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

2h2o2--> 2h2o + o2

this reaction is of the spontaneous decomposition of hydrogen peroxide down into water and oxygen.

add 2 molecules of hydrogen peroxide and 2 molecules of water. since oxygen is naturally diatomic, the total number of atoms of each element is now the same on both sides of the equation so it is balanced.

this may not be entirely correct but let me know.