Question

Consider the following equilibrium, for which Kp = 7.50×10⁻² at 470 ∘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)

Part B

Determine the value of Kp for the following reaction:

Cl2(g)+H2O(g)⇌2HCl(g)+12O2(g)

Part C

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

Answer 1

a)Answer is 13.3

b) The reaction Cl₂(g)+H₂O(g)⇌2HCl(g)+12O₂(g) is one half of the original. So you simply take the Kp value of the original, and raise it to an exponent of .5:
(7.50x10^-2)^.5 = .274

c) The equation used to find Kc from Kp is Kp=Kc(RT)^delta n, with R being the constant .0821, T being the temperature in Kelvin (923 K), and delta n being the difference in moles of the equation from part b (2.5-2=.5). So:
.274 = Kc(.0821 x 923)^.5
Kc = 3.45 x 10^-2