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

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

Since 1 mole of each SO3 and So2 are produced.

So,

P _(SO3) = P _(SO2) = x

After the second reaction:

SO3(g)      SO2(g) + ½ O2(g)

As given: P (O2) = 0.0275 atm

So,

P _(SO3) = x - 0.0275

P _(SO2) = x + 0.0275

Now, 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

So, for first reaction:

P _(SO3) = P _(SO2) = 0.40425 atm

and, for second reaction:

P _(SO3) = 0.40425 - 0.0275 = 0.37675 atm

P _(SO2) = 0.40425 + 0.0275 = 0.43175 atm

P (O2) = 0.0275 atm

K_p for the first reaction is:

K_p = P(_SO3) * P(_SO2)

= 0.40425 * 0.40425

K_p = 0.163 atm

K_p for the second reaction is:

K_p = [ P(_SO2) * P(_O2)^1/2 ] / P(_SO3)

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

= 0.0716 / 0.37675

K_p = 0.190