Question

How do you calculate ΔHF° given two reactions and their respective ΔHrxn° values?

Hello! Would anybody be so kind and help me figure out what I have to do for a problem like this? Thank you!

Calculate ΔHF° (SO3) from the following information.

S(s) + O2(g)  SO2(g) ΔHrxn° = ‐296.8 kJ
SO2(g) + 1/2 O2(g)  SO3(g) ΔHrxn° = ‐98.9 kJ

Answer 1

`Δ_text(f)H^@ = "-395.7 kJ"`

Explanation:

In this problem, you must use Hess' Law, which states that the energy involved in a chemical process is the same whether the process takes place in one or several steps.

You are given two equations:

`bb"(1)"color(white)(m) "S(s)" + "O"_2"(g)" → "SO"_2"(g)"; color(white)(mmll)Δ_text(rxn)H^@ = "-296.8 kJ"`
`bb"(2)"color(white)(m) "SO"_2"(g)" + "½O"_2"(g)"color(white)(l) → "SO"_3"(g)"; Δ_text(rxn)H^@ = color(white)(l)"-98.9kJ"`

From these, you must devise the target equation for the formation of `"SO"_3` from its elements:

`bb"(3)"color(white)(m)"S(s)" + "³/₂O"_2"(g)" → "SO"_3"(g)"; Δ_text(f)H^@ = ?`

The target equation has `"1S(s)"` on the left, so you write equation (1).

`bb"(4)"color(white)(m) "S(s)" + "O"_2"(g)" → "SO"_2"(g)"; color(white)(mmll)Δ_text(rxn)H^@ = "-296.8 kJ"`

Equation (4) has `"1SO"_2"(g)"` on the right, and that is not in the target equation.

You need an equation with `"1SO"_2"(g)"` on the left.

Write Equation (2).

`bb"(5)"color(white)(m) "SO"_2"(g)" + "½O"_2"(g)"color(white)(l) → "SO"_3"(g)"; Δ_text(rxn)H^@ = "-98.9kJ"`

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) "S(s)" + "O"_2"(g)" → color(red)(cancel(color(black)("SO"_2"(g)"))); color(white)(mmll)Δ_text(rxn)H^@ = "-296.8 kJ"`
`bb"(5)"ul(color(white)(m)color(red)(cancel(color(black)("SO"_2"(g)"))) + "½O"_2"(g)"color(white)(l) → "SO"_3"(g)"; Δ_text(rxn)H^@ =color(white)(ll) "-98.9kJ")`
`color(white)(mmm)"SO"_2"(g)" + "³/₂O"_2"(g)" → "SO"_3"(g)"; color(white)(m)Δ_text(f)H^@ = "-395.7 kJ"`

`Δ_text(f)H^@ = "-395.7 kJ"`