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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).

Respuesta :

Answer : The equilibrium pressure of [tex]H_2(g)[/tex] is, 288 torr

Explanation :  Given,

Initial pressure of [tex]AsH_3[/tex] = 392.0 torr

Total pressure = 488.0 torr

The balanced equilibrium reaction is,

                              [tex]AsH_3(g)\rightleftharpoons 2As(s)+3H_2(g)[/tex]

Initial pressure     392.0                            0

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

The expression of equilibrium constant [tex]K_p[/tex] for the reaction will be:

[tex]K_p=\frac{(p_{H_2})^3}{(p_{AsH_3})}[/tex]

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 [tex]H_2(g)[/tex] = 3p = 3(96) = 288 torr

Therefore, the equilibrium pressure of [tex]H_2(g)[/tex] is, 288 torr

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