Using the table below, what is the change in enthalpy for the following reaction?
CH4 (g) + 2O2 (g) arrow CO2 (g) + 2H2O (l)

Substance ΔHf (kJ/mol)
CO2 (g) -393.5
CO (g) -110.5
O2 (g) 0.0
O3 (g) 142.0
H2O (g) -241.8
H2O (l) -285.8
CH4 (g) -74.86

−890.2 kJ

−604.4 kJ

560.4 kJ

−802.2 kJ

Question

Using the table below, what is the change in enthalpy for the following reaction?
CH4 (g) + 2O2 (g) arrow CO2 (g) + 2H2O (l)

Substance	ΔHf (kJ/mol)
CO2 (g)	-393.5
CO (g)	-110.5
O2 (g)	0.0
O3 (g)	142.0
H2O (g)	-241.8
H2O (l)	-285.8
CH4 (g)	-74.86


−890.2 kJ
		
−604.4 kJ
		
560.4 kJ
		
−802.2 kJ

anonymous · Answer

I'm not totally sure, i m also learning this stuff...
so i think you should use this formula:
$$\sum H= \sum nH_{f\ products}-\sum nH_{f\ reactans}$$

anonymous · Answer

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