Hey Guys so I have this problem with ICE tables that is extremely confusing for me so I would love some help on it!

At a Certain Temperature, the equilibrium constant for the reaction of NO with Cl2 is 6250. If the initial concentration of NOCl is 1.0M and that of NO 0.10M what is the concentration of Cl2 at equilibrium?

2NO(g)+ Cl2(g)<-->2NOCl(g)

a) 1.0M
b) 0.062 M
c) 0.0016 M
d) 0.016 M
Anything Helps!

What's the (Cl2) initially. I've used 0 below.

......2NO(g)+ Cl2(g)<-->2NOCl(g)
I.....0.1M.....0........1.0M
C.....2x +x -2x
E.....0.1+2x...x........1-2x

Substitute the E line into Kc expression and solve for x. This is a cubic equation.

I think you made a typo for answer D. Are you sure that isn't 0.0106

By the way, instead of solving a cubic equation you can substitute each of the answers into the expression an see which comes out with K of 6250

Sure, I can help you with that! To solve this problem, we can use the ICE table approach. ICE stands for Initial, Change, and Equilibrium. Here's how you can approach this problem:

Step 1: Write the balanced chemical equation:
2NO(g) + Cl2(g) ⇌ 2NOCl(g)

Step 2: Set up the ICE table, filling in the initial concentrations and leaving the change and equilibrium concentrations unknown:

| NO(g) | Cl2(g) | NOCl(g)
---------------------------------------------
Initial | 0.10 M | ? | 1.0 M
Change | -2x | -x | +2x
Equilibrium | 0.10 - 2x | ? - x | 1.0 + 2x

We have to consider that the initial concentration of NOCl is 1.0 M, but this will change as the reaction progresses. Hence, we use "1.0 + 2x" for the equilibrium concentration.

Step 3: Write the expression for the equilibrium constant, Kc:
Kc = [NOCl]^2 / ([NO]^2 x [Cl2])

Given that Kc is 6250, we can plug in the values from the ICE table and solve for x:

6250 = (1.0 + 2x)^2 / [(0.10 - 2x)^2 x x]

Step 4: Simplify the equation and solve for x. This requires some algebraic manipulation, which I'll do for you.

6250 = (1 + 4x + 4x^2) / (0.01 - 0.4x + 4x^2)x

Cross-multiplying:

6250(0.01 - 0.4x + 4x^2) = 1 + 8x + 8x^2

62.5 - 250x + 2500x^2 = 1 + 8x + 8x^2

2500x^2 - 258x + 61.5 = 0

Now, you can solve this quadratic equation for x. Using a quadratic formula or factoring, you'll find two possible solutions for x.

Once you find the value(s) of x, substitute it/them back into the ICE table to find the concentration of Cl2 at equilibrium.

For this specific problem, x will have a very small value since the equilibrium constant Kc is quite large (6250). Therefore, we can make an approximation and assume that (0.10 - 2x) is approximately 0.10 and (1.0 + 2x) is approximately 1.0.

So, substituting x = 0 into our ICE table, we get the following equilibrium concentrations:

| NO(g) | Cl2(g) | NOCl(g)
----------------------------------------------
Initial | 0.10 M | ? | 1.0 M
Equilibrium | 0.10 M | 0 M | 1.0 M

From the ICE table, we can see that the concentration of Cl2 at equilibrium is 0 M. Therefore, the answer is option:

b) 0.062 M