Question

The gas arsine, AsH3, decomposes as follows. 2 AsH3(g) equilibrium reaction arrow 2 As(s) + 3 H2(g) In an experiment at a certain temperature, pure AsH3(g) was placed in an empty, rigid, sealed flask at a pressure of 392.0 torr. After 48 hours the pressure in the flask was observed to be constant at 488.0 torr. (a) Calculate the equilibrium pressure of H2(g).

Answer 1

Answer : The equilibrium pressure of
`H_2(g)` is, 288 torr

Explanation : Given,

Initial pressure of
`AsH_3` = 392.0 torr

Total pressure = 488.0 torr

The balanced equilibrium reaction is,

`AsH_3(g)\rightleftharpoons 2As(s)+3H_2(g)`

Initial pressure 392.0 0

At eqm. (392.0-2p) (3p)

The expression of equilibrium constant
`K_p` for the reaction will be:

`K_p=((p_(H_2))^3)/((p_(AsH_3)))`

As,

Total pressure at equilibrium = (392.0-2p) + (3p) = 488.0 torr

(392.0-2p) + (3p) = 488.0

392.0 + p = 488.0

p = 488.0 - 392.0

p = 96

Thus, the equilibrium pressure of
`H_2(g)` = 3p = 3(96) = 288 torr

Therefore, the equilibrium pressure of
`H_2(g)` is, 288 torr