### LeChatelier's Principle

In 1888, Henri LeChatelier gave a succinct statement of the principle he had announced 4 years prior. It is:

Every change of one of the factors of an equilibrium occasions a rearrangement of the system in such a direction that the factor in question experiences a change in a sense opposite to the original change.

Before going on to a more recent restatement, let's examine some important points gleaned from LeChatelier's words.

1. For LeChatelier's Principle to work, the chemical system MUST be at equilibrium first. Remember this means that there is no change in any of the concentrations of the substances involved.

2. Some type of change is involved in a factor that affects equilibrium. There are several factors and they are discussed below. The changes will involve increasing or decreasing something which affects the equilibrium.

3. In response to this change, the old equilibrium will rearrange itself. This means changing the old equilibrium concentrations to a new set of values without changing the Keq value. However, we will find that the Keq does change in value when the change involves temperature.

4. The last point is about the word "opposite." By this we mean that if something is increased, the induced change will be for something to decrease. Of course, the reverse case is also true.

Two examples of modern definitions are:

If a stress is applied to a system at equilibrium, then the system readjusts, if possible, to reduce the stress.

If a system at equilibrium is subjected to a stress, the equilibrium will shift in attempt to reduce the stress.

The four points discussed above are incorporated into each definition, it's just that sometimes it is implied, like the idea of "opposite." Most modern defnitions are very similar to these two in their wording.

Before going into a mess of examples, here are the factors that affect a chemical reaction at equilibrium.

Changes in:

1) concentration change
a) up
b) down
2) temperature
a) up
b) down
3) pressure
a) lower the pressure by increasing the volume
b) raise the pressure by decreasing the volume
c) raise the total pressure by adding an inert (non-reacting) gas

Please keep in mind that the above changes would be imposed on a system already at equilibrium.

Example #1: Which way will the equilibrium shift if more H2 is added to this reaction at equilibrium:

N2 + 3H2 ⇌ 2NH3

Solution:

The H2 amount goes up (by adding it), therefore according to LeChatelier's Principle, the reaction will try and use up the added H2. It does so by shifting the position of equilibrium to the right. This makes more NH3 by using up N2 and H2

Example #2: Using the same reaction in Example #1, which way will the equilibrium shift if some NH3 is removed from the reaction when it is at equilibrium.

Solution:

According to LeChatelier's Principle, the chemical system will attempt to replace the lost NH3. The stress was to remove NH3, so the opposite is to replace it. The equilibrium position will shift to the right in order to replace some of the lost NH3.

Example #3: Which way will the equilibrium shift if the system temperature goes up (heat is added):
2 SO2 + O2 ⇌ 2 SO3 + heat

Solution:

Even though heat is not a chemical substance, for the purposes of LeChatelier's Principle, you can treat it as if it has physical existence. Since heat is added, the reaction will shift to try and use up some of the added heat. In order to do this, the reaction must shift to the left.

Note that, since heat is not part of the equilibrium expression, the value of Keq would change when the chemical system is heated or cooled.

Example #4: Using the same reaction, which way will the equilibrium shift if heat is removed (that is, the temperature goes down).

Solution:

The reaction will attempt to do the opposite of what the stress was. Since the stress was to remove heat, the reaction will shift to the right to generate more heat (replacing only a part of what was lost).

Once again, the value of Keq would change, whereas in the other examples, its value WOULD NOT change.

Pressure Changes and their Effect on Equilibrium.

Remember that a pressure change potentially affecting the position of equilibrium can be accomplished three different ways.

(1) Increasing the volume of the reaction container will reduce the pressure.
(2) Reducing the volume of the reaction container will send the pressure up.
(3) Introducing an inert gas like argon into the reaction container will increase the total pressure.

Important point: the volume changes would affect ONLY the substances in a gaseous state. Liquid and solids would be unaffected by any volume changes. Since the below examples are all 100% gas phase, the position of the equilibrium will be changed, except in one particular circumstance, demonstrated in example #6.

Example #5: The container holding the following reaction (already at equilibrium) has its volume suddenly reduced by half. Which way will the equilibrium shift to compensate?

PCl3 + Cl2 ⇌ PCl5

Solution:

Since the volume went down, this means the pressure went up. The reaction will try to lessen the pressure by shifting to the side with the lesser number of gas molecules. This means a shift to the right because for every PCl5 molecule made, two molecules are used up. The lesser the total number of gas phase molecules in the container, the lesser the pressure.

Example #6: The container holding the following reaction (already at equilibrium) has its volume suddenly increased. Which way will the equilibrium shift to compensate?

H2 + Cl2 ⇌ 2HCl

Solution:

Neither side is favored over the other since both sides have the same number of total molecules (two). No matter which way the reaction shift, the total number of molecules would remain unchanged.

In cases like this, where there is an equal number of molecules on each side, the equilibrium would remain unchanged by the change in pressure (in either direction).

Example #6 is a favorite question to ask on a test. All other examples used would involve a change in the number of gas molecules, therefore showing a shift in the position of the equilibrium. In the above example, the number of gas molecules is the same on each side, so no shift.

You have been warned!

Example #7: The system below is already at equilibrium when some neon is added to the system. What happens to the position of the equilibrium? Does it shift right, left, or no change?

H2 + Cl2 ⇌ 2HCl

Solution:

The neon DOES NOT participate in the chemical reactions (forward and reverse) which make the equilibrium. Therefore, the presence of the inert gas has NO EFFECT on the position of the equilibrium. The [H2], [Cl2] and [HCl] would all remain unchanged.

Example #8: The system below is already at equilibrium when a catalyst is added to the system. What happens to the position of the equilibrium? Does it shift right, left, or no change?

PCl3 + Cl2 ⇌ PCl5

Solution:

There will be no change in the equilibrium. BOTH (with emphasis on both) the forward and the reverse reactions are speeded up. A catalyst just gets you to equilibrium faster, it doesn't affect the final position of equilibrium like changing the concentration would.

Example #9: Which way would the equilibrium shift if the pressure is increased?

N2 + 3H2 ⇌ 2NH3

Solution:

You should think of the pressure being increased by reducing the volume, not by adding an inert substance or heating the system up.

The position of the equilibrium would shift to the side with the lesser number of molecules. The shift would be to the right.

Example #10: An equilibrium mixture contains 0.810 mol HI, 0.500 mol I2, and 0.290 mol H2 in a 1.00 L flask.

(a) What is the equilibrium constant for the following reaction?
2HI(g) ⇌ H2(g) + I2(g)

(b) How many moles of I2 must be removed in order to double the number of moles of H2 at equilibrium?

Solution:

1) Write the equilibrium expression, insert values and solve:

 [H2] [I2] Kc = ––––––– [HI]2

 (0.290) (0.500) Kc = –––––––––––– (0.810)2

Kc = 0.221

2) Some discussion concerning (b):

The [H2] doubles to 0.580.

In order to gain 0.29 more H2, the HI must have 0.580 mol decompose, dropping it to 0.230.

The [I2] is x and we will compare it to 0.790 (the 0.500 already there plus the 0.290 made when the HI decomposed) after we determine its value.

3) Insert values into the equilibrium expression and solve:

 (0.580) (x) 0.221 = ––––––– (0.230)2

x = 0.020 M

0.790 − 0.020 = 0.770 mol must be removed.

4) Note: a real popular wrong answer would be 0.480 mol. However, remember that, as you remove I2, more H2 will be produced as the equilibrium shifts to the right. However, more I2 also gets made. So, you have to keep removing until arriving at an I2 value that will give the Kc value of 0.221.

Example #11: What would be the effect of reducing the volume of a container in which the following system was at equilibrium?

2NO(g) + O2(g) ⇌ 2NO2(g)

(a) decrease the rate   (b) decrease Kc   (c) decrease [NO2]   (d) increase [NO2]   (e) none of these

Solution:

When the volume of a container is decreased (at constant temperature), the pressure increases (remember Boyle's Law)

When the pressure increases, the system will try to reduce the pressure.

It does so by shifting the equilibrium towards the side with the lesser number of (gas phase only) molecules.

This direction is to the right, causing an increase in the [NO2].

Example #12: How would you increase the yield of product in the following reaction?

SO2(g) + NO2(g) ---> SO3(g) + NO(g)   $\text{ΔH}{\text{}}_{}^{o}$ = −42 kJ

(a) increase temperature   (b) decrease temperature   (c) increase volume   (d) decrease volume

Solution:

1) When the energy term is included, I like to rewrite the equation to include the heat as if it were a reactant/product. Like this:

SO2(g) + NO2(g) ---> SO3(g) + NO(g) + 42 kJ

The negative sign tells us the reaction is exothermic, meaning heat is produced/given off. Hence, its placement as a product.

Increasing the temperature means adding heat. The reaction will attempt to use up some of that added heat by shifting the point of equilibrium to the left. This would cause a decrease in the product yield.

This is the correct answer. Lowering the temperature is done by removing heat. The reaction attempts to replace the lost heat by shifting the equilibrium to the right, thus increasing the amount of product produced.

4) Answer choices (c) and (d):

There are two molecules on the reactant side and two on the product side.The reaction cannot shift to counteract an increase in volume (causing a drop in pressure) or a decrease in volume (causing an increase in pressure). This is because, for every two molecules removed from either side, two molecules are added in by the other side. There is no net change in the number of molecules inside the reaction chamber.

Example #13: What effect with the addition of a small amount of NaOH(s) have on the following system?

CH3COOH(aq) + H2O(ℓ) ⇌ H3O+(aq) + CH3COO¯(aq)

(a) it changes the value of Ka of CH3COOH
(b) it lowers the [CH3COOH]
(c) it increases the [H3O+]
(d) it favors the reverse reaction

Solution:

The hydroxide that was added will dissolve and then react with some H3O+ (until the hydroxide is used up), causing the hydronium ion concentration to decrease.

The system will react by trying to replace the H3O+ that reacted.

This is done by some CH3COOH reacting with the water to form some H3O+ and some CH3COO¯.

The position of the equilibrium is described as having been shifted to the right.

Example #14: In the following closed system:

CO(g) + 12O2 ⇌ CO2(g)   ΔH° = −198 kJ

which one(s) of the following statements is (are) correct?

I. Increasing the pressure would cause the quantity of CO2 to increase
II. Decreasing the volume would cause the quantity of CO2 to decrease
III. Decreasing the temperature would cause the quantity of CO2 to increase

(a) I and II
(b) I and III
(c) II and III
(d) I, II, and III
(e) none of these

Solution:

Statement I. is a true statement. In this case, the pressure is increased by reducing the volume. The equilibrium will respond by shifting to the right. This replaces 1.5 molecules of reactants with one molecule of product, thereby reducing the total number of gas phase molecules in the system. This reduction in amount of molecules causes a lowering of the pressure.

Statement II. is the exact reverse of statement I. That makes Statement II. false.

Answers (a), (c), and (d) are wrong because they contain statement II. The decision between (b) and (e) depends on the truth or falsity of statement III. The determination of it being a true statement is left to the student. Hint: think of the heat as a product and determine which way the equilibrium shifts when heat is removed by decreasing the temperature.

Example #15: A system at 700 °C contains N2, O2 and NO in equilibrium. If the volume of the system is decreased, what is the chemical effect?

(a) NO yield increases
(b) NO yield decreases
(c) N2 is consumed
(d) N2 is produced
(e) none of these

Solution:

Write the chemical equation: N2 + O2 ⇌ 2NO

A volume decrease means a pressure increase. The reaction would shift to the side with the lesser number of molecules, thereby lowering the pressure.

There are two molecules in total on the reactant side and two molecules total on the product side.

Since the number of molecules is the same on each side, neither direction would be favored by a volume decrease.

By the way, the temperature plays no role in the solution to this problem. It is there to confuse you and cause you distress.

Example #16: The fraction of NO2 present at equilibrium in the following reaction:

2NO(g) + O2 + heat ⇌ 2NO2(g)

is greater at which of the following conditions?

(a) low pressure and low temperature
(b) low pressure and high temperature
(c) high pressure and low temperature
(d) high pressure and high temperature
(e) the fraction is independent of T and P

Solution:

1) Pressure argument:

There are three molecules on the reactant side and two on the product side. As the pressure goes up, this favors the equilibrium being pushed to the right. The chemical system would try to counteract higher pressure by lessening the number of molecules, thereby lessening the pressure.

Our answer choices are narrowed to either (c) or (d).

2) Temperature argument:

The reaction is endothermic. As the temperature goes it, the system would attempt to counteract the added heat by shifting the point of equilibrium to the right, making more NO2 and lowering the temperature.