# According to Le Chatelier's Principle what will happen to this equilibrium: C2H4O (g) ---> CH4 (g) + CO (g) if we add CH4?

Jan 9, 2015

Le Chatelier's principle states that if a change in volume, pressure, temperature, or concentration takes place for a chemical system at equilibrium, the system will react in such a manner that will counteract the change made and establish a new equilibrium.

The chemical equilibrium for the decomposition of etylene oxide into methane and carbon monoxide is

${C}_{2} {H}_{4} {O}_{\left(g\right)} r i g h t \le f t h a r p \infty n s C {H}_{4 \left(g\right)} + C {O}_{\left(g\right)}$

If you were to add more $C {H}_{4}$, the equilibrium would have to shift in such a manner as to offset this increase in concentration. The only way this can happen is if more ${C}_{2} {H}_{4} O$ is produced, which is equivalent to more $C {H}_{4}$ being consumed.

The equilibrium will shift towards the reactant in order to counteract the increase in the concentration of one of the products.

LIkewise, an increase in the concentration of ${C}_{2} {H}_{4} O$ would shift the equilibrium towards the products, since the only way to offset the additional ${C}_{2} {H}_{4} O$ would be to produce more $C {H}_{4}$ and $C O$.

Here are some links to other possible scenarios addressed for this particular reaction by other contributors:

http://socratic.org/questions/according-to-le-chatelier-s-principle-what-will-happen-to-this-equilibrium-c2h4o-1

http://socratic.org/questions/according-to-le-chatelier-s-principle-what-will-happen-to-this-equilibrium-c2h4o-2