# Question 3c8f0

Apr 30, 2014

In Hess's Law calculations, you write equations to make unwanted substances cancel out.

Sometimes you have to reverse an equation to do this, and you reverse the sign of ΔH.

Sometimes you have to multiply or divide a given equation, and you do the same thing to the ΔH.

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 have to reverse equation 3 and its ΔH to put the CS₂ on the left. We 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 to get 2S(s). We also double 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