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 SO3(g) are introduced into an 8.00 dm3 container at 1105 K. At equilibrium, 0.58 mol of O2(g) has been formed. For the reaction:
2SO3(g) 2SO2(g) + O2(g)
at 1105 K
(a) Calculate Kc
(b) Calculate the initial partial pressure of SO3(g) and the equilibrium partial pressure of O2(g). Now calculate a value for Kp.
(c) Use the equation Kp = Kc(RT)Dn to calculate Kp for this equilibrium.
Set out your working in an 'equilibrium table' like this:
|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|
The volume of the container is fixed at 8.00 dm3. This reaction involves a change in gaseous volume on reaching equilibrium (an increase in this case) and so the total pressure will change.
For help with this, read about The Ideal Gas Law.
|Initial partial pressures||34||-||-||atm|
|Equilibrium partial pressures||(34 - 13.2)||13.2||6.6||atm|
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 dm3 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.
SO2(g) and O2(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 SO2(g) had been converted to SO3(g).
2SO2(g) + O2(g) 2SO3(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.)
|Initial moles||2 mol||1 mole|