✔ 最佳答案
C(s) + 2H₂(g) ⇌ CH₄(g)
A.
There is no effect.
This is because the effective concentration of solid is constant even more solid is added.
Both Q and K are unchanged, and Q = K.
B.
The equilibrium position shifts to the right and more CH₄ is formed.
This is because H₂ is removed to counteract the addition of H₂, according to Le Chatelier's principle.
K is unchanged. Before H₂ is added, Q = K. When H₂ is added, Q decreases immediately. Then Q increases and finally Q = K.
C.
The equilibrium position shifts to left and less CH₂ is formed.
This is because the raise in temperature favors the endothermic reaction, i.e. the backward reaction.
Before raising temperature, Q = K. In raising temperature, both Q and K increase. Finally, Q = K.
D.
The equilibrium position shifts to the right and more CH₄ is formed.
This is to lower the number of gaseous molecules to counteract the increase in pressure due to the decrease in volume.
K is unchanged. Originally, Q = K. When volume is lowered, Q decreases immediately. Then Q increases and finally Q = K.
E.
There is no effect because catalyst has no effect on the equilibrium position.
Q = K and both of them are unchanged.