# What is the relationship between the constant of equilibrium (Kc) and the position of equilibrium? [duplicate]

I understand that the constant of equilibrium is a ratio of the concentration of the products to the reactants and that this value allows one to estimate whether the position of equilibrium lies to the right (in which there will be more product at equilibrium) or lies to the left (where there will be more reactant at equilibrium). I'm also aware that Kc is affected only by a change in temperature and not a change in concentration, pressure etc.

I know the position of equilibrium is affected by changes in concentration, pressure and temperature (unlike Kc which is only affected by change in temperature). I understand Le Chatelier's Principle and how the position of equilibrium shifts in order to minimise the change, and that this maintains the equilibrium constant.

My problem arises when trying to relate the two. The position of equilibrium shifts so that the equilibrium constant will be maintained however I don't understand how this works or why! When we say that the position of equilibriums shifts (to the right for example), what does this actually mean? I know that the amount of product will increase but what is actually happening. Is the rate of the forwards reaction increasing? In other words, what actually is the position of equilibrium? If Kc is the ratio of product concentration to reactant concentration then what actually is the position of equilibrium?

• What is actually happening is that some molecules react to form some other molecules. As for the position of equilibrium, this is not a thing at all. – Ivan Neretin Jun 26 '18 at 16:39

What you are probably referring to is the degree of dissociation $\alpha$ when you ask about the position of equilibrium. At a given temperature the equilibrium constant is just that, constant, but as the pressure changes, say for a gaseous reaction then $\alpha$ changes.An example shows this.

In the gas phase reaction $\ce{A2 <=> 2A}$ with partial pressures $p_{A_2}$ and $p_{A}$ respectively, the equilibrium constant is $\displaystyle K_p=\frac{p_A^2}{P_{A_2}}$ and is defined at standard conditions of 1 atm. If $P$ is the total pressure then the two partial pressures are $\displaystyle p_A=\frac{2\alpha}{1+\alpha}P$ and $\displaystyle p_{A_2}=\frac{1-\alpha}{1+\alpha}P$ where and so after some simplification $\displaystyle K_p=\frac{4\alpha^2}{1-\alpha^2}P$.

Thus, as the equilibrium constant is constant at any total pressure $P$, the degree of dissociation must change as the pressure changes to compensate, thus the 'position of equilibrium' changes.

[Notes: The amount of A$_2$ reacted is $1-\alpha$ and the amount of A produced is $2\alpha$, thus the mole fractions of each are $(1-\alpha)/(1+\alpha)$ and $(2\alpha)/(1+\alpha)$ and so the partial pressure are as given above. The total amount needed for the mole fraction is $1-\alpha+2\alpha=1+\alpha$. You can derive related expressions in terms of concentrations.]

As long as the "equilibrium" shifts it is - by definition - no equilibrium. The equilibrium is reached as soon as nothing changes anymore. In theory this only occurs after an infinite amount of time.

As soon as the equilibrium is reached, the position of the equilibrium position is described by $K_c$.

And the system does not shift in order for $K_c$ to stay constant. The system shifts in order to reach the point of minimal energy. $K_c$ is only a way of thermodynamically describing that point.