I've had a look online and I think I now understand why the Kc value remains constant when adding/removing reactants/products, but I still don't understand why it does so upon dilution or changing pressure.
So if you dilute or decrease the pressure of a system at equilibrium, the concentration of all the species decreases. But, according to Le Chatelier's principle, this means the system shifts towards the side of the reaction with the higher number of moles to increase the concentration. However, wouldn't this mean that whilst the side of the reaction with higher moles increases, the other side of the reaction (with lower moles) decreases as a result - so one part of the concentration fraction increases but the other part decreases even more... wouldn't the Kc value then change? (as both parts of the concentration fraction are not both increasing to try to get back to the original Kc value, despite both parts being decreased as a result of the initial dilution/decrease in pressure)
Here I would suggest remembering exactly what K
c is, that
it is a fraction and a constant. You can get K
c of 2 with a concentration fraction of 6/3 and 3/1.5. 3M and 1.5M are both lower but still yield the same number 2.
Yes ultimately dilution lowers the concentration of all species in a solution, and according to the classic Le chat
one part of the fraction will oppose this change by increasing and the other decreasing, but it does this to establish, or even return to K
c not create a new Kc value.
The underlying theory of Kc is that it is a constant at a particular temperature. Since diluting a solution doesn't change the temperature of the system it remains the same value Straight from the OP heinemann textbook, I quote "For the general equation aW + bX cY + dZ at equilibrium
at a particular
temperature, the equilibrium expression can be written as [insert equation] where Kc is the equilibrium constant."
This also explains why temperature does change the value of Kc as it is dependent on temperature
Following this sweet knowledge, lets spice the math up and reinforce this concept with 3 species in a solution (1 product and 2 reactants) with still a K
c value of 2
X(aq) + Y(aq) <-> Z(aq)
Lets say [X] = 2M, [Y]= 1.5 and [Z]=6 in a 500mL solution
I dilute it by adding 500mL of water which halves the concentrations so now [X]=1. [Y]= 0.75 and [Z] = 3
Now the concentration fraction gives a value of 4, which is not equilibrium
Since Q
c>K
c it will favour a net reverse reaction which endorses Le chat's view of shifting towards the side with more particles.
So the concentration of Z will decrease and X and Y will increase.
Hypothetically [Z] will decrease from 3 to 2.5M and [X] will increase to 1.5M and [Y] will increase to roughly 0.833M which will yield a concentration ratio that is equal to equilibrium. this shows that despite the numerator decreasing and denominator of the fraction increasing it can still achieve the same value of equilibrium.
Hopefully this ungodly length of an answer helps