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?

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

Explanation:

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 gas , = (0.600 atm - 0.384 atm)=0.216 atm

Partial pressure of the gas, = 2(0.384 atm)=0.768 atm

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

it forms an anion that has a larger radius. anions have a negative charge and are larger than the original atom.