At 920 k, kp = 0.40 for the following reaction. 2 so2(g) + o2(g) equilibrium reaction arrow 2 so3(g) calculate the equilibrium partial pressures of so2, o2, and so3 produced from an initial mixture in which the partial pressures of so2 and o2 = 0.52 atm and the partial pressure of so3 = 0 (exactly).

The equilibrium partial pressure of sulfur dioxide, oxygen and sulfur trioxide is 0.366 atm, 0.443 atm and 0.154 atm respectively.

Explanation:

We are given:

Initial partial pressure of sulfur dioxide = 0.52 atm

Initial partial pressure of oxygen = 0.52 atm

Initial partial pressure of sulfur trioxide = 0 atm

For the given chemical reaction:

Initial:                 0.52      0.52

At eqllm:         0.52-2x    0.52-x        2x

The expression of for above equation follows:

We are given:

Putting values in above equation, we get:

Neglecting the negative values because partial pressure cannot be negative.

So, x = 0.077

Equilibrium partial pressure of sulfur dioxide =

Equilibrium partial pressure of oxygen =

Equilibrium partial pressure of sulfur trioxide =

Hence, the equilibrium partial pressure of sulfur dioxide, oxygen and sulfur trioxide is 0.366 atm, 0.443 atm and 0.154 atm respectively.