Say I have the following reactions:

$\ce{CH4 + H2O <=> CO + 3H2}$

$\ce{C2H6 + 2H2O -> 2CO + 5H2}$

Which undergo these processes in the same container. The first reaction goes to equilibrium and the second to completion. Clearly the reactions are entangled as the products are the same. The equilibrium concentration of methane will intuitively depend on the initial amount of ethane introduced. I'm not sure how to go about calculating the equilibrium concentration of methane however. My first thought was to find the amount of carbon monoxide and hydrogen gas generated by the second reaction and use this as initial product in the first reaction and to calculate equilibrium concentration from there but I'm having trouble doing this.


1 Answer 1



Suppose that initially, before any reaction, these are the species and amounts present:

$$\ce{CH4}: m \textrm{ moles} $$ $$\ce{H2O}: w \textrm{ moles} $$ $$\ce{C2H6}: e \textrm{ moles} $$ $$\ce{H2}: h \textrm{ moles} $$ $$\ce{CO}: c \textrm{ moles} $$

After the second reaction completes

Let's assume that the second reaction is limited by the amount of ethane present, not the amount of water. (This must be true, otherwise the first reaction could not be at "equilibrium" because one of the equilibrating species would head to a concentration of zero.) If the reaction is limited by ethane, then when it completes, $2e$ moles of carbon monoxide and $5e$ moles of hydrogen are formed.

$$\ce{CH4}: m \textrm{ moles} $$ $$\ce{H2O}: w - 2 e \textrm{ moles} $$ $$\ce{C2H6}: 0 \textrm{ moles} $$ $$\ce{H2}: h + 5e\textrm{ moles} $$ $$\ce{CO}: c + 2e\textrm{ moles} $$

After the first reaction equilibrates

The first reaction doesn't go to completion. Instead, let's assume the extent of the reaction from our current state to the equilibrated state is $\xi$ moles. That means that the species inventory becomes:

$$\ce{CH4}: m - \xi \textrm{ moles} $$ $$\ce{H2O}: w - 2 e - \xi \textrm{ moles} $$ $$\ce{C2H6}: 0 \textrm{ moles} $$ $$\ce{H2}: h + 5e + 3 \xi \textrm{ moles} $$ $$\ce{CO}: c + 2e + \xi \textrm{ moles} $$

These values can now be plugged into an equilibrium equation:

$$ K = \frac{[\ce{H2}]^3 [\ce{CO}]}{[\ce{CH4}][\ce{H2O}]}$$

$$ K = \frac{(h + 5e + 3 \xi)^3(c + 2e + \xi)}{(m-\xi)(w - 2 e - \xi)} $$

Now "all" you have to do is solve that equation for $\xi$.


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