Question

Many portable gas heaters and grills use propane, C3 Hs (g).
Using enthalpies of formation, calculate the quantity of heat produced when 13.0 g of propane is completely combusted in air under standard conditions. is forming.

Answer 1

To calculate the quantity of heat produced when 13.0 g of propane (C₃H₈) is combusted in air under standard conditions, we need to follow these steps:

1. **Write the balanced combustion reaction for propane**:

The complete combustion of propane is represented by the following reaction:

\[
C_3H_8(g) + 5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(g)
\]

2. **Find the standard enthalpy of formation (\( \Delta H_f^\circ \)) values**:

- For \(C_3H_8(g)\): \(\Delta H_f^\circ = -103.85 \, \text{kJ/mol} \)
- For \(CO_2(g)\): \(\Delta H_f^\circ = -393.5 \, \text{kJ/mol} \)
- For \(H_2O(g)\): \(\Delta H_f^\circ = -241.8 \, \text{kJ/mol} \)
- For \(O_2(g)\): \(\Delta H_f^\circ = 0 \, \text{kJ/mol} \) (since it's a standard state element)

3. **Calculate the standard enthalpy change (\( \Delta H_{rxn} \)) for the reaction**:

Using the equation:

\[
\Delta H_{rxn} = \sum (\Delta H_f^\circ \text{ of products}) - \sum (\Delta H_f^\circ \text{ of reactants})
\]

- Products:
3 moles of \(CO_2\) and 4 moles of \(H_2O\):
\[
\Delta H_{products} = [3(-393.5) + 4(-241.8)] \text{ kJ/mol}
\]

\[
\Delta H_{products} = -1180.5 - 967.2 = -2147.7 \text{ kJ/mol}
\]

- Reactants: 1 mole of \(C_3H_8\) and 5 moles of \(O_2\):
\[
\Delta H_{reactants} = [(-103.85) + 5(0)] \text{ kJ/mol} = -103.85 \text{ kJ/mol}
\]

Now we can find \( \Delta H_{rxn} \):

\[
\Delta H_{rxn} = -2147.7 - (-103.85) = -2043.85 \text{ kJ/mol}
\]

4. **Determine the number of moles of propane in 13.0 g**:

The molar mass of propane (C₃H₈) is:

\[
\text{C: } 3 \times 12.01 \text{ g/mol} + \text{H: } 8 \times 1.008 \text{ g/mol} = 36.046 \text{ g/mol}
\]

Now, calculate the number of moles of propane in 13.0 g:

\[
\text{moles of } C_3H_8 = \frac{13.0 \text{ g}}{36.046 \text{ g/mol}} \approx 0.360 \text{ moles}
\]

5. **Calculate heat produced from the combustion of 13.0 g of propane**:

Since we have \( \Delta H_{rxn} \) for 1 mole, we multiply by the number of moles of propane combusted:

\[
\text{Heat produced} = \Delta H_{rxn} \times \text{moles of } C_3H_8
\]

\[
\text{Heat produced} = -2043.85 \text{ kJ/mol} \times 0.360 \text{ moles} \approx -737.8 \text{ kJ}
\]

6. **Final Answer**:

The quantity of heat produced when 13.0 g of propane is completely combusted in air under standard conditions is approximately **737.8 kJ** (with the sign indicating it’s an exothermic reaction).