[Linh_]

this is the question and the beginning of my answer
https://imgur.com/a/Xw0AM7I

I just don't know what to do from here, I tried substituting but it doesn't help and I don't have any other idea.
Can someone help me please?

[Hunter2]

First is CH4 = k2*CH3°*H2/H° and CH4 = k3*C2H6*H°/CH3° ?

Only this two equations have to be combined.
Substitute H° in first equation with second equation and solve for CH4.

Initiation and Termination is in my opinion not necessary. It's not part of the chain reaction.
Termination have more possibilities like 2 H° => H2 or CH3° + H° => CH4

[Linh_]

First is CH4 = k2*CH3°*H2/H° and CH4 = k3*C2H6*H°/CH3° ?

CH4 = k2*CH3°*H2 + k3*C2H6*H°, from the two propagation reactions.
Also the suggested mechanism is part of the question and not something I chose.

Can you explain more please? substituting here doesn't lead to anything and I don't know what to do

[Hunter2]

No , you should not add, you should substitute.

CH4 = k2*CH3°*H2/H°

CH4 = k3*C2H6*H°/CH3°

H° = CH4*CH3°/(k3*C2H6)

Fill in this  for H° in first equation.

Solve for CH4 .

[Linh_]

Why is it possible to separate the expression for CH4 to 2 different expressions?
Shouldn't it be one that depends on both reactions?

[Hunter2]

It's because two reactions take place which create CH4.

The System is feed with C2H6 and H2

C2H6 will be cracked in two CH3° radicals.

This react with hydrogen according
CH3° + H2 => CH4 + H°
1. Methane source and a hydrogen radical
And H° react again with ethane according
C2H6 + H° => CH4 + CH3°
2. Methane source and again a methyl radical

Then the game starts again until all Ethane and hydrogen is consumed or the Termination reaction takes place.

[Linh_]

I just don't understand why can you separate the equations for CH4.

And also why do you add the products to the concentration equation?

Thanks for your help