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 : The equilibrium pressure of is, 288 torr

Explanation :  Given,

Initial pressure of = 392.0 torr

Total pressure = 488.0 torr

The balanced equilibrium reaction is,

Initial pressure     392.0                            0

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

The expression of equilibrium constant for the reaction will be:

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 = 3p = 3(96) = 288 torr

Therefore, the equilibrium pressure of is, 288 torr

i think   the answer is c hope i'm right

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml