Use bond energy data to find enthalpy of reaction

Question

Here is another one that got me. It goes by bonds broken and formed? Which are these? And I know it is reactant values added minus sum of product values

Using bond-energy data, what is ∆H° for the following reaction?

CH3OH(g) H2S(g) → CH3SH(g) H2O(g)

Bond Bond Energy (kJ/mol)

C-H 411

C-O 358

O-H 459

C-S 272

S-H 363

A. 1883 kJ

B. 10 kJ

C. –1883 kJ

D. –10 kJ

E. 593 kJ

Answer 1

Just sum all of the energies for breaking all of the bonds of the reactants and all of the energies for making the bonds of the products.  Breaking a bond is endothermic and making a bond is exothermic

CH3OH(g) + H2S(g) → CH3SH(g) + H2O(g)

DH = 3(C-H) + (C-O) + (O-H) + 2(S-H) + -3(C-H) + -(C-S) +  -(S-H) + 2(H-O)

DH = 3(411) + 358 + 459 + 2(363) - 3(411) - 272 - 363 - 2(459)

DH = -10 kJ

The answer is (D)