consider the following equilibrium

Question

2NOCl(g)--- 2 NO(g) + Cl2(g)
Initally 1.00 mol of NO and 1.00 mol of Cl2 are placed in a 5.00L container.
Calculate the initial concentrations of NOCl, NO and Cl2
NO= 0.20 mol/L
Cl2= 0.20 mol/l
NOCl=0
At equilibrium it is found that the NOCl = 0.120 mol/L calculate the value of k eq for the reaction
using ice table
change
NOCl= 0.120 mol/L
NO= -0.120 mol/L
Cl2= -0.06 mol/L
equilibrium
NOCl= 0.120 mol/L
NO= 0.08 mol/L
Cl2 = 0.14 mol/L
k eq= 0.08^2 0.14/ 0.120^2= 6.2x10^-2
if you could check this question out and point out any errors. I appreciate your time. thanks

Answer 1

it looks ok to me.