[lykzomg]

This is a question regarding a past Chemistry AP Exam

http://i57.tinypic.com/jqm1qs.jpg

Specifically #1 (f)
From (e) ii, I was able to obtain the concentrations of Isobutane and n-butane at equilibrium which were 0.0071 and 0.003M respectively.

In f, they ask what the concentrations would be if we started out with isobutane instead of n-butane.

Why is it that we don't inverse the Kc and get 0.003 and 0.0071 for n-butane and Isobutane respectively, and instead the equilibrium concentrations remain the same?

The equation we were given was n-butane <-> Isobutane, and the Kc for that was 2.5, which is based on the equation: Kc = [isobutane]/[n-butane] so if we started out with isobutane, wouldn't the equation for Kc be: Kc = [n-butane]/[isobutane]?

Thanks in advance!

[mjc123]

Are you sure you've posted the right image?
If the conditions are the same, the position of equilibrium is the same whichever end you start from. That's the point of the equilibrium CONSTANT.

Why is it that we don't inverse the Kc and get 0.003 and 0.0071 for n-butane and Isobutane respectively, and instead the equilibrium concentrations remain the same?

Those are the same concentrations. (Did you mean "isobutane and n-butane respectively"?)

if we started out with isobutane, wouldn't the equation for Kc be: Kc = [n-butane]/[isobutane]?

There isn't a different Kc for the forward and back reactions. Kc is for the equilibrium, in which forward and back reactions are occurring simultaneously.
If we defined a Kc' = [n-butane]/[isobutane], its value would be the reciprocal of the other Kc, i.e. 0.4, and you would get the same answer for the equilibrium concentrations. You can define Kc either way, but pick one and be consistent with it.