Question

A system at equilibrium contains I2(g) at a pressure of 0.18 atm and  I(g) at a pressure of 0.21 atm . The system is then compressed to half its volume.

Part A

Find the pressure of I2 when the system returns to equilibrium.

Part B

Find the pressure of I when the system returns to equilibrium.

Answer 1

The idea here is that you 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

P_total=P_I2+P_I

P_total=0.18+0.21=0.39 atm

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

So, for the initial state of the system, you have

P₁⋅V₁=n⋅R⋅T

for the final state of the system, you have

P₂⋅V₂=n⋅R⋅T

Divide these wto equations to get the equation for Boyle's Law

P₁V₁/P₂V₂=n⋅R⋅T/n⋅R⋅T ⇒ P₁V₁=P₂V₂

But you also know that V₂=V₁/2, since the volume is halved. This means that you 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 0.39 atm = 0.78 atm

The equilibrium constant will be

Kp = I²/I₂ = 0.21²/0.18 = 0.245

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

Since you're 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 a result 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.

P_i = _i⋅P_total , where

P_i - the partial pressure of gas i
_i - the mole fraction of gas i in the mixture;
P_total - the total pressure of the mixture.

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

K_p = (_I²⋅P₂²)/(_I2⋅P₂) = (_I²/_I2) P₂

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

_I + _I2 = 1

This means that you have

_I2 = 1 − _I

and so

K_p = 0.245 = [_I²/(1-_I)] 0.78

_I² = [0.245/0.78] (1−_I)

_I² + 0.314⋅_I − 0.314 = 0

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

_I = 0.425

The mole fraction of I₂ will thus be

_I² = 1−0.425 = 0.575

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

P_I = _I⋅P₂ = 0.425 0.78 atm = 0.33 atm

P_I2 = _I2⋅P₂ = 0.575 0.78 atm = 0.45 atm