Question

Consider the equilibrium I2(g) ⇌ 2I(g) at 860 K. Suppose 0.054 mol I2(g) is placed in a rigid 0.48-L container at 860 K. At equilibrium, I2(g) is 2.52% dissociated into its products. What is Kp at this temperature?

Answer 1

Let's write the equilibrium reaction once again:

I₂ <-----> 2I

With this reaction, we know that the expression for K is:

Kc = [I]² / [I₂]

and Kp = Kc (RT)

So, first we calculate Kc and then Kp. For Kc:

[I2] = 0.054 / 0.48 = 0.1125 M

The concentration for I is the 2.52% of dissociation: 0.1125 * 0.0252 = 0.002835 M

I₂ <--------------------> 2I

i. 0.1125 0

e. 0.1125 - 0.002835 (2* 0.002835)

Kc = (2*0.002835)² / (0.1125 - 0.002835)

Kc = 3.21x10^-5 / 0.109665

Kc = 2.93x10^-4

now for Kp, let's calculate first :

= 2-1 = 1

Kp = 2.93x10^-4 * 0.0821 * 860

Kp = 0.02067

If you need more explanations, or have some doubts, tell me in a comment and I'll gladly help you out.

Hope this helps