The purpose of the catalyst is not to react with the mechane (CH4) or the products of the cracking process (H2 and C). The only purpose of the catalyst is to lower the activation energy of this cracking reaction and, thus, the reaction can happen at a lower temperature than without the catalyst, Hence, the weight of the catalyst before and after the cracking reaction will be the same.

Hope this helps answer your question.

Professor Yehia Khalil

Yale University

USA

I'm synthesising carbon nanotubes through catalytic methane decomposition, carbon yield is therefore required to know carbon deposition

Dear Vidisha,

It may be assumed that decomposition of methane produces hydrogen and carbon black by the following single reaction:

  CH4 + 75.3kJ / mol → C + 2H2

In above reaction one mole of methane produce one mole of carbon plus 2 moles of Hydrogen, If Methane & Hydrogen known then Carbon can be find out.

 Methane is a preferred raw material for the production of hydrogen from a hydrocarbon because of its high H to C ratio (H/C = 4), availability and low cost. Furthermore, the C produced can be sold as a co-product into the carbon black market (ink, paints, tires, batteries, etc).

As Yehia F. Khalil wrote, a catalyst lowers the activation energy (more correctly the activation free energy) but is not consumed in the reaction. Hence it has no effect on yield, though it will allow yield to be obtained faster. The maximum possible yield of carbon is (atomic weight of carbon)/(molecular weight of methane) = 12/16 = 3/4 of the weight of the methane. Of course the yield can be less if some of the carbon is lost due to inefficiency.

I'm assuming that this is being done on a laboratory setup, so yes you could probably get away with assuming minimal carbon deposition on your catalyst. Generally, if your throughput is less than 1 kg/hr you can get away with that assumption. Please just note that this is more a rule of thumb for process engineering than a hard and fast rule.

Hope this helps