PCl5 --> PCl3 + Cl2 (Kp = 630 at 546 K)

Question

A system is prepared by placing equimolar amounts of the three gases shown in the equation above in a suitable rigid container held at constant volume. Equilibrium is established at 546 K.

(A) When equilibrium is established, how does [PCl3] compare to [Cl2]? Explain.

(B) When equilibrium is established, how does [PCl5} compare to [PCl3]? Explain.

(C) When equilibrium is established, how does the rate of the forward reaction compare to the rate of the reverse reaction? Explain.

(D) If the volume of the container is increased at constant temperature, then:
(1) What effect is observed on the number of moles of PCl5 in the system? Explain.
(2) Does the value of Kp increase, decrease, or remain the same? Explain.

Answer 1

The way you do this problem is to calculate Q and compare with K. Here is a good site.
http://www.ausetute.com.au/qquotient.html

Answer 2

So basically for the first 3 answers (Part (A)(B)(C)):
(A)and(B) - the concentrations would be equal since the mole ratio is equal and Q = K at equilibrium.
(C) - the rate of both reactions would be equal also, as Q = k, right?

(D)(1) - If the volume is increased at constant temperature their will be a shift to the right, if i am correct, because as the pressure decreases due to the increase in volume the shift will be towards the side which contains the most moles of gas. Thus the number of moles of PCl5 would decrease.
(2) - I am sort of confused for number 2, but I predict the Kp value would increase since the pressure of PCl5 decreases and it is causing the denominator of Kp equation (Kp = Kp(prod)/Kp(react)) to decrease.

So, I was wondering if I am thinking this clearly and if I am on the right path?

Answer 3

for d(2) K would remain the same because k only depends on temperature