Enthalpy of Formation

Question

Substance	H2O(l)	H2O(g)	CO2 (aq)	CO2 (g)	CO(g)	CH4 (g)	C2H6(g)
Hfo (kJ/mole)	-285.83	-241.82	-413.82	-393.51	-110.53	-74.81	-84.68

Use the enthalpy of formation data in the table to calculate the enthalpy of the reactions below:

 

CH_{4 (g)} + 2O_{2 (g)}  2H₂O_(g) + CO_{2 (g)}   ΔH = 

2C₂H_{6 (g)} + 7O_{2 (g)}  6H₂O_(g) + 4CO_{2 (g)} ΔH = 

CH_{4 (g)} + O_{2 (g)}  2H₂O_(g) + C_(s)   ΔH = 

Answer 1

Enthalpy of reaction = sum of enthalpy of formation of products - sum of enthalpy of formation of reactants

CH_{4 (g)} + 2O_{2 (g)}  2H₂O_(g) + CO_{2 (g)}   ΔH = [(2)(-241.82) + (-393.51)] - [(-74.81) + (2)(0)] = -802.34 kJ

You should be able to do the others using the same process. Remember that ∆Hformation of an element in its standard form is zero, so, e.g., ∆H_formation O₂(g) = 0; ∆H_formation C(s) = 0.