Question

The combustion reaction of propane is as follows.
C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l)
Using Hess's law and the reaction enthalpies given below, find the change in enthalpy for this reaction. Answer in KJ/mol

reaction (1):     C(s) + O2(g) → CO2(g)    
ΔH = −393.5 kJ/mol
reaction (2):     H2(g) + 1/2 O2(g) → H2O(l)    
ΔH = −285.8 kJ/mol
reaction (3):     3 C(s) + 4 H2(g) → C3H8(g)    
ΔH = −103.8 kJ/mol

Answer 1

Reactions:

C(s) + O2(g) --> CO2(g) HRxn = -393.5 kJ/mol

H2(g) + 1/2O2(g) --> H2O(l) HRxn = -285.8 kJ/mol

3C(s) + 4H2(g) --> C3H8(g) HRxn = -103.8 kJ/mol

since we need C3H8 in the reactants, consider inverting equation (3)

C3H8(g) -->3C(s) + 4H2(g) HRxn = +103.8 kJ/mol (change sign for inversion)

also... we need CO2 in the right, so keep reaction 1 as it is

C(s) + O2(g) --> CO2(g) HRxn = -393.5 kJ/mol

we need 3CO2 , so multiply equation 1 by 3

3C(s) + 3O2(g) --> 3CO2(g) HRxn = 3*(-393.5) = -1180.5 kJ/mol

H2O(l) must remain in the products

multiply equation (2) by 4

4H2(g) + 2O2(g) --> 4H2O(l) HRxn = 4*(-285.8) = -1143.2 kJ/mol

now...

the equations

C3H8(g) -->3C(s) + 4H2(g)   HRxn = +103.8 kJ/mol (change sign for inversion)

3C(s) + 3O2(g) --> 3CO2(g) HRxn = 3*(-393.5) = -1180.5 kJ/mol

4H2(g) + 2O2(g) --> 4H2O(l) HRxn = 4*(-285.8) = -1143.2  kJ/mol}

Add all

C3H8(g) + 3C(s) + 3O2(g) + 4H2(g) + 2O2(g) -->3C(s) + 4H2(g) + 3CO2(g) + 4H2O(l) Htotal = 103.8 -1180.5 -1143.2 = -2219.9 kJ/mol

cancel common terms

C3H8(g) + 3C(s) + 3O2(g) + 4H2(g) + 2O2(g) -->3C(s) + 4H2(g) + 3CO2(g) + 4H2O(l)

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

which is what we wanted

so

Hrxn =  -2219.9 kJ/mol