Question

The gas arsine, AsH3, decomposes as follows: 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). b. Calculate Kp for this reaction

Answer 1

The equilibrium pressure of $H_{2}$ is 288 torr. and $K_{p}$ for this reaction is 0.786 atm.

from the reaction:-        $2AsH_{3}$ ⇄ $2As + 3H_{2}$

initial concentration     392 torr.              0

at equilibrium.              392 - $2x$.               $3x$

and the final pressure in the flask = 488 torr.

Hence,

$( 392 - 2x ) + 3x = 488\\ x = 488-392\\x = 96 torr.$

The partial  pressure of $H_{2}$ is 3 × 96 =  288 torr.

and $AsH_{3}$ is 392 - ( 2 × 96 ) = 392 - 192 = 200 torr.

Now, to find $K_{p}$ for this reaction, we will use $K_{p} = \frac{(P_{H_{2} })^{2} }{( P_{AsH_{3} } )^{2}}$

putting all the values, we get,

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

     = 597.1968 torr.

     = 0.786 atm.  ( 1 atm = 760 torr. )

what do you mean by equilibrium?

Equilibrium in chemistry is the phase that exists when a chemical reaction and its opposite reaction happen at the same rates. This word's Latin origin dates back to the prefix aequi-, which means equal, and lbra, that indicates scale or balance.

Learn more about equilibrium reaction here:-

brainly.com/question/15118952

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