What is the delta-H of reaction for the conversion of C2H5OH(l) to CH3OCH3(l) if..?

Question

Given that[C2H5OH(l) + 3O2(g) = 2CO2(g) + 3H2O(g); delta-H of reaction = -1234 kJ] and [CH3OCH3(l) + 3O2(g) = 2CO2(g) + 3H2O(g); delta-H of reaction = -1328 kJ]???

Please show steps to solving this, thanks!!!!

Answer 1

You want to make the equation C2H5OH-->CH3OCH3 from the 2 equations given. So, start with the first one, leave it as is because C2H5OH is on the left side, where you want it. However, the second equation must be flipped to force CH3OCH3 to be on the right side. When you flip an equation around, you negate the H value, meaning if you flip the second equation, you get H=1328kJ for the flipped reaction:

C2H5OH+3O2-->2CO2 + 3H2O deltaH= -1234kJ
2CO2 + 3H2O--> CH3OCH3 + 3O2 deltaH= +1328kJ

Obvioulsy, there's alot more than C2H5OH-->CH3OCH3 going on in those 2 equations, but everything is on the correct side. We're going to add these equations to make the one you want. One important rule of adding equations is to have everything on the correct side for the result you wish to attain, which we did already. You are allowed to multiply and divide equations by numbers (you'd do the same operation to the deltaH, you don't need to do this in this case) and reverse equations (negate the deltaH value, you did that in one instance here). Another rule is that things that are exactly the same on one side as the other when you add the equations will cancel out. When you add equations, you add deltaH values too. So adding your two new equations gives:

{}inside of these means cancelled out
In this case 2CO2, 3H2O, and 3O2 appear on both sides when added, so they cancel out because they stayed the same on both sides of the equation.

C2H5OH+{3O2}-->{2CO2+3H2O} dH=-1234kJ
{2CO2+3H2O}-->CH3OCH3+{3O2} dH=1328kJ
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C2H5OH -->CH3OCH3 dH=94KJ