Question

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 1

Answer :

(a) The equilibrium pressure of $H_2$ gas is, 288 torr

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

Solution :

The given equilibrium reaction is,

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

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,

$(392.0-2x)+3x=488.0$

$x=96torr$

The partial pressure of $H_2$ at equilibrium = 3x = 3 × 96 = 288 torr

The partial pressure of $AsH_3$ at equilibrium = (392.0-2x) = [392.0-2(96)] = 200 torr

Now we have to calculate the $K_p$ for the reaction.

The expression of $K_p$ for the reaction will be:

$K_p=\frac{(p_{H_2})^3}{(p_{AsH_3})^2}$

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

$K_p=\frac{(288)^3}{(200)^2}$

$K_p=597.1968torr=\frac{597.1968}{760}=0.786atm$

Conversion used : (1 atm = 760 torr)

Therefore, the value $K_p$ for this reaction is, 0.786 atm