Question

A sample of HI (9.30 x 10-3 mol) was placed in an empty 2.00 L container at 1000 K. After equilibrium was reached, the concentration of I2 was 6.29 x 10- 4 M. Calculate the value of Kc at 1000 K for the reaction H2(g) + I2(g) 2HI(g).

Answer 1

A sample of HI (9.30 x 10-3 mol) was placed in an empty 2.00 L container at 1000 K.

[HI] initial = mol / V = (9.3*10^-3)/(2) = 0.00465 M

[I2] = [H2] = 0

After equilibrium was reached, the concentration of I2 was 6.29 x 10- 4 M.

then, we need to get the equilbirium concentration,

[HI] eq= 0.00465 - 2x

[I2] eq = [H2] eq = 0 + x

and we know that:  I2 was 6.29 x 10- 4 M.

x = [I2] = 6.29*10^-4 M

so

[HI] eq= 0.00465 - 2x =0.00465 -2*6.29*10^-4 = 0.003392 M

Now... in order to be able to :

Calculate the value of Kc at 1000 K for the reaction H2(g) + I2(g) 2HI(g).

We need

Kc expression of our known data:

2HI <-> H2 + I2

Kc = [H2][I2]/[HI]^2

Kc = (6.29*10^-4)(6.29*10^-4) /(0.003392^2)

Kc = 0.034386

Now, we need the reverse reaction

H2(g) + I2(g)<-> 2HI(g).

So Kc required = 1/Kc original

Kc required = 1/(0.034386) = 29.081