A couple of chemistry questions needing answering.....PLEASE HELP!!!?

Question

Use bond energies to estimate the enthalpy of formation of HBr(g).

H2(g) + Br2(g) 2HBr(g)

BE(H—H) = 436 kJ/mol
BE(Br—Br) = 192 kJ/mol
BE(H—Br) = 366 kJ/mol
a. +262 kJ
b. –52 kJ
c. –104 kJ
d. +104 kJ
e. +52 kJ

Answer 1

This is a Hess's Law problem: Products - Reactants

On the product side, you have 2 H-Br bonds

On the reactant side, you have 1 H-H bond and 1 Br-Br bond

So, the math expression would be 366x2 - 436 -192 = 104 kJ

This reaction produced 2 mol HBr, so 104 kJ = 2 mol

The enthalpy of formation = 52 kJ/mol