Equilibrium questions?

Question

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).

torr

(b) Calculate Kp for this reaction.

atm

huh? where do i even start?

Answer 1

by reading your textbook :)

i could have answered this 8 years ago  but not now!

Answer 2

Pressure . .2AsH3(g)  <==>  2As(s) + 3H2(g)

initial . . . . . .392 . . . . . . . . . . . . . . . . .0

change . . . . -2x . . . . . . . . . . . . . . . . +3x

final . . . . .392 - 2x . . . . . . . . . . . . . . .3x . .

The final pressure in the flask is P AsH3 + P H2 after the reaction has completed = 488 torr.

So (392 - 2x) + 3x = 488

392 + x = 488

x = 96 torr; 3x = P H2 = 3(96) = 288 torr

P AsH3 = 392 - 2x = 392 - 2(96) = 200 torr

Kp = (P H2)^3 / (P AsH3)^2 where P is in atm

P H2 = 288/760 atm = 0.379 atm

PAsH3 = 200/760 = 0.263 atm

Kp = (0.379)^3 / (0.263)^2 = 0.787

Answer 3

1.observe:from 2 mole of gas(AsH3) is formed 3mole of gas (H2)

2. initial: 392torr final 392-2x+3x ,x=advancement

3. 392+x=488,x=96torr,

a.eql. pressure H2 :3x=288torr;AsH3=200torr

b.Kp=(288torr)^3/(200torr)^2=597.2torr :yes,Kp has unit of measurement