Question

Calculate the standard enthalpy of formation of C₂H₆ from the following data:
2C₂H₆ (g) + 7 O₂ (g) → 4CO₂ (g) + 6 H₂O (I),
ΔH° = -3119 kJ
ΔfH° (CO₂) = -393.5 kJ mol⁻¹
ΔfH° (H₂O) = -285.8 kJ mol⁻¹
(-84.9 kJ mol⁻¹)

Answer 1

In thermodynamics we are interested in initial and final states.

You need to construct a Hess Cycle using the information given:


You can see that, in energy terms, the blue route is equal to the red route as the arrows start and finish in the same place. This is in accordance with Hess' Law.

So we can write:

(4×157.3)+ΔH=292

∴ΔH=−337.2kJ

Enthalpy of formation refers to the formation of 1 mole of a substance from its elements in their standard states under standard conditions.

We have found ΔH for:

4Cu+2O2→2Cu2O+O2

Which is the same as:

4Cu+O2→2Cu2O

This refers to the formation of 2 moles of copper oxide. We need the enthalpy change for the formation of 1 mole of copper oxide.

∴ΔHf[Cu2O]=ΔH2=−337.22=−168.6kJ/mol