I can't get my head around equilibrium or more specifically, the 'equilibrium shift'.
Suppose we have:
A + C -><- C + D
the principle states that if we add some A, there will be more products formed, so ultimately the amount of A will decrease and the amount of C and D will increase. This makes the equilibrium shift to the right.
Well, I can't help myself but take the idea further - if the equilibrium shifts to the right and the amount of C and D increases, surely the reaction will counter-act that, so that when there's more C and D, they react more and so in turn - their concentration decreases?
1. Concentration of A is increased
2. A and C react to produce more C + D (equilibrium shifts to the right)
3. C + D react (because there is more of it), forming A + C
4. Equilibrium shifts back to the left?
From my understanding, the equilibrium would just shift back to its initial position so that the concentrations are back to 50/50. A bit like on this diagram:
Where is the fault in my understanding? I am aware that something's not working out, whether it's my definition of equilibrium or equilibrium shifting. Or perhaps I am understanding it, just not connecting the dots with the information from the textbook?
I've been sitting here looking throughout the internet for some answers, but it seems that I can't comprehend the concept myself. Help please?
Thanks in advance.
Equilibrium issues with understanding Watch
- Thread Starter
Last edited by frostyy; 14-11-2016 at 18:35.
- 14-11-2016 18:34
- 14-11-2016 19:08
Do some research on a thing called Kc and get back to us if you don't get it.
If A + C <-> C + D, then nothing appears to be happening to C, so the equilibrium is just A <-> D. In which case, your understanding is fine.
- 14-11-2016 21:03
As already mentioned, eqm constant is the missing component in bsic level understanding.
What is worth noting also is that while overall amount at eqm can change after a "disturbance"/"perturbation", the ratio of conc of etc....ie eqm constant..remain fixed as long as T is constant.