Why do we say the position of equilibrium is shifted even when K is the same?

Question

I realize that there are similar questions that have already been asked here, however none of them answer the question. The answers all say that (when K is the same) you "you haven't really shifted the equilibrium position in the way it sounds like" or something similar.

Many people also say that the position of equilibrium is roughly equivalent to Q. However, going by that definition, the position of equilibrium would be the same if you only change concentrations. Yet, many textbooks explicitly say things like "increasing the concentration of a reactant shifts the position of equilibrium to the products". I understand that more products are produced to cause Q to reach K again, but once you are at equilibrium Q is the same as before, Q = K.

Is there any specific definition of position of equilibrium? Or is the term only used when you are saying "the position of equilibrium is shifting right/left" to mean that the forward/reverse reaction is happening at a higher rate for the moment?

The second option seems to be what people answer with, but it just does not make sense to me that the position of equilibrium would only be a term used to refer to a behavior of a reaction that is disturbed from equilibrium, instead of referring a quality at equilibrium that can be compared with other equilibrium states.

Answer 1

Is there any specific definition of position of equilibrium?

No, it is a fuzzy concept and not very helpful. When a reaction is at equilibrium, there is a certain composition of the equilibrium mixture. However, this composition is not unique. There are many other compositions (sets of concentrations) that are also at equilibrium.

I will use a simple reaction to illustrate this:

The concentration of the reactant $\ce{N2O4}$ is plotted on the y-axis and the concentration of the product $\ce{NO2}$ is plotted on the x-axis. The blue curve shows reaction mixtures that are at equilibrium. If I start with pure $\ce{N2O4}$ at a partial pressure or concentration of 10 (arbitrary units, just for illustration), the reaction will proceed on the red straight line (slope -1/2 because of the stoichiometry). The reaction will proceed net forward until it reaches equilibrium, marked with the circle. However, if I initially have 8 (purple straight line) or 12 (green straight line), I will reach different equilibrium compositions. I can't reach the circled composition because I don't have the right number of atoms.

Or is the term only used when you are saying "the position of equilibrium is shifting right/left" to mean that the forward/reverse reaction is happening at a higher rate for the moment?

"The position of equilibrium is shifting right" should be replaced by "the reaction is going in a net forward direction" until it reaches equilibrium. I will use the same example to illustrate this:

In this example, we start out at equilibrium on the red line, and add some product. We are pretending the reaction is turned off as we add the product, so the reactant concentration stays the same. Then (like in a cartoon where the character notices they stepped off the cliff), we pretend the reaction is turned on again. It is not at equilibrium, and we see a net reverse reaction. You could talk about this net reverse reaction as a shift to the left, but it is describing the reaction, not the equilibrium. Equilibrium states are defined by the blue curve, which does not shift.

You could also add reactant (half as much), as the next panel shows:

This time, there is a net forward reaction, and you end up at the same equilibrium composition (because the reaction moves along on the same green line, constrained by stoichiometry).

Below are the other two cases involving removal of reactant or product. Again, I chose the example such that both cases have the reaction composition constrained to the same line, purple in this case.

Why do we say the position of equilibrium is shifted even when K is the same?

I don't know why, but we shouldn't. We should say that when we disturb an equilibrium by adding or removing a single reactant or product, the reaction will go in the direction of re-establishing an equilibrium (net forward or reverse, depending on the details). Notice I say establish "an equilibrium" because there are multiple compositions (or sets of concentrations) that satisfy the equilibrium condition. After adding or removing a single species, you can not return to the same equilibrium state (because the number of atoms changed).

Answer 2

Maybe you understand better with a simple numerical example.

Suppose we are dealing with an equilibrium reaction A + B <=> C + D, where the equilibrium constant $K$ is $K = \frac{[C][D]}{[A][B]}$ = $1$. Suppose the volume of the system (in liters), the equilibrium constant and all concentrations (in mole/L) are all equal to $1$.

Now add one supplementary mole A to the system. As a consequence, Q = $2$. The system is out of equilibrium. To obtain a new equilibrium, a certain amount $x$ of A and B is destroyed producing $x$ new molecules of C and D. The new equilibrium state will be described by [A] = $2$ - $x$, [B] = $1$ - $x$, [C] = [D] = $1$ + $x$. This equilibrium has the same constant $K$ = $1$, and is described by :

$$K = \frac{[C][D]}{[A][B]} = \frac{(1 + x)^2}{(2-x)(1-x)} = 1 $$

With a little algebra, you find $x$ = $0.2$. The final system contains $1.8$ mole A, $0.8$ mole B, $1.2$ mole C and D. This correspond exactly to the sentence the position of equilibrium is shifting right that you were discussing in the question. There is no change of kinetics.

Answer 3

Is there any specific definition of position of equilibrium? Or is the term only used when you are saying "the position of equilibrium is shifting right/left" to mean that the forward/reverse reaction is happening at a higher rate for the moment?

Yes, there is a specific definition. It is the condition at which there is no net driving force for the (macroscopic) system to change. An imperfect mechanical analogue for this is a ball rolling in a frictionless basin. When it reaches the basin bottom the net force on the ball is null (the ball however has inertia which makes the analogy flawed; if friction occurred between basin and ball, however, the latter could exchange (lose) kinetic energy (as heat). The motion in the basin can be described as motion within a gravitational potential energy surface. The slope of the surface is proportional to the gravitational force driving motion.

In the molecular world the classical mechanical worldview must be supplemented by a statistical treatment. You can still describe reaction paths in terms of chemical potential (the analogue of gravitational potential) and even of chemical forces (although this is a weaker analogy since the vectorial property of a force, direction, has to be interpreted as the direction of a process such as a chemical reaction (what you call left or right).

In any case, at equilibrium the free energy of the system is at a minimum (and the net chemical force is null). Karsten's diagrams could be modified to show free energy (the sum of chemical potentials). The blue line would represent a potential energy minimum isoline.

Note that K is the equilibrium constant, not "the equilibrium". A condition of chemical reaction equilibrium is that Q=K. Equivalently that the chemical potentials of products and reactants are equal.

Answer 4

Think carefully about what is happening at equilibrium. First everything is constant T, P, numbers of moles of reactants and products etc. [Lets minimize such things as tidal forces from the time of day etc.] Everything is Rock Solid the universe is frozen in place! That seems a perfectly reasonable approach and some think that is the eventual end of the Universe, everything at equilibrium. Fortunately we have not arrived yet and There is continuous energy and matter flux that can perturb any [micro] system that attains equilibrium.

How is this possible? Why do reactions not proceed to completion and the entire universe fall into a minimum energy rock state? Because of the quantized properties of matter and energy and the constant interplay between them. Matter is constantly absorbing and emitting energy and this energy is contained in random molecular velocities and QM states of the molecules. This thermal energy or heat content is conveniently contained in temperature. This means that at equilibrium all the molecules of reactants and products have kinetic energy and their concentrations [chemical activities] satisfy the equilibrium constant.

All molecules are a relative energy minimum; an activation energy is necessary for a reaction to happen. Only an elementary reaction, a reaction where the products do not further react more quickly, can reach equilibrium. This means that the mechanism is identical in the forward and reverse directions. The direction of the reaction depends on the concentrations of the components, the two activation energies, and the number of molecules with sufficient energy to react; the latter is determined by the temperature. At equilibrium the reactions do not cease; the activities and activation energies are such that the forward and reverse reactions have the same rates. Changing a concentration will change a rate and the equilibrium will change to reestablish equal rates and equilibrium. The stoichiometry of the reaction determines whether the ratios change. A change in temperature changes the equilibrium position [the free energy changes] and results in a different equilibrium constant because the different activation energies respond differently to temperature.

This is why yields of reactions can be changed by manipulating reactant ratios or removing products as formed or by changing temperatures.