### LeChatelier's PrincipleTen Problems

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Problem #1: 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].

Problem #2: 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.

Problem #3: 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.

Problem #4: 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.

Problem #5: 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.

Problem #6: 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.

Problem #7: Chlorine is a useful oxidizing agent, often used in the treatment of wastewater. It can be produced by this reaction:

4HCl(g) + O2(g) ⇌ 2Cl2(g) + 2H2O(g) + 117 kJ

What changes in conditions will increase the yield of chlorine in this reaction at equilibrium?

Solution:

1) Potential change involving pressure:

The reactant side has five molecules total whereas the product side has four. Note: only the gas phase is considered when determining numbers of molecules on each side. Liquid and solid are excluded because they do not appear in the equilibrium expression.

Increasing chlorine means we must shift the equilibrium to the right.

This is done by increasing the pressure on the system (usually accomplished by reducing the volume).

The reaction will respond by shifting the equilibrium to the right, favoring the production of four molecules of products (including chlorine) while consuming five molecules of reactants.

This reduction of the total number of reactants and products will lower the pressure inside the container while increasing the number of chlorine molecules in the new equilibrium position.

2) Comment on changing pressure:

Sometimes, a question will propose increasing the pressure by adding an inert gas, such as argon. This addition does not shift the position of the equilibrium because the inert gas does not appear in the equilibrium expression.

3) Potential change involving temperature:

The reaction is exothermic.

Remember, our goal is to move the position of the equilibrium to the right.

If we remove heat, the position of the equilibrium will shift so as to replace the lost heat.

Removing of heat is caused by lowering the temperature of the system.

4) Potential change involving reactants:

If we add more of the reactants to the system, LeChatelier's Principle tells us that the reaction will shift away from the stress.

In that case, that means the equilibrium will shift to the right, increasing the amount of Cl2 produced.

5) Potential change involving products:

If we remove products from the system, LeChatelier's Principle says that the equilibriun position will shift so as to replace that which was lost.

That means a shift of the equilibrium position to the right, adding some H2O and some Cl2, attempting to replace that which was lost.

Problem #8: What will happen to the number of moles of SO3 in equilibrium with SO2 and O2 in each of the following cases?

2SO3(g) ⇌ 2SO2(g) + O2(g)
(b) The pressure is increased by decreasing the volume of the reaction container.
(c) In a rigid reaction container, the pressure is increased by adding argon gas.
(d) The temperature is decreased (the reaction is endothermic).
(e) Gaseous sulfur dioxide is removed.

Problem #9: Methanol (CH3OH) can be made by the reaction of CO with H2: CO(g) + 2H2(g) ⇌ CH3OH(g)

(a) Use the following thermochemical data to calculate ΔH° for this reaction:
 CO(g) −110.5 kJ/mol CH3OH(g) −201.0 kJ/mol

(b) To maximize the equilibrium yield of methanol, would you use a high or low temperature?
(c) To maximize the equilibrium yield of methanol, would you use a high or low pressure?

Problem #10: How does each of the following affect the yield of NO at equilibrium (increase, decrease, no effect)?

 4NH3(g) + 5O2(g) ⇌ 4NO(g) + 6H2O(g) ΔH° = −904.4 kJ

 (a) Increase in [NH3] (d) Addition of a catalyst (b) Increase in [H2O] (e) Increase in temperature (c) Decrease in [O2] (f) Decrease in the volume of the container in which the reaction occurs

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