Question

At very high temperature can following reaction occur.

2 CrSO₄(s) ↔ Cr₂O₃(s) + SO₂(g) + SO₃(g)

Rate constant (K_p) is 0,36 with some temperature. In some experiment a CrSO₄ is put in some container were partial pressure of SO₃ is 1,00 atm. What is partial pressure of SO₂ at equilibrium?

Answer 1

Given the value of rate constant, Kp = 0.36

For the above reaction, we have :

Kp = p_SO2 * p_SO3

Here, p_SO2 denotes partial pressure of SO₂ and similar is the case for SO₃.

We assume that when we add CrSO₄, only SO₃ is present in the container.

Given equation :

                              2 CrSO₄(s) ↔ Cr₂O₃(s) + SO₂(g) + SO₃(g)

Initial                             0                 0             0            1 atm

Eqb                               0                 0             x atm    1+x atm

Total pressure = x + 1 + x = 1 + 2x

Thus, pSO2 = x/(1+2x)

and pSO3 = (1+x)/(1+2x)

According to the equilibrium constant

Kp = 0.36 = [ x/(1+2x) ] * [ (1+x)/(1+2x) ]

Solving for x from the above equation, we get an imaginary value for x, which shows that there is some problem with the data. Kindly correct the data. The value of Kp is given incorrectly as 0,36 which I have assumed to be 0.36. Also there are no units given, while the units should be atm². Put the correct data in the above solution to get the answer.