using standard heats of formation

Question

Using the standard heats of formation that follow, calculate the standard enthalpy change for the following reaction.

CH₄(g) + 2O₂(g)CO₂(g) + 2H₂O(g)     H°_rxn = fill in the blank  kJ

Species	H°f (kJ/mol)
CH4(g)	-74.8
CO2(g)	-393.5
H2O(g)	-241.8

Answer 1

CH4(g)  +  2O2(g) ==> CO2(g)  +  2H2O(g) 

-74.8...........0.................-393.5........-241.8 ........∆H values in kJ per mole

To find ∆Hº_rxn, we use summation of the products - summation of reactants as follows:

sum of products = -393.5 + (2)(-241.8) = -393.5 + (-483.6) = -877.1 kJ

sum of reactants = -74.8 kJ

products - reactants = -877.1 kJ - (-74.8 kJ) = - 951.9 kJ

∆Hº_rxn = -951.9 kJ