Consider the following equilibrium.
2 NOCl(g) ⇌ 2 NO(g) + Cl2(g)
Determine Kc if 1.00 mol of NOCl(g) is initially placed in an empty 1.00 L flask, and 90% of the NOCl is consumed when the reaction reaches equilibrium.
So we know that Kc = [Cl2][NO]^2 / [NOCl]^2 based on the formula. But this is at equilibrium, so you must find the concentrations at equilibrium.
What you have to do here is calculate the intial molarity of NOCl, and then calculate the molarity of all three at equilibrium. You can use an ICE chart if that makes it easier:
2NOCl 2NO Cl2
Initial 1 0 0
Change -0.9 +0.9 +0.9
Equilibrium
so it will be: Kc = [Cl2][NO]^2 / [NOCl]^2
=0.9x0.9^2 / 0.9-1?
I don't agree with the ICE table. Also the denominator is 1-0.9 and not 0.9 -1. You can't have a negative number as a starting concentration. I would do the ICE chart like this.
............ 2NOCl(g) ⇌ 2 NO(g) + Cl2(g)
I..............1..................0...............0
C..........-0.9...............+0.9.........+0.45
To determine the equilibrium constant, Kc, for the given equilibrium, we need to find the concentration of each species at equilibrium.
We are given that initially, 1.00 mol of NOCl(g) is placed in a 1.00 L flask, and 90% of NOCl is consumed at equilibrium. This means that 0.90 mol of NOCl has reacted, and 0.1 mol of NOCl remains.
Since the reaction produces 2 mol of NO(g) and Cl2(g) for every 2 mol of NOCl(g) reacted, the amount of NO and Cl2 formed will be equivalent to the amount of NOCl reacted. Thus, there will be 0.90 mol of both NO(g) and Cl2(g) at equilibrium.
Now, we can calculate the concentrations of each species by dividing the number of moles by the volume:
[NOCl] = 0.1 mol / 1.00 L = 0.1 M
[NO] = 0.9 mol / 1.00 L = 0.9 M
[Cl2] = 0.9 mol / 1.00 L = 0.9 M
Finally, we can write the expression for the equilibrium constant, Kc, using the concentrations of the species at equilibrium:
Kc = ([NO]^2 * [Cl2]) / [NOCl]^2
= (0.9 M)^2 * (0.9 M) / (0.1 M)^2
= 7.29
Therefore, the value of Kc for the given equilibrium is 7.29.