Equilibrium Constant

 
 

Equilibrium

In a closed system, reversible reactions can reach dynamic equilibrium. At dynamic equilibrium, reactants are being formed just as fast as the products are being formed, resulting in the concentrations of the reactants and products staying the same.

At equilibrium:

  • The forward and reverse reactions are still happening

  • The rate of the forward and reverse reactions is the same

  • The concentration of reactants and products remains constant

The equilibrium constant, Kc

The equilibrium constant, Kc, gives a numerical value to the equilibrium position and therefore the reaction yield.

  • The more the forward reaction is favoured — the position of equilibrium lying over to the right-hand side — the larger the value of Kc.

  • The more the reverse reaction is favoured — the position of equilibrium lying over to the left-hand side — the smaller the value of Kc.

The formula for Kc is:

 
 

Square brackets are used to represent concentration, in mol dm-3.

The units of Kc will differ depending on the number of moles of reactants or products. We have to work out the units each time we calculate Kc.

Worked example – writing Kc expressions and determining units

Nitrogen and hydrogen react to form ammonia, as shown in the equation below:

 
 

To write the Kc expression, we put the products on the top and the reactants on the bottom, to the powers of their coefficients. This gives us the following expression:

 
 

To work out the units, we just need to write out the expression again but replace each substance in the equation with the units for concentration, mol dm-3.

 
 

Now we can cancel out anything that is the same on the top and bottom of our expression.

 
 

This leaves mol dm-3 x mol dm-3 on the bottom of our expression. Multiplying these units gives us mol2 dm-6. Then we bring this to the top of our expression, giving us mol-2 dm6.

Homogeneous and heterogeneous reactions

Reactions are homogeneous if all of the reactants and products are in the same state. For these reactions, all substances are included in the Kc expression. For example, in the reaction between nitrogen and hydrogen to form ammonia, all substances are in the gaseous state, so all components are included in the Kc expression.

 
 
 
 

Reactions are heterogeneous if the reactants and products are in different states. For heterogeneous reactions, only gaseous and aqueous components are included in the Kc expression with everything else left out. This is because the concentrations of solids and liquids will not affect the position of equilibrium.

For example, the thermal decomposition of calcium carbonate (limestone) into solid calcium oxide and gaseous carbon dioxide is a heterogeneous reaction. Since we exclude solids from the Kc expression, gaseous carbon dioxide is the only component that is included.

 
 

Using Kc expressions to calculate concentration

If you know the value of Kc for a particular reaction, you can calculate the concentration of a particular reactant by rearranging the Kc expression.

Worked example – using Kc to calculate concentration

Let’s look again at the reaction between hydrogen and nitrogen to form ammonia, as represented in the equation:

 
 
 
 

*At 500 °C, Kc = 8.00 x 10–2 dm6 mol–2.

At equilibrium, the concentration of N2 is 1.20 moldm–3 and the concentration of H2 is 2.00 moldm–3 . Calculate the equilibrium concentration of ammonia under these conditions.*

Simply substitute the numbers that we have been given into our equilibrium expression. We can then rearrange this to find the concentration of ammonia.

 
 

Using initial concentrations to calculate equilibrium concentrations

If you know the concentration of the reactants that you started with, and the concentration of the products at equilibrium, we can use this to work out the concentration of the reactants at equilibrium.

Worked example – calculating equilibrium concentrations from initial concentrations

The preparation of hydrogen iodide, HI (g), from hydrogen and iodine gases is a reversible reaction which reaches equilibrium at constant temperature.

 
Kc HI.jpg
 

A student mixed together 0.30 mol H2(g) with 0.20 mol I2(g) and the mixture was allowed to reach equilibrium. At equilibrium, 0.14 mol H2(g) was present. Use this information to work out the equilibrium concentrations of iodine and hydrogen iodide.

To answer this question, we are going to use the ICE method. ICE stands for initial, change and equilibrium. The best way to set this out is in a table, as shown below. Initial concentrations of products will always be zero.

 
 

If we started with 0.3 mol of hydrogen and have 0.14 left at equilibrium, this means that 0.16 mol has reacted. Since hydrogen and iodine react in a 1:1 ratio, this means that 0.16 mol of iodide has also reacted, leaving 0.04 mol at equilibrium.

 
 

There is a 1:2 ratio between hydrogen and hydrogen iodide, meaning that two moles of HI are formed for every 1 mol of hydrogen that reacts. So if 0.16 mol of H2 has reacted, 0.32 mol of HI are formed.

 
 

Partial pressure

In a gas mixture, each gas exerts its own pressure. This is known as partial pressure and will differ depending on the gas. The total pressure of a gas mixture is the sum of all of the partial pressures of the individual gases.

Worked example – total pressure

When nitric oxide and oxygen are placed in a sealed container, they react to form nitrogen dioxide, as shown in the equation below.

 
 

The equilibrium mixture contains nitric oxide (NO) at a partial pressure of 36 kPa. Total pressure inside the container is 99 kPa. Calculate the partial pressure of nitrogen dioxide at equilibrium.

  • Total pressure is the sum of the individual gas pressures, so total pressure = p(NO) + p(O2) + p(NO2)
  • If NO has a partial pressure of 36 kPa, that means oxygen has a partial pressure of 18 kPa, since they exist in a 2:1 ratio
  • Now we can just rearrange the equation to find the partial pressure of nitrogen dioxide.
  • 99 = 36 + 18 + p(NO2)
  • P(NO2) = 99 – 54 = 45
  • Partial pressure of NO2 = 45 kPa

Mole fractions

A mole fraction is the amount of a particular gas relative to the whole gas mixture.

For example, if I have 8 moles of nitrogen gas in a 10 mol mixture of air, the mole fraction of nitrogen is 8/10 or 4/5. You can calculate the mole fraction of a gas using the following equation:

 
 

If you know the mole fraction of a gas, as well as the total pressure of the gas mixture, then you can calculate the partial pressure of the gas. To do so, you use the equation:

 
Partial pressure.jpg
 

Worked example – mole fractions

In the manufacture of ammonia for fertilisers, nitrogen gas and hydrogen gas are reacted together in a reaction vessel, as shown in the equation below.

 
 
 
 

When 3.0 mol of SO2Cl2 is heated at 700 K, the equilibrium mixture contains 1.75 mol of chlorine. If the total pressure is 515 kPa, what is the partial pressure of SO2?

  • There is a 1:1 ratio between SO2 and Cl2. This means that if there are 1.75 mol of Cl2 at equilibrium, there must also be 1.75 mol of SO2.
  • If we started with 3 mol of SO2Cl, as 1.75 mol has reacted, then we are left with 1.25 mol of SO2Cl at equilibrium.
  • Total moles = 1.75 + 1.75 + 1.25 = 4.75
  • Mole fraction of SO2 = 1.75 / 4.75 = 0.37
  • Partial pressure = mole fraction of a gas x total pressure
  • Partial pressure of SO2 = 0.37 x 515 = 190 kPa

The equilibrium constant, Kp

The equilibrium constant, Kp, is similar to Kc except we use partial pressures instead of concentrations. The Kp expression is written just like the Kc expression, with the products on the top of the equation and the reactants on the bottom.

For example, the Kp expression for the reaction below would be:

 
 

If the reaction contains reactants and products in different states (i.e. it is a heterogeneous reaction), you only include the things that are in the gaseous state in the Kp expression.

If you know the partial pressures of each of these gases, you can pop them into your Kp expression to find the numerical value for Kp.

Let’s say the partial pressure of nitric oxide, NO, is 36 kPa, the partial pressure of oxygen is 18 kPa and the partial pressure of nitrogen dioxide, NO2, is 54 kPa. We just place those values into the Kp expression. Remember to square any gas pressures in which there are 2 moles in the balanced symbol equation (or cube any gas pressures which are present in 3 moles etc).

Our Kp expression now looks like this:

  • Kp = 542 / 362  x 18

  • Kp = 0.125

We work out units in the same way that we calculated the units for Kc. Remember it will be different for different reactions so you’ll have to work it out each time. In the example above, the units will be:

 
 

kPa2 cancels out on the top and the bottom, leaving just kPa on the bottom. We bring that to the top of the equation by dividing by 1, giving us kPa-1.

So Kp = 0.125 kPa-1

At equilibrium, there are 6.2 mol of NH3, 3.6 mol of H2 and 8.9 mol of N2. Calculate the mole fraction for nitrogen.

  • Total moles = 6.2 + 3.6 + 8.9 = 18.7 mol
  • Mole fraction for N2 = 8.9 / 18.7 = 0.48

Worked example – using mole fractions to calculate partial pressure

At high temperatures, SO2Cl2 dissociates into SO2 and Cl2, as shown in the equation below.

Changes to Kp

Just like Kc, the only thing that changes the value of Kp is temperature. Changing the concentration or pressure of a reaction will have no effect on Kp.

This is because, even though the position of equilibrium may shift to the right after increasing the pressure, the equilibrium moles of product would increase but the equilibrium moles of reactant would decrease. Therefore the value for Kp would stay the same.

Adding a catalyst does not affect the value of Kp either, since it only speeds up the rate of reaction without changing the yield.

We can predict how temperature changes will affect the value of Kp, depending on whether a reaction is exothermic or endothermic.

  • For exothermic reactions, increasing the temperature will cause the position of equilibrium to shift to the left-hand side, favouring the reactants. The value of Kp (or Kc) will decrease.

  • For endothermic reactions, increasing the temperature will cause the position of equilibrium to shift to the right-hand side, favouring the reactants. The value of Kp (or Kc) will increase.