CHEM 321 - ANALYTICAL CHEMISTRY Le Chatelier's principle and equilibrium shifts Question Bank Set 10

CHEM 321 - ANALYTICAL

CHEMISTRY - Le Chatelier’s principle

and equilibrium shifts

Question Bank - Set 10

Liberty University

Question 1

Question

A gaseous reaction involving nitrogen dioxide is at equilibrium according to the

following equation:

2NO2(g)⇌2NO(g) + O2(g)

If the pressure of nitrogen dioxide is increased, predict how the equilibrium

will shift and explain why.

Solution

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

for the reaction is:

2NO2(g)⇌2NO(g) + O2(g)

Step 2: Analyze the effect of increasing the pressure of nitrogen dioxide.

Increasing the pressure of nitrogen dioxide will cause the equilibrium to shift in

a direction that reduces the total number of gas moles to relieve the stress of

increased pressure.

Step 3: Determine the direction of the equilibrium shift. In this reaction,

the total number of gas moles decreases from 3 moles on the left side to 2 moles

on the right side. Therefore, increasing the pressure will cause the equilibrium

to shift to the right to reduce the total number of moles of gas.

Step 4: Explain the equilibrium shift. By shifting to the right, the equilib-

rium will consume more nitrogen dioxide and produce more nitrogen monoxide

and oxygen gas to decrease the total number of gas moles and relieve the pres-

sure increase.

Step 5: Conclusion Increasing the pressure of nitrogen dioxide will cause the

equilibrium to shift to the right, favoring the formation of nitrogen monoxide

and oxygen gas. This shift is in accordance with Le Chatelier’s principle, which

states that a system at equilibrium will adjust in response to a stress applied to

it.

Question 2

Question

Consider the reaction:

2CO(g) + O2(g)⇌2CO2(g)

If the pressure is increased in the system at equilibrium, predict the direction

in which the equilibrium will shift and explain why using Le Chatelier’s principle.

Solution

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

subjected to a stress (such as a change in pressure), the system will shift the

equilibrium position to counteract the imposition of the stress.

Step 2: In this reaction, the total number of gas molecules on the left side of

the equation is 3 (2CO + O2), while on the right side, there are 2 gas molecules

(2CO2).

Step 3: When the pressure is increased in the system, the system will shift

the equilibrium position to reduce the total number of gas molecules present.

Step 4: To do this, the system will shift towards the side with fewer gas

molecules. In this case, it will shift to the right (towards the products) to

consume some of the reactants and produce more products.

Step 5: Therefore, when the pressure is increased in the system, the equilib-

rium will shift to the right to decrease the total number of gas molecules and

alleviate the pressure increase.

Question 3

Question

In a closed system at equilibrium, the reaction N2(g) + 3H2(g)⇌2NH3(g)is

established. If the volume of the container is suddenly decreased, predict the

direction in which the equilibrium will shift and explain why.

Solution

Step 1: Write the expression for the equilibrium constant, Keq. The equilibrium

constant for the given reaction is defined as:

Keq =[NH3]2

[N2][H2]3

Step 2: Predict the direction of the equilibrium shift. When the pressure

is increased by decreasing the volume, the system will respond by shifting the

equilibrium to reduce the pressure. Since there are four moles of gas on the left

side and two moles of gas on the right side of the reaction, the equilibrium will

shift to the side with fewer moles of gas to decrease the pressure. Therefore, the

equilibrium will shift to the left, favoring the reactants.

Step 3: Justify the equilibrium shift. Shifting the equilibrium to the left

will decrease the concentration of ammonia and increase the concentrations of

nitrogen and hydrogen, thereby decreasing the overall pressure in the system.

This shift is in accordance with Le Chatelier’s principle, which states that a

system at equilibrium will respond to a stress by shifting in a direction that

reduces the stress.

Therefore, in response to the decrease in volume (and increase in pressure),

the system will shift to the left to relieve the pressure by favoring the reactants.

Question 4

Question

Consider the following equilibrium reaction:

2SO2(g) + O2(g)⇌2SO3(g)

If an external pressure is applied to the system at equilibrium, predict how

the position of the equilibrium will shift. Justify your answer.

Solution

To determine how the position of the equilibrium will shift when an external

pressure is applied to the system, we can consider the effect of the pressure on

the number of moles of gas on each side of the reaction. Le Chatelier’s principle

states that when a system at equilibrium is subjected to a stress, the system

shifts to relieve that stress and establish a new equilibrium.

Step 1: Determine the change in the number of moles of gas on each side

of the reaction.

In the given reaction, the number of moles of gas before external pressure is

applied: - Left side: 2SO2+O2→3moles of gas - Right side: 2SO3→2moles

of gas

Step 2: Analyze the effect of increased pressure on the equilibrium.

When an external pressure is applied to the system, the equilibrium will

shift to minimize the total number of moles of gas. In this case, an increase in

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

to decrease the pressure.

Step 3: Predict the direction of the equilibrium shift.

Since the left side of the reaction has more moles of gas, the equilibrium

will shift in the direction of the right side where there are fewer moles of gas.

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

formation of more SO3gas.

Thus, when an external pressure is applied to the system at equilibrium, the

position of the equilibrium will shift to the right in order to decrease the total

moles of gas and relieve the pressure.

Question 5

Question

For the reaction N2O4(g)⇌2NO2(g)at a certain temperature, the equilibrium

constant Kcis 0.056. If the initial concentration of N2O4is 0.30 M, calculate

the equilibrium concentrations of both N2O4and NO2.

Solution

Step 1: Write the expression for the equilibrium constant Kcusing the concen-

trations of products and reactants.

Kc=[NO2]2

[N2O4]= 0.056

Step 2: Substitute the initial concentration of N2O4into the equilibrium

constant expression and simplify.

0.056 = (2x)2

0.30 −x

Step 3: Expand and solve the equation for x.

0.056 = 4x2

0.30 −x

0.056(0.30 −x) = 4x2

0.0168 −0.056x= 4x2

4x2+ 0.056x−0.0168 = 0

Step 4: Solve the quadratic equation to find the value of x, which represents

the change in concentration of N2O4.

x=−0.056 ±√(0.056)2−4(4)(−0.0168)

2(4)

Step 5: After finding the value of x, calculate the equilibrium concentrations

of N2O4and NO2.

[N2O4]eq = 0.30 −x

[NO2]eq = 2x

Question 6

Question

For the reaction:

2 A(g) + B(g)⇌C(g)

The equilibrium constant, Kc, is 0.50. If the initial concentrations of A,

B, and C are all 1.00 M, what will be the concentrations of A, B, and C at

equilibrium if the volume of the system is doubled at constant temperature?

Solution

Step 1: Write the equilibrium expression for the reaction:

Kc=[C]1

[A]2[B]1= 0.50

Step 2: Calculate the initial Qc (reaction quotient) using the initial concen-

trations of A, B, and C:

Qc=[1]1

[1]2[1]1= 1

Step 3: Compare Qc to Kc to determine the direction in which the reaction

must shift in order to reach equilibrium. Since Qc > Kc, the system must shift

to the right to form more products.

Step 4: If the volume of the system is doubled at constant temperature,

the total pressure remains constant. According to Le Chatelier’s principle, an

increase in volume favors the side of the reaction with more gas molecules. In

this case, the right side of the reaction has 1 mole of gas while the left side has

2 moles of gas. Therefore, the equilibrium will shift to the right to counteract

the volume increase.

Step 5: Let xbe the change in concentration for C. At equilibrium, the

concentrations will be:

[A] = 1.00 −2x

[B] = 1.00 −x

[C] = x

Step 6: Substitute the equilibrium concentrations into the equilibrium ex-

pression and solve for x:

0.50 = x

(1.00 −2x)2(1.00 −x)

Step 7: After solving for x, substitute back into the equilibrium expressions

for A, B, and C to find their concentrations at equilibrium.

Question 7

Question

Consider the following equilibrium reaction:

3A(g) + B(g) ⇌2C(g) + D(g)

If the concentration of substance Ais decreased, predict how the equilibrium

will shift. Justify your answer.

Solution

Step 1: Write the equilibrium expression for the reaction:

K=[C]2[D]

[A]3[B]

Step 2: Understand Le Chatelier’s principle which states that when a system

at equilibrium is subjected to a stress, the system will adjust in a way that

minimizes the effect of that stress.

Step 3: If the concentration of substance Ais decreased, the system will try

to counteract this change by shifting the equilibrium position to the side that

produces more of substance A. This means the reaction will shift to the left

(towards A).

Step 4: This can be further explained with Le Chatelier’s principle as follows:

- When [A]decreases, the system will try to increase it by moving to the left,

i.e., towards reactants. - In order to increase [A],[B]will also increase because

of the stoichiometry of the reaction. - Consequently, [C]and [D]will decrease

as the reaction shifts left.

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

more substance A.

Question 8

Question

Consider the following reaction at equilibrium:

2 CO(g) + O2(g)⇌2 CO2(g)

If more CO gas is added to the system at constant temperature and pressure,

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, K.

The balanced chemical equation is:

2 CO(g) + O2(g)⇌2 CO2(g)

The equilibrium constant expression is given by:

K=[CO2]2

[CO]2[O2]

Step 2: Determine the initial reaction quotient (Q) and compare it to the

equilibrium constant (K) to predict the direction of the equilibrium shift.

Initially, when more CO is added, the concentration of CO will increase

while the concentrations of CO2 and O2 will remain constant. Therefore, the

reaction quotient (Q) will increase.

Since Q will be greater than Kafter more CO is added, the system will shift

to the left to decrease Q and reach a new equilibrium.

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

CO and O2 from CO2.

Question 9

Question

For the reaction 2SO2(g) + O2(g)⇌2SO3(g)at equilibrium at a certain tem-

perature, the concentration of SO2is increased. Predict the direction in which

the equilibrium will shift and explain your reasoning.

Solution

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

2SO2(g) + O2(g)⇌2SO3(g)

Step 2: Le Chatelier’s principle states that if a system at equilibrium is

disturbed by a change in temperature, pressure, or the concentration of a com-

ponent, the system will shift its equilibrium position so as to counteract the

effect of the disturbance.

Step 3: In this reaction, if the concentration of SO2is increased, the sys-

tem will try to counteract this change by shifting the equilibrium position in a

direction that uses up SO2and produces more SO3.

Step 4: Therefore, the equilibrium will shift to the right to produce more

SO3.

Step 5: The equilibrium expression for this reaction is given by:

K=[SO3]2

[SO2]2[O2]

Step 6: By increasing the concentration of SO2, the concentrations of SO3

and O2will also increase due to the right shift of the equilibrium.

Step 7: Overall, the equilibrium position will shift to the right, favoring the

formation of more SO3until a new equilibrium is reached.

Question 10

Question

For the reaction N2(g) + 3H2(g)⇌2NH3(g), the equilibrium constant, Kc, is

7.24 ×10−3at a certain temperature. If 0.60 moles of N2, 1.80 moles of H2, and

1.20 moles of NH3are placed in a 2.00 L container at equilibrium, calculate the

equilibrium concentrations of all species.

Solution

Step 1: Write the expression for Kc:

Kc=[NH3]2

[N2][H2]3

Step 2: Substitute the given values into the expression for Kc:

7.24 ×10−3=(1.20)2

(0.60)(1.80)3

Step 3: Solve for the equilibrium concentrations of NH3:

(1.20)2= (0.60)(1.80)3×7.24 ×10−3

1.44 = (0.60)(1.8)3×7.24 ×10−3

1.44 = (0.60)(5.832) ×10−3

1.44 = 3.4992 ×10−3

Step 4: Calculate the equilibrium concentrations of NH3:

[NH3] = √3.4992 ×10−3

[NH3] = 0.0592

Step 5: Calculate the equilibrium concentrations of N2and H2:

[N2] = 0.60 −(1.20 ×2) = −1.80

[H2] = 1.80 −(1.20 ×3) = −2.40

Step 6: Since concentrations cannot be negative, all equilibrium concentra-

tions are zero.

Therefore, at equilibrium, the concentrations of N2, H2, and NH3are 0.00

mol/L.

Question 11

Question

For the reaction: C(s) + H2O(g) ⇋CO(g) + H2(g), which way will the equilib-

rium shift if the pressure of the system is increased by decreasing the volume?

Solution

Step 1: Identify the effect of the volume change on the system’s pressure.

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

will increase.

Step 2: Determine the change in the number of gas moles on each side of

the reaction.

Initially, there are 1 mole of gas on the reactant side (C(s)) and 1 mole of

gas on the product side (CO(g) +H2(g)).

Step 3: Apply Le Chatelier’s principle to predict the direction of the equi-

librium shift.

Since the pressure of the system is increased by decreasing the volume, the

system will respond by shifting the equilibrium to minimize the impact of the

pressure change.

According to Le Chatelier’s principle, the equilibrium will shift in the direc-

tion that reduces the total number of moles of gas.

Step 4: Determine which side of the reaction has fewer gas moles.

On the reactant side, there is 1 mole of gas, and on the product side, there

are 2 moles of gas.

Step 5: Predict the direction of the equilibrium shift.

To minimize the increase in pressure caused by the decrease in volume, the

equilibrium will shift to the side with fewer gas moles. Therefore, the equilibrium

will shift to the left (towards the reactants).

Therefore, when the volume is decreased, causing an increase in pressure,

the equilibrium will shift to the left: C(s) + H2O(g) ⇋CO(g) + H2(g).

Question 12

Question

A gaseous equilibrium is established according to the reaction:

2NO(g)+O2(g)⇌2NO2(g)

For this reaction, the equilibrium constant, Kc, is 8.00 ×104at a certain

temperature.

If the pressure of NO(g)is increased, predict the direction in which the

equilibrium will shift and explain why.

Solution

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

reaction:

Kc=[NO2]2

[NO]2[O2]= 8.00 ×104

Step 2: Assume the initial concentrations of NO(g),O2(g), and NO2(g)are

[NO]0,[O2]0, and [NO2]0respectively.

Step 3: When the pressure of N O(g)is increased (and thus its concentration),

the system will try to counteract this change by shifting the equilibrium position

in a direction that reduces the pressure of NO(g).

Step 4: Since the forward reaction results in a decrease in the number of

moles of gas (2NO(g)+O2(g)→2NO2(g)), the equilibrium will shift to the right

to consume more NO(g)and O2(g).

Step 5: This means the concentration of NO(g)will decrease while the con-

centrations of O2(g)and NO2(g)will increase.

Step 6: Consequently, the equilibrium constant Kcwill remain constant since

the temperature is held constant, but the concentrations of the species involved

will change to favor the reverse reaction.

Therefore, the equilibrium will shift to the right to consume more NO(g)

and O2(g)when the pressure of NO(g)is increased.

Question 13

Question

For the reaction:

2 NOBr (g) ⇌2 NO (g) +Br2(g)

which is at equilibrium, if the temperature is increased, predict how the

equilibrium will shift. Justify your answer.

Solution

Step 1: Le Chatelier’s Principle states that if a system at equilibrium is dis-

turbed, the system will shift to counteract the disturbance.

Step 2: When the temperature of a system at equilibrium is increased, the

system will shift in the direction of the endothermic reaction to absorb the

excess heat.

Step 3: In the given reaction, the forward reaction is endothermic (since it

is absorbing energy to break the bonds in NOBr).

Step 4: Therefore, if the temperature is increased, the equilibrium will shift

to the right (towards the products) to counteract the increase in temperature.

Step 5: Therefore, an increase in temperature will result in an increase in

the concentrations of NO and Br2while decreasing the concentration of NOBr.

Question 14

Question

A reaction mixture initially contains equal concentrations of A and B, which

react to form products C and D according to the equation:

A+B⇌C+D

If C is removed from the reaction mixture, predict the direction in which the

equilibrium will shift. Justify your answer.

Solution

Step 1: When C is removed, the equilibrium will shift to counteract the loss of

C. This can be explained using Le Chatelier’s principle.

Step 2: In the given reaction, C is on the product side. By removing C from

the mixture, the equilibrium will shift to produce more C in order to replace

what was lost.

Step 3: To produce more C, the reaction must favor the forward reaction,

which is the formation of C and D from A and B.

Step 4: Therefore, the equilibrium will shift to the right, increasing the

concentration of C and D in the reaction mixture.

Step 5: In conclusion, when C is removed from the reaction mixture, the

equilibrium will shift towards the products, favoring the formation of C and D.

Question 15

Question

For the reaction N2O4(g)⇌2NO2(g)at equilibrium, if the pressure of N2O4is

increased, predict the direction in which the equilibrium will shift and explain

your reasoning.

Solution

Step 1: Write the balanced equilibrium equation for the reaction: The balanced

equilibrium equation for the reaction is:

N2O4(g)⇌2NO2(g)

Step 2: Define the equilibrium expression: The equilibrium expression for

the reaction is:

K=[NO2]2

[N2O4]

Step 3: Analyze the effect of increasing the pressure of N2O4: According

to Le Chatelier’s principle, if the pressure of one of the reactants or products

is increased, the equilibrium will shift in the direction that reduces the total

pressure. In this case, increasing the pressure of N2O4will cause the equilibrium

to shift towards the right to decrease the pressure by consuming some of the

N2O4gas.

Step 4: Predict the direction in which the equilibrium will shift: Therefore,

if the pressure of N2O4is increased, the equilibrium will shift to the right to

decrease the pressure.

Step 5: Justification: By consuming some of the N2O4gas and producing

more NO2, the total number of gas molecules will decrease, thus reducing the

total pressure and bringing the system back to equilibrium.

Question 16

Question

A reaction mixture initially contains 0.20 M of NO2and 0.30 M of O2. The

following equilibrium is established:

2NO2(g) + O2(g)⇌2NO2(g)

If the equilibrium constant, Kc, for the reaction at a certain temperature is

2.50 ×103, calculate the new equilibrium concentration of NO2if the concen-

tration of O2is increased to 0.40 M.

Solution

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

equilibrium constant, Kc. The balanced equilibrium equation is:

2NO2(g) + O2(g)⇌2NO2(g)

The expression for the equilibrium constant, Kc, is given by:

Kc=[NO2]2

[NO2]2[O2]=[NO2]2

[O2]

Step 2: Calculate the initial value of Qc. Given initial concentrations:

[NO2]initial = 0.20 M and [O2]initial = 0.30 M

Substitute the initial concentrations into the expression for Qc:

Qc=0.202

0.30 = 0.1333

Step 3: Determine the direction of the reaction shift. Since Qcis less than

Kc, the reaction will shift to the right to reach equilibrium.

Step 4: Determine the changes in concentrations. Let xrepresent the change

in concentration of NO2.

After the reaction reaches equilibrium: [NO2] = 0.20−2xand [O2] = 0.40−x

Step 5: Set up the equation using the equilibrium constant Kc. Substitute

the equilibrium concentrations into the expression for Kc:

2.50 ×103=(0.20 −2x)2

(0.40 −x)

Step 6: Solve for x.

2.50 ×103=(0.20 −2x)2

0.40 −x

2.50 ×103=(0.04 −0.80x+ 4x2)

0.40 −x

0.10 −2x= 4x2

4x2+ 2x−0.10 = 0

Using the quadratic formula to solve for x:

x=−2±√(−2)2−4(4)(−0.10)

2(4)

x≈0.0783

Step 7: Calculate the new equilibrium concentration of NO2.

[NO2]final = 0.20 −2(0.0783)

[NO2]final ≈0.0434 M

Therefore, the new equilibrium concentration of NO2is approximately 0.0434

M when the concentration of O2is increased to 0.40 M.

Question 17

Question

A reaction mixture initially contains 0.20 M nitrogen dioxide (NO2) and 0.10

M dinitrogen tetroxide (N2O4) at equilibrium according to the reaction:

2NO2(g)⇌N2O4(g)

If the concentration of NO2is increased to 0.30 M, predict the direction in

which the equilibrium will shift (towards the products, towards the reactants,

or no shift) and explain why.

Solution

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

reaction:

The equilibrium constant expression for the reaction 2NO2(g)⇌N2O4(g)

is:

Kc=[N2O4]2

[NO2]2

Step 2: Calculate the initial equilibrium constant, Kc,initial, using the initial

concentrations of NO2and N2O4:

Given: Initial concentration of NO2,[NO2]initial = 0.20 M Initial concentra-

tion of N2O4,[N2O4]initial = 0.10 M

Substitute the initial concentrations into the equilibrium constant expres-

sion:

Kc,initial =(0.10)2

(0.20)2= 0.25

Step 3: Determine the direction in which the equilibrium will shift when the

concentration of NO2is increased to 0.30 M:

Given: New concentration of NO2,[NO2]new = 0.30 M

Since the new [NO2]is greater than the initial [NO2], the reaction will shift

to consume some of the added NO2. This means that the equilibrium will

shift towards the products to alleviate the stress caused by the increase in NO2

concentration.

Therefore, the equilibrium will shift towards the products when the concen-

tration of NO2is increased to 0.30 M.

Question 18

Question

For the reaction 2NO(g) + O2(g)⇌2NO2(g)at equilibrium, if the concentra-

tion of O2is increased, predict how the equilibrium will shift and explain your

reasoning.

Solution

Step 1: Write down the balanced chemical equation and the equilibrium expres-

sion. The balanced chemical equation for the given reaction is:

2NO(g)+O2(g)⇌2NO2(g)

The equilibrium expression for this reaction is:

Kc=[NO2]2

[NO]2[O2]

Step 2: Determine the initial direction of the equilibrium shift. Adding more

O2will increase the concentration of a reactant. According to Le Chatelier’s

principle, when a reactant is added, the equilibrium will shift to the right to

consume the added reactant and re-establish equilibrium.

Step 3: Re-write the equilibrium expression with increased O2. Since O2

was added, the concentration of O2in the equilibrium expression will increase.

Step 4: Predict the shift in equilibrium. The equilibrium will shift to the

right to consume the added O2until a new equilibrium is established.

Therefore, an increase in the concentration of O2will cause the equilibrium

to shift to the right.

Question 19

Question

For the reaction

2A+ 3B⇌C+D,

the equilibrium constant, Kc, is 10. If 0.1 moles of A, 0.2 moles of B, and

0.3 moles of Care mixed in a 1 L flask, determine the direction in which the

reaction will proceed in order to establish equilibrium. Justify your answer.

Solution

Step 1: Calculate the initial concentrations of A,B,C, and D. Given: - Initial

moles of A: 0.1 mol - Initial moles of B: 0.2 mol - Initial moles of C: 0.3 mol -

Volume of the flask: 1 L

Calculate initial concentrations:

[A]0=0.1mol

1L= 0.1M

[B]0=0.2mol

1L= 0.2M

[C]0=0.3mol

1L= 0.3M

[D]0= 0 M

(since initially, no moles of Dare present)

Step 2: Calculate the reaction quotient, Qc, using the initial concentrations.

Qc=[C]0[C]0

[A]2

0[B]3

=0.3∗0

0.12∗0.23= 0

Step 3: Compare Qcand Kc. Since Qc< Kc, the reaction will proceed in

the forward direction (to the right) to establish equilibrium. This means more

products will be formed until Qcequals Kc.

Therefore, the reaction will proceed to the right to establish equilibrium.

Question 20

Question

For the reaction:

2SO2(g) + O2(g)⇌2SO3(g)

at equilibrium, an external pressure is applied to the system. Explain how

Le Chatelier’s principle can be used to predict the direction of the shift in

equilibrium and the effect on the concentrations of the reactants and products.

Solution

Step 1: According to Le Chatelier’s principle, when a stress is applied to a

system at equilibrium, the system will shift in a direction that minimizes the

effect of the stress.

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

in the direction that produces fewer molecules of gas to reduce the pressure.

Step 3: Since there are 3 moles of gas on the left side and only 2 moles of

gas on the right side, increasing the pressure would cause the system to shift to

the right to reduce the total number of gas molecules.

Step 4: As a result, the concentration of SO2and O2would decrease, while

the concentration of SO3would increase, returning the system to a new equi-

librium position.

Step 5: Therefore, applying external pressure would shift the equilibrium

towards the formation of more SO3, leading to a decrease in the concentrations

of SO2and O2and an increase in the concentration of SO3in order to relieve

the applied stress.

Question 21

Question

For the reaction

2SO2(g) + O2(g)⇌2SO3(g)

at equilibrium, the concentration of SO2is increased. Predict qualitatively, us-

ing Le Chatelier’s principle, what effect this change will have on the equilibrium

position of the reaction.

Solution

Step 1: According to Le Chatelier’s principle, if we increase the concentration of

a reactant, the equilibrium will shift to the right to consume some of the added

reactant.

Step 2: In this reaction, increasing the concentration of SO2will cause the

system to react by consuming some of the added SO2.

Step 3: Consequently, the equilibrium position of the reaction will shift to

the right in order to produce more SO3and consume some of the excess SO2

added.

Therefore, the equilibrium position of the reaction

2SO2(g) + O2(g)⇌2SO3(g)

will shift to the right when the concentration of SO2is increased based on Le

Chatelier’s principle.

Question 22

Question

For the reaction represented by the equation below, predict whether an increase

in pressure will shift the equilibrium to the left, to the right, or will not shift

the equilibrium:

2 SO2(g) + O2(g)⇌2SO3(g)

Solution

Step 1: First, let’s determine the effect of an increase in pressure on the reaction.

An increase in pressure will shift the equilibrium in the direction that reduces

the total number of gas molecules (moles) involved in the reaction.

Step 2: Examine the balanced equation to determine the total number of

gas molecules on each side of the equation:

- Left side: 2 moles of SO� gas + 1 mole of O� gas = 3 moles of gas - Right

side: 2 moles of SO� gas = 2 moles of gas

Step 3: Compare the total number of gas molecules on each side of the

equation.

Since there are more gas molecules on the left side of the equation, an increase

in pressure will cause the system to shift in the direction that reduces the total

number of gas molecules.

Step 4: Therefore, an increase in pressure will shift the equilibrium to the

right to form more SO� gas in order to reduce the total number of gas molecules

and relieve the pressure increase.

Question 23

Question

For the reaction:

2NOBr(g)⇌2N O(g) + Br2(g)

, which of the following changes will shift the equilibrium to the right: a) Increas-

ing the temperature, b) Decreasing the volume of the container, or c) Adding

more NOBr gas to the container? Justify your answer.

Solution

Step 1: Let’s analyze each possible change: a) Increasing the temperature will

favor an endothermic reaction to absorb the extra heat. The reaction in ques-

tion is endothermic because it involves breaking bonds. Thus, increasing the

temperature will shift the equilibrium to the right. b) Decreasing the volume

of the container will favor the side of the reaction with fewer gas molecules to

reduce the pressure. In this case, the right side of the reaction has fewer gas

molecules (1 mol of Br2 gas vs. 2 mol of NOBr gas), so decreasing the volume

will shift the equilibrium to the right. c) Adding more NOBr gas to the con-

tainer will increase the concentration of the reactant NOBr. According to Le

Chatelier’s principle, an increase in reactants will shift the equilibrium to the

right to offset the change and consume some of the added NOBr.

Step 2: Conclusion: Both increasing the temperature and decreasing the

volume of the container will shift the equilibrium to the right. However, adding

more NOBr gas to the container will not shift the equilibrium to the right.

Question 24

Question

For the reaction

2NOCl(g)⇌2NO(g) + Cl2(g)

at equilibrium, increasing the pressure by decreasing the volume of the container

will result in which of the following changes to the system: I. Increase in the

concentration of NOCl II. Decrease in the concentration of NO III. Increase in

the concentration of Cl2

Solution

Step 1: First, we need to determine the effect of increasing the pressure by

decreasing the volume of the container on the reaction system. According to

Le Chatelier’s principle, when the pressure is increased, the system will shift

towards the side with fewer moles of gas molecules to counteract the increase in

pressure.

Step 2: Analyzing the given reaction, we can see that there are 3 moles of

gas molecules on the left side (2NOCl) and 3 moles of gas molecules on the

right side (2NO +Cl2). Therefore, changing the pressure will not cause the

system to shift to either side to reduce the number of gas molecules.

Step 3: Since there is no net effect on the total gas molecules in the system,

the concentrations of NOCl, NO, and Cl2will remain unchanged when the

pressure is increased by decreasing the volume of the container.

Step 4: Therefore, the correct answers are: I. No change in the concentra-

tion of NOCl II. No change in the concentration of NO III. No change in the

concentration of Cl2

Question 25

Question

A gaseous equilibrium is established at 300 K with a total pressure of 2.50 atm.

The reaction is described by the equation:

2A(g) + B(g) � 3C(g)

The equilibrium concentrations are found to be [A] = 0.40 M, [B] = 0.60 M,

and [C] = 1.20 M. Determine the value of the equilibrium constant Kpfor this

reaction at 300 K.

Solution

Step 1: Write the expression for the equilibrium constant Kpin terms of the

partial pressures:

Kp=(PC)3

(PA)2·(PB)

Step 2: Calculate the partial pressures of each gas using the ideal gas law

P V =nRT :

For A:

PA= (0.40 M)·(0.0821 L·atm/mol ·K)·300 K= 0.984 atm

For B:

PB= (0.60 M)·(0.0821 L·atm/mol ·K)·300 K= 1.476 atm

For C:

PC= (1.20 M)·(0.0821 L·atm/mol ·K)·300 K= 2.952 atm

Step 3: Substitute the partial pressures into the expression for Kpand solve

for its value:

Kp=(2.952)3

(0.984)2·(1.476)

Kp=24.80

1.39 ·1.48 ≈24.80

2.06 ≈12.04

Therefore, the equilibrium constant Kpfor the reaction at 300 K is approx-

imately 12.04 atm.

Step 5: Conclusion Increasing the pressure of nitrogen dioxide will cause the

equilibrium to shift to the right, favoring the formation of nitrogen monoxide

and oxygen gas. This shift is in accordance with Le Chatelier’s principle, which

states that a system at equilibrium will adjust in response to a stress applied to

it.

Question 2

Question

Consider the reaction:

2CO(g) + O2(g)⇌2CO2(g)

If the pressure is increased in the system at equilibrium, predict the direction

in which the equilibrium will shift and explain why using Le Chatelier’s principle.

Solution

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

subjected to a stress (such as a change in pressure), the system will shift the

equilibrium position to counteract the imposition of the stress.

Step 2: In this reaction, the total number of gas molecules on the left side of

the equation is 3 (2CO + O2), while on the right side, there are 2 gas molecules

(2CO2).

Step 3: When the pressure is increased in the system, the system will shift

the equilibrium position to reduce the total number of gas molecules present.

Step 4: To do this, the system will shift towards the side with fewer gas

molecules. In this case, it will shift to the right (towards the products) to

consume some of the reactants and produce more products.

Step 5: Therefore, when the pressure is increased in the system, the equilib-

rium will shift to the right to decrease the total number of gas molecules and

alleviate the pressure increase.

Question 3

Question

In a closed system at equilibrium, the reaction N2(g) + 3H2(g)⇌2NH3(g)is

established. If the volume of the container is suddenly decreased, predict the

direction in which the equilibrium will shift and explain why.

Solution

Step 1: Write the expression for the equilibrium constant, Keq. The equilibrium

constant for the given reaction is defined as:

Keq =[NH3]2

[N2][H2]3

Step 2: Predict the direction of the equilibrium shift. When the pressure

is increased by decreasing the volume, the system will respond by shifting the

equilibrium to reduce the pressure. Since there are four moles of gas on the left

side and two moles of gas on the right side of the reaction, the equilibrium will

shift to the side with fewer moles of gas to decrease the pressure. Therefore, the

equilibrium will shift to the left, favoring the reactants.

Step 3: Justify the equilibrium shift. Shifting the equilibrium to the left

will decrease the concentration of ammonia and increase the concentrations of

nitrogen and hydrogen, thereby decreasing the overall pressure in the system.

This shift is in accordance with Le Chatelier’s principle, which states that a

system at equilibrium will respond to a stress by shifting in a direction that

reduces the stress.

Therefore, in response to the decrease in volume (and increase in pressure),

the system will shift to the left to relieve the pressure by favoring the reactants.

Question 4

Question

Consider the following equilibrium reaction:

2SO2(g) + O2(g)⇌2SO3(g)

If an external pressure is applied to the system at equilibrium, predict how

the position of the equilibrium will shift. Justify your answer.

Solution

To determine how the position of the equilibrium will shift when an external

pressure is applied to the system, we can consider the effect of the pressure on

the number of moles of gas on each side of the reaction. Le Chatelier’s principle

states that when a system at equilibrium is subjected to a stress, the system

shifts to relieve that stress and establish a new equilibrium.

Step 1: Determine the change in the number of moles of gas on each side

of the reaction.

In the given reaction, the number of moles of gas before external pressure is

applied: - Left side: 2SO2+O2→3moles of gas - Right side: 2SO3→2moles

of gas

Step 2: Analyze the effect of increased pressure on the equilibrium.

When an external pressure is applied to the system, the equilibrium will

shift to minimize the total number of moles of gas. In this case, an increase in

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

to decrease the pressure.

Step 3: Predict the direction of the equilibrium shift.

Since the left side of the reaction has more moles of gas, the equilibrium

will shift in the direction of the right side where there are fewer moles of gas.

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

formation of more SO3gas.

Thus, when an external pressure is applied to the system at equilibrium, the

position of the equilibrium will shift to the right in order to decrease the total

moles of gas and relieve the pressure.

Question 5

Question

For the reaction N2O4(g)⇌2NO2(g)at a certain temperature, the equilibrium

constant Kcis 0.056. If the initial concentration of N2O4is 0.30 M, calculate

the equilibrium concentrations of both N2O4and NO2.

Solution

Step 1: Write the expression for the equilibrium constant Kcusing the concen-

trations of products and reactants.

Kc=[NO2]2

[N2O4]= 0.056

Step 2: Substitute the initial concentration of N2O4into the equilibrium

constant expression and simplify.

0.056 = (2x)2

0.30 −x

Step 3: Expand and solve the equation for x.

0.056 = 4x2

0.30 −x

0.056(0.30 −x) = 4x2

0.0168 −0.056x= 4x2

4x2+ 0.056x−0.0168 = 0

Step 4: Solve the quadratic equation to find the value of x, which represents

the change in concentration of N2O4.

x=−0.056 ±√(0.056)2−4(4)(−0.0168)

2(4)

Step 5: After finding the value of x, calculate the equilibrium concentrations

of N2O4and NO2.

[N2O4]eq = 0.30 −x

[NO2]eq = 2x

Question 6

Question

For the reaction:

2 A(g) + B(g)⇌C(g)

The equilibrium constant, Kc, is 0.50. If the initial concentrations of A,

B, and C are all 1.00 M, what will be the concentrations of A, B, and C at

equilibrium if the volume of the system is doubled at constant temperature?

Solution

Step 1: Write the equilibrium expression for the reaction:

Kc=[C]1

[A]2[B]1= 0.50

Step 2: Calculate the initial Qc (reaction quotient) using the initial concen-

trations of A, B, and C:

Qc=[1]1

[1]2[1]1= 1

Step 3: Compare Qc to Kc to determine the direction in which the reaction

must shift in order to reach equilibrium. Since Qc > Kc, the system must shift

to the right to form more products.

Step 4: If the volume of the system is doubled at constant temperature,

the total pressure remains constant. According to Le Chatelier’s principle, an

increase in volume favors the side of the reaction with more gas molecules. In

this case, the right side of the reaction has 1 mole of gas while the left side has

2 moles of gas. Therefore, the equilibrium will shift to the right to counteract

the volume increase.

Step 5: Let xbe the change in concentration for C. At equilibrium, the

concentrations will be:

[A] = 1.00 −2x

[B] = 1.00 −x

[C] = x

Step 6: Substitute the equilibrium concentrations into the equilibrium ex-

pression and solve for x:

0.50 = x

(1.00 −2x)2(1.00 −x)

Step 7: After solving for x, substitute back into the equilibrium expressions

for A, B, and C to find their concentrations at equilibrium.

Question 7

Question

Consider the following equilibrium reaction:

3A(g) + B(g) ⇌2C(g) + D(g)

If the concentration of substance Ais decreased, predict how the equilibrium

will shift. Justify your answer.

Solution

Step 1: Write the equilibrium expression for the reaction:

K=[C]2[D]

[A]3[B]

Step 2: Understand Le Chatelier’s principle which states that when a system

at equilibrium is subjected to a stress, the system will adjust in a way that

minimizes the effect of that stress.

Step 3: If the concentration of substance Ais decreased, the system will try

to counteract this change by shifting the equilibrium position to the side that

produces more of substance A. This means the reaction will shift to the left

(towards A).

Step 4: This can be further explained with Le Chatelier’s principle as follows:

- When [A]decreases, the system will try to increase it by moving to the left,

i.e., towards reactants. - In order to increase [A],[B]will also increase because

of the stoichiometry of the reaction. - Consequently, [C]and [D]will decrease

as the reaction shifts left.

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

more substance A.

Question 8

Question

Consider the following reaction at equilibrium:

2 CO(g) + O2(g)⇌2 CO2(g)

If more CO gas is added to the system at constant temperature and pressure,

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, K.

The balanced chemical equation is:

2 CO(g) + O2(g)⇌2 CO2(g)

The equilibrium constant expression is given by:

K=[CO2]2

[CO]2[O2]

Step 2: Determine the initial reaction quotient (Q) and compare it to the

equilibrium constant (K) to predict the direction of the equilibrium shift.

Initially, when more CO is added, the concentration of CO will increase

while the concentrations of CO2 and O2 will remain constant. Therefore, the

reaction quotient (Q) will increase.

Since Q will be greater than Kafter more CO is added, the system will shift

to the left to decrease Q and reach a new equilibrium.

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

CO and O2 from CO2.

Question 9

Question

For the reaction 2SO2(g) + O2(g)⇌2SO3(g)at equilibrium at a certain tem-

perature, the concentration of SO2is increased. Predict the direction in which

the equilibrium will shift and explain your reasoning.

Solution

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

2SO2(g) + O2(g)⇌2SO3(g)

Step 2: Le Chatelier’s principle states that if a system at equilibrium is

disturbed by a change in temperature, pressure, or the concentration of a com-

ponent, the system will shift its equilibrium position so as to counteract the

effect of the disturbance.

Step 3: In this reaction, if the concentration of SO2is increased, the sys-

tem will try to counteract this change by shifting the equilibrium position in a

direction that uses up SO2and produces more SO3.

Step 4: Therefore, the equilibrium will shift to the right to produce more

SO3.

Step 5: The equilibrium expression for this reaction is given by:

K=[SO3]2

[SO2]2[O2]

Step 6: By increasing the concentration of SO2, the concentrations of SO3

and O2will also increase due to the right shift of the equilibrium.

Step 7: Overall, the equilibrium position will shift to the right, favoring the

formation of more SO3until a new equilibrium is reached.

Question 10

Question

For the reaction N2(g) + 3H2(g)⇌2NH3(g), the equilibrium constant, Kc, is

7.24 ×10−3at a certain temperature. If 0.60 moles of N2, 1.80 moles of H2, and

1.20 moles of NH3are placed in a 2.00 L container at equilibrium, calculate the

equilibrium concentrations of all species.

Solution

Step 1: Write the expression for Kc:

Kc=[NH3]2

[N2][H2]3

Step 2: Substitute the given values into the expression for Kc:

7.24 ×10−3=(1.20)2

(0.60)(1.80)3

Step 3: Solve for the equilibrium concentrations of NH3:

(1.20)2= (0.60)(1.80)3×7.24 ×10−3

1.44 = (0.60)(1.8)3×7.24 ×10−3

1.44 = (0.60)(5.832) ×10−3

1.44 = 3.4992 ×10−3

Step 4: Calculate the equilibrium concentrations of NH3:

[NH3] = √3.4992 ×10−3

[NH3] = 0.0592

Step 5: Calculate the equilibrium concentrations of N2and H2:

[N2] = 0.60 −(1.20 ×2) = −1.80

[H2] = 1.80 −(1.20 ×3) = −2.40

Step 6: Since concentrations cannot be negative, all equilibrium concentra-

tions are zero.

Therefore, at equilibrium, the concentrations of N2, H2, and NH3are 0.00

mol/L.

Question 11

Question

For the reaction: C(s) + H2O(g) ⇋CO(g) + H2(g), which way will the equilib-

rium shift if the pressure of the system is increased by decreasing the volume?

Solution

Step 1: Identify the effect of the volume change on the system’s pressure.

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

will increase.

Step 2: Determine the change in the number of gas moles on each side of

the reaction.

Initially, there are 1 mole of gas on the reactant side (C(s)) and 1 mole of

gas on the product side (CO(g) +H2(g)).

Step 3: Apply Le Chatelier’s principle to predict the direction of the equi-

librium shift.

Since the pressure of the system is increased by decreasing the volume, the

system will respond by shifting the equilibrium to minimize the impact of the

pressure change.

According to Le Chatelier’s principle, the equilibrium will shift in the direc-

tion that reduces the total number of moles of gas.

Step 4: Determine which side of the reaction has fewer gas moles.

On the reactant side, there is 1 mole of gas, and on the product side, there

are 2 moles of gas.

Step 5: Predict the direction of the equilibrium shift.

To minimize the increase in pressure caused by the decrease in volume, the

equilibrium will shift to the side with fewer gas moles. Therefore, the equilibrium

will shift to the left (towards the reactants).

Therefore, when the volume is decreased, causing an increase in pressure,

the equilibrium will shift to the left: C(s) + H2O(g) ⇋CO(g) + H2(g).

Question 12

Question

A gaseous equilibrium is established according to the reaction:

2NO(g)+O2(g)⇌2NO2(g)

For this reaction, the equilibrium constant, Kc, is 8.00 ×104at a certain

temperature.

If the pressure of NO(g)is increased, predict the direction in which the

equilibrium will shift and explain why.

Solution

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

reaction:

Kc=[NO2]2

[NO]2[O2]= 8.00 ×104

Step 2: Assume the initial concentrations of NO(g),O2(g), and NO2(g)are

[NO]0,[O2]0, and [NO2]0respectively.

Step 3: When the pressure of N O(g)is increased (and thus its concentration),

the system will try to counteract this change by shifting the equilibrium position

in a direction that reduces the pressure of NO(g).

Step 4: Since the forward reaction results in a decrease in the number of

moles of gas (2NO(g)+O2(g)→2NO2(g)), the equilibrium will shift to the right

to consume more NO(g)and O2(g).

Step 5: This means the concentration of NO(g)will decrease while the con-

centrations of O2(g)and NO2(g)will increase.

Step 6: Consequently, the equilibrium constant Kcwill remain constant since

the temperature is held constant, but the concentrations of the species involved

will change to favor the reverse reaction.

Therefore, the equilibrium will shift to the right to consume more NO(g)

and O2(g)when the pressure of NO(g)is increased.

Question 13

Question

For the reaction:

2 NOBr (g) ⇌2 NO (g) +Br2(g)

which is at equilibrium, if the temperature is increased, predict how the

equilibrium will shift. Justify your answer.

Solution

Step 1: Le Chatelier’s Principle states that if a system at equilibrium is dis-

turbed, the system will shift to counteract the disturbance.

Step 2: When the temperature of a system at equilibrium is increased, the

system will shift in the direction of the endothermic reaction to absorb the

excess heat.

Step 3: In the given reaction, the forward reaction is endothermic (since it

is absorbing energy to break the bonds in NOBr).

Step 4: Therefore, if the temperature is increased, the equilibrium will shift

to the right (towards the products) to counteract the increase in temperature.

Step 5: Therefore, an increase in temperature will result in an increase in

the concentrations of NO and Br2while decreasing the concentration of NOBr.

Question 14

Question

A reaction mixture initially contains equal concentrations of A and B, which

react to form products C and D according to the equation:

A+B⇌C+D

If C is removed from the reaction mixture, predict the direction in which the

equilibrium will shift. Justify your answer.

Solution

Step 1: When C is removed, the equilibrium will shift to counteract the loss of

C. This can be explained using Le Chatelier’s principle.

Step 2: In the given reaction, C is on the product side. By removing C from

the mixture, the equilibrium will shift to produce more C in order to replace

what was lost.

Step 3: To produce more C, the reaction must favor the forward reaction,

which is the formation of C and D from A and B.

Step 4: Therefore, the equilibrium will shift to the right, increasing the

concentration of C and D in the reaction mixture.

Step 5: In conclusion, when C is removed from the reaction mixture, the

equilibrium will shift towards the products, favoring the formation of C and D.

Question 15

Question

For the reaction N2O4(g)⇌2NO2(g)at equilibrium, if the pressure of N2O4is

increased, predict the direction in which the equilibrium will shift and explain

your reasoning.

Solution

Step 1: Write the balanced equilibrium equation for the reaction: The balanced

equilibrium equation for the reaction is:

N2O4(g)⇌2NO2(g)

Step 2: Define the equilibrium expression: The equilibrium expression for

the reaction is:

K=[NO2]2

[N2O4]

Step 3: Analyze the effect of increasing the pressure of N2O4: According

to Le Chatelier’s principle, if the pressure of one of the reactants or products

is increased, the equilibrium will shift in the direction that reduces the total

pressure. In this case, increasing the pressure of N2O4will cause the equilibrium

to shift towards the right to decrease the pressure by consuming some of the

N2O4gas.

Step 4: Predict the direction in which the equilibrium will shift: Therefore,

if the pressure of N2O4is increased, the equilibrium will shift to the right to

decrease the pressure.

Step 5: Justification: By consuming some of the N2O4gas and producing

more NO2, the total number of gas molecules will decrease, thus reducing the

total pressure and bringing the system back to equilibrium.

Question 16

Question

A reaction mixture initially contains 0.20 M of NO2and 0.30 M of O2. The

following equilibrium is established:

2NO2(g) + O2(g)⇌2NO2(g)

If the equilibrium constant, Kc, for the reaction at a certain temperature is

2.50 ×103, calculate the new equilibrium concentration of NO2if the concen-

tration of O2is increased to 0.40 M.

Solution

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

equilibrium constant, Kc. The balanced equilibrium equation is:

2NO2(g) + O2(g)⇌2NO2(g)

The expression for the equilibrium constant, Kc, is given by:

Kc=[NO2]2

[NO2]2[O2]=[NO2]2

[O2]

Step 2: Calculate the initial value of Qc. Given initial concentrations:

[NO2]initial = 0.20 M and [O2]initial = 0.30 M

Substitute the initial concentrations into the expression for Qc:

Qc=0.202

0.30 = 0.1333

Step 3: Determine the direction of the reaction shift. Since Qcis less than

Kc, the reaction will shift to the right to reach equilibrium.

Step 4: Determine the changes in concentrations. Let xrepresent the change

in concentration of NO2.

After the reaction reaches equilibrium: [NO2] = 0.20−2xand [O2] = 0.40−x

Step 5: Set up the equation using the equilibrium constant Kc. Substitute

the equilibrium concentrations into the expression for Kc:

2.50 ×103=(0.20 −2x)2

(0.40 −x)

Step 6: Solve for x.

2.50 ×103=(0.20 −2x)2

0.40 −x

2.50 ×103=(0.04 −0.80x+ 4x2)

0.40 −x

0.10 −2x= 4x2

4x2+ 2x−0.10 = 0

Using the quadratic formula to solve for x:

x=−2±√(−2)2−4(4)(−0.10)

2(4)

x≈0.0783

Step 7: Calculate the new equilibrium concentration of NO2.

[NO2]final = 0.20 −2(0.0783)

[NO2]final ≈0.0434 M

Therefore, the new equilibrium concentration of NO2is approximately 0.0434

M when the concentration of O2is increased to 0.40 M.

Question 17

Question

A reaction mixture initially contains 0.20 M nitrogen dioxide (NO2) and 0.10

M dinitrogen tetroxide (N2O4) at equilibrium according to the reaction:

2NO2(g)⇌N2O4(g)

If the concentration of NO2is increased to 0.30 M, predict the direction in

which the equilibrium will shift (towards the products, towards the reactants,

or no shift) and explain why.

Solution

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

reaction:

The equilibrium constant expression for the reaction 2NO2(g)⇌N2O4(g)

is:

Kc=[N2O4]2

[NO2]2

Step 2: Calculate the initial equilibrium constant, Kc,initial, using the initial

concentrations of NO2and N2O4:

Given: Initial concentration of NO2,[NO2]initial = 0.20 M Initial concentra-

tion of N2O4,[N2O4]initial = 0.10 M

Substitute the initial concentrations into the equilibrium constant expres-

sion:

Kc,initial =(0.10)2

(0.20)2= 0.25

Step 3: Determine the direction in which the equilibrium will shift when the

concentration of NO2is increased to 0.30 M:

Given: New concentration of NO2,[NO2]new = 0.30 M

Since the new [NO2]is greater than the initial [NO2], the reaction will shift

to consume some of the added NO2. This means that the equilibrium will

shift towards the products to alleviate the stress caused by the increase in NO2

concentration.

Therefore, the equilibrium will shift towards the products when the concen-

tration of NO2is increased to 0.30 M.

Question 18

Question

For the reaction 2NO(g) + O2(g)⇌2NO2(g)at equilibrium, if the concentra-

tion of O2is increased, predict how the equilibrium will shift and explain your

reasoning.

Solution

Step 1: Write down the balanced chemical equation and the equilibrium expres-

sion. The balanced chemical equation for the given reaction is:

2NO(g)+O2(g)⇌2NO2(g)

The equilibrium expression for this reaction is:

Kc=[NO2]2

[NO]2[O2]

Step 2: Determine the initial direction of the equilibrium shift. Adding more

O2will increase the concentration of a reactant. According to Le Chatelier’s

principle, when a reactant is added, the equilibrium will shift to the right to

consume the added reactant and re-establish equilibrium.

Step 3: Re-write the equilibrium expression with increased O2. Since O2

was added, the concentration of O2in the equilibrium expression will increase.

Step 4: Predict the shift in equilibrium. The equilibrium will shift to the

right to consume the added O2until a new equilibrium is established.

Therefore, an increase in the concentration of O2will cause the equilibrium

to shift to the right.

Question 19

Question

For the reaction

2A+ 3B⇌C+D,

the equilibrium constant, Kc, is 10. If 0.1 moles of A, 0.2 moles of B, and

0.3 moles of Care mixed in a 1 L flask, determine the direction in which the

reaction will proceed in order to establish equilibrium. Justify your answer.

Solution

Step 1: Calculate the initial concentrations of A,B,C, and D. Given: - Initial

moles of A: 0.1 mol - Initial moles of B: 0.2 mol - Initial moles of C: 0.3 mol -

Volume of the flask: 1 L

Calculate initial concentrations:

[A]0=0.1mol

1L= 0.1M

[B]0=0.2mol

1L= 0.2M

[C]0=0.3mol

1L= 0.3M

[D]0= 0 M

(since initially, no moles of Dare present)

Step 2: Calculate the reaction quotient, Qc, using the initial concentrations.

Qc=[C]0[C]0

[A]2

0[B]3

=0.3∗0

0.12∗0.23= 0

Step 3: Compare Qcand Kc. Since Qc< Kc, the reaction will proceed in

the forward direction (to the right) to establish equilibrium. This means more

products will be formed until Qcequals Kc.

Therefore, the reaction will proceed to the right to establish equilibrium.

Question 20

Question

For the reaction:

2SO2(g) + O2(g)⇌2SO3(g)

at equilibrium, an external pressure is applied to the system. Explain how

Le Chatelier’s principle can be used to predict the direction of the shift in

equilibrium and the effect on the concentrations of the reactants and products.

Solution

Step 1: According to Le Chatelier’s principle, when a stress is applied to a

system at equilibrium, the system will shift in a direction that minimizes the

effect of the stress.

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

in the direction that produces fewer molecules of gas to reduce the pressure.

Step 3: Since there are 3 moles of gas on the left side and only 2 moles of

gas on the right side, increasing the pressure would cause the system to shift to

the right to reduce the total number of gas molecules.

Step 4: As a result, the concentration of SO2and O2would decrease, while

the concentration of SO3would increase, returning the system to a new equi-

librium position.

Step 5: Therefore, applying external pressure would shift the equilibrium

towards the formation of more SO3, leading to a decrease in the concentrations

of SO2and O2and an increase in the concentration of SO3in order to relieve

the applied stress.

Question 21

Question

For the reaction

2SO2(g) + O2(g)⇌2SO3(g)

at equilibrium, the concentration of SO2is increased. Predict qualitatively, us-

ing Le Chatelier’s principle, what effect this change will have on the equilibrium

position of the reaction.

Solution

Step 1: According to Le Chatelier’s principle, if we increase the concentration of

a reactant, the equilibrium will shift to the right to consume some of the added

reactant.

Step 2: In this reaction, increasing the concentration of SO2will cause the

system to react by consuming some of the added SO2.

Step 3: Consequently, the equilibrium position of the reaction will shift to

the right in order to produce more SO3and consume some of the excess SO2

added.

Therefore, the equilibrium position of the reaction

2SO2(g) + O2(g)⇌2SO3(g)

will shift to the right when the concentration of SO2is increased based on Le

Chatelier’s principle.

Question 22

Question

For the reaction represented by the equation below, predict whether an increase

in pressure will shift the equilibrium to the left, to the right, or will not shift

the equilibrium:

2 SO2(g) + O2(g)⇌2SO3(g)

Solution

Step 1: First, let’s determine the effect of an increase in pressure on the reaction.

An increase in pressure will shift the equilibrium in the direction that reduces

the total number of gas molecules (moles) involved in the reaction.

Step 2: Examine the balanced equation to determine the total number of

gas molecules on each side of the equation:

- Left side: 2 moles of SO� gas + 1 mole of O� gas = 3 moles of gas - Right

side: 2 moles of SO� gas = 2 moles of gas

Step 3: Compare the total number of gas molecules on each side of the

equation.

Since there are more gas molecules on the left side of the equation, an increase

in pressure will cause the system to shift in the direction that reduces the total

number of gas molecules.