Question

Consider the following equilibrium, for which Kp = 7.55×10⁻² at 500 ∘C:

2Cl₂(g)+2H₂O(g)⇌4HCl(g)+O₂(g)

Part A

Determine the value of Kp for the following reaction:

4HCl(g)+O₂(g)⇌2Cl₂(g)+2H₂O(g)

Express the equilibrium constant to three significant digits.

Part B

Determine the value of Kp for the following reaction:

Cl₂(g)+H₂O(g)⇌2HCl(g)+1/2O₂(g)

Express the equilibrium constant to three significant digits.

Part C

What is the value of Kc for the reaction in Part B?

Answer 1

2Cl₂(g)+2H₂O(g) <-------------------> 4HCl(g)+O₂(g)              Kp1 = 7.55×10⁻²

Kp1 = [HCl]^4 [O2]/[Cl2]^2 [H2O]^2

part A )

4HCl(g)+O₂(g) <-----------------> 2Cl₂(g)+2H₂O(g)

Kp = [Cl2]^2 [H2O]^2 / [HCl]^4 [O2]

      = (1/Kp1)

      = 1 / 7.55×10⁻²

      = 13.2

Kp = 13.2

Part B :

Cl₂(g) + H₂O(g) <-----------------> 2HCl(g) + 1/2O₂(g)

Kp = [HCl]^2 [O2]^(1/2) /[Cl2] [H2O]

       = sqrt (Kp1)

       = sqrt (7.55×10⁻²)

       = 0.274

Kp = 0.274

Part C:

Dn = 5/2 - 2 = 1/2 = 0.5

Kp = Kc (RT)^Dn

0.274 = Kc (0.0821 x 773)^0.5

Kc = 0.0344