Question

At 500 degree C, F_2 gas is stable and does not dissociate, but at 840 degree C, some dissociation occurs: F_2 (g) 2 F(g). A flask filled with 0.600 atm of F_2 at 500 degree C was heated to 840 degree C, and the pressure at equilibrium was measured to be 0.984 atm. What is the equilibrium constant K_p for the dissociation of F_2 gas at 840 degree C?

Answer 1

2.73 is the equilibrium constant for the dissociation of
`F_2` gas at 840 degree Celsius.

Step-by-step explanation:

`F_2(g)\rightleftharpoons 2F(g)`

Initial

0.600 atm 0

Equilibrium

(0.600 atm - p) 2p

Total pressure at equilibrium = P = 0.984 atm

P= 0.600 atm - p)+2p=0.984 atm

p = 0.384 atm

Partial pressure of the
`F_2` gas ,
`p_(f_2)`= (0.600 atm - 0.384 atm)=0.216 atm

Partial pressure of the
`F` gas,
`p_(f)` = 2(0.384 atm)=0.768 atm

`K_p=((p_(F))^2)/(p_(F_2))`

`K_p=((0.768 atm)^2)/(0.216 atm)=2.73`

2.73 is the equilibrium constant for the dissociation of
`F_2` gas at 840 degree Celsius.