Question

Hess's Law

Given the following data:

Br2(l) + 5F2(g) → 2BrF5(l)	ΔH°=-918.0 kJ
BrF3(l) + Br2(l) → 3BrF(g)	ΔH°=125.2 kJ
Br2(l) + F2(g) → 2BrF(g)	ΔH°=-117.2 kJ

calculate ΔH° for the reaction:

BrF₅(l) → BrF₃(l) + F₂(g)

Answer 1

Br2(l) + 5F2(g) = 2BrF5(l)

Reverse the reaction and divided by 2

BrF5(l) = 0.5Br2(l) + 2.5F2(g)

H1 = 918/2 = 459 kJ

BrF3(l) + Br2(l) = 3BrF(g)

Reverse the reaction

3BrF(g) = BrF3(l) + Br2(l)

H2 = - 125.2 kJ

Br2(l) + F2(g) = 2BrF(g)

Multiply by 1.5

1.5Br2(l) + 1.5F2(g) = 3BrF(g)

H3 = - 1.5*117.2 = - 175.8 kJ

Now add these reactions

BrF5(l) + 3BrF(g) + 1.5Br2(l) + 1.5F2(g) = 3BrF(g) + BrF3(l) + Br2(l) + 0.5Br2(l) + 2.5F2(g)

BrF5(l) = BrF3(l) + F2(g)

From the Hess law

H = H1 + H2 + H3

= 459 - 125.2 - 175.8

= 158 kJ