[Boxxxed]

Determine standard enthalpy change of the following reaction

CH4 + 1/2 O2 ------> CH3OH

1. CH4 + 2 O2 -------> CO2 + 2 H2O

2. CH3OH + 3/2 O2 ------->CO2 + H2O


---------------------
All I have is

CO2 + H2O -------->CH3OH + 3/2 O2

If I cancel out the compounds now I get

CH4 + 1/2 O2 --------->CH3OH + H20

How do I get rid of the water?

[sinthreck]

If you have written equation 2 straight from the textbook, then the textbook has made a mistake. The equation is not balanced.

2. CH₃OH + 3/2 O₂ -------> CO₂ + H₂O

check if the number of hydrogens & oxygens on the LHS equals that of the RHS

Once you balance this equation, you can solve this problem.

You should easily be able to work out which compound you need to change because you were having some problems with it (as you said in an earlier post).

[Boxxxed]

The heat of reaction value for equation 2 is - 676 KJ/M

How does this change when I balance the equation? Or is it the value for the balanced equation?