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

I₂(g) + F₂(g) <==> 2 IF (g)                ; K_p = P_IF² / 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_IF² / P_I2.P_F2

= 3.66²/(1.46*1.46)

= 6.28 < K_p

As the system is yet to reach the equilibrium, concentration of I₂ and F₂ will further decrease to form IF so that the value of (P_IF² / 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 I₂ and F₂ will be less than 1.46 atm.

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