This page offers just enough to cover the requirements of one of the UK A level Exam Boards to show that reactions with large negative values of ΔG° have large values for their equilibrium constants, while those with large positive values of ΔG° have very small values of their equilibrium constants.
ΔG° = -RT ln K
R is the gas constant with a value of 8.314 J K
T is the temperature of the reaction in Kelvin.
It is important to realise that we are talking about | |

Note: In fact, under the conditions that a reaction is in a state of dynamic equilibrium, ΔG (as opposed to the free energy change under standard conditions, ΔG°) is zero. | |

If you look up or calculate the value of the standard free energy of a reaction, you will end up with units of kJ mol You must convert your standard free energy value into joules by multiplying the kJ value by 1000.
ln K (that is a letter L, not a letter I) is the natural logarithm of the equilibrium constant K. For the purposes of A level chemistry (or its equivalents), it doesn't matter in the least if you don't know what this means, but you must be able to convert it into a value for K. How you do this will depend on your calculator. Once you have calculated a value for ln K, you just press the e
Suppose you have a fairly big negative value of ΔG° = -60.0 kJ mol The first thing you have to do is remember to convert it into J by multiplying by 1000, giving -60000 J mol And let's suppose that we are interested in the equilibrium constant for the reaction at 100°C - which is 373 K.
That is a huge value for an equilibrium constant, and means that at equilibrium the reaction has almost gone to completion. In the equilibrium constant expression, there must be lots of products at the top and hardly any reactants at the bottom.
Now let's repeat the same exercise with a fairly big positive value of ΔG° = +60.0 kJ mol And we will keep the same temperature as before - 373 K.
That is a tiny value for an equilibrium constant, and there has been virtually no reaction at all at equilibrium. In the equilibrium constant expression, there must be hardly any products at the top and lots of reactants at the bottom.
It is up to you now to play around with your own examples until you are confident of the mechanics of getting an answer. What happens if you change the value of ΔG°? Try the calculations again with values closer to zero, positive and negative. What happens if you change the temperature? **Where would you like to go now?**
© Jim Clark 2017 |