I used to think that the equilibrium constant was a quantitative measure of the equilibrium. However, changing concentration and pressure in a system at equilibrium that shifts the equilibrium doesn't affect the equilibrium constant, which is solely temperature dependent. This means that my definition was wrong. What is the real relationship between the two, or what do they actually mean? (I haven't been able to find any suitable definition for "equilibrium position," and the only definition I know how "equilibrium constant" is the value of the reaction quotient when a system has reached equilibrium).


3 Answers 3


Here's a more simplistic response.

Equilibrium position: a particular set of concentrations of reactant and product species.
Equilibrium constant: a number that describes how far "forward" equilibrium lies for a particular system (as described by a particular equation).

Example: for the system $\ce{A <=> B}$, the mass action expression is $\ce{\frac{[B]}{[A]}}$. Let's say $K = 200$. This means, at equilibrium, $\ce{[B]}$ will be 200 times greater than $\ce{[A]}$. That's what the equilibrium constant tells you - equilibrium for this system lies to the right, and the K value is greater than 1.

There are infinitely many different equilibrium positions that satisfy K. We could have $\ce{[B]} = 20~\mathrm{M}$ and $\ce{[A]} = 0.1~\mathrm{M}$. Or we could have $\ce{[B]} = 1~\mathrm{M}$ and $\ce{[A]} = 0.005~\mathrm{M}$. Those are two different equilibrium positions that are both at equilibrium. In both cases, $Q = K$ and the system is at equilibrium.

Some notes:

  1. If we describe the exact same system with a different equation, $\ce{B <=> A}$, then the mass action expression is $\ce{\frac{[A]}{[B]}}$. We would find that $K = 0.005$ (that is, $1/200$) and we would say that equilibrium lies to the left. Of course, the molecules don't behave any differently, but the equation we use to describe the system is different, and that difference is reflected in a different $K$.

  2. If we have a system with more than one product or reactant or with stoichiometric coefficients other than 1, then it's a bit more complicated than just saying something like "product concentrations are 200x greater than reactant concentrations". Instead, equilibrium may be achieved at many different equilibrium positions. You have to look at the mass action expression to interpret the meaning of $K$.

  3. The equilibrium constant $K$ is just one way of expressing how far "forward" a particular reaction will proceed. Gibbs free energy change ($\Delta G$) is another way. And, for redox systems, cell potential is yet another way.


Equilibrium position:

A reaction reaches equilibrium position when it has no further tendency to change; that is, the reaction does remain 'spontaneous in neither direction' .

This happens when the reaction Gibbs energy becomes zero viz. $\Delta_r G= 0\;.$

Equilibrium constant:

The reaction quotient $Q$ is given by

$$\Delta_r G= \Delta_r G^{\Theta} + RT\ln Q\;\tag 1 $$

where $^\Theta$ denotes the term is measured under standard conditions.

Now, in equilibrium, $\Delta_r G= 0$ which implies

$$\Delta_r G^\Theta = -RT\ln Q_\textrm{equilibrium}\tag 2$$

From $(2)$ we can define $$Q_\textrm{equilibrium}= e^{-\frac{\Delta_rG^\Theta}{RT}}=K\equiv \textrm{equilibrium constant}\;.$$

Response of Equilibrium constant to the shifting of Equilibrium position:

Shifting of equilibrium position implies a change in the conditions the reaction is carried under. This may include change in temperature or pressure.

That would definitely influence the standard reaction Gibbs energy $\Delta_r G^\Theta$ and hence the value of $K\;.$

Note that the equilibrium position is still the condition when $\Delta_r G$ becomes zero; but the value of $K,$ the equilibrium constant, does not necessarily need to remain the same (for instance, $K$ doesn't change for a change in pressure since $\Delta_rG^\Theta$ remains the same irrespective of the change in pressure; nevertheless the individual components can change without altering the value of $K$ in whole).


I will try to give a simpler answer!

$Q,$ the reaction quotient, basically tells you the ratio of products:reactants right now. When you add $\ce{HF}$ to pure water $Q,$ is $0$ at the start because in the reaction $$\ce{H2O + HF <=> H3O+ + F-}$$ the ratio of products to reactants is $0$. $$\ce{Q=\dfrac{[H3O+][F^{-}]}{[HF]}=0}$$

However, the equilibrium constant of HF is $$K=2.7\times 10^{-2}~\rm M$$

$K$ is the value of $Q$ at equilibrium. The expression for $Q$ and $K$ are the same however $Q$ describes the state of a system and $K$ describes the system at equilibrium. Thus, as the reaction proceeds, more and more $\ce{HF}$ dissociates. This causes a larger $\ce{[H3O+][F^{-}]}$ and hence $Q$ increases and approaches K. Le Chatelier's principle tells us that a system that has $Q\neq K$ will approach $K.$ When $Q$ is not $K,$ the system is not at equilibrium.

Hope this helps!

  • $\begingroup$ Q is the reaction quotient not the equilibrium constant. K is the equilibrium constant and it is equal to the reaction quotient at equilibrium. $\endgroup$
    – bon
    Commented May 25, 2016 at 21:24

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