C3H8(g)+5O2(g)→3CO2(g)+4H2O(g). Calculate ΔrH∘ for this reaction using standard enthalpies of formation. (The standard enthalpy of formation of gaseous propane is -103.85 kJmol−1 .)

Question

C3H8(g)+5O2(g)→3CO2(g)+4H2O(g). Calculate ΔrH∘ for this reaction using standard enthalpies of formation. (The standard enthalpy of formation of gaseous propane is -103.85 kJmol−1 .)

Answer 1

C₃H₈(g) + 5O₂(g) ==> 3CO₂(g) + 4H₂O(g)

∆Hº_rxn = [ ∑∆Hº_f CO2g)x3 + ∆Hº_f H2O(g)x4 ] - (-103.85)

Look up the ∆Hº_f for CO2 gas and multiply that by 3. Look up the ∆Hº_f for H2O gas and multiply that by 4. Add these together and then subtract the value of -103.85 to get your answer.