Asample of so3 is introduced into an evacuated sealed container and heated to 600 k. the following equilibrium is established: 2 so3( g) ∆ 2 so2( g) + o2( g) the total pressure in the system is 3.0 atm and the mole fraction of o2 is 0.12. find kp

The value of  Kp is 0.050.

As per this law, the vapor pressure of a component at a given temperature should be equivalent to the mole fraction of that component and then it should be multiplied by the vapor pressure of that component in the pure state.

Since mole fraction of oxygen is 0.12

Now the partial pressure should be = 0.12(3) = 0.36

The partial pressure of SO_2 is = 2(0.36) = 0.72

Now the partial pressure of SO_3 is  [3 - (0.36+0.720)] atm = 1.92 atm

Now Kp is

= 0.050

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The value of is 0.050.

Explanation:

According to Raoult's law, the vapor pressure of a component at a given temperature is equal to the mole fraction of that component multiplied by the vapor pressure of that component in the pure state.

As we know the mole fraction of is 0.12

The partial pressure of

The partial pressure of Thus the partial pressure of is = [3 - (0.36+0.720)] atm = 1.92 atm

= 1.92 atm

The value of is 0.050.

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