CHEM 321 - ANALYTICAL CHEMISTRY Le Chatelier's principle and equilibrium shifts Question Bank Set 9

CHEM 321 - ANALYTICAL

CHEMISTRY - Le Chatelier’s principle

and equilibrium shifts

Question Bank - Set 9

Liberty University

Question 1

Question

A reaction mixture at equilibrium is composed of 0.20 M N2, 0.30 M H2, and

0.25 M NH3. The equilibrium constant, Kc, for the reaction N2(g) + 3H2(g)

⇌2NH3(g) is 0.34. If the equilibrium mixture is disturbed by increasing the

concentration of N2to 0.30 M, predict the direction in which the equilibrium

will shift. Justify your answer.

Solution

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

for the equilibrium constant.

The balanced chemical equation for the reaction is:

N2(g) + 3H2(g) ⇌2NH3(g)

The expression for the equilibrium constant, Kc, for this reaction is:

Kc =[NH3]2

[N2][H2]3

Step 2: Calculate the initial value of the equilibrium constant, Qc, using the

initial concentrations provided.

Substitute the initial concentrations into the expression for Qc:

Qc =(0.25)2

(0.20)(0.30)3

Qc ≈0.69

Step 3: Compare Qc and Kc to determine the direction of the equilibrium

shift.

Since Qc = 0.69 and Kc = 0.34, Qc is larger than Kc. This means that the

system is not at equilibrium.

Step 4: Predict the direction of the equilibrium shift.

Since Qc is larger than Kc, the reaction will shift to the left to establish a

new equilibrium.

Therefore, with the increased concentration of N2, the equilibrium will shift

towards the reactants N2and H2.

Question 2

Question

A gaseous equilibrium is established in the reaction below at a certain temper-

ature:

2 NOCl(g) ⇌2 NO(g) +Cl2(g)

When the pressure of NOCl is increased, predict how the equilibrium will

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

Solution

Step 1: Initially, in the reaction system, there are 2 moles of NOCl for every 2

moles of NO and 1 mole of Cl2.

Step 2: When the pressure of NOCl is increased, Le Chatelier’s principle

states that the system will shift to counteract this change. Since the reaction

produces 2 moles of gas for every 1 mole of NOCl consumed, increasing the

pressure will cause the system to shift to the side with fewer moles of gas to

alleviate the pressure increase.

Step 3: In this case, the forward reaction results in the production of 3 moles

of gas (2 moles of NO and 1 mole of Cl2) for every 2 moles of NOCl consumed.

Therefore, the system will shift to the right to decrease the total number of gas

molecules and reduce the pressure.

Step 4: As a result, increasing the pressure of NOCl will cause the equilib-

rium to shift towards the products, favoring the formation of NO and Cl2.

Step 5: In conclusion, an increase in the pressure of NOCl will cause the

equilibrium to shift to the right, favoring the formation of NO and Cl2according

to Le Chatelier’s principle.

Question 3

Question

Consider the following equilibrium reaction:

N2O4(g)⇌2NO2(g)

If the initial concentration of N2O4is 0.20 M, and the equilibrium constant,

Kc, for the reaction at a certain temperature is 0.050, what will be the concen-

tration of NO2at equilibrium?

Solution

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

Kc=[NO2]2

[N2O4]= 0.050

Step 2: Since the initial concentration of N2O4,[N2O4]initial = 0.20 M,

and assuming the concentration of NO2at the beginning is 0M, we denote

[NO2]initial = 0 M and [NO2]at equilibrium as x.

Step 3: Set up an ICE table (Initial, Change, Equilibrium) to represent the

changes in concentration:

Species N2O4NO2

Initial (M) 0.20 0

Change (M) −x+2x

Equilibrium (M) 0.20 −x2x

Step 4: Substitute the equilibrium concentrations into the equilibrium con-

stant expression and solve for x:

Kc=(2x)2

0.20 −x= 0.050

4x2= 0.050(0.20 −x)

4x2= 0.01 −0.05x

4x2+ 0.05x−0.01 = 0

Step 5: Solve the quadratic equation for xusing the quadratic formula:

x=−0.05 ±√0.052−4(4)(−0.01)

2(4)

x=−0.05 ±√0.0025 + 0.16

x=−0.05 ±√0.1625

x=−0.05 ±0.4034

Step 6: Calculate the possible values for x:

x=0.3534

8≈0.0442 M (rejected) or x=−0.4534

8≈ −0.0567 M

Since concentration cannot be negative, the concentration of NO2at equi-

librium is approximately 0.044 M.

Question 4

Question

For the reaction:

N2O4(g)⇌2NO2(g)

Which of the following changes would result in an increase in the concentra-

tion of N2O4 at equilibrium? Justify your answer with Le Chatelier’s principle.

A. Decreasing the volume of the container B. Increasing the temperature

C. Decreasing the pressure by adding an inert gas D. Removing some NO2 gas

from the system

Solution

Step 1: Write the balanced chemical equation for the given reaction and identify

the direction of the reaction. The reaction is:

N2O4(g)⇌2NO2(g)

In this reaction, N2O4 is in the gas phase and decomposes into NO2 gas.

This is an endothermic reaction, meaning it absorbs heat.

Step 2: Apply Le Chatelier’s principle to the given changes: A. Decreasing

the volume of the container: If the volume decreases, the system will shift to

the side with fewer gas molecules to relieve the increase in pressure. Therefore,

it will shift towards more N2O4 to decrease the total pressure of the system,

resulting in an increase in the concentration of N2O4 at equilibrium.

B. Increasing the temperature: Since the reaction is endothermic, increasing

the temperature will favor the endothermic reaction to absorb the additional

heat. Thus, the equilibrium will shift towards the products, resulting in a

decrease in the concentration of N2O4.

C. Decreasing the pressure by adding an inert gas: The addition of an inert

gas does not affect the concentrations of the reactants and products in the

equilibrium mixture, as they do not participate in the reaction. Therefore, this

change will not increase the concentration of N2O4.

D. Removing some NO2 gas from the system: If NO2 gas is removed, the

system will shift towards the formation of more NO2 to replace what was lost.

This will result in a decrease in the concentration of N2O4 at equilibrium.

Therefore, the correct answer is A. Decreasing the volume of the con-

tainer, as it will lead to an increase in the concentration of N2O4 at equilibrium.

Question 5

Question

For the reaction below at a certain temperature, 1.00 mol of solid carbon dioxide

is placed in a 4.00 L container. The reaction that occurs is:

CO2(s)⇌CO2(g)

The reaction reaches equilibrium, and it is found that 0.25 mol of carbon

dioxide is present in the gaseous state.

What is the value of the equilibrium constant, Kc, for this reaction at the

given temperature?

Solution

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

temperature.

Kc=[CO2(g)]

[CO2(s)]

Step 2: Determine the molar concentration of CO2(g)at equilibrium. Given

that 0.25 mol of CO2(g)is present in a 4.00 L container:

[CO2(g)] = 0.25 mol

4.00 L= 0.0625 M

Step 3: Determine the molar concentration of CO2(s)at equilibrium. Since

1.00 mol of CO2(s)was placed in the 4.00 L container to start with:

[CO2(s)] = 1.00 mol

4.00 L= 0.25 M

Step 4: Substitute the molar concentrations into the equilibrium constant

expression to find Kc.

Kc=0.0625

0.25 = 0.25

Therefore, the equilibrium constant, Kc, for the reaction at the given tem-

perature is 0.25.

Question 6

Question

Consider the following equilibrium reaction at a certain temperature:

3A(g) + 2B(g)↔C(g)

1. If the concentration of substance A is increased, predict the direction in

which the equilibrium will shift. 2. Provide a reason for your prediction.

Solution

1. When the concentration of substance A is increased, according to Le Chate-

lier’s principle, the equilibrium will shift to oppose this change.

2. The equilibrium will shift to the right because the system will try to

consume the excess A added to bring the system back to equilibrium. This

shift will result in an increase in the concentration of C and a decrease in the

concentrations of A and B.

Question 7

Question

A reaction is at equilibrium with the following equilibrium expression:

2CO(g) + O2(g)⇌2CO2(g)

If the concentration of CO2is increased, predict the direction in which the

equilibrium will shift and explain why.

Solution

Step 1: Write the balanced equilibrium equation and the expression for the

equilibrium constant (K).

2CO(g) + O2(g)⇌2CO2(g)

K=[CO2]2

[CO]2[O2]

Step 2: Increase in CO2concentration will disturb the equilibrium. Accord-

ing to Le Chatelier’s principle, the system will shift in a direction that opposes

the change.

Step 3: Since an increase in CO2concentration will drive the equilibrium

towards the reactants, in this case towards 2CO(g) + O2(g). This shift occurs

in order to consume some of the excess CO2that was added.

Therefore, the equilibrium will shift to the left, in the direction of the reac-

tants.

Question 8

Question

A student is performing an experiment where they determine the concentration

of ammonia gas at equilibrium in a sealed container at 500 K according to the

following reaction:

2NH3(g)⇌N2(g)+ 3H2(g)

Initially, the student adds 0.10 mol of ammonia gas in a 1.0 L container at 500

K. The equilibrium constant, Kc, for this reaction at 500 K is 0.10. Calculate

the concentration of ammonia gas at equilibrium. (Assume the volume remains

constant and there are no other species present in the container.)

Solution

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

reaction:

Kc=[N2]×[H2]3

[NH3]2

Step 2: Initially, the concentration of ammonia, [N H3]initial, is 0.10 mol.

Let [NH3]eq be the concentration of ammonia gas at equilibrium.

Step 3: The change in concentration of ammonia gas, [NH3], will be:

[NH3]initial = [NH3]eq + 2x

where ’x’ is the change in concentration.

Step 4: The concentrations of N2and H2can also be expressed in terms of

’x’:

[N2] = x

[H2] = 3x

Step 5: Substitute the expressions for [N2],[H2], and [NH3]into the equi-

librium constant expression:

0.10 = (x)(3x)3/(0.10 −2x)2

Step 6: Solve for ’x’:

27x4

(0.10 −2x)2= 0.10

Step 7: Simplify the equation and solve for ’x’. This is a challenging algebraic

step.

Step 8: Calculate the concentration of ammonia gas at equilibrium, [NH3]eq,

using the value of ’x’ obtained.

Step 9: Finally, convert the concentration of ammonia gas at equilibrium to

mol/L.

Question 9

Question

For the reaction

N2O4(g)⇌2NO2(g)

at a certain temperature, a student initially adds 0.10 mol of N2O4(g)toa1.0Lflask.T hemixturewasallowedtoreachequilibriumandwasf oundtocontain0.060molof N2O4

at equilibrium. Calculate the equilibrium constant Kcfor the reaction.

Solution

Step 1: Write the expression for the equilibrium constant Kc. The equilibrium

constant Kcis defined as:

Kc=[NO2]2

[N2O4]

Step 2: Set up an ICE table to determine the changes in concentration for

each species.

Species N2O4NO2

Initial (mol/L) 0.10 0

Change (mol/L) −x+2x

Equilibrium (mol/L) 0.10 −x2x

Step 3: Write the equilibrium constant expression using the equilibrium

concentrations from the ICE table. Substitute the equilibrium concentrations

into the equilibrium constant expression:

Kc=(2x)2

0.10 −x

Step 4: Use the given information to find the value of x. Given: At equilib-

rium, N2O4= 0.060 mol. So, 0.10 −x= 0.060 ⇒x= 0.04 mol.

Step 5: Calculate the equilibrium constant Kcusing the value of x. Substi-

tute x= 0.04 into the equilibrium constant expression:

Kc=(2(0.04))2

0.10 −0.04

Kc=0.16

0.06

Kc= 2.67

Therefore, the equilibrium constant Kcfor the reaction is 2.67.

Question 10

Question

For the reaction below at equilibrium, determine the effect on the equilibrium

position if the volume of the container is decreased.

N2(g)+3H2(g)⇌2NH3(g)

Solution

Step 1: Write the balanced equation including the phase of each chemical

species.

N2(g)+3H2(g)⇌2NH3(g)

Step 2: Identify the initial effect of decreasing the volume on the concen-

tration of each species. When the volume of the container is decreased, the

pressure will increase. According to Le Chatelier’s principle, the system will

shift in a direction that opposes the imposed change.

Step 3: Determine the effect on the equilibrium position. In this reaction,

the total number of moles of gas on the left side is 1 (from N2) + 3 (from H2) =

4, and on the right side is 2 (from NH3). Therefore, by increasing the pressure,

the system will shift to the side with fewer moles of gas to relieve the stress. In

this case, the system will shift to the left to decrease the total pressure, favoring

the formation of N2 and H2 and decreasing NH3.

Step 4: Write the new equilibrium expression to reflect the change. The equi-

librium position will shift to the left, resulting in an increase in N2andH2concentrationsandadecreaseinNH3concentration.

Step 5: Write the updated equilibrium equation.

N2(g)+3H2(g)⇌2NH3(g)

Therefore, decreasing the volume of the container will result in a leftward

shift of the equilibrium position.

Question 11

Question

For the reaction 2SO2(g) + O2(g) ⇌2SO3(g), a student adds an inert gas at

constant volume, causing the pressure to increase. Explain how this change will

affect the equilibrium position according to Le Chatelier’s principle.

Solution

Step 1: According to Le Chatelier’s principle, when a system experiencing equi-

librium is subjected to a change in concentration, pressure, or temperature, the

system will adjust to counteract that change and reach a new equilibrium.

Step 2: In this reaction, adding an inert gas at constant volume will only

increase the total pressure of the system without affecting the concentrations of

the reactants and products.

Step 3: Since the inert gas does not participate in the reaction, the reaction

will shift in a direction that reduces the total pressure.

Step 4: To decrease the total pressure, the reaction will shift in the direction

that decreases the number of gas molecules present. In this case, the reaction

will shift to the right (towards the formation of more SO3) to consume some of

the excess pressure.

Step 5: Therefore, adding an inert gas at constant volume will cause the

equilibrium position to shift to the right, favoring the formation of more SO3.

Question 12

Question

A mixture of hydrogen gas and iodine gas is placed in a closed container at a

certain temperature. The following equilibrium is established:

H2(g) + I2(g)⇌2HI(g)

If the container is suddenly compressed to half its volume, explain in detail

the direction in which the equilibrium will shift. Justify your answer using Le

Chatelier’s principle.

Solution

Step 1: Identify the effect of the volume change on the system. When the

volume of the container is reduced, the system will try to alleviate the change

in pressure by shifting the equilibrium in a direction that reduces the number

of gas molecules.

Step 2: Determine the initial impact of reducing the volume on the equi-

librium. Since there is an increase in pressure due to volume reduction, the

equilibrium will shift in the direction that reduces the total moles of gas.

Step 3: Analyze the stoichiometry of the reaction. From the balanced equa-

tion, we can see that on the left side, there are 1 + 1 = 2 moles of gas, and on

the right side, there are 2 moles of gas. Hence, there is no change in the total

moles of gas.

Step 4: Apply Le Chatelier’s principle to determine the direction of the

equilibrium shift. Since the total moles of gas in the system are the same before

and after the volume change, the equilibrium will not shift in either direction

as a response to the change in volume.

Step 5: Conclusion. In this case, reducing the volume of the container will

not cause a shift in the equilibrium position of the reaction because the total

moles of gas remain unchanged.

Question 13

Question

At a certain temperature, the equilibrium reaction involving nitrogen dioxide

(NO2) can be represented as:

2NO2(g)⇌N2O4(g)

If the concentration of NO2is increased, predict how the equilibrium will

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

Solution

Step 1: According to Le Chatelier’s principle, if the concentration of a reactant

is increased, the system will shift to consume some of the increase, favoring the

formation of products.

Step 2: In this reaction, when the concentration of NO2is increased, the

system will shift to the right to consume some of the additional NO2.

Step 3: As a result, the equilibrium will shift towards the products, N2O4,

to partially offset the increase in NO2concentration.

Therefore, the equilibrium will shift to the right, favoring the formation of

N2O4, when the concentration of NO2is increased.

Question 14

Question

A student is conducting an experiment to study the equilibrium between nitro-

gen dioxide (NO2) and dinitrogen tetroxide (N2O4). The student starts with a

reaction mixture containing a brown gas. The balanced chemical equation for

the reaction is:

2NO2(g)⇌N2O4(g)

During the experiment, the student increases the temperature of the reaction

mixture. Predict how the color of the gas mixture will change and explain this

observation using Le Chatelier’s principle.

Solution

Step 1: The color change of the gas mixture will be from brown to colorless.

Step 2: According to Le Chatelier’s principle, when a system at equilibrium

is subjected to a change in temperature, it will respond in a way that opposes

the change in order to restore equilibrium.

Step 3: In this case, by increasing the temperature of the reaction mixture,

the system will shift in the endothermic direction to absorb the excess heat.

Step 4: The forward reaction (2NO2→N2O4) is endothermic, meaning it

absorbs heat.

Step 5: Therefore, as the system shifts to the right to consume heat, the

concentration of nitrogen dioxide decreases and the concentration of dinitrogen

tetroxide increases.

Step 6: Since the brown color is associated with nitrogen dioxide and the

colorless is with dinitrogen tetroxide, the change in color will be from brown to

colorless as the equilibrium shifts in response to the increased temperature.

Step 7: Thus, the gas mixture will change from brown to colorless when the

temperature is increased, in accordance with Le Chatelier’s principle.

Question 15

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 of the shift in equilibrium and provide a justification for

your answer.

Solution

Step 1: Write the balanced equilibrium equation:

2SO2(g) + O2(g)⇌2SO3(g)

Step 2: Apply Le Chatelier’s principle - If the concentration of SO2is in-

creased, the reaction will shift to the right to consume the added SO2and

establish a new equilibrium.

Step 3: Justification - According to Le Chatelier’s principle, when the con-

centration of a reactant is increased, the system will shift in the direction that

consumes that reactant to relieve the stress and restore equilibrium. In this

case, as SO2is a reactant, the equilibrium will shift to the right to consume the

added SO2and form more SO3until a new equilibrium is established.

Question 16

Question

Consider the following equilibrium reaction:

2SO2(g) + O2(g)⇌2SO3(g)

If the concentration of SO2is increased by adding more of it to the reaction

vessel at constant temperature, predict the direction in which the equilibrium

will shift and explain your reasoning.

Solution

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

Kc=[SO3]2

[SO2]2[O2]

Step 2: Analyze the impact of increasing the concentration of SO2on the

equilibrium. According to Le Chatelier’s principle, if the concentration of a

reactant is increased, the equilibrium will shift to consume some of that reactant

and move the reaction towards the products to re-establish equilibrium.

Step 3: Since the concentration of SO2is increased, the reaction will shift

to the right to consume some of the excess SO2to form more SO3and O2until

a new equilibrium is reached.

Step 4: Therefore, the equilibrium will shift to the right in response to the

increase in SO2concentration in order to relieve the stress on the system.

Thus, the equilibrium will shift to the right when the concentration of SO2

is increased.

Question 17

Question

A reaction is at equilibrium with the following equilibrium constant:

Kc= 3.5×10−3

Given the balanced chemical equation below, which of the following changes

will NOT cause a shift in the equilibrium position to the left?

2A(g) + B(g) ⇌3C(g)

Options:

• A decrease in the volume of the reaction container

• An increase in the concentration of B

• An increase in the concentration of C

• An increase in the pressure by applying an external force

Solution

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

changing the conditions, the position of the equilibrium will shift to counteract

the disturbance. Let’s analyze each option to determine which change will NOT

cause a shift in the equilibrium position to the left.

Option A: A decrease in the volume of the reaction container

•Explanation: If the volume is decreased, the pressure will increase. Ac-

cording to Le Chatelier’s principle, the equilibrium will shift in a direction

that reduces the pressure. Since the reaction has 2 moles of gas on the left

side and 3 moles of gas on the right side, reducing the volume will favor

the side with fewer moles of gas (the left side), leading to a shift to the

left.

•Conclusion: This change will cause a shift in the equilibrium position to

the left.

Option B: An increase in the concentration of B

•Explanation: Adding more B will increase the concentration of B. Ac-

cording to Le Chatelier’s principle, the equilibrium will shift in a direction

that consumes the added component. In this case, the added B will be

consumed by the reaction to form more C, leading to a shift to the right

(towards the product side).

•Conclusion: This change will cause a shift in the equilibrium position to

the right.

Option C: An increase in the concentration of C

•Explanation: Adding more C will increase the concentration of C. Fol-

lowing Le Chatelier’s principle, the equilibrium will shift in a direction

that reduces the excess. In this case, the excess C will be used up in the

reverse reaction to form more A and B, shifting the equilibrium to the

left.

•Conclusion: This change will cause a shift in the equilibrium position to

the left.

Option D: An increase in the pressure by applying an external

force

•Explanation: Increasing the pressure by applying an external force will

cause the equilibrium to shift in a direction that reduces the pressure.

Since the reaction has more moles of gas on the product side (3 moles of

gas), increasing the pressure will favor the side with fewer moles of gas

(the reactant side), leading to a shift to the left.

•Conclusion: This change will cause a shift in the equilibrium position to

the left.

Conclusion: The change that will NOT cause a shift in the equilibrium

position to the left is an increase in the concentration of B (Option B).

Question 18

Question

A reaction mixture initially contains 0.20 mol of solid calcium carbonate, CaCO3,

in a 1.0 L container. The system is in equilibrium with the reaction:

CaCO3(s)⇌CaO(s) + CO2(g)

Calculate the equilibrium concentration of carbon dioxide, CO2, when the

system is heated to increase the temperature. Assume that the volume of the

container remains constant. The equilibrium constant, Kc, for this reaction at

the given temperature is 3.0.

Solution

Step 1: Write the equilibrium expression for the reaction:

Kc=[CaO][CO2]

[CaCO3]= 3.0

Step 2: Since the initial concentration of CaCO3is 0.20 mol in a 1.0 L

container, the initial concentration of CaCO3is:

[CaCO3] = 0.20 mol

1.0L= 0.20 M

Step 3: Let x mol/L be the equilibrium concentration of CO2. At equilib-

rium, the concentrations of CaO and CO2will both be x mol/L.

Step 4: Substitute the initial and equilibrium concentrations into the equi-

librium expression:

3.0 = x2

0.20

Step 5: Solve for x, the equilibrium concentration of CO2:

x=√3.0×0.20 = √0.60 ≈0.77 M

Therefore, the equilibrium concentration of carbon dioxide, CO2, when the

system is heated to increase the temperature is approximately 0.77 M.

Question 19

Question

A reaction mixture initially contains 0.20 M NO, 0.40 M H2, and 0.10 M H2O.

The following equilibrium is established:

2 NO(g) + 2 H2(g)⇌N2(g) + 2 H2O(g)

If the concentration of N2is found to be 0.05 M at equilibrium, calculate

the equilibrium constant, Kc, for the reaction.

Solution

Step 1: Write the expression for the equilibrium constant, Kc, for the reaction.

Kc=[N2][H2O]2

[NO]2[H2]2

Step 2: Given that the concentration of N2is 0.05 M at equilibrium, we plug

in this value into the equilibrium constant expression.

0.05 = (0.10)2

(0.20)2(0.40)2

Step 3: Simplify the expression and solve for Kc.

0.05 = 0.01

0.16(0.16)

0.05 = 0.01

0.0256

0.05 = 0.390625 ×10−3

Therefore, the equilibrium constant, Kc, for the reaction is 0.390625 ×10−3.

Question 20

Question

A closed vessel initially contains a gaseous mixture of 0.1 mol of NO, 0.2 mol of

H2, and 0.3 mol of H2O at equilibrium:

2NO(g)+2H2(g)⇌2NH3(g) + H2O(g)

If the volume of the vessel is suddenly increased, predict the direction in which

the equilibrium shifts and explain why.

Solution

Step 1: Write the expression for the equilibrium constant (Kc) for the given

reaction.

Kc=[NH3]2[H2O]

[NO]2[H2]2

Step 2: Analyze the effect of increasing the volume of the vessel on the

equilibrium. - When the volume of the vessel is increased, the total pressure in

the container decreases. - According to Le Chatelier’s principle, the equilibrium

will shift in a direction that counteracts the change.

Step 3: Determine the impact of the volume increase on the equilibrium. -

Increasing the volume of the vessel will lead to a decrease in pressure. - The

side of the reaction with the greater number of gas molecules will be favored to

increase pressure.

Step 4: Count the number of gas molecules on each side of the reaction. -

Left side: 2 moles of NO gas + 2 moles of H2gas = 4 moles of gas - Right side:

2 moles of NH3gas + 1 mole of H2O gas = 3 moles of gas

Step 5: Determine the direction in which the equilibrium will shift. - As the

total pressure decreases (pressure increases on the right side), the equilibrium

will shift to the side with a higher number of gas molecules. - Therefore, the

equilibrium will shift to the left to increase the total pressure inside the vessel.

Step 6: Conclusion The equilibrium will shift to the left (towards the reac-

tants) to counteract the increase in volume, favoring the formation of more NO

and H2gas molecules.

Question 21

Question

A reaction has an equilibrium constant, Kc, of 0.25 at a certain temperature. If

the concentration of one of the products is increased by a factor of 2, how will

the equilibrium shift? Justify your answer.

Solution

Step 1: Write the balanced chemical reaction and the expression for the equi-

librium constant. Let’s assume the reaction is:

aA +bB ⇌cC +dD

The equilibrium constant expression for this reaction would be:

Kc =[C]c[D]d

[A]a[B]b

Step 2: Given that the equilibrium constant, Kc, is 0.25, we have:

Kc = 0.25 = [C]c[D]d

[A]a[B]b

Step 3: If the concentration of one of the products, let’s say [C], is increased

by a factor of 2, then the new equilibrium constant expression can be written

as:

Kc′=(2[C′])c[D]d

[A]a[B]b= 2c([C]c[D]d

[A]a[B]b)

Step 4: Since the equilibrium constant, Kc, does not change with tempera-

ture, Kc’ must equal Kc. Thus, we have:

2c×0.25 = 0.25

Step 5: Solving the equation, we find:

2c= 1

c= 0

Step 6: The equilibrium will shift to the right if c is positive, to the left if

c is negative, and remain unchanged if c is zero. In this case, since c is 0, the

equilibrium will remain unchanged.

Question 22

Question

For the following reaction at equilibrium, predict the shift in equilibrium when

the volume of the container is decreased:

CO(g)+3H2(g)⇌CH4(g) + H2O(g)

Solution

1. Determine the initial reaction quotient, Q, using the given equilibrium con-

centrations/volumes. Let’s assume the reaction at equilibrium has the following

concentrations: [CO] = 0.1M, [H2] = 0.2M, [CH4] = 0.5M, and [H2O] = 0.3M.

2. Calculate Qby substituting the initial concentrations into the expression

for Q:

Q=[CH4][H2O]

[CO][H2]3

3. Substituting the values, we get:

Q=(0.5)(0.3)

(0.1)(0.2)3= 37.5

4. Determine the equilibrium constant, K, for the reaction. Let’s assume

K= 50.

5. Compare Qand Kto determine the direction of the reaction. Since

Q < K, the reaction will shift to the right to reach equilibrium.

6. When the volume of the container is decreased, the system will try to

counteract the change. According to Le Chatelier’s principle, the system will

shift in the direction that reduces the total number of gas molecules (moles) to

help relieve the increased pressure.

7. In this reaction, the total number of moles on the left side of the reaction

is 4 (1 mole of CO and 3 moles of H2) while on the right side, the total number

of moles is 2 (1 mole of CH4and 1 mole of H2O).

8. Therefore, by decreasing the volume of the container, the system will shift

to the right (towards the side with fewer moles) to reduce the pressure.

Thus, the equilibrium will shift to the right when the volume of the container

is decreased for the given reaction.

Question 23

Question

For the reaction:

2A +2B ⇌C+D

If the concentration of reactants A and B are both increased, predict the effect

on the equilibrium position of the reaction. Justify your answer.

Solution

1. When the concentration of reactants A and B are both increased, accord-

ing to Le Chatelier’s principle, the equilibrium will shift in a direction

that reduces the concentrations of A and B, and favors the formation of

products C and D.

2. The increase in the concentrations of A and B will cause the reaction to

shift to the right in order to consume some of the additional reactants.

3. Consequently, the concentrations of C and D will increase, while the con-

centrations of A and B will decrease until a new equilibrium is established.

This new equilibrium will have higher concentrations of both C and D

compared to the initial equilibrium.

Question 24

Question

For the following reaction at equilibrium:

2A(g) + B(g)⇌3C(g)

If the concentration of A is increased, predict the direction in which the equi-

librium will shift and explain your reasoning. Assume no volume or temperature

changes occur.

Solution

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

chemical equation for the given reaction is:

2A(g) + B(g)⇌3C(g)

Step 2: Determine the effect of increasing the concentration of A. If the

concentration of A is increased, according to Le Chatelier’s principle, the equi-

librium will shift to counteract this change. This means the reaction will shift

in the direction that consumes A to alleviate the increase.

Step 3: Predict the direction in which the equilibrium will shift. Since the

forward reaction consumes A and the reverse reaction produces A, an increase

in the concentration of A will cause the equilibrium to shift to the right (towards

C) to consume the excess A.

Therefore, the equilibrium will shift to the right (towards the products) when

the concentration of A is increased.

Question 25

Question

A student is performing an experiment to study the equilibrium reaction:

N2O4(g)⇌2NO2(g)

Initially, the student adds 0.15 mol of N2O4to a 1.0 L flask at 25°C. The

equilibrium constant, Kc, for the reaction is 0.050. If the student also adds

0.10 mol of NO2to the flask at equilibrium, calculate the new equilibrium

concentrations of N2O4and NO2.

Solution

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

reaction.

Kc=[NO2]2

[N2O4]= 0.050

Step 2: Define the initial amounts and changes in concentration for the

reactants and products. - Initial: [N2O4]0= 0.15 mol/L, [NO2]0= 0 mol/L -

Change: −2x(as 2 moles of NO2are formed for every mole of N2O4consumed)

- New equilibrium: [N2O4] = 0.15 −2x,[NO2] = 0 + 2x

Step 3: Substitute the expressions for equilibrium concentrations into the

equilibrium constant expression.

Kc=(2x)2

0.15 −2x= 0.050

Step 4: Solve for x.

4x2= 0.05(0.15 −2x)

4x2= 0.0075 −0.1x

4x2+ 0.1x−0.0075 = 0

Step 5: Solve the quadratic equation using the quadratic formula.

x=−b±√b2−4ac

Step 6: Plug in the values a= 4,b= 0.1, and c=−0.0075 into the formula

to get the values of x.

Step 7: Calculate the new equilibrium concentrations of N2O4and NO2.

[N2O4] = 0.15 −2x

[NO2] = 2x

Step 8: Plug in the calculated value of xto find the new equilibrium con-

centrations.

Step 3: Compare Qc and Kc to determine the direction of the equilibrium

shift.

Since Qc = 0.69 and Kc = 0.34, Qc is larger than Kc. This means that the

system is not at equilibrium.

Step 4: Predict the direction of the equilibrium shift.

Since Qc is larger than Kc, the reaction will shift to the left to establish a

new equilibrium.

Therefore, with the increased concentration of N2, the equilibrium will shift

towards the reactants N2and H2.

Question 2

Question

A gaseous equilibrium is established in the reaction below at a certain temper-

ature:

2 NOCl(g) ⇌2 NO(g) +Cl2(g)

When the pressure of NOCl is increased, predict how the equilibrium will

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

Solution

Step 1: Initially, in the reaction system, there are 2 moles of NOCl for every 2

moles of NO and 1 mole of Cl2.

Step 2: When the pressure of NOCl is increased, Le Chatelier’s principle

states that the system will shift to counteract this change. Since the reaction

produces 2 moles of gas for every 1 mole of NOCl consumed, increasing the

pressure will cause the system to shift to the side with fewer moles of gas to

alleviate the pressure increase.

Step 3: In this case, the forward reaction results in the production of 3 moles

of gas (2 moles of NO and 1 mole of Cl2) for every 2 moles of NOCl consumed.

Therefore, the system will shift to the right to decrease the total number of gas

molecules and reduce the pressure.

Step 4: As a result, increasing the pressure of NOCl will cause the equilib-

rium to shift towards the products, favoring the formation of NO and Cl2.

Step 5: In conclusion, an increase in the pressure of NOCl will cause the

equilibrium to shift to the right, favoring the formation of NO and Cl2according

to Le Chatelier’s principle.

Question 3

Question

Consider the following equilibrium reaction:

N2O4(g)⇌2NO2(g)

If the initial concentration of N2O4is 0.20 M, and the equilibrium constant,

Kc, for the reaction at a certain temperature is 0.050, what will be the concen-

tration of NO2at equilibrium?

Solution

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

Kc=[NO2]2

[N2O4]= 0.050

Step 2: Since the initial concentration of N2O4,[N2O4]initial = 0.20 M,

and assuming the concentration of NO2at the beginning is 0M, we denote

[NO2]initial = 0 M and [NO2]at equilibrium as x.

Step 3: Set up an ICE table (Initial, Change, Equilibrium) to represent the

changes in concentration:

Species N2O4NO2

Initial (M) 0.20 0

Change (M) −x+2x

Equilibrium (M) 0.20 −x2x

Step 4: Substitute the equilibrium concentrations into the equilibrium con-

stant expression and solve for x:

Kc=(2x)2

0.20 −x= 0.050

4x2= 0.050(0.20 −x)

4x2= 0.01 −0.05x

4x2+ 0.05x−0.01 = 0

Step 5: Solve the quadratic equation for xusing the quadratic formula:

x=−0.05 ±√0.052−4(4)(−0.01)

2(4)

x=−0.05 ±√0.0025 + 0.16

x=−0.05 ±√0.1625

x=−0.05 ±0.4034

Step 6: Calculate the possible values for x:

x=0.3534

8≈0.0442 M (rejected) or x=−0.4534

8≈ −0.0567 M

Since concentration cannot be negative, the concentration of NO2at equi-

librium is approximately 0.044 M.