This page provides two separate problems which together relate a number of ideas.

**Problem 1** involves calculating some equilibrium concentrations and then a value for Kc. The equation pV = nRT is used to calculate some corresponding (partial) pressures, from which a value for Kp can be calculated. In this calculation the volume remains constant and so the total pressure changes on reaching equilibrium. Finally, the equation Kp = Kc(RT)^{Dn} is used to calculate a value for Kp again, but this time from the value of Kc.

**Problem 2** involves calculating the equilibrium molar amounts of reactants and products, and then their partial pressures, from which a value for Kp can be obtained. In this calculation the total pressure remains constant and so the volume changes on reaching equilibrium.

See if you can follow through these examples.

3.00 moles of pure SO_{3}(g) are introduced into an 8.00 dm^{3} container at 1105 K. At equilibrium, 0.58 mol of O_{2}(g) has been formed. For the reaction:

2SO_{3}(g) 2SO_{2}(g) + O_{2}(g)

at 1105 K

(a) Calculate Kc

(b) Calculate the initial partial pressure of SO_{3}(g) and the equilibrium partial pressure of O_{2}(g). Now calculate a value for Kp.

(c) Use the equation **Kp = Kc(RT) ^{Dn}** to calculate Kp for this equilibrium.

- Decide on the initial amounts in moles of reactants and products.
- Decide on the change in moles of each reactant and product for equilibrium to be reached.
- Work out the number of moles of each reactant and product at equilibrium (what is the value of y below?).
- Work out the concentration in mol dm
^{-3}of each reactant and product at equilibrium.Set out your working in an 'equilibrium table' like this:

2SO _{3}(g)2SO _{2}(g)+ O _{2}(g)Initial moles 3.00 - - mol Change -2y +2y +y Equilibrium moles (3.00 - 1.16) 1.16 0.58 mol Equilibrium concentrations ^{ }(3.00 - 1.16)/8.00 ^{ }1.16/8.00 ^{ }0.58/8.00 ^{ }mol dm ^{-3}

- Write the equilibrium expression for the reaction in terms of Kc, and substitute the equilibrium concentrations as appropriate to calculate its value.

The volume of the container is fixed at 8.00 dm^{3}. This reaction involves a change in gaseous volume on reaching equilibrium (an increase in this case) and so the total pressure will change.

- Use the equation PV = nRT to calculate the initial pressure of the SO
_{3}(g). - Use the equation pV = nRT to calculate the equilibrium partial pressure of the O
_{2}(g).

For help with this, read about The Ideal Gas Law.

- Decide on the change in partial pressure of each reactant and product for equilibrium to be reached. (What is the value of y above in terms of partial pressure?)
- Work out the equilibrium partial pressure of each reactant and product.
2SO _{3}(g)2SO _{2}(g)+ O _{2}(g)Initial partial pressures 34 - - atm Change -2y +2y +y Equilibrium partial pressures (34 - 13.2) 13.2 6.6 atm - Write an equilibrium expression for the reaction in terms of Kp, substitute in the equilibrium partial pressures and calculate the value of Kp.

There are two common values of the gas constant, R. Kc has been calculated using concentrations in mol dm^{-3}, and so the appropriate value for R is 0.0820575 atm dm^{3} K^{-1} mol^{-1}. (More often you will be given the SI units value for R which is 8.3142 J K^{-1} mol^{-1}. See The Ideal Gas Law for more help with this.

SO_{2}(g) and O_{2}(g) in the ratio **2 moles : 1 mole** were mixed at constant temperature and a constant pressure of 9 atmospheres in the presence of a catalyst. At equilibrium, one third of the SO_{2}(g) had been converted to SO_{3}(g).

2SO_{2}(g) + O_{2}(g) 2SO_{3}(g)

In this calculation the total pressure remains constant at 9 atm and so the volume will change on reaching equilibrium. You will need to know that the partial pressure of a chemical component is equal to its mole fraction of the total pressure. (Dalton's Law of Partial Pressures says that the total pressure is equal to the sum of tha partial pressures.)

- Set out an 'equilibrium table', part of which is shown below.
- Work out the amount in moles of each reactant and product at equilibrium. (What is the value of y?)
2SO _{2}(g)+ O _{2}(g)2SO _{3}(g)Initial moles 2 mol 1 mole Change -2y -y +2y Equilibrium moles 1.33 0.66 0.66 mol - Work out the equilibrium partial pressure of each component.
- Write the equilibrium expression for Kp and calculate its value.