Question

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?

Answer 1

Answer:

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