K=1.6x10^-5 mol/L for the following reaction

Question

2NOCl(g).... 2 NO(g) + Cl2(g)
Calculate the concentrations of all species at equilibrium for each of the following original mixtures

E) 2.4 mol of NOCl, 2.4 mol of NO, and 1.2 mol Cl2 in a 1.0 L flask

F)1.9 mol/L concentration of all three gases

Answer 1

In order to try and get this spacing right, I will write the equation from top to bottom instead of horizontally.
1st are the equatios. Next column is starting molarityies (mols/L). Third column is change. Fourth column is final concentration. This is just an ICE chart.
2NOCl(g) 2.4 +2x 2.4+2x
|
|
v
2NO(g) 2.4 -2x 2.4-2x
+
Cl2(g) 1.2 -x 1.2-x

You know the reaction will shift to the left BECAUSE Q = (NOCl)^2/(NO)^2*(Cl2) = (2.4)^2/(2.4)^2*(1.2) = 1.2 and that is larger than Keq listed in the problem which means Q is too large and it will sift to the left to reach equilibrium.
The columns indicate changes taking place.
Now substitute all of that (the last column) into Keq expression and solve for x.
Keq = 1.6 x 10^-5 = (NOCl)^2/(NO)^2*(Cl2)^2.
post your work if you get stuck.