Question

Estimate the enthalpy of combustion of methane in kJ.mol^-1

CH4(g) + 2O2(g) ------> CO2(g) + 2H2O(g)

Bond Dissociation Enthalpies /kJ.mol^-1
Single Bonds Double Bonds
C C 350 C O 732
C H 410 O O 498
O H 460
C O 350
O O 180

A. 668
B. 540
C. –540
D. –668 (CORRECT ANSWER)
E. none of the above

I get a different answer when I attempt to solve this question.

Answer 1

If you had shown your work I could have found the error. You probably used the O-O of 180 instead of the O=O of 498.

Answer 2

This is what I did

deltaHc(reaction)=deltaHc(products)-deltaHc(reactants)
deltaHc(reaction)=(1mol(2(732kJ/mol))+1mol(2(460kJ/mol)))-(1mol(4(410kJ/mol))+1mol(498kJ/mol))=246kJ/mol

I also did this
deltaHc(reaction)=-deltaHc(products)+deltaHc(reactants)
deltaHc(reaction)=-(1mol(2(732kJ/mol))+1mol(2(460kJ/mol)))+
(1mol(4(410kJ/mol))+1mol(498kJ/mol))=-246kJ/mol

Answer 3

I see several problems.
You want dHrxn = (n*dHreact) - (n*dHproducts). Using dHo formations is not quite the same as dHbond energies.
One problem is the H2O or H-O-H-H.
Each OH is 460, you have two of those for 1 mol H2O = 920 and you have 2 mol H2O which is 920*2 - ?
The correct value to use for O2 is 2*498.
The -688 answer is correct.
(4*410)+ (2*498) - (2*732) - (4*460)