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

1 Answer
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_2H_4O_((g)) rightleftharpoons CH_(4(g)) + CO_((g))#

If you were to add more #CH_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_2H_4O# is produced, which is equivalent to more #CH_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_2H_4O# would shift the equilibrium towards the products, since the only way to offset the additional #C_2H_4O# would be to produce more #CH_4# and #CO#.

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