The enthalpy change for the combustion of one mole of propane under those conditions is approximately −2.26 × 10³ kJ mol⁻¹.

Quick walkthrough

  1. Given data
    • 2.00 g of propane releases −102.5 kJ of heat.
  2. 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

  1. 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