Hess's Law Problem. PLEASE HELP

Question

4. For the reaction (at 298 K):

2 AgI (s) + Br2 (g) → 2 AgBr (s) + I2 (s) ΔHrxn = -108.0 KJ

If the following are also true:

ΔHof for AgBr (s) = -100.4 KJ/mol
and,
ΔHof for Br2 (g) = +30.9 KJ/mol

then, calculate the value of ΔHof for AgI (s).

Answer 1

These sort of problems are all solved the same way.

General equation: ΔHrxn = Σ(n*ΔHf products) - Σ(n*ΔHf reactants)

 

-108.0 kJ = (2*ΔHf AgBr s + 1*ΔHf I2 s) - (2*ΔHf AgI s + 1*ΔHf Br2 g)

-108 kJ = (2mol*-100.4 kJ/mol) + (1mol*0 kJ/mol) - (2mol*ΔHf AgI s) - (1mol*30.9 kJ/mol)

         Please note that at STP (298K and 1 atm), ΔHf of any element in its natural state is zero. Iodine is a diatomic solid both in this reaction and at STP, so its ΔHf is 0 kJ/mol.

 

-108 kJ = -200.8 kJ + 0 kJ - (2mol*ΔHf AgI s) - 30.9 kJ

(-108 + 200.8 + 30.9)kJ = -2mol*ΔHf AgI s

(123.7 kJ)/(-2mol) = ΔHf AgI s

 

-61.85 kJ/mol = ΔHf AgI s