3
$\begingroup$

I know that equilibrium constant depends on temperature and pressure. But according to Le Chatelier's principle when we change pressure of reagents the equilibrium is the same. Why?

$\endgroup$
1
$\begingroup$

Good question here - the equilibrium constant is all a little bit abstract, and therefore tough for some of us chemists, at least initially :) I have always thought about the concept of 'equilibrium' in terms of a balloon.

Imagine a balloon, and you've blown the balloon up until it is as full as can be. At this stage the balloon is in one of two 'equilibrium' states:

balloonDeflated $\rightleftharpoons$ balloonInflated

Pressure:

Initially, while blowing up the balloon, you and the balloon are involved in a delicate dance. You want the balloon to be the biggest balloon in the world! But, the balloon doesn't want to be too fat. You and the balloon dance a delicate equilibrium.

So, while you try to blow the balloon up as big as possible, the balloon says: 'No!', until it finally explodes in your face :( Too much pressure from one side. Blown out the equilibrium, and lost the battle.

Temperature:

You've huffed and puffed just as much as you can - (or as much as you can be bothered) - and the balloon is just the right size, determined by the effort you can be bothered putting into it. The balloon size is now in equilibrium with your effort. This is the same with temperature, you can push a reaction faster and further with higher temperatures (more puff).

Equilibrium State

Of course, once you and the balloon have agreed on the optimal size, and you have tied a knot in the balloon, you have reached a happy equilibrium. Yay! Now the fun can begin.

What will happen if you squeeze on one side of the balloon?

roundShape $\rightarrow$ distortedShape

In this analogy, you have added more reagent to the left hand side of the equilibrium. But, by definition, this is no longer an equilibrium, there is more on the left than the right. So, the balloon bulges out to the right, to even up the pressure (i.e. you get more right hand reactants produced to counter the influx of left hand reagent).

I think, the concept of the 'pressure' of reagents is a reference to the concentration of reagents on either side of the equilibrium. A greater concentration on the left, causes a flow to the right and vice versa. Note that this is distinct from the concept of pressure in the normal terms (pounds per square or force).

And this whole process is fluid, a change to one side, affects the other side. Temperature / environmental pressure will affect the initial equilibrium state (your effort / lung power and the potential size of the balloon will affect the initial equilibrium state) and after this, the pressures imposed on either sides of the equilibrium will be countered by a shift to the low-pressure side.

$\endgroup$
3
$\begingroup$

The position of equilibrium depends on temperature and pressure, as Nick Burns describes in his answer. The equilibrium constant depends on temperature only. The equilibrium constant is not a measure of the position of equilibrium, but rather which positions (representing different compositions) along the the reaction coordinate surface are equilibrium positions. Changing the temperature changes the shape of the reaction coordinate surface. Changing the pressure just moves you to a different point on the surface.

Example: consider a hypothetical reaction in which two molecules of gas react to form a third.

$$\ce{A(g) + B(g)<=>C(g)}$$

The reaction quotient $Q$ is simply the ratio of the product of the partial pressure(s) of the product(s) to the product of the partial pressure(s) of the reactant(s). $Q$ can have any value representing different points on the reaction coordinate surface.

$$Q=\frac{P_C}{P_A P_B}$$

The equilibrium constant represents special values of Q in which the reaction coordinate surface has a free energy minimum. For the sake of our example, let's make the equilibrium constant an easy value to work with, say $K=2$.

$$K=Q_{eq}=\frac{P_{C,eq}}{P_{A,eq} P_{B,eq}}=2$$

Since we are talking about gases, we also have the constraint the sum of the partial pressures of the reagents must equal the total pressure of the system. Again, for ease, let us set this value to an easy value to work with, say $P_T = 7 \text{ kPa}$.

$$\frac{P_{C}}{P_{A} P_{B}}=2$$ $$P_A + P_B + P_C = 7$$

Now we have two equations of three variables, which means there are an infinite number of possible values of the partial pressures of the reactions components that fit these two conditions. One is $P_A = 1$, $P_B =2$, and $P_C =4$. Another equally valid combo is $P_A = 2$, $P_B=1$, and $P_C=4$. The position of the equilibrium in this case is better given by a ratio of the sum of the partial pressure(s) of the product(s) to the sum of the partial pressure(s) of the reactant(s):

$$\frac{P_C}{P_A + P_B}=\frac{4}{3}$$

If we change the total pressure to, say $P_T = 4$, we change the allowable combinations of $P_A$, $P_B$, and $P_C$ that can get us to the equilibrium condition of $Q=2$. One allowable combination (of an infinite number) is $P_A=1$, $P_B=1$, and $P_C=2$. The position of the equilibrium is now:

$$\frac{P_C}{P_A + P_B}=\frac{2}{2}=1$$

Le Chatelier's principle suggests that When pressure decreases, the equilibrium shifts toward the side of the reaction with more equivalents of gas.

If we consider changing the situation from $P_A = 1$, $P_B =2$, $P_C =4$, and $P_T=7$ to a situation with $P_T=2$, we can do some math to figure out new values of $P_A$, $P_B$, and $P_C$ that fit this new situation. First, we set the partial pressures of the compoents to $\frac{2}{7}$ of their original value, and calculate Q. $$P_A =1\times \frac{2}{7}=\frac{2}{7}$$ $$P_B =2\times \frac{2}{7}=\frac{4}{7}$$ $$P_A =4\times \frac{2}{7}=\frac{8}{7}$$ $$Q=\frac{P_C}{P_A P_B}=\frac{\frac{8}{7}}{\frac{2}{7} \times \frac{4}{7}}=\frac{\frac{8}{7}}{\frac{8}{49}}=49$$

Since $C>K$ the equilibrium will shift to decrease $Q$. This can happen by decreasing $P_C$ and increasing $P_A$ and $P_B$. Since there is an reaction equation relating these three components in a 1:1:1 stoichiometric ratio, the change must have the same value vut in different directions:

$$K=2= \frac{\frac{8}{7}-x}{\left(\frac{2}{7}+x \right)\left( \frac{4}{7}+x\right)}$$

Rather than goof up the algebra, I let WolframAlpha do it. The solution for $x$ is:

$$x=\frac{1}{128}\left(-19 \pm \sqrt{681} \right)\approx \{+0.2534..., -1.6106...\}$$

The first answer for $x$ is nonsense, since it would put one partial pressure less than $0$. If we change the partial pressures appropriately, we get new values of

$$P_A = \frac{2}{7}+\frac{1}{128}\left(-19 + \sqrt{681} \right)=0.5391...$$ $$P_B = \frac{4}{7}+\frac{1}{128}\left(-19 + \sqrt{681} \right)=0.8248...$$ $$P_C = \frac{8}{7}-\frac{1}{128}\left(-19 + \sqrt{681} \right)=0.8894...$$

The position of equilibrium is

$$\frac{0.8894...}{0.5391...+0.8248...}\approx 0.65$$

which is less than the value of $1.33$ from when $P_T=7$ and different from the value of $1$, when the partial pressures were different (but still met the criteria than $P_T=2$ and $Q=2$).

Changing the temperature changes the values of $Q$ which define the equilibrium state:

$$K=e^{-\Delta G^\circ/RT}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.