For the equilibrium: O2(g) + 2 F2(g) 2 OF2(g) ; Kp = 2.3 x 10–15 Which ONE of th
ID: 947395 • Letter: F
Question
For the equilibrium: O2(g) + 2 F2(g)
2 OF2(g) ; Kp = 2.3 x 10–15
Which ONE of the following statements is true?
A) If the reaction mixture initially contains only OF2(g), then at equilibrium, the reaction mixture will consist of essentially only O2(g) and F2(g).
B) For this equilibrium, Kc = Kp.
C) If the reaction mixture initially contains only OF2(g), then the total pressure at equilibrium will be less than the total initial pressure.
D) If the reaction mixture initially contains only O2(g) and F2(g), then at equilibrium, the reaction mixture will consist of essentially only OF2(g).
E) If the reaction mixture initially contains only O2(g) and F2(g), then the total pressure at equilibrium will be greater than the total initial pressure.
Explanation / Answer
The equation is
O2 + 2F2 --> 2OF2
A) If the reaction mixture initially contains only OF2(g), then at equilibrium, the reaction mixture will consist of essentially only O2(g) and F2(g).
True : As the Kp for reverse reaction will be high and it will go towards formation of O2 and F2 only
B) For this equilibrium, Kc = Kp.
FAlse : We know that the relation between them is
Kp = Kc (RT)^ng
Where ng = Number of moles of products - Numer of moles of reactants
So here ng = 2-3 = -1
so Kp < Kc
C) If the reaction mixture initially contains only OF2(g), then the total pressure at equilibrium will be less than the total initial pressure.
False : The pressure will remain constant
D) If the reaction mixture initially contains only O2(g) and F2(g), then at equilibrium, the reaction mixture will consist of essentially only OF2(g).
False : The reaction will hardly go toward formation of OF2 as the Kp is very very low
E) If the reaction mixture initially contains only O2(g) and F2(g), then the total pressure at equilibrium will be greater than the total initial pressure.
False