Questions: O mylabmastering.pearson.com Post-Lecture - Chapter 5 - Homework Problem 5.72 - Enhanced - with Feedback 14 of 18 Reviex I Constants I PeriodicTable The following table contains some standard enthalpies of formation, ΔH°f, at 298 K. Substance Formula ΔH°f (kJ / mol) --------- Carbon dioxide CO2(g) -393.5 Carbon monoxide CO(g) -110.5 Methane CH4(g) -74.80 Propane C3H8(g) -103.85 Water H2O(l) -285.8 Water vapor H2O(g) -241.8 Part A Many portable gas heaters and grills use propane, C3H8(g) Using enthalpies of formation, calculate the quantity of heat produced when 11.0 g of propane is completely combusted in air under standard conditions. Assume that liquid water is forming. Express the heat in kilojoules to three significant digits.

Solution Steps

The balanced chemical equation for the combustion of propane (\(\mathrm{C}_3\mathrm{H}_8\)) is:

\[ \mathrm{C}_3\mathrm{H}_8(\mathrm{g}) + 5\mathrm{O}_2(\mathrm{g}) \rightarrow 3\mathrm{CO}_2(\mathrm{g}) + 4\mathrm{H}_2\mathrm{O}(\mathrm{l}) \]

The molar mass of propane (\(\mathrm{C}_3\mathrm{H}_8\)) is:

\[ 3 \times 12.01 + 8 \times 1.008 = 44.094 \, \mathrm{g/mol} \]

Given 11.0 g of propane, calculate the moles:

\[ \text{Moles of } \mathrm{C}_3\mathrm{H}_8 = \frac{11.0 \, \mathrm{g}}{44.094 \, \mathrm{g/mol}} = 0.2494 \, \mathrm{mol} \]

Using the standard enthalpies of formation:

\[ \Delta H^\circ_{\text{reaction}} = \left[3(-393.5) + 4(-285.8)\right] - \left[(-103.85)\right] \]

\[ = \left[-1180.5 - 1143.2\right] + 103.85 \]

\[ = -2223.7 + 103.85 = -2119.85 \, \mathrm{kJ/mol} \]

Multiply the moles of propane by the enthalpy change per mole:

\[ \text{Heat produced} = 0.2494 \, \mathrm{mol} \times (-2119.85 \, \mathrm{kJ/mol}) = -528.8 \, \mathrm{kJ} \]

Final Answer