Question

Question #3c8f0

Answer 1

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