Question

At 400K, the equilibrium constant for the reaction I2(g) + F2(g)
<==> 2 IF (g) Kp = 7.0. A closed vessel at 400 K is charged
with 1.46 atm I2, 1.46 atm F2, and 3.66 atm of IF. Which of the
following statements is true? **CAN YOU PLEASE EXPLAIN WHY?
THANK YOU!**

A. The equilibrium partial pressures of I2 F2, and IF will not change.

B. At equilibrium, the total pressure of IF wil 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 than1.46 atm.

Answer #1

I_{2}(g) + F_{2}(g) <==> 2 IF (g)
; K_{p} = P_{IF}^{2} /
P_{I2}.P_{F2}

Given that at equilibrium, the K_{p} is 7.

Let us check out if the system is currently at equilibrium or not.

P_{IF}^{2} / P_{I2}.P_{F2}

= 3.66^{2}/(1.46*1.46)

= 6.28 < K_{p}

As the system is yet to reach the equilibrium, concentration of
I_{2} and F_{2} will further decrease to form IF so
that the value of (P_{IF}^{2} /
P_{I2}.P_{F2} ) reaches to the value of 7.

So, at equilibrium pressure of IF will be greater than 3.66 atm
as more IF will produce and pressure of I_{2} and
F_{2} will be less than 1.46 atm.

Ans: (B) At equilibrium, the total pressure of IF wil be greater than 3.66 atm.

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