The gas arsine, AsH3, decomposes as follows. 2 AsH3(g) 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).

96 torr

Explanation:

After 48 hours, we are told, the pressure is 488 torr. Since we started with 392 torr of arsine, we can conclude there is an increase in pressure cause by the arsine producing hydrogen and the solid As.

We can calculate the equilibrium pressure of H₂ if we account for the total pressure in terms of the stoichiometry of the equilibrium equation:

                          2 AsH₃(g )     ⇆    2 As(s)        +      3 H2(g)

Initial p (atm)          392                                                  0

Change                   -2x                                                  3x

Equilibrium           392 - 2x                                           3x

and the total pressure at equilibrium is the sum of the pressure of the gasses arsine and hydrogen:

488 = (392 - 2x)+ 3x

488 - 392 = x

x = 96

Therefore the equilibrium pressure of H₂ is 3 x 96 = 288 torr

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