This is a question involving Le Chatelier's Principle. That Principle says it so beautifully if you write it out but most students don't get it. So here is a country boy's take on it in very simple words.
In a system at equilibrium, the equilibrium will shift so as to undo what we do to the reaction in order to reach a new equilibrium under the new conditions. CaO (s) + H2O (l) ==> Ca(OH)2 (s) + 65.2 kJ
Now let's rewrite it since 65.2kJ is heat to
CaO (s) + H2O (l) ==> Ca(OH)2 (s) + heat.
I'll do the first two. You do the last two.
a. The mixture is heated.
So the reaction gives off heat. If we heat the mixture at equilibrium, it will shift (move to the right or to the left) so as to undo what we did. So if we are adding heat it will move to the left because it doesn't want to add more heat So solid CaO will increase. The problem doesn't ask but liquid water will increase also and solid Ca(OH)2 will decrease.
b. H2O (l) is added to the mixture.
Adding H2O will make the equilibrium shift to the right so CaO will decrease, Ca(OH)2 will increase and more heat will be generated.
c. Ca(OH)2 is removed from the mixture.
d. The mixture is cooled.
I shall be happy to check your answers for c and d.
Consider the following reaction. CaO (s) + H2O (l) Ca(OH)2 (s) + 65.2 kJ
If you start with a container holding a mixture of the CaO (s), H2O (l) and Ca(OH)2 (s) at equilibrium, describe what happens to the amount of CaO if each of the following changes was made to the original sample?
a. The mixture is heated.
b. H2O (l) is added to the mixture.
c. Ca(OH)2 is removed from the mixture.
d. The mixture is cooled.
1 answer