CHEM 321 - ANALYTICAL CHEMISTRY Le Chatelier's principle and equilibrium shifts Question Bank Set 7

CHEM 321 - ANALYTICAL

CHEMISTRY - Le Chatelier’s principle

and equilibrium shifts

Question Bank - Set 7

Liberty University

Question 1

Question

A chemical reaction is represented by the equation:

2A(g) + 3B(g)⇌C(g)

If the concentration of substance A is increased, predict the effect on the equi-

librium position of the reaction based on Le Chatelier’s principle. Justify your

answer.

Solution

Step 1: Le Chatelier’s principle states that if a system in equilibrium is subjected

to a stress, the system will react in a way that partially offsets the stress to

restore equilibrium.

Step 2: If the concentration of substance A is increased, the reaction will

shift to the right to help consume the excess A and restore equilibrium.

Step 3: As the reaction shifts to the right, more of substance C will be

formed, while substances A and B will be consumed.

Step 4: Therefore, the equilibrium position of the reaction will shift to the

right when the concentration of substance A is increased, according to Le Chate-

lier’s principle.

Question 2

Question

For the reaction CO(g)+3H2O(g)⇌CO2H2O(l), which is at equilibrium, predict

the direction of each of the following changes and state how each change will

affect the equilibrium position:

1. Addition of more CO(g)

2. Removal of some H2O(g)

3. Increasing the volume of the container at constant temperature

Solution

Step 1: Addition of more CO(g)When more CO(g)is added to the reaction,

the reaction will shift to the right to consume the additional reactant. This

is in accordance with Le Chatelier’s principle, which states that a system at

equilibrium will respond to a stress by shifting in the direction that helps to

relieve that stress. Consequently, the equilibrium position will shift towards the

products in this case.

Step 2: Removal of some H2O(g)If some H2O(g)is removed from the reaction,

the equilibrium will shift to the left to replace the lost water molecules. This

will result in an increase in the concentration of the reactants, thus favoring the

reactants at equilibrium.

Step 3: Increasing the volume of the container at constant temperature

When the volume of the container is increased at constant temperature, the

equilibrium will shift in the direction that produces more gas molecules to oc-

cupy the additional volume. In this reaction, there are more gas molecules on

the left side (4 moles of gas) compared to the right side (1 mole of liquid).

Therefore, the equilibrium position will shift to the left to increase the pressure

by favoring the side with more gas molecules.

Question 3

Question

Consider the following equilibrium reaction:

2A(g) + 3B(g)⇌C(g) + D(g)

If the concentration of substance A is increased, predict the effect on the

concentrations of substances B, C, and D at equilibrium. Justify your answer.

Solution

Step 1: Identify the stoichiometry of the reaction.

The stoichiometry of the reaction is 2:3:1:1 for A, B, C, and D respectively.

Step 2: Apply Le Chatelier’s Principle.

When the concentration of substance A is increased, the equilibrium will

shift to the right to consume some of the excess A. This is to counteract the

change and re-establish equilibrium.

Step 3: Determine the effects on the concentrations of substances B, C, and

Since A is involved in the consumption of reactants B, an increase in the

concentration of A will lead to an increase in the concentration of B.

For the product side, the equilibrium will shift to produce more C and D in

order to consume the excess A. Consequently, the concentrations of C and D

will increase as well.

In summary: - Concentration of B will increase. - Concentrations of C and

D will also increase.

Therefore, the increase in the concentration of substance A will result in an

increase in the concentrations of substances B, C, and D at equilibrium.

Question 4

Question

For the reaction 2SO2(g) + O2(g) ⇌2SO3(g) at equilibrium, Le Chatelier’s

principle predicts that an increase in temperature will cause the equilibrium to

shift towards the side with the

gasmolecules.

Solution

Step 1: First, let’s determine the change in the number of gas molecules when

the reaction shifts towards the product side (SO3). In this reaction, there are 3

moles of gas molecules on the reactant side (2 moles of SO2and 1 mole of O2)

and 2 moles of gas molecules on the product side (2 moles of SO3). So, the shift

to the product side increases the number of gas molecules by 1 mole.

Step 2: Next, let’s consider Le Chatelier’s principle and the effect of an

increase in temperature on this reaction. When the temperature is increased,

the system will try to counteract the change by absorbing the excess heat. This

can be accomplished by favoring the reaction that absorbs heat. In this case,

the forward reaction (formation of SO3) is exothermic, meaning it releases heat.

Therefore, increasing the temperature will shift the equilibrium towards the

reactants to absorb the excess heat.

Step 3: Combining the information from step 1 and step 2, we conclude that

an increase in temperature will cause the equilibrium to shift towards the side

with the fewer gas molecules. Therefore, the equilibrium will shift towards the

reactant side (towards the left side of the reaction) when the temperature

is increased.

Question 5

Question

A reaction is at equilibrium when the following equation is satisfied:

2A(g) + B(g) <=> 3C(g) + D(g)

If the concentration of B is increased, predict how the equilibrium will shift.

Justify your answer using Le Chatelier’s principle.

Solution

Step 1: Determine the effect of increasing the concentration of B on the equi-

librium.

When the concentration of B is increased, the reaction will shift to reduce

the added concentration of B in order to establish a new equilibrium.

Step 2: Apply Le Chatelier’s principle to predict the direction of the shift.

Increasing the concentration of B (a reactant) will cause the reaction to shift

to the right to consume B and form more product. This will result in an increase

in the concentrations of C and D, while decreasing the concentration of A. The

equilibrium will shift towards the products to partially offset the increase in B

concentration.

Step 3: Write the new equilibrium expression after the shift.

The new equilibrium expression will be:

2A(g) + B(g) <=> 3C(g) + D(g)

Step 4: Justify the predicted shift using Le Chatelier’s principle.

Le Chatelier’s principle states that a system at equilibrium will respond to

any stress by shifting the equilibrium position in a direction that minimizes the

effect of the stress. In this case, the increase in B concentration is a stress on

the system, so the equilibrium shifts to the right to consume B and produce

more product to relieve the stress.

Question 6

Question

For the reaction:

2A(g) + B(g)⇌C(g)

If the initial concentrations of A,B, and Care 0.2 M, 0.1 M, and 0.5 M,

respectively, and the equilibrium constant (Kc) for the reaction is 4.5, predict

the direction in which the reaction will shift if the pressure is increased by

decreasing the volume of the container. Justify your answer.

Solution

Step 1: Write the equilibrium expression for the given reaction.

Kc=[C]1

[A]2[B]= 4.5

Step 2: Substitute the initial concentrations into the equilibrium expression.

4.5 = 0.5

(0.2)2(0.1)

Step 3: Calculate the value of Kcusing the initial concentrations.

Kc=0.5

0.004 = 125

Step 4: Determine the reaction quotient, Qc, for the reaction with a de-

creased volume.

Qc=[C]′1

[A]′2[B]′

Given that the volume is decreased, the total pressure will increase, so the

reaction will shift to the side with fewer moles of gas to counteract the change.

Since there are 2 moles of gas on the reactant side and 1 mole of gas on the

product side, the reaction will shift to the right to decrease the total pressure.

Therefore, the reaction will shift towards the formation of more products.

In conclusion, the reaction will shift to the right to relieve the increase in

pressure by decreasing the volume of the container.

Question 7

Question

For the reaction:

2 CO(g) + O2(g)⇌2CO2(g)

how will the following changes affect the equilibrium composition? Justify your

answers using Le Chatelier’s principle.

1. Increasing the pressure by decreasing the volume of the container

2. Adding more CO(g)

3. Removing some CO2(g)

Solution

1. Increasing the pressure by decreasing the volume of the con-

tainer:

According to Le Chatelier’s principle, if the pressure is increased by de-

creasing the volume, the system will shift towards the side with fewer

moles of gas to alleviate the stress. In this case, there are 3 moles of

gas on the left side and 2 moles of gas on the right side. Therefore, the

equilibrium will shift to the right to decrease the total number of moles

of gas. As a result, the concentrations of CO and O2 will decrease, while

the concentration of CO2 will increase.

2. Adding more CO(g):

When more CO is added, the equilibrium will shift to the left to counter-

act the change. This is because adding more CO will increase the con-

centration of one of the reactants, causing the system to favor the reverse

reaction to consume the excess reactant. Therefore, the concentration of

CO2 will decrease, while the concentrations of CO and O2 will increase.

3. Removing some CO2(g):

If CO2 is removed from the system, the equilibrium will shift to the right

to replace the lost CO2. This is because removing CO2 causes a decrease

in the concentration of one of the products, leading the system to favor the

forward reaction to produce more CO2. Consequently, the concentrations

of CO and O2 will decrease, while the concentration of CO2 will increase.

Question 8

Question

A chemist is studying the following reaction at equilibrium:

2A(g) + B(g)⇌C(g)

If the chemist increases the pressure in the reaction vessel by decreasing the

volume, predict the direction in which the equilibrium will shift. Justify your

answer using Le Chatelier’s principle.

Solution

Step 1: When the pressure is increased by decreasing the volume, the system

will respond by shifting the equilibrium to reduce the pressure.

Step 2: Since the reaction involves gaseous reactants and products, we need

to consider the stoichiometry of the reaction to determine how the equilibrium

will shift.

Step 3: The reaction involves 3 moles of gas on the left side and 1 mole of

gas on the right side of the reaction.

Step 4: By decreasing the volume and increasing the pressure, the system

will shift the equilibrium to the side with fewer moles of gas to alleviate the

pressure increase.

Step 5: Therefore, the equilibrium will shift to the right to decrease the total

moles of gas and reduce the pressure in the system.

Step 6: In summary, by decreasing the volume and increasing the pressure,

the equilibrium will shift towards the products (C(g)) to reduce the pressure in

the reaction vessel, following Le Chatelier’s principle.

Question 9

Question

For the reaction:

2H2O(g)⇌2H2(g) + O2(g)

Calculate the equilibrium constant, Kc, given that at equilibrium, the concen-

trations of H2and O2are both 0.05 M and the concentration of H2O is 0.1 M.

If the volume of the container is then doubled, explain the direction in which

the equilibrium will shift.

Solution

Step 1: The equilibrium constant, Kc, can be calculated using the formula:

Kc=[H2]2[O2]

[H2O]2

Given that at equilibrium, the concentrations are:

[H2] = 0.05 M

[O2] = 0.05 M

[H2O] = 0.1M

Substitute these values into the formula:

Kc=(0.05)2(0.05)

(0.1)2=0.000125

0.01 = 0.0125

Therefore, the equilibrium constant, Kc, is 0.0125.

Step 2: When the volume of the container is doubled, the total pressure

on the system decreases. According to Le Chatelier’s principle, the system will

respond by shifting the equilibrium to minimize the effect of the volume change.

Since there are more moles of gas on the reactant side (2 moles of H2O)

compared to the product side (2 moles of H2and 1 mole of O2), the equilibrium

will shift to the right to decrease the total pressure on the system.

Therefore, the equilibrium will shift towards the products side, increasing

the concentrations of H2and O2while decreasing the concentration of H2O to

partially relieve the decrease in pressure caused by the doubling of the volume.

Question 10

Question

For the reaction below, indicate the direction in which the equilibrium will shift

if the concentration of NO2is increased. Justify your answer.

N2O4(g)⇌2NO2(g)

Solution

Step 1: Write the balanced chemical equation for the reaction. The balanced

chemical equation for the reaction is:

N2O4(g)⇌2NO2(g)

Step 2: Determine the effect of increasing the concentration of NO2on the

equilibrium position. When the concentration of NO2is increased, according to

Le Chatelier’s principle, the equilibrium will shift in the direction that helps to

consume the added species. In this case, increasing the concentration of NO2

will cause the equilibrium to shift to the left, consuming some of the NO2to

restore equilibrium.

Step 3: Write the new equilibrium expression after the shift. After the

equilibrium shift, the reaction quotient Qwill be increased and will favor the

reverse reaction.

Q=[NO2]2

[N2O4]

Step 4: Conclusion Therefore, if the concentration of NO2is increased, the

equilibrium will shift to the left to consume some of the additional NO2added

to the system and restore equilibrium.

Question 11

Question

Consider the following equilibrium reaction at 500 K:

2 H2(g) + O2(g)⇌2H2O(g)

If you were to increase the pressure of the system by decreasing the volume,

describe the direction in which the system will shift to re-establish equilibrium

based on Le Chatelier’s principle. Justify your answer.

Solution

Step 1: According to Le Chatelier’s principle, when the pressure of a system

at equilibrium is increased by decreasing the volume, the system will shift to

reduce the pressure.

Step 2: In this particular system, the total number of moles of gas on the

reactant side is 3 (2 moles of H2 and 1 mole of O2), and on the product side is

2 (2 moles of H2O).

Step 3: By decreasing the volume and thereby increasing the pressure, the

system will shift in the direction that reduces the number of moles of gas.

Step 4: In order to reduce the pressure, the system will favor the reaction

that leads to a decrease in the total number of moles of gas. Therefore, the

system will shift to the left, favoring the formation of H2 and O2 from H2O.

Step 5: To summarize, if the pressure of the system is increased by decreasing

the volume, the equilibrium will shift to the left, favoring the formation of H2

and O2 from H2O in order to reduce the pressure.

Question 12

Question

A reaction is at equilibrium with the following equilibrium constant:

Kc= 2.0×10−3

If the concentration of one of the reactants is increased by a factor of 10,

predict the direction in which the reaction will shift according to Le Chatelier’s

principle. Justify your answer.

Solution

Step 1: Determine the initial reaction quotient, Q.

Given that the equilibrium constant, Kc, is equal to 2.0×10−3, we assume

the reaction is at equilibrium when Q = K.

Step 2: Determine the initial concentrations of reactants and products.

Since the reaction is at equilibrium:

Qc=[C]c[D]d

[A]a[B]b=Kc= 2.0×10−3

Step 3: If one of the reactants is increased by a factor of 10, the new con-

centration will be:

Let’s assume the concentration of reactant A is increased by a factor of 10,

so:

[A]new = 10[A]initial

Thus, the new reaction quotient, Q’, will be:

Q′

c=[C]c[D]d

(10[A])a[B]b=1

10a×Qc

Step 4: Compare Q’ to K.

Since Q′

c=1

10a×Qc, and Qc=Kc= 2.0×10−3, if a > 0, then Q′

c< Kc,

and the reaction will shift to the right to reach a new equilibrium.

Therefore, in this case, the reaction will shift to the right to re-establish

equilibrium when the concentration of reactant A is increased by a factor of 10

according to Le Chatelier’s principle.

Question 13

Question

Consider the following reaction at equilibrium:

2SO2(g) + O2(g)⇌2SO3(g)

Which of the following changes will decrease the concentration of SO3(g)at

equilibrium? (A) Adding more SO2(g)(B) Removing some O2(g)(C) Increasing

the volume of the container (D) Adding a catalyst

Solution

Le Chatelier’s principle states that if a system at equilibrium is disturbed by

changing the conditions (like concentration, pressure, temperature), the system

will shift its equilibrium position to counteract the change. Let’s analyze how

each of the changes will affect the equilibrium in this particular reaction:

Step 1: Adding more SO2(g): If we add more SO2(g), the reaction will

shift to the right to consume the excess SO2(g). This will lead to an increase

in the concentration of SO3(g), not a decrease. Therefore, this change will not

decrease the concentration of SO3(g)at equilibrium.

Step 2: Removing some O2(g): If we remove some O2(g), the reaction

will shift to the left to replace the lost O2(g), resulting in an increase in the

concentration of SO2(g)and a decrease in the concentration of SO3(g). Thus,

this change will decrease the concentration of SO3(g)at equilibrium.

Step 3: Increasing the volume of the container: If the volume of the

container is increased, the system will try to decrease the pressure by shifting

to the side with more moles of gas. In this case, both sides have the same

number of moles of gas, so changing the volume will not have an effect on the

concentration of SO3(g)at equilibrium.

Step 4: Adding a catalyst: Adding a catalyst will only increase the rate

at which the reaction reaches equilibrium, but it will not affect the position of

the equilibrium. Therefore, adding a catalyst will not decrease the concentration

of SO3(g)at equilibrium.

Therefore, the correct answer is: (B) Removing some O2(g), which will

decrease the concentration of SO3(g)at equilibrium.

Question 14

Question

A reaction mixture initially contains SO2(g),O2(g), and SO3(g) at equilibrium

in a 5 L container. The equilibrium concentrations are found to be: [SO2] = 0.10

M, [O2] = 0.20 M, and [SO3] = 1.0M. If the volume of the container is suddenly

decreased to 2 L at constant temperature, predict the direction in which the

reaction will shift (towards the reactants or products) to reestablish equilibrium.

Justify your answer using Le Chatelier’s principle.

Solution

Step 1: Write the balanced chemical equation representing the reaction:

2 SO2(g) + O2(g) ←−→ 2 SO3(g)

Step 2: Calculate the initial equilibrium constant, K, using the initial equi-

librium concentrations:

K=[SO3]2

[SO2]2[O2]=(1.0)2

(0.10)2(0.20) = 250

Step 3: Calculate the new equilibrium concentrations after the volume

change: The new concentrations can be calculated using the ideal gas law,

since the temperature is constant:

[SO2]2=P1V1

P2V2

[SO2]1=5

2(0.10) = 0.25 M

[O2]2=P1V1

P2V2

[O2]1=5

2(0.20) = 0.50 M

[SO3]2=P1V1

P2V2

[SO3]1=5

2(1.0) = 2.50 M

Step 4: Calculate the new equilibrium constant, K2, using the new equilib-

rium concentrations:

K2=[SO3]2

[SO2]2

2[O2]2

=(2.50)2

(0.25)2(0.50) = 400

Step 5: Compare the initial equilibrium constant, K, with the new equilib-

rium constant, K2. Since K2> K, the reaction quotient, Q, is now less than

K and the reaction will shift to the right (towards the products) to reestablish

equilibrium. This is in accordance with Le Chatelier’s principle, which states

that a system will shift in the direction that relieves the stress imposed on it.

Question 15

Question

For the reaction:

2 CO(g) + O2(g) ⇌2 CO2(g)

which equilibrium shift occurs when the pressure of the system is increased by

decreasing the volume, assuming all species are ideal gases? Justify your answer.

Solution

Step 1: Write the balanced chemical equation for the reaction:

2 CO(g) + O2(g) ⇌2 CO2(g)

Step 2: Identify the total number of moles of gas on each side of the reaction:

Initially, we have 3 moles of gas on the left side of the reaction and 2 moles of

gas on the right side.

Step 3: Determine how the total pressure of the system will be affected

by the change: When the volume is decreased, the pressure of the system will

increase according to the ideal gas law:

P V =nRT

Since the number of moles and temperature remain constant, a decrease in

volume will lead to an increase in pressure.

Step 4: Apply Le Chatelier’s Principle to predict the equilibrium shift: Due

to the increase in pressure, the system will try to counteract this change by

shifting the equilibrium in the direction that reduces the total number of moles

of gas. In this case, the system will shift to the right, favoring the production

of CO2.

Therefore, when the pressure of the system is increased by decreasing the

volume, the equilibrium shift will favor the formation of CO2gas.

Question 16

Question

For the reaction 2SO3(g)⇌2SO2(g) + O2(g)at equilibrium, an increase in

pressure shifts the equilibrium to the left. Explain this observation using Le

Chatelier’s principle.

Solution

Step 1: According to Le Chatelier’s principle, when a system at equilibrium

is subjected to a change, it will react in a way that tends to counteract that

change.

Step 2: In this case, an increase in pressure causes the system to shift to the

side with fewer gas molecules in order to reduce the total pressure.

Step 3: The reaction 2SO3(g)⇌2SO2(g) + O2(g)has 3 moles of gas on the

left side and 3 moles of gas on the right side.

Step 4: By removing 1 mole of gas (O2) from the right side and moving

to the left side, the equilibrium shifts to the left to alleviate the increase in

pressure.

Step 5: As a result of the shift, more SO3molecules will react to form more

SO2and O2molecules until a new equilibrium is established.

Step 6: Therefore, the equilibrium shifts to the left (towards the side with

fewer gas molecules) when the pressure is increased for the given reaction.

Question 17

Question

For the reaction 2SO2(g) + O2(g)⇌2SO3(g)at a certain temperature, an

equilibrium mixture is found to contain 0.30 M of SO2, 0.40 M of O2, and

1.20 M of SO3. Calculate the equilibrium constant, Kc, for the reaction at this

temperature.

Solution

Step 1: Write the expression for the equilibrium constant, Kc, for the given

reaction.

Kc=[SO3]2

[SO2]2·[O2]

Step 2: Substitute the given equilibrium concentrations into the expression

for Kc.

Kc=(1.20)2

(0.30)2·(0.40)

Step 3: Calculate Kc.

Kc=1.44

0.09 ·0.40 =1.44

0.036 = 40

Step 4: Therefore, the equilibrium constant, Kc, for the reaction at the given

temperature is 40.

Question 18

Question

For the reaction below at a certain temperature, CO(g)+ 2H2O(g)⇌CO2H2g,

the equilibrium constant Kcis 1.2×10−3. If initially, the system contains

0.50 moles of CO, 0.20 moles of H2O, and 0.30 moles of CO2, predict in which

direction the system will shift to reach equilibrium when the total pressure is

reduced by halving the volume of the container.

Solution

Step 1: Determine the reaction quotient Qc. The reaction quotient Qcis given

by the formula:

Qc=[CO2][H2]2

[CO].

Substitute the initial concentrations into the formula to get Qc:

Qc=(0.30)(0.20)2

0.50 = 0.024.

Step 2: Compare Qcwith Kcto determine the direction of the reaction.

Since Qc= 0.024 is greater than Kc= 1.2×10−3, the system will shift to the

left to reach equilibrium. This means the concentration of CO2will decrease,

and the concentrations of CO and H2O will increase.

Step 3: Determine how decreasing the total pressure will affect the system.

When the total pressure is reduced by halving the volume of the container, the

system will respond by shifting in the direction that reduces the total moles

of gas. This means the system will shift to the left, towards the reactants, to

decrease the total pressure.

Therefore, the system will shift to the left (towards the reactants) to reach

equilibrium when the total pressure is reduced by halving the volume of the

container.

Question 19

Question

A reaction mixture at equilibrium contains nitrogen gas, hydrogen gas, and

ammonia gas according to the following balanced equation:

N2(g)+ 3H2(g)⇌2NH3(g)

If the volume of the container is suddenly decreased, predict the direction in

which the equilibrium will shift and explain why.

Solution

Step 1: Identify the change caused by decreasing the volume of the container.

When the volume of the container is decreased, the pressure inside the container

will increase. According to Le Chatelier’s principle, the equilibrium will shift in

such a way as to counteract this change.

Step 2: Determine the effect of increased pressure on the reaction. In the

given reaction, the total number of moles of gas on the left side is 1 (1 mol of

N2and 3 mol of H2), while the total number of moles of gas on the right side is

2 mol of NH3. Therefore, the reaction involves a decrease in the number of gas

moles.

Step 3: Predict the direction of equilibrium shift. To counteract the increase

in pressure caused by decreasing the volume, the equilibrium will shift in the

direction that leads to a decrease in the total number of gas moles. This means

the equilibrium will shift towards the side with fewer gas moles. In this case,

the equilibrium will shift to the left to decrease the total pressure inside the

container.

Step 4: Justification. When the equilibrium shifts to the left, the concentra-

tions of N2and H2will increase, while the concentration of NH3will decrease.

This shift helps to reduce the total number of gas moles and, therefore, decrease

the pressure.

Question 20

Question

For the reaction

2A+B⇌C+D

, which is at equilibrium, the system is then subjected to an increase in pressure

by decreasing the volume of the container. How will the equilibrium position

shift in response to this change according to Le Chatelier’s principle?

Solution

Step 1: When the volume of the container is decreased, the system responds

by trying to counteract the change by favoring the side of the reaction that

produces fewer moles of gas.

Step 2: In this reaction, since there are 3 moles of gas on the left side (2

moles of A and 1 mole of B) and only 2 moles of gas on the right side (1 mole

of C and 1 mole of D), decreasing the volume will shift the equilibrium position

towards the side of the reaction with fewer moles of gas.

Step 3: Therefore, in response to the increase in pressure caused by the

decreased volume, the equilibrium position will shift to the right, favoring the

formation of C and D.

Question 21

Question

Consider the following equilibrium reaction:

N2(g)+3H2(g)⇌2NH3(g)

If the concentration of N2in the system is increased, predict the effect on

the equilibrium position of the reaction according to Le Chatelier’s principle.

Justify your answer.

Solution

Step 1: Recall Le Chatelier’s principle, which states that when a system at

equilibrium is subjected to a change in concentration, temperature, or pressure,

the equilibrium position will shift to counteract the effect of that change.

Step 2: In this reaction, the forward reaction results in the formation of

ammonia from nitrogen and hydrogen gas. If the concentration of N2is in-

creased, Le Chatelier’s principle predicts that the reaction will shift to the right

to counteract this increase in N2.

Step 3: By shifting to the right, more nitrogen and hydrogen gas will react to

form ammonia until equilibrium is reestablished. This will result in a decrease

in the concentration of nitrogen gas.

Step 4: As a result of the shift to the right, the concentrations of nitrogen

and hydrogen gas will decrease while the concentration of ammonia will increase

until a new equilibrium is reached.

Step 5: Therefore, the equilibrium position of the reaction will shift to the

right if the concentration of N2is increased, in order to consume the excess

nitrogen added to the system.

Question 22

Question

For the equilibrium reaction:

2A +B⇌C+D

The equilibrium constant, Kc, is 5.0. If the concentration of substance A is

increased by a factor of 3, predict the direction in which equilibrium will shift

and briefly explain why.

Solution

Step 1: Write the reaction quotient, Qc, based on the change in concentration

of substance A.

Qc=[C][D]

[A]2[B]

Step 2: Since the concentration of substance A is increased by a factor of 3,

the new concentration of A will be 3[A]. The new reaction quotient becomes:

Qc=[C][D]

(3[A])2[B]=[C][D]

9[A]2[B]

Step 3: Compare the new reaction quotient, Qc, with the equilibrium con-

stant Kc. - If Qc< Kc, the reaction will shift to the right to reach equilibrium. -

If Qc> Kc, the reaction will shift to the left to reach equilibrium. - If Qc=Kc,

the reaction is at equilibrium and no shift will occur.

Step 4: Calculate the new Qc:

Qc=[C][D]

9[A]2[B]=1

9Kc

Step 5: Since Qc=1

9Kc, which is less than the original Kc, the reaction

will shift to the right to reach a new equilibrium position. This shift occurs to

relieve the stress caused by the increase in A concentration.

Question 23

Question

Consider the following equilibrium reaction:

2 NOBr (g) ⇌2 NO (g) +Br2(g)

How will each of the following changes affect the equilibrium position (shift

left, shift right, no shift)? Justify your answer using Le Chatelier’s principle.

1. Increase in temperature 2. Decrease in pressure 3. Addition of NO gas 4.

Removal of some NOBr gas

Solution

1. Increase in temperature:

• According to Le Chatelier’s principle, if the temperature of an equilibrium

reaction is increased, the system will shift in the endothermic direction to

counteract the change.

• In this case, the forward reaction is endothermic (+82 kJ/mol), so an

increase in temperature will shift the equilibrium to the right (towards

the products) to consume the added heat.

• Therefore, the equilibrium position will shift to the right.

2. Decrease in pressure:

• When the pressure is decreased, the system will shift in the direction that

produces more moles of gas to counteract the change.

• In this reaction, the total number of moles of gas decreases from 2 to 3 as

we move from left to right.

• As a result, a decrease in pressure will cause the equilibrium to shift to

the right to increase the total gas moles.

• Therefore, the equilibrium position will shift to the right.

3. Addition of NO gas:

• Adding more NO gas will increase the concentration of a reactant in the

equilibrium reaction.

• According to Le Chatelier’s principle, the system will shift in the direction

that consumes the added reactant to counteract the change.

• In this case, the added NO will be consumed by the forward reaction,

causing the equilibrium to shift to the right to consume the excess NO.

• Therefore, the equilibrium position will shift to the right.

4. Removal of some NOBr gas:

• Removing some NOBr gas will decrease the concentration of a reactant in

the equilibrium reaction.

• The system will therefore shift in the direction that produces more of the

removed species to counteract the change.

• In this case, removing NOBr will cause the equilibrium position to shift

to the right to produce more NO and Br2.

• Therefore, the equilibrium position will shift to the right.

Question 24

Question

For the reaction 2SO2(g) + O2(g)⇌2SO3(g), if the concentration of SO2is

increased while keeping the total pressure constant, predict the direction in

which the equilibrium will shift according to Le Chatelier’s principle. Justify

your answer.

Solution

Step 1: First, let’s consider the effect of increasing the concentration of SO2on

the equilibrium position. According to Le Chatelier’s principle, an increase in

the concentration of a reactant will cause the system to shift in the direction

that consumes some of that reactant.

Step 2: In this reaction, SO2is a reactant. By increasing its concentration,

the system will shift in the direction that consumes SO2, which is the forward

reaction.

Step 3: The forward reaction consumes SO2and produces SO3, causing the

concentration of SO2to decrease and the concentration of SO3to increase.

Step 4: As a result, the equilibrium will shift to the right to consume some

of the excess SO2added, and the concentrations of SO2and SO3will both

increase.

Step 5: In conclusion, when the concentration of SO2is increased while keep-

ing the total pressure constant, the equilibrium will shift to the right (towards

the formation of more SO3) according to Le Chatelier’s principle.

Question 25

Question

A reaction is at equilibrium according to the equation:

2A(g) + B(g) <=> C(g) + D(g)

If the concentration of B is increased, predict how the equilibrium will shift

according to Le Chatelier’s principle. Justify your answer.

Solution

Step 1: When the concentration of B is increased, the equilibrium will shift to

the right in order to counteract the increase in B. This is because according

to Le Chatelier’s principle, a system at equilibrium will respond to stress by

shifting the equilibrium position in a direction that tends to relieve the stress.

Step 2: As B is a reactant in the reaction, increasing its concentration will

drive the reaction in the forward direction to consume the excess B added.

This will result in an increase in the concentrations of C and D, while the

concentrations of A and B will decrease.

Step 3: Therefore, the equilibrium will shift to the right, favoring the for-

mation of C and D, in response to the increase in the concentration of B. This

shift will continue until a new equilibrium is established where the rates of the

forward and reverse reactions once again become equal.

Question 2

Question

For the reaction CO(g)+3H2O(g)⇌CO2H2O(l), which is at equilibrium, predict

the direction of each of the following changes and state how each change will

affect the equilibrium position:

1. Addition of more CO(g)

2. Removal of some H2O(g)

3. Increasing the volume of the container at constant temperature

Solution

Step 1: Addition of more CO(g)When more CO(g)is added to the reaction,

the reaction will shift to the right to consume the additional reactant. This

is in accordance with Le Chatelier’s principle, which states that a system at

equilibrium will respond to a stress by shifting in the direction that helps to

relieve that stress. Consequently, the equilibrium position will shift towards the

products in this case.

Step 2: Removal of some H2O(g)If some H2O(g)is removed from the reaction,

the equilibrium will shift to the left to replace the lost water molecules. This

will result in an increase in the concentration of the reactants, thus favoring the

reactants at equilibrium.

Step 3: Increasing the volume of the container at constant temperature

When the volume of the container is increased at constant temperature, the

equilibrium will shift in the direction that produces more gas molecules to oc-

cupy the additional volume. In this reaction, there are more gas molecules on

the left side (4 moles of gas) compared to the right side (1 mole of liquid).

Therefore, the equilibrium position will shift to the left to increase the pressure

by favoring the side with more gas molecules.

Question 3

Question

Consider the following equilibrium reaction:

2A(g) + 3B(g)⇌C(g) + D(g)

If the concentration of substance A is increased, predict the effect on the

concentrations of substances B, C, and D at equilibrium. Justify your answer.