Asked by Bob

The molar enthalpies of combustion of CH3COCOOH(l)CH3COOH(l) and CO(g) are respectively -1275kJ/mol, -875kJ/mol, and -283kJ/mol. What is the enthalpy change for the reaction below?
CH3COCOOH= CH3COOH +CO

a)1867kJ
b)-1867kJ
c)117kJ
d)-117kJ
e)-2433kJ

I think it's d)
Answered by DrBob222
Did you check your work as I suggested earlier? I still don't get -117 kJ. Post your work and let me see if I can find the error.
Answered by Bob
I thought what your suppose to do is
-1275 - ((-875) + (-283))= -117kj/mols
Don't u just subtract? To find the difference?
Answered by DrBob222
delta Hrxn = (sum DH products)-(sum DH reactants)
DHrxn = (-873-283)-(-1275)
-1158 + 1275 = +117 kJ.
Answered by Bob
Can you explain why you have to subtract the other way?