At a certain temperature, the equilibrium constant, 𝐾c, is 6.01×10−3 for the reaction

Question

Cl2(g)↽−−⇀2Cl(g)

A. If 3.21 g Cl2 is placed in a 2.50 L flask at this temperature, what are the equilibrium concentrations of Cl2 and Cl?
[Cl2]=
M
[Cl]=
M
B. Following the establishment of equilibrium in part A, the volume of the flask is suddenly increased to 4.00 L while the temperature is held constant. What are the new equilibrium concentrations of Cl2 and Cl?
[Cl2]=
M
[Cl]=
M
C. Following the establishment of equilibrium in part A, the volume of the flask is instead suddenly decreased to 1.00 L while the temperature is held constant. What are the new equilibrium concentrations of Cl2 and Cl?
[Cl2]=
M
[Cl]=
M

Answer 1

A. (Cl2) = M = moles/L where moles = g/molar mass = 3.21/71 = 0.045 mols, then 0.045/02.50 L = 0.018M.
..................Cl2(g)↽−−⇀2Cl(g)
I ..................0.018............0
C...................-x.................2x
E................0.018-x............2x
Substitute the E line into the Kc expression and solve for x = (Cl^-) and evaluate 0.0108-x = (Cl2)
For B: See the previous problem.
For C: See the previous problem.
Note: This is a Kc problem and the previous problem was a Kp. For this one decreasing volume increaes concn and increasing volume decreases concn.
Post your work if you get stuck.