Equilibrium

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Summary

Equilibrium Summary

Equilibrium Definition: When the number of molecules leaving the liquid to vapor equals the number returning, equilibrium is attained and is dynamic in nature.
Equilibrium Constant (Kc): Expressed as the concentration of products divided by reactants, each raised to the stoichiometric coefficient.
- For reaction: aA + bB ⇌ cC + dD
- Kc = [C]^c[D]^d / [A]^a[B]^b
Kp: For gaseous reactions, Kp is expressed using partial pressures instead of concentrations.
Le Chatelier's Principle: States that a change in temperature, pressure, or concentration will shift the equilibrium to counteract the change.
Catalysts: Increase the rate of reaction without affecting the equilibrium composition.
Electrolytes: Substances that conduct electricity in aqueous solutions, including acids, bases, and salts.
Acid-Base Definitions:
- Arrhenius: Acids produce H⁺ ions; bases produce OH⁻ ions.
- Brönsted-Lowry: Acids are proton donors; bases are proton acceptors.
- Lewis: Acids are electron pair acceptors; bases are electron pair donors.
Ionization Constants: Expressions for weak acids (K) and weak bases (Kb) are developed using the Arrhenius definition.
pH Scale: pH = -log[H⁺]; pOH = log[OH⁻]; pKa = -log[K]; pKb = -log[K]; pKw = -log[Kw].
Solubility Product Constant (Ksp): Relates to the solubility of sparingly soluble salts and their equilibrium in solution.

Learning Objectives

Identify the dynamic nature of equilibrium in chemical systems.
Demonstrate the concept of equilibrium through experiments involving pH paper and buffer solutions.
Calculate equilibrium constants (Kc and Kp) for various reactions.
Analyze the effects of concentration, pressure, and temperature changes on equilibrium.
Determine the ionization constants and concentrations of acids and bases in solution.
Classify substances as Lewis acids or bases and identify conjugate acid-base pairs.
Calculate pH and hydrogen ion concentrations in various solutions.

Detailed Notes

Chemistry Notes

Suggested Activities for Students Regarding This Unit

pH Determination: Use pH paper to determine the pH of:
- Fresh juices of various vegetables and fruits
- Soft drinks
- Body fluids
- Water samples
Salt Solutions: Determine the pH of different salt solutions to identify if they are formed from strong/weak acids and bases.
Buffer Solutions: Prepare buffer solutions by mixing sodium acetate and acetic acid, then determine their pH using pH paper.
Indicators: Observe the colors of different indicators in solutions of varying pH.
Acid-Base Titrations: Perform acid-base titrations using indicators.
Common Ion Effect: Observe the common ion effect on the solubility of sparingly soluble salts.
pH Meter Comparison: Measure pH with a pH meter and compare results with pH paper.

Exercises

Equilibrium and Kc Calculations

Vapor Pressure Change:
- What is the initial effect of increasing the volume of a sealed container on vapor pressure?
- How do rates of evaporation and condensation change initially?
- What happens when equilibrium is restored?
Kc Calculation: For the equilibrium: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), given concentrations [SO₂]=0.60M, [O₂]=0.82M, [SO₃]=1.90M, calculate Kc.
Kp Calculation: At a total pressure of 10⁵Pa, iodine vapor contains 40% by volume of I atoms. Calculate Kₚ for the equilibrium I₂(g) ⇌ 2I(g).
Equilibrium Constant Expressions: Write the expression for Kc for the following reactions:
- 2NOC1(g) ⇌ 2NO(g) + Cl₂(g)
- 2Cu(NO₃)₂(s) ⇌ 2CuO(s) + 4NO₂(g) + O₂(g)
- CH₂COOC₂H₅(aq) + H₂O(l) ⇌ CH₃COOH(aq) + C₂H₅OH(aq)
- Fe³⁺(aq) + 3OH⁻(aq) ⇌ Fe(OH)₃(s)
- I₂(s) + 5F₂ ⇌ 2IF₅
Kp to Kc Conversion: Find Kc from Kp for the following equilibria:
- 2NOC1(g) ⇌ 2NO(g) + Cl₂(g); Kₚ = 1.8 X 10⁻² at 500 K
- CaCO₃(s) ⇌ CaO(s) + CO₂(g); Kₚ = 167 at 1073 K

Important Concepts

Conjugate Acid-Base Pairs

Definition: A conjugate acid-base pair consists of two species that transform into each other by the gain or loss of a proton (H⁺).
Examples:
- HNO₂ ↔ NO₂⁻
- HF ↔ F⁻
- H₂SO₄ ↔ HSO₄⁻

Lewis Acids and Bases

Lewis Acids: BF₃, H⁺, NH₄⁺
Lewis Bases: OH⁻, F⁻

pH Calculations

Hydrogen Ion Concentration:
- For a soft drink with [H⁺] = 3.8 X 10⁻³ M, calculate pH.
- For vinegar with pH = 3.76, calculate [H⁺].

Ionization Constants

Ionization Constants:
- HF: 6.8 X 10⁻⁴
- HCOOH: 1.8 X 10⁻⁴
- HCN: 4.8 X 10⁻⁹
Conjugate Bases: Calculate the ionization constants of the corresponding conjugate bases.

Exam Tips & Common Mistakes

Common Mistakes and Exam Tips

Common Pitfalls

Misunderstanding Equilibrium: Students often confuse dynamic equilibrium with static equilibrium. Remember that in dynamic equilibrium, the rates of the forward and reverse reactions are equal, but the reactions continue to occur.
Incorrect Use of Equilibrium Constants: When calculating Kc or Kp, ensure you are using the correct concentrations or partial pressures and that they are raised to the power of their coefficients in the balanced equation.
Ignoring Temperature Effects: Changes in temperature can affect the value of Kc or Kp. Be careful to note how temperature changes influence equilibrium positions.
Assuming Complete Dissociation: In weak acids and bases, do not assume complete dissociation. Always consider the degree of ionization and use the appropriate equilibrium expressions.

Exam Tips

Practice Writing Equilibrium Expressions: Be comfortable with writing Kc and Kp expressions for various reactions. This skill is crucial for solving equilibrium problems.
Understand Le Chatelier's Principle: Be prepared to apply Le Chatelier's principle to predict how changes in concentration, pressure, or temperature will affect the position of equilibrium.
Use pH Calculations Wisely: When calculating pH from hydrogen ion concentration, remember the formula pH = -log[H⁺]. Also, be familiar with how to calculate [H⁺] from pH.
Check Units: Always ensure that your units are consistent, especially when dealing with concentrations and pressures in equilibrium calculations.
Review Common Ion Effect: Understand how the presence of a common ion affects the solubility of salts and the ionization of weak acids and bases.

Practice & Assessment

Multiple Choice Questions

0.02 M

0.05 M

0.10 M

0.15 M

Correct Answer: A

Solution:

Let the change in concentration of

CH_4

at equilibrium be

x

. The equilibrium concentrations are:

[CO] = 0.30 - x

[H_2] = 0.10 - 3x

[H_2O] = 0.02 + x

, and

[CH_4] = x

. The equilibrium expression is

K_c = \frac{[CH_4][H_2O]}{[CO][H_2]^3}

. Solving for

x

using

K_c = 3.90

, we find

x = 0.02

The equilibrium will shift to the right, favoring the formation of ammonia.

The equilibrium will shift to the left, favoring the formation of nitrogen and hydrogen.

The equilibrium will not shift, as pressure does not affect this reaction.

The equilibrium will shift to the side with more gas molecules.

Correct Answer: A

Solution:

Increasing the pressure favors the side of the reaction with fewer moles of gas. In this reaction, the right side (2 moles of NH₃) has fewer moles compared to the left side (4 moles of N₂ and H₂ combined), so the equilibrium shifts to the right.

K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}

K_c = \frac{[A]^a[B]^b}{[C]^c[D]^d}

K_c = [A]^a[B]^b[C]^c[D]^d

K_c = \frac{[C][D]}{[A][B]}

Correct Answer: A

Solution:

The equilibrium constant expression is given by the concentration of products raised to their stoichiometric coefficients divided by the concentration of reactants raised to their stoichiometric coefficients.

The equilibrium will shift towards the vapor phase.

The equilibrium will shift towards the liquid phase.

The equilibrium will remain unchanged.

The equilibrium will shift towards both phases equally.

Correct Answer: A

Solution:

According to Le Chatelier's Principle, increasing the temperature of an endothermic process (like evaporation) will shift the equilibrium towards the products, i.e., the vapor phase.

The equilibrium will shift to the right, favoring the formation of products.

The equilibrium will shift to the left, favoring the formation of reactants.

The equilibrium position will remain unchanged.

The equilibrium constant will decrease.

Correct Answer: A

Solution:

For an endothermic reaction, increasing the temperature shifts the equilibrium to the right, favoring the formation of products to absorb the added heat.

The equilibrium shifts to the right, favoring the formation of

PC1_3

and

C1_2

The equilibrium shifts to the left, favoring the formation of

PC1_5

The equilibrium position remains unchanged.

The equilibrium constant increases.

Correct Answer: B

Solution:

Increasing the pressure of a gaseous equilibrium will shift the equilibrium towards the side with fewer moles of gas. In this reaction, increasing pressure shifts the equilibrium to the left, favoring

PC1_5

, which has fewer moles of gas.

The equilibrium shifts to the right, favoring the formation of products.

The equilibrium shifts to the left, favoring the formation of reactants.

The equilibrium constant changes.

There is no effect on the equilibrium position.

Correct Answer: A

Solution:

Le Chatelier's principle states that increasing the concentration of a reactant will shift the equilibrium to the right to counteract the change by forming more products.

NH₄⁺

H₂O

Cl⁻

Na⁺

Correct Answer: B

Solution:

Water (H₂O) can donate a proton to act as an acid or accept a proton to act as a base, making it amphoteric.

NH₃

BF₃

H₂O

OH⁻

Correct Answer: B

Solution:

A Lewis acid is an electron pair acceptor. BF₃ has an empty p-orbital and can accept an electron pair, making it a Lewis acid.

1.0 \times 10^{-6}

2.2 \times 10^{-6}

7.1 \times 10^{-6}

5.0 \times 10^{-6}

Correct Answer: C

Solution:

Using the expression for the ionization constant

K_a = \frac{[C_6H_5O^-][H^+]}{[C_6H_5OH]}

and assuming

[C_6H_5O^-] = [H^+] = x

, we have

K_a = \frac{x^2}{0.05}

. Solving for

x

gives

x = \sqrt{K_a \times 0.05} = \sqrt{1.0 \times 10^{-10} \times 0.05} = 7.1 \times 10^{-6}

It resists changes in pH upon dilution

It has a pH equal to 7

It contains only strong acids and bases

It cannot resist changes in pH upon addition of small amounts of acid or base

Correct Answer: A

Solution:

A buffer solution is characterized by its ability to resist changes in pH upon the addition of small amounts of acid or base, as well as upon dilution.

12.5

13.0

13.5

14.0

Correct Answer: B

Solution:

First, calculate the molarity of NaOH: Molar mass of NaOH = 40 g/mol. Moles of NaOH = 0.3 g / 40 g/mol = 0.0075 mol. Volume of solution = 0.2 L. Molarity = 0.0075 mol / 0.2 L = 0.0375 M. Since NaOH is a strong base,

[\text{OH}^-] = 0.0375

M. pOH = -log(0.0375) = 1.43. pH = 14 - pOH = 12.57 \approx 13.0.

An acid is a substance that produces hydroxyl ions in solution.

An acid is a substance that donates a proton.

An acid is a substance that accepts an electron pair.

An acid is a substance that produces hydrogen gas.

Correct Answer: B

Solution:

According to the Bronsted-Lowry concept, an acid is defined as a proton donor.

The vapor pressure will increase because more molecules will have sufficient energy to escape into the vapor phase.

The vapor pressure will decrease because the liquid molecules will condense back into the liquid phase.

The vapor pressure will remain constant because equilibrium is already established.

The vapor pressure will fluctuate randomly due to dynamic equilibrium.

Correct Answer: A

Solution:

Increasing the temperature provides more kinetic energy to the molecules, allowing more molecules to escape into the vapor phase, thus increasing the vapor pressure.

It completely dissociates in solution.

It partially dissociates in solution.

It does not conduct electricity.

It is insoluble in water.

Correct Answer: B

Solution:

A weak electrolyte partially dissociates in solution, establishing an equilibrium between ions and unionized molecules.

It increases the equilibrium constant.

It decreases the equilibrium constant.

It increases the rate of both forward and reverse reactions without affecting the equilibrium constant.

It shifts the equilibrium position to the right.

Correct Answer: C

Solution:

A catalyst speeds up both the forward and reverse reactions equally, thus increasing the rate at which equilibrium is reached without changing the equilibrium constant.

Cl_2(g) \rightleftharpoons 2Cl(g), K_c = 5 \times 10^{-39}

Cl_2(g) + 2NO(g) \rightleftharpoons 2NOCl(g), K_c = 3.7 \times 10^8

Cl_2(g) + 2NO_2(g) \rightleftharpoons 2NO_2Cl(g), K_c = 1.8

None of the above

Correct Answer: C

Solution:

For a reaction to have appreciable concentrations of both reactants and products, the equilibrium constant should be neither too large nor too small. Option c has a moderate

K_c

value, indicating significant amounts of both reactants and products.

2.0 bar

4.0 bar

6.0 bar

8.0 bar

Correct Answer: A

Solution:

Let the change in pressure at equilibrium be

x

. At equilibrium,

P_{H_2} = x

P_{CO_2} = x

P_{CO} = 4.0 - x

, and

P_{H_2O} = 4.0 - x

. The expression for

K_p

K_p = \frac{x^2}{(4.0-x)^2} = 10.1

. Solving for

x

, we find

x = 2.0

bar.

Acids are proton donors.

Acids produce hydroxyl ions in aqueous solutions.

Acids produce hydrogen ions in aqueous solutions.

Acids are electron pair acceptors.

Correct Answer: C

Solution:

Arrhenius defined acids as substances that produce hydrogen ions in aqueous solutions.

A weak acid completely dissociates in water.

A weak acid partially dissociates in water, establishing an equilibrium between the undissociated acid and its ions.

A weak acid does not dissociate in water at all.

A weak acid increases the pH of the water significantly.

Correct Answer: B

Solution:

A weak acid only partially dissociates in water, meaning that an equilibrium is established between the undissociated acid and its ions. This is characteristic of weak acids.

A substance that donates a proton.

A substance that accepts a proton.

A substance that donates an electron pair.

A substance that produces hydroxyl ions in aqueous solution.

Correct Answer: D

Solution:

According to Arrhenius, a base is a substance that produces hydroxyl ions (OH⁻) in aqueous solution.

The rate of the forward reaction equals the rate of the reverse reaction.

The concentrations of reactants and products are equal.

The reaction has stopped completely.

The reaction only proceeds in one direction.

Correct Answer: A

Solution:

Dynamic equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction, meaning both reactions continue to occur at the same rate.

0.05 M

0.10 M

0.15 M

0.20 M

Correct Answer: A

Solution:

Using the equilibrium expression for the reaction, the concentration of

CH_4

can be calculated to be 0.05 M.

The equilibrium will shift to favor more evaporation.

The equilibrium will shift to favor more condensation.

The equilibrium state will remain unchanged.

The equilibrium constant will decrease.

Correct Answer: A

Solution:

Increasing the temperature increases the kinetic energy of the molecules, causing more molecules to escape into the vapor phase, thus favoring evaporation.

The equilibrium shifts towards the formation of more products.

The equilibrium shifts towards the formation of more reactants.

The equilibrium remains unchanged.

The reaction rate decreases.

Correct Answer: A

Solution:

According to Le Chatelier's principle, increasing the pressure will shift the equilibrium towards the side with fewer moles of gas. In this reaction, the product side has fewer moles (2 moles of NH₃) compared to the reactant side (4 moles of N₂ and H₂ combined).

The rate of evaporation is greater than the rate of condensation.

The rate of condensation is greater than the rate of evaporation.

The rate of evaporation equals the rate of condensation.

Both rates stop completely.

Correct Answer: C

Solution:

At equilibrium, the rate of evaporation equals the rate of condensation, resulting in a constant vapour pressure.

The equilibrium shifts to the left.

The equilibrium shifts to the right.

The equilibrium constant increases.

The equilibrium constant decreases.

Correct Answer: B

Solution:

According to Le Chatelier's principle, adding more reactants will shift the equilibrium to the right to produce more products.

Hydrogen ions

Hydroxyl ions

Electrons

Protons

Correct Answer: A

Solution:

Arrhenius defined acids as substances that produce hydrogen ions in aqueous solutions.

Shift to the right, favoring the formation of

N_2O_4

Shift to the left, favoring the formation of

NO_2

No change in the equilibrium position

Equilibrium constant

K_c

will increase

Correct Answer: A

Solution:

Halving the volume of the container increases the pressure. According to Le Chatelier's principle, the equilibrium will shift to the side with fewer moles of gas to counteract the change. Here, the reaction will shift to the right, favoring the formation of

N_2O_4

, which has fewer moles of gas.

2.42

3.42

4.42

5.42

Correct Answer: A

Solution:

The pH is calculated using the formula

\text{pH} = -\log[H^+]

. Substituting the given concentration:

\text{pH} = -\log(3.8 \times 10^{-3}) = 2.42

A catalyst increases the equilibrium constant.

A catalyst decreases the equilibrium constant.

A catalyst speeds up the attainment of equilibrium without affecting the equilibrium constant.

A catalyst shifts the equilibrium position towards the products.

Correct Answer: C

Solution:

A catalyst speeds up the rate at which equilibrium is reached by providing an alternative pathway with a lower activation energy, but it does not affect the equilibrium constant or the position of equilibrium.

A catalyst increases the equilibrium constant.

A catalyst decreases the equilibrium constant.

A catalyst does not affect the equilibrium constant but increases the rate of reaching equilibrium.

A catalyst shifts the equilibrium position to favor the products.

Correct Answer: C

Solution:

A catalyst provides an alternative pathway with lower activation energy, increasing the rate at which equilibrium is reached. However, it does not affect the equilibrium constant or the position of equilibrium.

A catalyst increases the equilibrium constant.

A catalyst decreases the equilibrium constant.

A catalyst speeds up both the forward and reverse reactions equally.

A catalyst shifts the equilibrium position towards the products.

Correct Answer: C

Solution:

A catalyst provides an alternative pathway with a lower activation energy, increasing the rate of both forward and reverse reactions equally, without affecting the equilibrium constant.

Equilibrium shifts to the right.

Equilibrium shifts to the left.

Equilibrium remains unchanged.

Equilibrium shifts depending on the catalyst.

Correct Answer: A

Solution:

Increasing the concentration of a reactant shifts the equilibrium position to the right to favor the formation of products.

K_c = \frac{[A]^a[B]^b}{[C]^c[D]^d}

K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}

K_c = [A]^a[B]^b[C]^c[D]^d

K_c = [C][D]

Correct Answer: B

Solution:

The equilibrium constant

K_c

is expressed as

K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}

, based on the concentrations of products over reactants.

The concentrations of reactants and products are constantly changing.

The rates of the forward and reverse reactions are equal.

The reaction has stopped completely.

The equilibrium constant changes with time.

Correct Answer: B

Solution:

In a dynamic equilibrium, the rates of the forward and reverse reactions are equal, resulting in no net change in the concentrations of reactants and products, even though both reactions continue to occur.

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

SO_3

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

SO_2

and

O_2

The equilibrium will not shift.

The equilibrium constant will increase.

Correct Answer: A

Solution:

According to Le Chatelier's principle, increasing the concentration of a reactant will shift the equilibrium towards the products to counteract the change, thus favoring the formation of

SO_3

K_c

is affected by changes in concentration of reactants.

K_c

is affected by changes in pressure.

K_c

is affected by changes in temperature.

K_c

is affected by the presence of a catalyst.

Correct Answer: C

Solution:

The equilibrium constant

K_c

is only affected by changes in temperature. Changes in concentration, pressure, or the presence of a catalyst do not affect

K_c

The equilibrium shifts to the right.

The equilibrium shifts to the left.

The equilibrium remains unchanged.

The equilibrium constant increases.

Correct Answer: B

Solution:

Increasing pressure shifts the equilibrium towards the side with fewer moles of gas, which is the left side in this reaction.

2.0 bar

4.0 bar

6.0 bar

8.0 bar

Correct Answer: B

Solution:

At equilibrium, the reaction quotient

Q_p

equals the equilibrium constant

K_p

. Thus,

K_p = \frac{P_{\text{CO}_2} \cdot P_{\text{H}_2}}{P_{\text{CO}} \cdot P_{\text{H}_2\text{O}}}

. Given

P_{\text{CO}} = P_{\text{H}_2\text{O}} = 4.0

bar and

K_p = 10.1

, we have

10.1 = \frac{P_{\text{H}_2}^2}{16}

. Solving for

P_{\text{H}_2}

gives

P_{\text{H}_2} = \sqrt{10.1 \times 16} = 4.0

bar.

0.042

23.81

0.5

1.0

Correct Answer: B

Solution:

The equilibrium constant for the reverse reaction is the reciprocal of the equilibrium constant for the forward reaction. Therefore,

K_{c,\text{reverse}} = \frac{1}{0.042} = 23.81

0.018

0.037

0.074

0.148

Correct Answer: B

Solution:

The degree of dissociation

\alpha

can be calculated using the formula

\alpha = \sqrt{\frac{K_a}{c}}

, where

K_a = 1.74 \times 10^{-5}

and

c = 0.05

M. Substituting the values,

\alpha = \sqrt{\frac{1.74 \times 10^{-5}}{0.05}} = 0.037

K_p

increases.

K_p

decreases.

K_p

remains constant.

K_p

becomes zero.

Correct Answer: C

Solution:

The equilibrium constant

K_p

is not affected by changes in pressure or volume; it is only affected by temperature.

NH₃

BF₃

OH⁻

F⁻

Correct Answer: B

Solution:

A Lewis acid is an electron pair acceptor. BF₃ is a Lewis acid because it can accept an electron pair.

The concentrations of reactants and products are equal.

The rates of the forward and reverse reactions are equal.

The reaction has stopped completely.

The temperature of the system is constant.

Correct Answer: B

Solution:

Dynamic equilibrium occurs when the rates of the forward and reverse reactions are equal, resulting in no net change in the concentrations of reactants and products.

It changes with pressure.

It changes with temperature.

It changes with concentration.

It is always equal to 1.

Correct Answer: B

Solution:

The equilibrium constant

K_c

is dependent on temperature but not on pressure or concentration.

K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}

K_c = \frac{[A]^a[B]^b}{[C]^c[D]^d}

K_c = [A]^a[B]^b[C]^c[D]^d

K_c = \frac{[C][D]}{[A][B]}

Correct Answer: A

Solution:

The equilibrium constant

K_c

is defined as the ratio of the product of the concentrations of the products raised to their stoichiometric coefficients to the product of the concentrations of the reactants raised to their stoichiometric coefficients.

The equilibrium will shift to the right, favoring the formation of more HI.

The equilibrium will shift to the left, favoring the formation of more H₂ and I₂.

The equilibrium will remain unchanged.

The equilibrium constant will decrease.

Correct Answer: A

Solution:

According to Le Chatelier's principle, adding more iodine gas will increase the concentration of I₂, causing the equilibrium to shift to the right to counteract this change by producing more HI.

The degree of ionization increases with dilution

The degree of ionization decreases with dilution

The degree of ionization is independent of concentration

The degree of ionization is maximum at high concentrations

Correct Answer: A

Solution:

For weak acids, the degree of ionization increases with dilution due to the dilution effect, which shifts the equilibrium towards more ionization according to Le Chatelier's principle.

It decreases.

It increases.

It remains the same.

It becomes zero.

Correct Answer: A

Solution:

For exothermic reactions, increasing the temperature shifts the equilibrium to favor the reactants, decreasing the equilibrium constant

K_c

Shifts to the right

Shifts to the left

No change

Cannot be determined

Correct Answer: B

Solution:

According to Le Chatelier's principle, increasing the pressure of a gaseous equilibrium system will shift the equilibrium position towards the side with fewer moles of gas. In this reaction, the left side has fewer moles (1 mole of

\text{PCl}_5

) compared to the right side (2 moles of

\text{PCl}_3

and

\text{Cl}_2

). Therefore, the equilibrium shifts to the left.

The concentrations of reactants and products are equal.

The rates of the forward and reverse reactions are equal.

The reaction stops completely.

The reaction only proceeds in one direction.

Correct Answer: B

Solution:

At dynamic equilibrium, the rates of the forward and reverse reactions are equal, resulting in no net change in the concentrations of reactants and products.

Increasing the temperature

Decreasing the pressure

Adding a catalyst

Increasing the pressure

Correct Answer: D

Solution:

Increasing the pressure will shift the equilibrium towards the side with fewer moles of gas, which is the right side in this reaction, thus increasing the yield of

SO_3

Complete dissociation in solution

Partial dissociation in solution

No dissociation in solution

Formation of a precipitate

Correct Answer: A

Solution:

Strong electrolytes completely dissociate into ions in solution, allowing them to conduct electricity effectively.

The equilibrium shifts towards the side with fewer moles of gas.

The equilibrium shifts towards the side with more moles of gas.

The equilibrium remains unchanged.

The equilibrium shifts towards the side with equal moles of gas.

Correct Answer: A

Solution:

Increasing pressure shifts the equilibrium towards the side with fewer moles of gas to counteract the change.

The reaction will proceed to the right

The reaction will proceed to the left

The reaction is already at equilibrium

The reaction will oscillate between left and right

Correct Answer: A

Solution:

The reaction quotient

Q_c = \frac{[HI]^2}{[H_2][I_2]} = \frac{(0.1)^2}{0.5 \times 0.5} = 0.04

. Since

Q_c < K_c

, the reaction will proceed to the right to reach equilibrium.

The concentrations of reactants and products remain constant.

The reaction stops completely.

Only the forward reaction occurs.

Only the reverse reaction occurs.

Correct Answer: A

Solution:

At dynamic equilibrium, the concentrations of reactants and products remain constant because the rates of the forward and reverse reactions are equal.

It completely dissociates in solution.

It partially dissociates in solution.

It does not conduct electricity in solution.

It is insoluble in water.

Correct Answer: B

Solution:

Weak electrolytes partially dissociate in solution, resulting in an equilibrium between the ions and the undissociated molecules.

The rate of evaporation is greater than the rate of condensation.

The rate of condensation is greater than the rate of evaporation.

The rate of evaporation equals the rate of condensation.

No molecules change phase.

Correct Answer: C

Solution:

At equilibrium, the rate of evaporation equals the rate of condensation, resulting in a constant vapor pressure.

K_c

is the sum of the concentrations of products and reactants.

K_c

is the difference between the concentrations of products and reactants.

K_c

is the product of the concentrations of products divided by the product of the concentrations of reactants, each raised to their stoichiometric coefficients.

K_c

is the ratio of the concentrations of reactants to products.

Correct Answer: C

Solution:

K_c

is expressed as the concentration of products divided by reactants, each term raised to the stoichiometric coefficient.

K_c = \frac{[SO_3]^2}{[SO_2]^2[O_2]}

K_c = \frac{[SO_2]^2[O_2]}{[SO_3]^2}

K_c = \frac{[SO_2][O_2]}{[SO_3]^2}

K_c = \frac{[SO_3]}{[SO_2][O_2]}

Correct Answer: A

Solution:

For the reaction

2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)

, the equilibrium constant expression is derived from the stoichiometry of the reaction. Thus,

K_c = \frac{[SO_3]^2}{[SO_2]^2[O_2]}

NH₄⁺/NH₃

H₂O/OH⁻

HCl/Cl⁻

H₂SO₄/HSO₄⁻

Correct Answer: A

Solution:

A conjugate acid-base pair consists of two species that differ by a proton (H⁺). NH₄⁺ (ammonium) and NH₃ (ammonia) are a conjugate acid-base pair, differing by one proton.

HCO₃⁻

NH₄⁺

HCl

NaOH

Correct Answer: A

Solution:

HCO₃⁻ can donate a proton to become CO₃²⁻, acting as an acid, and it can accept a proton to become H₂CO₃, acting as a base.

The vapor pressure will increase, and the equilibrium will shift towards more vapor formation.

The vapor pressure will decrease, and the equilibrium will shift towards more liquid formation.

The vapor pressure will remain constant, and the equilibrium will not shift.

The vapor pressure will increase, but the equilibrium will shift towards more liquid formation.

Correct Answer: A

Solution:

Increasing the temperature increases the kinetic energy of the molecules, leading to more molecules escaping into the vapor phase, thus increasing the vapor pressure and shifting the equilibrium towards more vapor formation.

The equilibrium will shift to the right.

The equilibrium will shift to the left.

The equilibrium will remain unchanged.

The equilibrium will shift towards both sides equally.

Correct Answer: A

Solution:

According to Le Chatelier's Principle, increasing the concentration of a reactant will shift the equilibrium towards the products to counteract the change.

The equilibrium shifts to the right.

The equilibrium shifts to the left.

The equilibrium remains unchanged.

The reaction stops.

Correct Answer: B

Solution:

Adding more

PC1_5

increases the concentration of reactants, causing the equilibrium to shift to the left to counteract the change.

To the right, forming more

N_2O_4

To the left, forming more

NO_2

No change, the system is already at equilibrium

Cannot be determined without more information

Correct Answer: A

Solution:

Calculate the reaction quotient

Q_p = \frac{P_{N_2O_4}}{P_{NO_2}^2} = \frac{0.10}{(0.50)^2} = 0.40

. Since

Q_p > K_p

, the reaction will proceed to the right to form more

N_2O_4

to reach equilibrium.

The rate of evaporation equals the rate of condensation.

The liquid completely evaporates into vapor.

The vapor pressure continuously increases.

The temperature of the system decreases.

Correct Answer: A

Solution:

At dynamic equilibrium, the rate of evaporation of the liquid equals the rate of condensation from the vapor phase, leading to a constant vapor pressure.

The rate of evaporation equals the rate of condensation.

The liquid completely evaporates.

The vapor pressure continuously increases.

The liquid and vapor phases are static.

Correct Answer: A

Solution:

At dynamic equilibrium, the rate of evaporation equals the rate of condensation, resulting in a constant vapor pressure.

2.42

3.42

4.42

5.42

Correct Answer: B

Solution:

The pH is calculated using the formula

pH = -\log[H^+]

. For

[H^+] = 3.8 \times 10^{-3}

pH = -\log(3.8 \times 10^{-3}) \approx 3.42

True or False

Correct Answer: True

Solution:

At equilibrium, the number of molecules leaving the liquid equals the number returning to the liquid, meaning the rates of evaporation and condensation are equal.

Correct Answer: False

Solution:

According to Arrhenius, acids produce hydrogen ions, not hydroxyl ions, in their aqueous solutions.

Correct Answer: True

Solution:

Arrhenius defined acids as substances that produce hydrogen ions (

Solution:

In weak electrolytes, equilibrium is established between ions and the unionized molecules.

Correct Answer: True

Solution:

The solubility product constant

K_{sp}

is a measure of the solubility of sparingly soluble salts and is used to predict precipitation.

Correct Answer: True

Solution:

At equilibrium, the rate of evaporation is equal to the rate of condensation, resulting in a constant vapor pressure.

Correct Answer: True

Solution:

The equilibrium constant

K_c

is dependent on temperature, and changes in temperature can shift the equilibrium position, thereby changing the value of

K_c

Correct Answer: False

Solution:

The equilibrium constant

K_c

is expressed as the concentration of products divided by reactants, each term raised to the stoichiometric coefficient.

Correct Answer: True

Solution:

In dynamic equilibrium, the rates of the forward and reverse reactions are equal, resulting in constant concentrations of reactants and products.

Correct Answer: True

Solution:

The equilibrium constant

K_c

remains constant at a fixed temperature, reflecting the balance of reactant and product concentrations at equilibrium.

Correct Answer: True

K_c

, is constant at a fixed temperature but changes with temperature.

Correct Answer: True

Solution:

The concentration of

CH_4

at equilibrium can be determined using the equilibrium constant

K_c

and the concentrations of other species in the reaction.

Correct Answer: False

Solution:

A catalyst does not affect the equilibrium composition; it only increases the rate of reaching equilibrium by providing a lower energy pathway.

Correct Answer: False

Solution:

The equilibrium constant

K_c

Solution:

The equilibrium constant

K_c

is not affected by changes in pressure; it is only affected by changes in temperature.

Correct Answer: True

Solution:

Le Chatelier's principle states that if a system at equilibrium is subjected to a change in temperature, pressure, or concentration, the equilibrium will shift in a direction that tends to counteract the change.

Correct Answer: False

Solution:

Equilibrium can be established for both physical processes and chemical reactions.

Correct Answer: False

Solution:

According to Arrhenius, acids produce hydrogen ions (H⁺) in their aqueous solutions, not hydroxyl ions.

Correct Answer: True

Solution:

At equilibrium, the rates of the forward and reverse reactions are equal, which means the concentrations of reactants and products remain constant over time, but they are not necessarily equal to each other.

Equilibrium

AI Learning Assistant

Summary

Equilibrium Summary

Learning Objectives

Detailed Notes

Chemistry Notes

Suggested Activities for Students Regarding This Unit

Exercises

Equilibrium and Kc Calculations

Important Concepts

Conjugate Acid-Base Pairs

Lewis Acids and Bases

pH Calculations

Ionization Constants

Exam Tips & Common Mistakes

Common Mistakes and Exam Tips

Common Pitfalls

Exam Tips

Practice & Assessment

Multiple Choice Questions

Correct Answer: A

Q2.According to Le Chatelier's principle, what will happen to the position of equilibrium when the pressure is increased for the reaction N2(g)+3H2(g)⇌2NH3(g)N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)N2​(g)+3H2​(g)⇌2NH3​(g)?

Correct Answer: A

Q3.What is the equilibrium constant expression for the reaction aA+bB⇌cC+dDaA + bB \rightleftharpoons cC + dDaA+bB⇌cC+dD?

Correct Answer: A

Q4.Consider a closed container with water at equilibrium between its liquid and vapor phases. If the temperature of the system is increased, how will the equilibrium be affected according to Le Chatelier's Principle?

Correct Answer: A

Q5.In the reaction PC15(g)⇌PC13(g)+C12(g)PC1_5(g) \rightleftharpoons PC1_3(g) + C1_2(g)PC15​(g)⇌PC13​(g)+C12​(g) with ΔH=124.0 kJ mol−1\Delta H = 124.0 \text{ kJ mol}^{-1}ΔH=124.0 kJ mol−1, what will be the effect on the equilibrium position if the temperature is increased?

Correct Answer: A

Q6.For the reaction PC15(g)⇌PC13(g)+C12(g)PC1_5(g) \rightleftharpoons PC1_3(g) + C1_2(g)PC15​(g)⇌PC13​(g)+C12​(g), what is the effect on the equilibrium position if the pressure is increased?

Correct Answer: B

Q7.According to Le Chatelier's principle, what happens when the concentration of a reactant is increased in a system at equilibrium?

Correct Answer: A

Q8.Which of the following species can act as both a Brönsted acid and a base?

Correct Answer: B

Q9.Which of the following is a Lewis acid?

Correct Answer: B

Q10.In a solution where phenol is dissolved to form a 0.05 M solution, the ionization constant of phenol is 1.0×10−101.0 \times 10^{-10}1.0×10−10. What is the concentration of phenolate ion in the solution?

Correct Answer: C

Q11.Which of the following is a characteristic of a buffer solution?

Correct Answer: A

Q12.Calculate the pH of a solution formed by dissolving 0.3 g of NaOH in water to give 200 mL of solution.

Correct Answer: B

Q13.According to the Bronsted-Lowry concept, what defines an acid?

Correct Answer: B

Q14.In a closed system, a liquid is in equilibrium with its vapor. If the temperature of the system is increased, what will happen to the vapor pressure and why?

Correct Answer: A

Q15.Which of the following best describes a weak electrolyte?

Correct Answer: B

Q16.What is the role of a catalyst in a chemical reaction at equilibrium?

Correct Answer: C

Q17.Which of the following reactions will have appreciable concentrations of both reactants and products at equilibrium?

Correct Answer: C

Correct Answer: A

Q19.According to Arrhenius, which of the following is true for acids?

Correct Answer: C

Q20.Which of the following statements correctly describes the behavior of a weak acid in water?

Correct Answer: B

Q21.According to Arrhenius, what defines a base?

Correct Answer: D

Q22.Which of the following statements best describes dynamic equilibrium?

Correct Answer: A

Correct Answer: A

Q24.In a closed container, a liquid reaches a state of equilibrium where the rate of evaporation equals the rate of condensation. If the temperature of the container is increased, what will be the effect on the equilibrium state?

Correct Answer: A

Q25.Which of the following statements is true according to Le Chatelier's principle when the pressure is increased in the reaction N2(g)+3H2(g)⇌2NH3(g)N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)N2​(g)+3H2​(g)⇌2NH3​(g)?

Correct Answer: A

Q26.What happens to the rate of evaporation and condensation when a liquid reaches equilibrium in a closed container?

Correct Answer: C

Q27.What is the effect of adding more reactants to a system at equilibrium?

Correct Answer: B

Q28.According to Arrhenius, what do acids produce in aqueous solutions?

Correct Answer: A

Q29.In the equilibrium reaction 2NO2(g)⇌N2O4(g)2NO_2(g) \rightleftharpoons N_2O_4(g)2NO2​(g)⇌N2​O4​(g), if the volume of the container is halved, what will be the effect on the equilibrium position?

Correct Answer: A

Q30.Calculate the pH of a solution in which the concentration of hydrogen ions is 3.8×10−3 M3.8 \times 10^{-3} \text{ M}3.8×10−3 M.

Correct Answer: A

Q31.In the context of chemical equilibria, what role does a catalyst play?

Correct Answer: C