“The overall enthalpy change of a reaction is independent of the route taken.”
Given that Δ H θ combustion of water is -286 KJ mol-1
Given that Δ H θ combustion of Carbon dioxide is -394 KJ mol-1
Given that Δ H θ combustion of Methanol is -726 KJ mol-1
Calculate the Δ H θ formation of Methanol.
Step 1 – Balance the equation.
C + H2 >> C8H18
C + 2 H2 >> C8H18
Step 2 – Make the route diagram and label the Δ H’s
Step 3 – realise the route.
In this case, we want Δ H4+ Δ H3 which is equal to Δ H1 + Δ H2
This is because, regardless of what route we take, we end up getting the same product, and therefore the enthalpy change will be the same.
Step 4 – Substitute the values
Δ H1 + Δ H2 = Δ H3 + Δ H4
-394 + (2x-286) = Δ H1 + -726
-966 = Δ H3 – 726
- make the wanted delta H the subject of the formulae, and add the unit. -
Δ H3 = -966 + 726
Δ H3 = -240 Kj Mol -1