The following equilibrium is set up in a glass syringe. 2(NO2) (brown gas) ‹-› N2O4 (colourless gas) ∆H = -58 kJmol-1. Using le Chatelier's principle, predict and explain how heating up the mixture would affect it's appearance.

At first this question can seem confusing, but by breaking it down we can better understand what we are being asked to do. Firstly, we need to recall le Chatelier's principle - that an equalibrium will shift to try and counteract an external change put upon it (in this case increasing temperature). Secondly, we need to undertand what the change in enthalpy (∆H = -58kJmol-1) means in this equasion. ∆H is negative for the forwards reaction, meaning that there is a decrease in enthalpy and stored energy in the molecules on the right hand side of the equalibrium. We know that energy cannot be created or destroyed and so the energy is released as heat in the forwards reaction, and taken in during the reverse reaction.

Now we can apply le Chatelier's principle. If the temperature is increased, we know the equalibrium will try to oppose that change by taking in more heat. We have established that the forwards reaction will release heat and the reverse will do the opposite, and so we know that the equalibrium will favour the backwards reaction and produce more NO2 in order to take in more heat and oppose the external temperature increase. We have been told in the question that NO2 is a brown gas, and so the mixture will turn a darker shade of brown as more is produced.