At 400K , the equilibrium constant for the reaction I2(g) + F2(g) <--> 2IF (g) K
ID: 948355 • Letter: A
Question
At 400K , the equilibrium constant for the reaction I2(g) + F2(g) <--> 2IF (g) Kp = 7.0. A closed vessel at 400k is charged with 1.46 atm I2, 1.46 atm F2, and 3.66 atm of IF. Which of the following statements is true?
a. the equilibrium partial pressures of I 2, F2, and IF will not change.
b. At equilibrium, the total pressure of IF will be greater than 3.66 atm.
c. At equilibrium, the pressures of I2 will decrease and F2 will increase.
d. The reaction will go to completion since there are equal amounts of I2 and F2.
e. At equilibrium, the total pressure of F2 will be greater than 1.46 atm.
THE ANSWER IS B. is it because of the (x)^2 when doing the rate constant?
Not sure how to answer this.
Explanation / Answer
I2(g) + F2(g) <--> 2IF (g) Kp = 7.0
Q = 3.66^2 / (1.46*1.46) = 6.28
Kp = pIF^2 / pI2*pF2 = 7
as Q<kp .the forward reaction is favourable.so that
At equilibrium, the total pressure of IF will be greater than 3.66 atm.
answer: B
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