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.

a. energy is absorbed in the reaction.

if in a given reaction the energy of products is greater than the energy of reactants it means that reaction is endothermic and during the reaction, the energy is absorbed by reactants. in an endothermic reaction, the products have more stored chemical energy than the reactants. and in this reaction the reactants are more stable than products before the compounds with less energy are always more stable.