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# Measurement of enthalpy change of formation for methane watch

1. Suggest why the enthalpy change of formation of methane cannot be measured directly. (and thus must be calculated)
2. (Original post by AbdullahAAK)
Suggest why the enthalpy change of formation of methane cannot be measured directly. (and thus must be calculated)
There are quite a few correct answers, what are you suggestions for any of them?
3. (Original post by Pigster)
There are quite a few correct answers, what are you suggestions for any of them?
I have no idea. A few will suffice.
4. (Original post by AbdullahAAK)
I have no idea. A few will suffice.
What happens when you mix C and H2 together?

And why?
5. (Original post by Pigster)
What happens when you mix C and H2 together?

And why?
I think nothing happens (At standard conditions) as there is insufficient heat in the atmosphere to allow breakage of bonds in the reactants.
Hence one must use Hess's Law and enthalpy changes of combustion for the products and reactants (already calculated and given to student) to calculate the enthalpy change of reaction for C and H2.

Is that correct?
6. (Original post by AbdullahAAK)
I think nothing happens (At standard conditions) as there is insufficient heat in the atmosphere to allow breakage of bonds in the reactants.
Hence one must use Hess's Law and enthalpy changes of combustion for the products and reactants (already calculated and given to student) to calculate the enthalpy change of reaction for C and H2.

Is that correct?
They simply don't react (as you say)
This is because the activation energy is so high (as you allude to)
Which makes any rate of reaction very slow
(which means any energy change would not be detectable since any heat gain/loss by the reaction would be lost to/gained from the surroundings before it could be detected)
And anyway, even if C did react with H2, what is the chance that 1 mol of C and 2 mol of H2 would form 1 mol of CH4 and nothing else, e.g. ethane, benzene etc. etc.

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