Can anyone find out if I did it right or not? Thanks

Question

A 0.682?g sample of ICl(g) is placed in a 625?mL reaction vessel at 682 K. When equilibrium is reached between the ICl(g) and I2(g) and Cl2(g) formed by its dissociation, 0.0383 g I2(g)is present.

What is Kc for this reaction?
0.682g / 162.35g/mol = 4.2x10^-3mol

4.2x10^-3mol / 0.625L = 6.72x10^-3 M

2ICl(g) <=> I2(g) + Cl(g)
I 6.72x10^-3 0 0
C -2x +x +x
E 6.72x10^-3-2x x x

I2 (0.0383g/253.8g/mol ) / 0.625L = 2.41x10^-4M

Kc = [I2(g)][Cl2(g)] / [ICl(g)]^2

Kc = (x)^2 / (6.72x10^-3-2x)^2

Kc = (2.41x10^-4)^2 / 6.72x10^-3-2(2.41x10^-4))^2

Kc = 8.64x10^-6

Answer 1

You are right on with everything, BUT the last line; the answer for Kc is not right. I also noticed a typo in the equation but that doesn't affect the work you did.

Answer 2

Kc = 0.00149

but the answer from the book is Kc = 9.31x10^-6

Answer 3

You are right and the book is wrong.
I played around with the numbers and this what I found.

If you don't subtract the 2x from the 6.72E-3 in the denominator AND you don't square the denominator, you gt 8.64E-6.

If you subtract the 2x from 6.72E-3 in the denominator but you don't square the denominator, you get 9.31E-6.

If you do it right you get 1.49E-3

Your Kc expression with numbers substituted is correct.
Kc = (2.41x10^-4)^2 / 6.72x10^-3-2(2.41x10^-4))^2
You just did my first example above to get the 8.64 number. The book answer comes from the second example above to get the 9.31 number.

Answer 4

Thanks a lot DrBob222