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

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

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.

Kc = 0.00149

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

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.

Thanks a lot DrBob222

To determine if you did the calculation correctly, we can go through the steps:

1. You correctly calculated the number of moles of ICl(g) using the given mass and molar mass: 0.682 g / 162.35 g/mol = 4.2x10^-3 mol.

2. Then, you divided the number of moles by the volume in liters to obtain the concentration in molarity: 4.2x10^-3 mol / 0.625 L = 6.72x10^-3 M. This step is correct.

3. You set up the reaction: 2ICl(g) ⇌ I2(g) + Cl2(g).

4. By using the initial concentration of ICl(g) and assuming the changes in concentration to be "-2x" for ICl(g) and "+x" for both I2(g) and Cl2(g), you correctly set up the expression for the equilibrium concentrations: 6.72x10^-3 - 2x, x, x.

5. Using the given mass of I2(g) and its molar mass, you calculated its concentration: 0.0383 g / 253.8 g/mol / 0.625 L = 2.41x10^-4 M. This step is correct.

6. Finally, you used the equilibrium concentrations to calculate Kc: Kc = [I2(g)][Cl2(g)] / [ICl(g)]^2 = (x)^2 / (6.72x10^-3 - 2x)^2. Using the value of x obtained from the concentration of I2(g), you correctly calculated Kc.

Therefore, based on the given information and your calculations, the value of Kc for this reaction is indeed 8.64x10^-6. Well done!