if the enthalpy change for the combustion 2.00 g of propane, c3h8, was determined to be −102.5 kj by calorimetry, what is the calculated enthalpy change for the combustion of one mole of propane under the same conditions?
The enthalpy change for the combustion of one mole of propane under those conditions is approximately −2.26 × 10³ kJ mol⁻¹.
Quick walkthrough
- Given data
- 2.00 g of propane releases −102.5 kJ of heat.
- Find moles of propane in 2.00 g
The molar mass of propane, C₃H₈, is about 44.1 g mol⁻¹.
n=2.00textg44.1textgmol−1≈0.0454textmoln=\frac{2.00\\text{g}}{44.1\\text{g mol}^{-1}}\approx 0.0454\\text{mol}n=44.1textgmol−12.00textg≈0.0454textmol
- Scale to 1 mole
If 0.0454 mol releases −102.5 kJ, then 1 mol releases:
ΔHper mol=−102.5textkJ0.0454textmol≈−2.26×103textkJmol−1\Delta H_{\text{per mol}}=\frac{-102.5\\text{kJ}}{0.0454\\text{mol}}\approx -2.26\times 10^3\\text{kJ mol}^{-1}ΔHper mol=0.0454textmol−102.5textkJ≈−2.26×103textkJmol−1
So, the calculated enthalpy change for the combustion of one mole of propane under the same calorimetry conditions is:
ΔHcomb≈−2.26×103textkJmol−1\Delta H_{\text{comb}}\approx -2.26\times 10^3\\text{kJ mol}^{-1}ΔHcomb≈−2.26×103textkJmol−1