Naked Science Forum

Non Life Sciences => Chemistry => Topic started by: Justin_Is_Here on 27/05/2014 05:25:26

Title: Decomposition of methane
Post by: Justin_Is_Here on 27/05/2014 05:25:26

At what temperature, in absence of air, does methane break up to form carbon and hydrogen?

Title: Re: Decomposition of methane
Post by: chiralSPO on 27/05/2014 18:15:48

Using the data presented by NIST (shown below; http://webbook.nist.gov/chemistry/form-ser.html) and the equations:
CH₄ → C + 2 H₂ and ΔG = ΔH – TΔS

If find the minimum temperature required for this reaction to be:
0 = 74870 – (261.36 + 5.6 – 188.66)T
T = 956.2 K = 683 °C

Bear in mind, this is the minimum temperature at which the reaction would be favorable, it may well require substantially higher temperatures for the reaction to happen at any reasonable rate.
----------------
Methane:
ΔfH°gas   -74.87   kJ/mol
S°gas   188.66 ± 0.42   J/mol*K

Hydrogen:
ΔfH°gas       0          kJ/mol
S°gas,1 bar   130.680 ± 0.003   J/mol*K

Carbon (graphite):
ΔfH°solid       0          kJ/mol
S°solid,1 bar   5.6 ± 0.5   J/mol*K   

Title: Re: Decomposition of methane
Post by: Bored chemist on 28/05/2014 20:41:37

That's the temperature at which the equilibrium constant is equal to 1 (at 1 atm).
Of course, even partial decomposition might be adequate from the OP's  point of view, in which case a lower temperature would do the job.
And the equilibrium will depend on the pressure so changing that would also affect the temperature you need.

And, it's likely that other reactions will take place too.

And then there's the issue of what the reaction rate is.
And that might be altered by the addition of a catalyst- which might also change the yield of different products.
So, the answer is, by looking stuff up in the data book and doing some maths, we still don't really know.

Title: Re: Decomposition of methane
Post by: CliffordK on 28/05/2014 22:37:10

How important is the phase of the end products.

In the case of Methane (CH₄), carbon (graphite/diamond) has very high melting and boiling points.  I would think that would reduce the reaction rate, but perhaps also help push it towards completion by removing reaction products.

Water, on the other hand, would decompose into H₂+O₂ with high temperature products all being gases (also increasing the moles of gases in a closed vessel).