# Question 86d62

Apr 14, 2014

WARNING. This is a long answer.

I think you are referring to Hess's Law calculations.

Hess's Law states that the enthalpy change of a reaction is the same whether you do it in one step or many steps.

This means that we can add up other reactions that give the same products and sum their energies instead.

EXAMPLE

Determine the heat of combustion, ΔH_c, of CS₂, given the following equations.

1. C(s) + O₂(g) → CO₂(g); ΔH_c = -393.5 kJ
2. S(s) + O₂(g) → SO₂(g); ΔH_c = -296.8 kJ
3. C(s) + 2S(s) → CS₂(l); ΔH_f = 87.9 kJ

Solution

Write down the target equation, the one you are trying to get.
CS₂(l) + 2O₂(g) → CO₂(g) + 2SO₂(g)

Start with equation 3. It contains the first compound in the target (CS₂). We reverse equation 3 and its ΔH to put the CS₂ on the left and get equation A below.

A. CS₂(l) → C(s) + 2S(s); -ΔH_f = -87.9 kJ

Now we eliminate C(s) and S(s) one at a time. Equation 1 contains C(s), so we write it as Equation B below.

B. C(s) + O₂(g) → CO₂(g); ΔH_c = -393.5 kJ

We use Equation 2 to eliminate the S(s), but we have to double it and its ΔH. We then get equation C below.

C. 2S(s) + 2O₂(g) → 2SO₂(g); ΔH_c = -593.6 kJ

Finally, we add equations A, B, and C to get the target equation. We cancel things that appear on opposite sides of the reaction arrows.

A. CS₂(l) → C(s) + 2S(s); -ΔH_f = -87.9 kJ
B. C(s) + O₂(g) → CO₂(g); ΔH_f = -393.5 kJ
C. 2S(s) + 2O₂(g) → 2SO₂(g); ΔH_f = -593.6 kJ

CS₂(l) + 3O₂(g) → CO₂(g) + 2SO₂(g); ΔH_c# = -1075.0 kJ