Question

A system at equilibrium contains I 2 (g) at a pressure of 0.19 atm and I(g) at a pressure of 0.22 atm . The system is then compressed to half its volume. Part A Find the pressure of I 2 when the system returns to equilibrium. Part B Find the pressure of I when the system returns to equilibrium.

Answer 1

First we need to use the initial equilibrium partial pressures to find the equilibrium constant, Kp.

After that, you will need to use Boyle's Law to find the total pressure of the system after the volume of the container is halved.

So, the equilibrium reaction looks like this

I2(g]⇌2I(g]

The total pressure of the system is equal to the sum of the partial pressure of its components

Ptotal=PI2+PI

         = 0.19 atm+ 0.22 atm

         = 0.41atm

When you decrease volume while keeping the number of moles and temperature constant, pressure will increase proportionally - this is known as Boyle's Law.

According to it: P1V1 = P2V2

But you also know that V2=V1/2, since the volume is halved. This means that we have

P1⋅V1=P2⋅ (V1/2)

⇒P2=2P1

The total pressure of the system after the volume is halved will be equal to

P2=2 * 0.41atm

   = 0.82 atm

The equilibrium constant will be

Kp= (I)² / (I2)

    = (0.22)² / (0.19)

    = 0.255

Try to predict what will happen to the partial pressures of the two gases after the volume is halved.

Since we are dealing with an equilibrium that involves gases, increasing the overall pressure of the system will favor the side will the smaller number of moles.

So you can expect the partial pressure of I2 to increase relative to that of I, given the fact that both partial pressures will increase as aresult of the reduction of the volume of the container.

Now, you can express the partial pressure of a gas that's part of a mixture by using its mole fraction and the total pressure of the mixture.

Pi=χi⋅Ptotal , where

Pi - the partial pressure of gas i;
χi - the mole fraction of gas i in the mixture;
Ptotal - the total pressure of the mixture.

The equilibrium constant for the second state of the system can thus be written as

Kp= χ²_i⋅P^₂2/χ_I2⋅P₂

     = χ_I² /χ_I2 ⋅P₂

Also, since you only have two components in the mixture, you know that

χ_I+χ_I2=1

This means that you have

χ_I2=1−χ_I

and so

Kp=0.255= χ_I2 /1−χ_I⋅ (0.82)

χ_I²=0.255/0.82⋅(1−χ_I)

χ_I²+0.311χ_I−0.311=0

The solution to this quadratic is - only take the positive solution

χ_I=0.425

The mole fraction of I2 will thus be

χI2=1−0.425=0.575

The partial pressures of the two gases after the new equilibrium is established will be

PI=χI⋅P2=0.425 * 0.82 atm=0.349atm

PI2=χI2⋅P2=0.575 * 0.82 atm=0.471 atm