Using the table below, what is the change in enthalpy for the following reaction when all reactants and products are in the gaseous state?
Carbon Monoxide + Oxygen Carbon Dioxide
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

A. −283.0 kJ

B. −708.0 kJ

C. −566.0 kJ

D. 566.0 kJ

Question

Using the table below, what is the change in enthalpy for the following reaction when all reactants and products are in the gaseous state?
Carbon Monoxide + Oxygen  Carbon Dioxide
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
 
		
A. −283.0 kJ
 		
B. −708.0 kJ
 		
C. −566.0 kJ
 		
D. 566.0 kJ

anonymous · Answer

@iolanaakau1

Enthalpy change = enthalpy of products - enthalpy of reactants

Now try on your own :D

jfraser · Answer

balance the reaction, add up all the pieces, profit.

jfraser · Answer

I get -566kJ, check your work

anonymous · Answer

@JFraser , I am getting A as the answer

jfraser · Answer

are you multiplying by the stoichiometric coefficients?

jfraser · Answer

The reaction has to balance before we can use the DH values

anonymous · Answer

You got  me ...  You are absolutely right . I am little sleepy now :)