Can Someone Explain Hess's Law?

How do you find enthalpy by combining equations?

For Example,

ΔH for the reaction

IF5(g) → IF3(g) + F2(g)

is ____ kJ, give the data below.

IF(g) + F2(g) → IF3(g) ΔH = -390 kJ

IF(g) + 2F2(g) → IF5(g) ΔH = -745 kJ

1 Answer
Dec 7, 2017

ΔH = "+355 kJ"

Hess's Law states that the energy involved in a chemical process is the same whether the process takes place in one or in several steps.

You are given two equations:

bb"(1)"color(white)(m) "IF(g)" + "F"_2"(g)" → "IF"_3"(g)"; color(white)(m)ΔH = "-390 kJ"
bb"(2)"color(white)(m) "IF(g)" + "2F"_2"(g)"color(white)(l) → "IF"_5"(g)"; ΔH = "-745 kJ"

From these, you must devise a third equation (the target equation).

bb"(3)"color(white)(m)"IF"_5"(g)" → "IF"_3"(g)" + "F"_2"(g)"; ΔH = ?

The target equation has "IF"_5"(g)" on the left, so you reverse equation (2).

bb"(4)"color(white)(m)"IF"_5"(g)"→ "IF(g)" + "2F"_2"(g)"; ΔH ="+745 kJ"

When you reverse an equation, you change the sign of its ΔH.

Equation (4) has "IF(g)" on the right, and that is not in the target equation.

You need an equation with "IF(g)" on the left.

Just rewrite Equation (1).

bb"(5)"color(white)(m) "IF(g)" + "F"_2"(g)" → "IF"_3"(g)"; color(white)(m)ΔH = "-390 kJ"

Now, you add equations (4) and (5), cancelling species that appear on opposite sides of the reaction arrows.

When you add equations, you add their ΔH values.

This gives us the target equation (6):

bb"(4)"color(white)(m)"IF"_5"(g)"→ color(red)(cancel(color(black)("IF(g)"))) + color(red)(cancel(color(black)(2)))"F"_2"(g)"; ΔH ="+745 kJ"
ul(bb"(5)"color(white)(m) color(red)(cancel(color(black)("IF(g)"))) + color(red)(cancel(color(black)("F"_2"(g)"))) → "IF"_3"(g)"; color(white)(ll)ΔH = color(white)(l)"-390 kJ")
bb"(6)"color(white)(m)"IF"_5"(g)" → "IF"_3"(g)" + "F"_2"(g)"; color(white)(m)ΔH = "+355 kJ"