Why should a system try to reverse any change that has been done to it?

I understood Le Chatelier's principle perfectly but my question is: Why should a system try to reverse any change that has been done to it?

I understand equilibrium constant, mole concept and partial pressures but I do not understand quantum mechanics yet. If you use any other technical terms in your answer, please also explain them. Answers with intuitive examples will be highly appreciated.

• At an abstract level, Le Chatelier's principle is a (not so obvious) consequence of the second law of thermodynamics. The second law, along with the fact that a small part of the system should be in equilibrium with the rest, leads to thermodynamic stability conditions, which can be summarized as the entropy being a concave function (or, equivalently, the internal energy being a convex function). Le Chatelier's principle follows from the concavity/convexity requirements. Dec 16, 2015 at 4:59
• Further reading: Section 22 of L. D. Landau and E. M. Lifshitz, "Statistital Physics Part I", 3rd revised and enlarged ed. (1980) and Section 8.5 of H. B. Callen, "Thermodynamics and an Introduction to Thermostatistics", 2nd ed. (1985). Dec 16, 2015 at 5:02
• Would be great if I could write a detailed answer exemplifying this point, but it seems difficult to produce something more readable than what is in the books I recommended. I might try when I have enough time and energy. Dec 16, 2015 at 5:12

Well, lets first write out the general equation of a chemical reaction:

$$\ce{aA + bB <=> cC + dD}$$

As you obviously know, since this is at equilibrium, not all the reactants react to form the products. Hence in a solution which is at equilibrium, all species of $\ce{A, B, C, D}$ will be present. However the special thing is that the ratio of the concentration of these species will usually be constant, no matter how much of each species you start with. This ratio is known as the equilibrium constant. The only thing that effects this ratio is temperature. The equilibrium constant is given by:

$$K_\mathrm{c} = \ce{\frac{[A]^a[B]^b}{[C]^c[D]^d}}$$

One more equation you need to know is the reaction quotient which gives the ratio of products to reactants at a certain time during a reaction and is given by:

$$Q_\mathrm{c} = \ce{\frac{[A]^a[B]^b}{[C]^c[D]^d}}$$ Note how this is similar to the equilibrium constant, but here, $\ce{[A]}$, $\ce{[B]}$, $\ce{[C]}$ and $\ce{[D]}$ are the concentration of the substance at a specific time during the reaction and doesn't necessary have to be when the system has reached equilibrium. This means that if $Q$ is smaller than $K$, the reactants will react to form more products so that $Q$ will increase so that it will eventually equal $K$ and when $Q$ is larger than $K$, vice versa. When $Q$ equals $K$, the system is at equilibrium.

So now lets apply this concept to help us explain Le Chatelier's Principle by considering different scenarios. Lets consider a mixture of hydrogen and nitrogen gas in a container. They react to form ammonia gas. The equation of this reaction is:

$$\ce{N2(g) + 3H2(g) <=> 2NH3(g) + heat}$$

Also the equilbrium constant is:

$$K_\mathrm{c} = \ce{\frac{[NH3]^2}{[N2][H2]^3}}$$

Scenario 1: Taking or adding more reactants or products

Lets say that this mixture of gases have finally reached equilibrium, so $K = Q$. Now I suddenly take some hydrogen has out.This would decrease the concentration of hydrogen gas and hence increase $Q$. So now $Q>K$. Therefore for the system to adjust itself, it needs to decrease the value of $Q$. It does this by creating more products and decreasing the amount of reactants. So in short, the equilibrium shifts to the left. This is what Le Chatelier's Principle predicts. Now by using the exact same logic, we can predict what happens if I were to change the concentration of ammonia, nitrogen, etc.

Scenario 2: Changing pressure or volume

Pressure and volume are inversely proportional to each other. Meaning that an increase is pressure will result in a decrease in volume and vice versa. This is due to the ideal gas law:

$$pV = nRT$$

Therefore, increasing the pressure of a system has the effect on the equilibrium as decreasing the volume.

Now, back to example of hydrogen, nitrogen and ammonia gas in a container. If I were to suddenly decrease the volume of the container by two (equivalent to doubling the pressure) this would mean that the concentration of each species would double. Now, how would this effect our $Q$ value, would it be larger or smaller than $K$? Well the numerator would increase by fourfold while the denominator would increase by sixteenfold. Hence $Q<K$. Therefore as mentioned above, the equilibrium will shift to the right.

This can be applied to if the pressure was decreased, volume was decreased, etc.

Scenario 3: Temperature is changed

Now this one isn't really explained by using equilibrium constant. Instead will use a more intuitive approach.

For endothermic reactions, heat is required for the reaction to go forward. So by increasing the temperature, there is an increase in heat energy available to the system. Therefore the forward reaction will occur more readily. Hence it is expected that the equilibrium will shift to the right. Meanwhile, a decrease in temperature means less heat energy, hence equilibrium will shift to the left. The same logic can be applied to exothermic reactions.

For a simple reaction, we can derive the free energy change as

$$\Delta G (T,p)=\Delta G^\circ(T)+k_\mathrm{B}T\ln K_\text{eq}$$

Here, $\Delta G (T,p)$ is the free energy change of the reaction, $\Delta G^\circ (T)$ is the free energy change at that temperature but 1 bar pressure or you can use any scale of absolute pressure as long as you are using $K_\text{eq}$ expression correctly. Now $K_\text{eq}$ can be expressed as (for a simple $\ce{A <=> C}$ reaction; here $p_\ce{A}$ and $p_\ce{C}$ are partial pressures of $\ce{A}$ and $\ce{C}$)

$$K_\text{eq}=\frac{p_\ce{C}}{p_\ce{A}}$$

So, at equilibrium $\Delta G(T,p)=0$ yields

$$K_\text{eq}=\exp\left[-\frac{\Delta G^\circ(T)}{k_\mathrm{B}T}\right]$$

So far everything is fine and also the effect of temperature, pressure on equilibrium has been explained nicely in the above answer but still the question is why $Q$ goes to $K$ at equilibrium is unexplained.

At any particular temperature $\Delta G(T,p)$ would be zero if and only if ratio of $p_\ce{C}/p_\ce{A}$ is such that value of $\Delta G^\circ(T)$ is cancelled out by $k_\mathrm{B}T\ln K_\text{eq}$ or in mathematical term

$$\Delta G^\circ(T)=-k_\mathrm{B}T\ln K_\text{eq}$$

So it's a single point on thermodynamic surface. Now if you change one pressure the other pressure must change itself to keep the ratio constant. On the hand if you change temperature your pressure ratio will also change to keep $\Delta G(T,p)$ zero.

Now the final question is, why is $\Delta G(T,p)$ is zero at equilibrium. This is something like gravitational surface. In gravitational surface every object will try to go as down as possible to make it's center of mass as close as possible to the gravitational field. On a macroscopic scale it can be seen as equilibriation of chemical potential. If you keep two different temperature body together they will exchange heat between each other and eventually end up being at the same temperature. In the same way if you mix two chemical compound they will try to exchange chemical potential between themselves (The currency here is conversion of one molecule into another) as long as their chemical potential is not equal. The above formulation can be written as

For $\ce{A}$, chemical potential is

$$\mu_\ce{A} (T,p)=\mu_\ce{A}^\circ(T)+k_\mathrm{B}T\ln\frac{p_\ce{A}}{p^\circ}$$

similarly for $\ce{C}$

$$\mu_\ce{C} (T,p)=\mu_\ce{C}^\circ(T)+k_\mathrm{B}T\ln\frac{p_\ce{C}}{p^\circ}$$

At equilibrium

$$\mu_\mathrm{A}(T,p)=\mu_\mathrm{C}(T,p)$$

And you will get

$$K_\text{eq}= \exp\left[-\frac{\Delta G^\circ(T)}{k_\mathrm{B}T}\right]$$