Adrian Dingle’s Chemistry Blog

Consider the two equations we introduced today;

Equation 1:

Delta G STANDARD = – RT ln K

Since K is the equilibrium constant, we are AT equilibrium, the amounts of products and reactants in the mixture are fixed, and the SIGN of Delta G STANDARD can be thought of as a guide to the ratio of the amount of products to the amount of reactants at equilibrium, and should NOT be thought of as a predictor of the feasibility of the reaction.

IF it so happens that products and reactants are equally favored at equilibrium, then Delta G STANDARD is zero, BUT Delta G STANDARD is not *necessarily* ZERO at equilibrium (I think this is the key).

Equation 2:

Delta G = Delta G STANDARD + RT ln Q

Since Q is NOT the K, and we are NOT necessarily at the equilibrium position, the SIGN of Delta G can be thought of as a prediction about which way the reaction (that has reactants and products defined by Q), will go.

If Delta G STANDARD is negative at equilibrium, then we will have lots of products at equilibrium, meaning Q  needs to be big (greater than 1). As Q gets larger (i.e. as we get more products), the term ‘RT ln Q’ gets increasingly positive and eventually adding IT to a negative Delta G STANDARD will make Delta G = 0 and equilibrium will be established and no further change occurs.

It is possible that Q could already be too large and therefore Delta G is positive. IF so, then the reaction will need to from more reactants, reduce the value of Q, and allow Delta G to reach zero, i.e. allow equilibrium to be established.

If Delta G STANDARD is positive at equilibrium, then we will have lots of reactants at equilibrium, meaning Q needs to be small (i.e. less than 1). As Q gets smaller (i.e. as we get more reactants), the term ‘RT ln Q’ gets increasingly negative and eventually adding IT to a positive Delta G STANDARD will make Delta G = 0 and equilibrium will be established and no further change occurs.

It is possible that Q could already be too small and therefore Delta G is negative, IF so, then the reaction will need more products, decrease the value of Q, and allow Delta G to reach zero, i.e. allow equilibrium to be established.

IN short, it is Delta G (NOT Delta G STANDARD) that will be zero at equilibrium and the sign of IT (in combination with Delta G STANDARD and RT ln Q in Equation #2.), will define which way the reaction proceeds.

This is messy – really messy, and don’t even get me started on work, internal energy and all that nonsense (which is a catastrophic minefield of bizarre ‘conventions’), but I thought it was time to crystallize a few thoughts on this.

As usual, when studying Thermodynamics with my AP classes, I did a lab that involves a couple of reactions; one endothermic and one exothermic followed by the application of q = m c Delta T. The calculations associated with this lab and these reactions are very messy in two ways. Firstly the problem of deciding what actually constitutes ‘m’ in the equations is tricky. Gases are given off, single replacement reactions lead to the depositing of solids and reactants are in excess – what do we add together to get the correct ‘m’? Secondly, the problem of assigning signs correctly in order to end up with a positive sign for the endothermic reaction and a negative sign for the exothermic reaction is a prickly one. Most texts make much too difficult a song and dance about this IMO, and frankly you often run into the sign convention problem.

Frankly, the ‘mass issue’ is really a non-issue when it comes to the AP exam (as explained at the end of the document below), but the sign issue has the potential to be much less easy to clean up. This year, for the first time, I decided to create a short document that attempts to clarify things once and for all – I think it worked very well, and it seemed to be the clearest for the kids in many years.