H2 (g) + I2 (g) ⇌2HI (g) Initially, only H2 and I2 were present at concentrations of [H2]=3.65M and [I2]=2.70M. The equilibrium concentration of I2 is 0.0100 M. What is the equilibrium constant, Kc, for the reaction at this temperature?

Kc for this equilibrium is 1076

Explanation:

The equilibrium reaction is:

H₂ (g) + I₂ (g) ⇌ 2HI (g)

We propose the first situation:

Initially 3.65 m 2.70m -

React x x 2x

In the equilibrium [I₂] is 0.01 mol/L so, we can determine the x which is the amount that has reacted

Eq (3.65 - x) (2.70-x) = 0.01 2x

2.70-x = 0.01 → x = 2.69 moles, therefore the [C] in the equilibrium are:

3.65 - 2.69 = 0.96 M = [H₂]

0.01 M = [I₂]

2.69. 2 = 5.38 M = [HI]

Let's make the expresison for Kc → [HI]² / [H₂]. [I₂]

Kc = 5.38² / 2.69. 0.01 = 1076