When K=1

[Amelia Vernon 2L]

I know that if K>1, it slightly favors the products, and if K<1, it slightly favors the reactants. However, I forgot what Dr. Lavelle said about K=1. Does this mean that the reaction is no longer occurring? Or is this just telling us something about the concentrations of the reactants and products?

[Fariha Hameed 1D]

Hi! When K=1 the reaction is definitely still occurring but has reached equilibrium. In the equilibrium equation aA+bB<-->cC+dD, the value of [C]^cx[D]^d would equal the value of [A]^ax[B]^b, and neither the reactant nor product would be favored. Hope this helps!

[Hannah Markovic 3C]

The previous answer is wrong. K=1 does not give any information about whether the reaction is at equilibrium. Q=K will tell you that the reaction is at equilibrium (rates of forward and reverse reactions are equal). K=1 just means that the concentration of reactants and of products are equal at equilibrium. Both the forward and reverse reactions are still occurring, but they are occurring at the same rate, so no change is happening in the concentrations of the species. Also, to clarify, K>1 means that there is a greater concentration of products than reactants at equilibrium, while Q<K means the formation of products is favored when the species are at the concentrations that you used to calculate Q. K<1 means that there is a greater concentration of reactants than products at equilibrium, while Q>K means the formation of reactants is favored. Make sure not to confuse Q and K! Hope this helps :)

PS if you need a trick to remember how Q relates to K, all you have to do is write the inequality with K on the left (alphabetical order) and then draw a line to the inequality which makes an arrow that shows you how the reaction will proceed :) for example if K<Q, you get an arrow pointing left, so the reaction will form reactants. You can also check this because if K<Q, Q>K which means the concentration of products when you measured Q is bigger than the concentration at equilibrium (since products in the numerator of the equation for K/Q) and so the reaction will shift left to form more reactants in order to reach equilibrium.

[Fariha Hameed 1D]

Oh yes sorry about that! We know that neither products nor reactants are favored at K=1 , but we would not know if it is at equilibrium. We would have to be given value of the equilibrium constant for that particular reaction to find out.