OOOHH i get it now thanks!
Theres also one other thing i don't get- its about equilibrium. I understand the whole lcp stuff about temperature and pressure but i don't get everything about concentration, i.e. when you add things to the system.
For q9 i know that the answer is b, i was just wondering if part c and d have any effect on the equilibrium.
Also for 10, this is what i was talking about the concentration stuff, how do you do it
With Q9, C and D would end up shifting the equilibrium to the left because you introduce more product, not reactant.
So.
If there is an excess of either reactant or product, then the equilibrium has been disturbed (or it was never at equilibrium)
. Therefore, according to LCP, the system wants to get rid of that EXCESS.Conversely, if there is a
shortage of either reactant or product, then LCP predicts that the system wants to
bring in more of what's lacked to address the shortage. This is how equilibrium is (re-)established.
So for Q10, where the effect is not immediately obvious, you must analyse what happens.
You introduced hydrochloric acid. Obviously it's not going to react with water or the ammonium ion here. Does it react with ammonia?
No. The HCl will have a tendency to react with that hydroxide ion, which is clearly a base.
HCl + OH
- -> Cl
- + H
2O
So since we take out OH
- through the introduction of the acid, the equilibrium will shift to the right.
Now, immediately I can tell this question is beyond the HSC's scope, however it's not unreasonable. This is because of unnecessarily introducing option C: Equation is driven to completion.
A few drops of HCl is probably insufficient to do so. In practicality? Yes, this is possible if you had an unlimited amount of HCl. This is because you end up introducing so much acid that all the base ends up neutralised, and you have excess acid instead.
So for this question? It should be A.