Write a balanced equation for the combustion of C3H8(g) (propane) – i.e. its reaction with O2(g) forming the products CO2(g) and H2O(l). Given the following standard heats of formation: ΔHf° of CO2(g) is -393.5 kJ/mol ΔHf° of H2O(l) is -286 kJ/mol ΔHf° of C3H8(g) is -103.8 kJ/mol What is the standard heat of reaction (ΔH°) for the combustion reaction of C3H8(g)? Calculate the difference, ΔH-ΔE=Δ(PV) for the combustion reaction of 1 mole of propane. (Assume standard state conditions and 298 K for all reactants and products.)
The Correct Answer and Explanation is:
Balanced Combustion Reaction of Propane (C₃H₈):
The balanced chemical equation is:C3H8(g)+5O2(g)→3CO2(g)+4H2O(l)\text{C}_3\text{H}_8(g) + 5\text{O}_2(g) \rightarrow 3\text{CO}_2(g) + 4\text{H}_2\text{O}(l)C3H8(g)+5O2(g)→3CO2(g)+4H2O(l)
Step 1: Calculate the Standard Heat of Reaction (ΔH°)
Use the standard enthalpy of formation (ΔHf°) values:
- ΔHf° [C₃H₈(g)] = -103.8 kJ/mol
- ΔHf° [CO₂(g)] = -393.5 kJ/mol
- ΔHf° [H₂O(l)] = -286 kJ/mol
Apply the formula:ΔH∘=∑ΔHf∘(products)−∑ΔHf∘(reactants)\Delta H^\circ = \sum \Delta H_f^\circ \text{(products)} – \sum \Delta H_f^\circ \text{(reactants)}ΔH∘=∑ΔHf∘(products)−∑ΔHf∘(reactants)=[3(−393.5)+4(−286)]−[1(−103.8)+5(0)]= \left[3(-393.5) + 4(-286)\right] – \left[1(-103.8) + 5(0)\right]=[3(−393.5)+4(−286)]−[1(−103.8)+5(0)]=[−1180.5−1144]−[−103.8]= [-1180.5 – 1144] – [-103.8]=[−1180.5−1144]−[−103.8]=−2324.5+103.8=−2220.7 kJ/mol= -2324.5 + 103.8 = -2220.7 \text{ kJ/mol}=−2324.5+103.8=−2220.7 kJ/mol
So, the standard heat of combustion of propane is:ΔH∘=−2220.7 kJ/mol\boxed{\Delta H^\circ = -2220.7 \text{ kJ/mol}}ΔH∘=−2220.7 kJ/mol
Step 2: Calculate ΔH – ΔE = Δ(PV)
Use the ideal gas law relationship:Δ(PV)=Δngas⋅RT\Delta(PV) = \Delta n_{\text{gas}} \cdot RTΔ(PV)=Δngas⋅RT
At standard conditions (T = 298 K, R = 8.314 J/mol·K = 0.008314 kJ/mol·K):
- Reactant gas moles: 1 C₃H₈ + 5 O₂ = 6 mol
- Product gas moles: 3 CO₂ (H₂O is liquid, not gas) = 3 mol
- Δn = n_products – n_reactants = 3 – 6 = -3 mol
Now calculate:Δ(PV)=−3⋅0.008314⋅298≈−7.43 kJ\Delta(PV) = -3 \cdot 0.008314 \cdot 298 \approx -7.43 \text{ kJ}Δ(PV)=−3⋅0.008314⋅298≈−7.43 kJ
So,ΔH−ΔE=Δ(PV)=−7.43 kJ\boxed{\Delta H – \Delta E = \Delta(PV) = -7.43 \text{ kJ}}ΔH−ΔE=Δ(PV)=−7.43 kJ
Explanation)
The combustion of propane (C₃H₈) is a chemical reaction in which propane reacts with oxygen (O₂) to form carbon dioxide (CO₂) and water (H₂O). This reaction is highly exothermic and releases a significant amount of energy in the form of heat.
To begin, we write and balance the chemical equation to ensure conservation of mass. Each molecule of propane contains three carbon atoms and eight hydrogen atoms, so we need three CO₂ molecules and four H₂O molecules to balance carbon and hydrogen, respectively. Balancing oxygen then requires five O₂ molecules.
To find the standard enthalpy change (ΔH°), we use standard heats of formation. This is done by subtracting the total enthalpies of the reactants from those of the products. Since elements in their standard form, like O₂(g), have a ΔHf° of zero, they do not contribute directly to the calculation. The calculation shows that burning one mole of propane releases 2220.7 kJ of heat.
We also explore the difference between enthalpy (ΔH) and internal energy (ΔE) by computing the change in pressure-volume work (Δ(PV)). This depends on the change in moles of gas. Since the number of gas molecules decreases (from 6 to 3), the system compresses, and the volume decreases, resulting in a negative Δ(PV) value of -7.43 kJ. This small difference reflects the work done by the system due to volume change, which slightly reduces the energy released as internal energy.
Overall, this analysis highlights how thermodynamic quantities relate to molecular changes in combustion and helps us understand energy conservation in chemical processes.
