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

Answer:

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

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

$K_p=\frac{(p_{F})^2}{p_{F_2}}$

$K_p=\frac{(0.768 atm)^2}{0.216 atm}=2.73$

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