The gas arsine (ash3) decomposes as follows: 2ash3(g) < > 2as(s) + 3h2(g) in an experiment pure ash3(g) was placed in an empty, rigid, sealed flask at a pressure of 392.0 torr. after 48 h the pressure in the flask was observed to be constant at 488.0 torr. a. calculate the equilibrium pressure of h2(g). b. calculate kp for this reaction.

Answer :

(a) The equilibrium pressure of gas is, 288 torr

(b) The value for this reaction is, 0.786 atm

Solution :

The given equilibrium reaction is,

Initially pressure        392.0                           0

At equilibrium         (392.0-2x)                     3x

The final pressure in the flask after the reaction has completed is 488.0 torr. That means,

The partial pressure of at equilibrium = 3x = 3 × 96 = 288 torr

The partial pressure of at equilibrium = (392.0-2x) = [392.0-2(96)] = 200 torr

Now we have to calculate the for the reaction.

The expression of for the reaction will be:

Now put all the given values in this expression, we get:

Conversion used : (1 atm = 760 torr)

Therefore, the value for this reaction is, 0.786 atm