\text{1}\text{.45}\text{mole}\right)\text{=}\text{1}\text{.04}\text{mole}\end{array}$

The heat released during the combustion is calculated as,

$\begin{array}{l}\text{Heat}\text{released}\text{=}\left(\text{0}\text{.412}\text{mole}{\text{H}}_{\text{2}}\text{×}\frac{\text{571}\text{.6}\text{kJ}}{\text{2}\text{mole}{\text{H}}_{\text{2}}}\right)\text{+}\left(\text{1}\text{.04}\text{mole}{\text{CH}}_{\text{4}}\text{×}\frac{\text{890}\text{.3}\text{kJ}}{\text{1}\text{mole}{\text{CH}}_{\text{4}}}\right)\\ \\ \text{Heat}\text{released}\text{=}\text{1}{\text{.04×10}}^{\text{3}}\text{kJ}\end{array}$

The heat released is calculated as $\text{1}{\text{.04×10}}^{\text{3}}\text{kJ}\text{.}$

Conclusion

The heat released was calculated as $\text{1}{\text{.04×10}}^{\text{3}}\text{kJ}\text{.}$