Calculate the standard enthalpy of combustion of propane

Question

Given the following data (1) C(s) + O2(g) ==> CO2(g) ΔHθ = -393 kJ/mol-1 (2) H2(g) + 1/2O2(g) ==> H2O(l) ΔHθ = -286 kJ/mol-1 (3) 3C(s) + 4H2(g) ==> C3H8(g) ΔHθ = -104 kJ/mol-1 Calculate the standard enthalpy of combustion of propane

Answer 1

Target equation is combustion of propane: C3H8(g) + 5O2(g) ==> 3CO2(g) + 4H2O(l)

(1) C(s) + O2(g) ==> CO2(g) ... ∆H = -393 kJ/mol

(2) H2(g) + 1/2 O2(g) ==> H2O(l) ... ∆H = -286 kJ/mol

(3) 3C(s) + 4H2(g) ==> C3H8(g) ...ΔH = -104 kJ/mol

Using Hess' Law we can perform the following operations:

copy eq. (1) and multiply by 3: 3C(s) + 3O2(g) ==> 3CO2(g) ... ∆H = 3x-393 = -1179 kJ

copy eq. (2) and multiply by 4: 4H2(g) + 2 O2(g) ==> 4H2O(l) ... ∆H = -1144 kJ

reverse eq. (3): C3H8(g) ===> 3C(s) + 4H2(g) ... ∆H = +104 kJ

Add these three equations and cancel and combine where appropriate to obtain...

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

∆H = -1179 kJ + -1144 kJ + 104 kJ = -2219 kJ/mol