Consider the following system, which is at equilibrium,
CO(g) + 3 H2(g) <--------> CH4(g) + H2O(g).
The result of removing some CH4(g) and H2O(g) from the system will be
A. more CH4 is consumed to restore the equilibrium
B. Kc decreases
C. more H2O(g) is consumed to restore the equilibrium
D. more CO(g) is produced
E. more CH4(g) and H2O(g) are produced to replace that which is removed
The answer is E
I understand why the answer is E but why isn't the answer also be b?
Tbe ONLY time the value of Kc changes is with temperature.
The answer is not only B because removing some CH4(g) and H2O(g) from the system will disturb the equilibrium by reducing the concentration of these reactants. According to Le Chatelier's principle, the system will try to counteract this change and restore the equilibrium by favoring the formation of more CH4(g) and H2O(g) to replace what was removed. So, in addition to B, option E is also correct.
The answer is not B because the removal of CH4(g) and H2O(g) from the system will not directly affect the value of Kc, which is the equilibrium constant. Kc is determined by the ratio of the concentrations of the products and reactants at equilibrium, and removing some of the products and reactants does not change this ratio.
In this particular reaction, Kc is determined by the concentrations of CH4(g) and H2O(g) in the numerator, and the concentrations of CO(g) and H2(g) in the denominator. Removing some CH4(g) and H2O(g) will not change these concentrations directly, so Kc will remain the same.
Therefore, the correct answer is E, as removing some CH4(g) and H2O(g) from the system will cause the system to shift towards the products to replace what has been removed. This means that more CH4(g) and H2O(g) will be produced to restore the equilibrium.