Question

Many portable gas heaters and grills use propane, \(\mathrm{C}_{3} \mathrm{H}_{8}(g),\) as a fuel. Using standard enthalpies of formation, calculate the quantity of heat produced when \(10.0 \mathrm{~g}\) of propane is completely combusted in air under standard conditions.

Short answer

The total heat produced during the complete combustion of 10.0 g of propane under standard conditions is -243.34 kJ.

Step-by-step solution

Identify the standard enthalpies of formation

Standard enthalpies of formation (\(\Delta H_\mathrm{f}^\circ\)) are given as follows (all values are in kJ/mol): - \(\Delta H_\mathrm{f}^\circ(C_3H_8) = -103.85\) - \(\Delta H_\mathrm{f}^\circ(O_2) = 0\) - \(\Delta H_\mathrm{f}^\circ(CO_2) = -393.5\) - \(\Delta H_\mathrm{f}^\circ(H_2O) = -285.8\)

Calculate the standard enthalpy change of the reaction

Using the standard enthalpies of formation, we can calculate the enthalpy change of the reaction as follows: \[\Delta H_\mathrm{rxn}^\circ = [3 \times \Delta H_\mathrm{f}^\circ(CO_2) + 4 \times \Delta H_\mathrm{f}^\circ(H_2O)] - [\Delta H_\mathrm{f}^\circ(C_3H_8) + 5 \times \Delta H_\mathrm{f}^\circ(O_2)]\] \[\Delta H_\mathrm{rxn}^\circ = [3 \times (-393.5) + 4\times (-285.8)] - [(-103.85) + 5 \times 0]\] \[\Delta H_\mathrm{rxn}^\circ = (-1180.5) - (-103.85) = -1076.65 \,\mathrm{kJ/mol}\] The standard enthalpy change of the reaction is -1076.65 kJ/mol.

Calculate the number of moles of propane

Given that the mass of propane is 10.0 g, we can determine the number of moles (n) by using the molar mass of propane: \[n = \frac{m}{M}\] Propane has a molar mass of \(3 \times 12.01 + 8 \times 1.01 = 44.1 \,\mathrm{g/mol}\). So: \[n = \frac{10.0}{44.1} = 0.226 \,\mathrm{mol}\] There are 0.226 moles of propane.

Calculate the total heat produced

Now, we can calculate the total heat produced during the complete combustion of 10.0 g of propane using the moles of propane (0.226 mol) and the standard enthalpy change of the reaction (-1076.65 kJ/mol): \[q = n \times \Delta H_\mathrm{rxn}^\circ\] \[q = 0.226 \times (-1076.65) = -243.34 \,\mathrm{kJ}\] Therefore, the total heat produced during the complete combustion of 10.0 g of propane is -243.34 kJ (negative sign indicating that the heat is released).

Key concepts

Standard Enthalpies of Formation

Standard enthalpies of formation are crucial for calculating the heat change in chemical reactions. These values, often symbolized as \(\Delta H_\text{f}^\circ\), represent the heat change when one mole of a compound is formed from its elements in their most stable forms under standard conditions (which is 1 atmosphere and 298 K). These values have been determined through experimental measurements and are compiled in tables for your convenience.

• For propane \( (C_3H_8) \), the standard enthalpy of formation is \(-103.85 \text{ kJ/mol}\).
• For oxygen \( (O_2) \), since it is elemental in its most stable form, \(\Delta H_\text{f}^\circ = 0 \text{ kJ/mol}\).
• For carbon dioxide \( (CO_2) \), it is \(-393.5 \text{ kJ/mol}\).
• For water \( (H_2O) \), it is \(-285.8 \text{ kJ/mol}\).

By using these values, you can determine the enthalpy changes in chemical reactions involving these substances. Remember, the values for elements in their most stable form (like \( O_2 \) and \( H_2 \)) are always zero.

Combustion Reaction

A combustion reaction typically involves a hydrocarbon, like propane, reacting with oxygen to produce carbon dioxide and water. It releases energy in the form of heat, making it an exothermic reaction.
In the context of propane:

• The balanced chemical equation for its combustion is: \[(C_3H_8) + 5(O_2) \rightarrow 3(CO_2) + 4(H_2O)\]
• This tells us that one mole of propane requires five moles of oxygen to produce three moles of carbon dioxide and four moles of water.
• The enthalpy change for this reaction, as calculated, is \(-1076.65 \text{ kJ/mol}\) of propane.

Because the reaction releases heat, it is important to account for this energy change when assessing the energy efficiency and environmental impact of using propane as a fuel.

Moles of Propane

Understanding the concept of moles is essential for calculating how much heat is produced in reactions like propane combustion. A mole is a fundamental unit in chemistry used to express amounts of a chemical substance. The amount is based on Avogadro's number, roughly \(6.022 \times 10^{23}\) entities, which could be atoms, molecules, or ions.
To determine how many moles of propane are in a certain mass, you divide the mass (in grams) by the molar mass of propane. Propane, which has a chemical formula of \( C_3H_8 \), has a molar mass calculated as:

• Carbon: \(3 \times 12.01 = 36.03\)
• Hydrogen: \(8 \times 1.01 = 8.08\)
• Total Molar Mass = \(36.03 + 8.08 = 44.1 \text{ g/mol}\)

So, if you take 10.0 grams of propane, convert it to moles: \( n = \frac{10.0}{44.1} = 0.226 \text{ mol}\). These moles are then used to calculate the total heat produced during combustion by multiplying by the enthalpy change of the reaction.