Question

A sample of iron(II)sulfate was heated in an evacuated container to 920 K, where the following reactions occur:

               2FeSO₄(s) ↔ Fe₂O₃(s) + SO₃(g) + SO₂(g)

               SO₃(g) ↔ SO₂(g) + ½ O₂(g)

At equilibrium the total pressure is 0.836 atm and the partial pressure of oxygen was 0.0275 atm. Calculate K_p for both reactions.

Answer 1

I answer this before, so I'm gonna answer again. Step by step.

1) Based on the 1: 1 stoichiometry of the first reaction,

P_SO3 = P_SO2

call this value 'x'

2) After the second reaction finishes, at equilibrium we know the following:

the P_SO3 value has gone down from that produced by reaction 1 (because some SO₃ produced by reaction 1 was used up)
the P_SO2 has gone up from that produced in reaction 1 (because some SO₂ was made, in addition to that made in reaction 1)

3) Since we know P_O2 at equilibrium to be 0.0275 atm:

P_SO3 = x - 0.0275
P_SO2 = x + 0.0275

4) The sum of all three equilibrium pressures is 0.836 atm:

(x - 0.0275) + (x + 0.0275) + 0.0275 = 0.836
x = 0.40425 atm

5) K_p for the first reaction is:

K_p = (P_SO3) (P_SO2)

K_p = (0.40425) (0.40425) = 0.163

5) K_p for the second reaction is:

K_p = [ (P_SO2) (P_O2)^1/2 ] / (P_SO3)

K_p = [(0.43175) (0.0275)^1/2] / (0.37675) = 0.190

Hope this helps