I thought that the value of Keq changes with temperature only...but is this only the case if the equation itself does not change?
So for example, is it true that...
X + Y --> A + B
has a different Keq than
2X + 2Y --> 2A + 2B
?
If so, why is this? Technically you are just increasing the reactant and product concentrations and you are not increasing temperature, right?
You are absolutely correct that the numerical value of K
eq only changes with temperature. However, that value applies for a reaction exactly as written. If you change the phases of any component or as you have done, double all of the stoichiometric terms, then it's a new reaction and thus a new K
eq. For instance, if we reverse a reaction, then then we need the recipricol of the original K
eq for the new reaction (reverse reaction).
You should note that other changes (pressure, volume, moles of reactant, and moles of product) can shift the equilibrium, but aslong as the temperature remains constant, then the numerical value of K
eq does not change.
Are you talking about the equilibrium constant Keq, or the rate constant k?
There shouldn't be too much confusion about Keq. Anything that upsets the balance of a reaction in equilibrium (changing concentrations, temperature, pressure, etc) will change Keq. The point is that different reaction conditions establish a different Keq.
I think you might be mixing up the equilibrium conditions with the numerical value of K
eq. A great example involves pH of an aqueous solution at 298K. No matter what we do, add acid or add base, the sum of pH + pOH = 14, as long as the temperature does not change from 298K. This is because the autoionization of water has a K
w = 10^
-14 at that temperature.
Adding base will shift the equilibrium so that the [H
3O+] will go down and the [OH-] will go up, but the new values after the addition will still multiply out to be 10^
-14.