Question

Enthalpy of combustion of CH4 , C,HE at 298 k are -890.3 KJ/mol, -393.5 KJ/mol and
-285.8 KJ/mol respectively . Calculate the enthalpy of formation of methane CH4
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Answer 1

Answer:

This is one of those fun little Hess’ Law problems. You’ve got several reactions whose enthalpies you know, and one (the target) whose enthalpy you need to find. Hess’ Law says that you can add enthalpies and reactions and the result will be the enthalpy of the summation reaction.

The trick is finding a way to add up the reactions whose enthalpies you know to get to the target reaction.

So what do we know?

C+O2→CO2  … ΔH = -393 kJ/mol

H2+12O2→H2O  … ΔH = -286 kJ/mol

CH4+2 O2→CO2+2 H2O  … ΔH = -892 kJ/mol

And our target, the heat of formation of  CH4  is, by definition:

C+2 H2→CH4  

We know our target has  CH4  on the right side, so let’s reverse equation 3 (and change the sign of its ΔH).

(A)  CO2+2 H2O→CH4+2 O2  … ΔH = +892 kJ/mol

Our target does not have any  CO2  in it, so let’s add in equation 1, which has  CO2  on the right side. It will cancel out the  CO2  on the left side of the reversed equation 3 and one of the  O2  molecules on the right side. It will also provide us with the  C  we need on the left side of the target equation. We’re not changing equation 2 at all, so its ΔH remains as is.

(B)  C+O2→CO2  … ΔH = -393 kJ/mol

We also need two  H2  molecules on the left side, so let’s double equation 2 (and double ΔH, too) (this also provides another  O2  to cancel the second  O2  in the reversed equation 3, as well as two  H2O  molecules in the reversed equation 3):

(C)  2 H2+O2→2 H2O  … ΔH = -572 kJ/mol

Now if we add up the three equations I’ve labelled (A), (B), and (C), and cancel anything that appears on both sides, we should have the target. If we do, we know we’ve found the right combination to satisfy Hess’ Law. And we can then add up the enthalpies of (A) + (B) + (C) to get the final answer:

+892 kJ/mol + (-393 kJ/mol) + (-572 kJ/mol) = -73 kJ/mol

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