Does it mean that if one side of a reaction is favoured, the reaction will shift to the other side? As I was reading up on the van't Hoff equation, it says that an increase in temperature of an endothermic reaction will favour the products. It also showed a graph where lnK increases with temperature (T). But I thought what would happen in this case is the reactants will be favoured and then the equilibrium will shift to the right to de-stress the equilibrium?

So in summary, I'm just quite confused about what it means by favouring a side of an equilibrium.

  • $\begingroup$ Doesn't it mean that the favoured side is the side to where the equilibrium shifts? $\endgroup$ – Kenny Lau May 15 '16 at 10:17
  • $\begingroup$ Favoring side of equilibrium is nothing but the side whose concentration or pressure in the mixture will be higher on reaching equilibrium. Increasing temperature of endothermic reaction leads to the increase in yield. This can be understood from Le Chatelier's prinicple, which states that the reaction will proceed in a direction which will oppose the change enforced on the system. $\endgroup$ – ShankRam May 15 '16 at 10:49
  • $\begingroup$ Thermodynamics and kinetics play a huge role in deciding the favored side. $\endgroup$ – ShankRam May 15 '16 at 10:51

In general, a equilibrium will shift in a direction that opposes the stress that was placed on the system. For example, if you add heat to an endothermic reaction, you will receive more products, as this will lower the temperature of the solution. When you have the dissociation of an acid, the acid will dissociate more (act like an acid) when you make the solution more basic. An equilibrium will shift to minimize any change in a system, so when evaluating what will happen after a system is stressed, look to see which side of the equilibrium relives that stress, and that side will be favored.

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