does reaction quotient After equilibrium has been achieved make any sense? At equilibrium point rates of forward and backward reaction are same and the system is in dynamic equilibrium.The graph of concentration vs time shows a straight line parallel to x axis after equilibrium point. Reaction quotient refers to ratio of concentration at any stage.This means that after equilibrium point reaction Quotient value will same as equilibrium constant as no more products are formed.So how can it have a value more than that of K ?! I have been thinking about this for hours but I get more confused with time.
$\begingroup$
$\endgroup$
1
asked Oct 14, 2014 at 18:08
-
3$\begingroup$ These two answers provide some insight into the relationship between $K$ and $Q$, but the main thrust is that at equilibrium $Q=K$. chemistry.stackexchange.com/questions/4410/k-eq-vs-k-c/… and chemistry.stackexchange.com/questions/4977/… $\endgroup$– Ben NorrisCommented Oct 14, 2014 at 18:48
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
$K$ is just the specific value of $Q$ for which $\Delta_\mathrm r G = 0$. In that case: $$\begin{align} \Delta_\mathrm r G &= \Delta_\mathrm r G^\circ + RT \ln Q \\ 0 &= \Delta_\mathrm r G^\circ + RT \ln K \\ \Delta_\mathrm r G^\circ &= -RT \ln K \end{align}$$
For any system not at the equilibrium it is possible that $Q < K$ or $Q > K$. What is the "forward direction" is just a convention, it depends on how you decide to write the equilibrium. There is nothing special in one direction or the other. The system will "move" toward the direction that leads to equilibrium in both cases. In the first case the reaction will consume reactants to generate products, while in the second case the opposite will happen. You can express this idea also in terms of the ratio $\frac Q K$. If it is > 1 then the reaction will consume reactants. This is also often referred to as the "Le Chatelier principle".
