Consider the equilibrium between PCl 5 , PCl 3 , and Cl 2 . PCl 5 (g) PCl 3 (g)
ID: 1088181 • Letter: C
Question
Consider the equilibrium between PCl5, PCl3, and Cl2.
PCl5(g) PCl3(g) + Cl2(g) K = 7.71×10-2 at 535 K
The reaction is allowed to reach equilibrium in a 17.6-L flask. At equilibrium, [PCl5] = 0.126 M, [PCl3] = 9.84×10-2 M and [Cl2] = 9.84×10-2 M.
(a) The equilibrium mixture is transferred to an 8.80-L flask. In which direction will the reaction proceed to reach equilibrium?
_________to the right to the left
(b) Calculate the new equilibrium concentrations that result when the equilibrium mixture is transferred to 8.80-L flask.
Explanation / Answer
pcL5(G) <-----> pcL3(g) + Cl2(g)
initial 0.126 M 0.0984 M 0.0984 M
after changing
volume 0.126*17.6/8.8 0.0984*17.6/8.8 0.0984*17.6/8.8
= 0.252 M = 0.1968 M = 0.1968 M
chnage + x -x -x
equilbrium 0.252+x 0.1968-x 0.1968-x
from equation
Kc = [PCl3][Cl2]/[pcL5]
0.0771 = ((0.1968-x)^2/(0.252+x))
x = 0.0454
at equilbrium,
[PCl5] = 0.252+0.0454 = 0.2974 M
[PCl3] = 0.1968-0.0454 = 0.1514 M
[Cl2] = 0.1968-0.0454 = 0.1514 M