Equilibrium MCQ Quiz - Objective Question with Answer for Equilibrium - Download Free PDF

CONCEPT:

Solubility Product (K_sp) and Solubility (S)

• The solubility product constant (K_sp) is the equilibrium constant for a solid substance dissolving in an aqueous solution.
• For a salt dissolving into its constituent ions, the relation between solubility (S) and K_sp depends on the stoichiometry of the dissociation.
• The general method is to:
  • Write the dissociation equation
  • Express ion concentrations in terms of solubility S
  • Substitute into the K_sp expression and solve for S

EXPLANATION:

• A - II: AB type (e.g. AgCl)
  • Dissociation: AB ⇌ A⁺ + B⁻
  • K_sp = S × S = S² ⇒ S = √K_sp
• B - I: AB₂ type (e.g. PbI₂)
  • Dissociation: AB₂ ⇌ A²⁺ + 2B⁻
  • K_sp = (2S)² × S = 4S³ ⇒ S = ∛(K_sp/4)
• C - III: AB₃ type (e.g. Al(OH)₃)
  • Dissociation: AB₃ ⇌ A³⁺ + 3B⁻
  • K_sp = S × (3S)³ = 27S⁴ ⇒ S = (K_sp/27)^1/4
• D - IV: A₃B₂ type (e.g. Ca₃(PO₄)₂)
  • Dissociation: A₃B₂ ⇌ 3A²⁺ + 2B³⁻
  • K_sp = (3S)³ × (2S)² = 108S⁵ ⇒ S = ∛(K_sp/108)

Therefore, the correct answer is: Option 1) A - I, B - II, C - III, D - IV

CONCEPT:

Effect of Temperature and Concentration on Chemical Equilibrium

• The given reaction is:

  N₂(g) + O₂(g) → 2NO(g)

• This reaction is endothermic because ΔH = +180.7 kJ mol⁻¹.
• According to Le Chatelier's Principle:
  • For an endothermic reaction (ΔH > 0), increasing the temperature shifts the equilibrium toward the products (higher yield of NO).
  • Increasing the concentration of reactants (N₂ and O₂) also shifts the equilibrium toward the products, increasing the yield of NO.

EXPLANATION:

• Higher temperature: Since the reaction is endothermic, increasing the temperature supplies more energy to drive the reaction forward.
• Higher concentration of N₂: Adding more N₂ shifts the equilibrium toward the products.
• Higher concentration of O₂: Adding more O₂ also shifts the equilibrium toward the products.

Therefore, the correct answer is A, C, D only.

CONCEPT:

Relation Between K_C and K_P

• The equilibrium constant Kp is related to Kc by the equation:

  K_p = K_C (RT)^Δn_g

• Where:
  • K_C is the equilibrium constant in terms of concentration.
  • R is the gas constant (0.0831 L atm mol^-1 K^-1).
  • T is the temperature in Kelvin (1000 K in this case).
  • Δn_g is the change in the number of moles of gas (products - reactants).
• For the given reaction, the backward rate constant is 2500 times the forward rate constant, and we are asked to calculate Kp at 1000 K.

EXPLANATION:

• The relationship between the rate constants for the forward and backward reactions is:

  \(KC = \frac{k{\text{f}}}{k{\text{b}}}\)

  Given that the backward rate constant is 2500 times the forward rate constant, we have:

  \(KC = \frac{k{\text{f}}}{2500 k{\text{f}}} = \frac{1}{2500}\)

• Next, we use the equation to find Kp :

  K_p = K_C (RT)^Δn_g

  Given that Δn_g = 1 (because there is a change in the number of moles from 1 mole of reactants to 2 moles of products), we substitute the known values:

  \(Kp = \frac{1}{2500} (0.0831 × 1000)^1\)

• Solving the equation:

  \(Kp = \frac{1}{2500} \times 83.1 = 0.033\)

Therefore, the correct value of K_p is 0.033.

CONCEPT:

Phosphoric Acid Ionization and Acid Strength

• Phosphoric acid (H₃PO₄) ionizes in three steps, leading to the formation of H₂PO₄^-, HPO₄^2-, and PO₄^3- respectively.
• Each ionization step has an associated ionization constant, denoted as Kₐ₁, Kₐ₂, and Kₐ₃.
• The overall ionization constant (K) is the product of the individual ionization constants: K = Kₐ₁ × Kₐ₂ × Kₐ₃.
• The logarithm of the overall ionization constant is the sum of the logarithms of the individual constants: log K = log Kₐ₁ + log Kₐ₂ + log Kₐ₃.
• The strength of an acid decreases as it ionizes further (i.e., H₃PO₄ is stronger than H₂PO₄^-, which is stronger than HPO₄^2-).

EXPLANATION:

• Statement A: log K = log Kₐ₁ + log Kₐ₂ + log Kₐ₃ is correct because the overall ionization constant is the product of the individual ionization constants, and the logarithm of a product is the sum of the logarithms of the individual factors.
• Statement B: H₃PO₄ is stronger than H₂PO₄^- and HPO₄^2- is correct because the acid strength decreases with successive ionizations. The first ionization constant (Kₐ₁) is the largest, making H₃PO₄ the strongest acid among its ionized forms.
• Statement C: Kₐ₁ > Kₐ₂ > Kₐ₃ is correct because the ionization constants decrease sequentially as the molecule loses protons, reflecting decreasing acid strength with each step.
• Statement D: K = (Kₐ₁ + Kₐ₂)/2 is incorrect because the overall ionization constant is the product of Kₐ₁, Kₐ₂, and Kₐ₃, not an average of the first two constants.

Therefore, the correct answer is A, B, and C only.

CONCEPT:

Neutralisation Reaction using N₁V₁ = N₂V₂ Method

• In acid-base neutralisation, the milliequivalents of acid = milliequivalents of base.
• The formula used is:

  N_acid × V_acid = N_base × V_base

• Normality (N) = Molarity × n-factor
• For H₃PO₃ (phosphorous acid), only 2 hydrogen ions are ionisable (n-factor = 2).
• For Ca(OH)₂, each molecule provides 2 OH^– ions (n-factor = 2).

EXPLANATION:

• Given:
  • Molarity of H₃PO₃ = 0.2 M
  • Volume of H₃PO₃ = 0.25 L
  • Molarity of Ca(OH)₂ = 0.1 M
• Step 1: Calculate Normalities
  • N_acid = 0.2 × 2 = 0.4 N
  • N_base = 0.1 × 2 = 0.2 N
• Step 2: Apply formula:

  N_acid × V_acid = N_base × V_base

  0.4 × 250 mL = 0.2 × V_base

  V_base = (0.4 × 250) / 0.2 = 500 mL

Therefore, the volume of Ca(OH)₂ required to neutralise the acid is 500 mL.

Concept:

The pH Scale

• The pH is the measure of the concentration of acid in the solution. 
• It is measured from the scale of 1 to 14. 
• If the solution is acidic or basic, depends upon the following parameters. 
• pH < 7 - Acidic
• pH > 7 - Basic
• pH = 7 - Neutral

​Conclusion:

• So, from the above explanation, it is clear that the Higher the pH, the weaker is the acid, and the Lower the pH, the weaker is the base.
• So, Statements 2 and 4 are correct. ​

Additional Information

• The pH of water at 25°c is 7. When it is heated to 100°C, the pH of water decreases to 6.14.
• On increasing the temperature of the water, the equilibrium would again move to lower the temperature.
• It would happen by absorbing the extra heat. 
• Hydrogen ions and hydroxide ions will remain at the same concentration in pure water and the water remains neutral even if its pH changes.

Concept:

pH of acidic buffer - 

• Buffer solution - Buffer solution is a mixture of a weak acid and its conjugate base or a weak base and its conjugate acid. It is a solution highly known for maintaining its pH on dilution with a small amount of strong acid or strong base by maintaining the H⁺ ion concentration inside it. 
• Two types of buffer solutions are there
• Acidic buffer - Highly specific for pH<7 and is a mixture of a weak acid and its conjugate base.
• Weak buffer - Highly specific for pH>7 and is a mixture of a weak base and its conjugate acid.
• pH of the acidic buffer is given by the formula -

pH = pKa + log ([salt]/[acid])

where pK_a = negative logarithms of the acid dissociation constant.

As acetic acid is a weak acid and sodium acetate is its conjugate salt with a strong base(NaOH), they form an acidic buffer whose pH is calculated by the formula -

pH = pKa + log ([salt]/[acid])

Chemical reaction - CH₃COOH + NaOH \(\rightarrow\) CH₃COONa +H₂O

Calculation -

Given,

• pKa = 4.57 (negative logarithms of the acid dissociation constant of acetic acid)
• [acid] = 0.01 M (concentration of acetic acid)
• [salt] = 0.10 M (concentration of salt)

Put all these values in the formula,

pH = pKa + log ([salt]/[acid])

=4.57 + log(0.10 / 0.01)

=4.57 + log(10)

= 4.57 + 1   -- (∵ log10 = 1)

=5.57

So, the pH of the buffer solution is = 5.57

Hence, the correct answer is option 2.

Chemical Equilibrium: Consider a general case of a reversible reaction

A + B ⇔ C + D

• With respect of time, the rate of forward reaction decreased, and the rate of backward reaction increased.
• With an instant of time, a stage is reached at which the rate of forward and reverse reaction become equal and the concentration of reactant and product becomes constant.
• The equilibrium is dynamic in 

The correct answer is Kolbe.

Key Points

• Synthesis of Acetic Acid
  • Acetic acid was first made or created from inorganic components in 1845 by German scientist Hermann Kolbe.
  • He created carbon disulfide with chlorine, and the end result was carbon tetrachloride.
  • After that, tetrachloroethylene was produced through thermal decomposition, trichloroacetic acid was created through aqueous chlorination, and acetic acid was finally produced through an electrolytic reduction.
  • He exclusively used inorganic materials made of hydrocarbons.
  • Carbon, hydrogen, and oxygen are inorganic elements that are used to create acetic acid while acetic acid is an organic one.
  • Acetic acid has a chemical formula and is a weak acid.
  • Kolbe's work was groundbreaking since he synthesised organic chemicals from inorganic ones.
  • In addition to bacterial fermentation, acetic acid is also made synthetically.
  • Some fruits naturally contain it.

Additional Information

• Kolbe's work was groundbreaking since he synthesised organic chemicals from inorganic ones. 
• In addition to bacterial fermentation, acetic acid is also made synthetically.
•  some fruits naturally contain it.

The correct answer is Less than 7

Key Points 

• When a salt is formed from the neutralization of a strong acid and a weak base, the resulting solution is acidic. This is because the strong acid completely dissociates in water, providing a high concentration of (H+) ions, while the weak base does not dissociate completely and therefore provides fewer (OH^-) ions to neutralize the (H+). As a result, the (H⁺) concentration in the solution is higher than the (OH-) concentration, leading to an acidic solution.

Therefore, the pH value of a salt made up of a strong acid and a weak base is Less than 7

• The cations of strong acids do not hydrolyze and therefore the solutions of salts formed by strong acids and strong bases are neutral, i.e. their pH is 7, whereas the solutions of salts formed by strong bases and weak acids are alkaline, i.e. their pH>7.
• Salts that are from strong bases and weak acids do hydrolyze, which gives them a pH greater than 7.
• The pH value range of a weak base is 7.1 to 10 whereas a strong base is 10.1 to 14.
• All substances can be classified as neutral (with a pH of about 7), basic (pH greater than 7), or acidic (pH less than 7), and the pH tells us how strong or weak that substance is as well. For example, a substance with pH = 8 is a very weak base, but a substance with pH = 3 is a strong acid.

The correct answer is II Only.

Key Points List of Acid and Natural Resources:

Explanation:

A reversible reaction is a chemical reaction where the reactants form products that, in turn, react together to give the reactants back. Reversible reactions will reach an equilibrium point where the concentrations of the reactants and products will no longer change.

According to equilibrium constant, K_c

\({{\rm{K}}_{\rm{c}}} = \frac{{{{\left[ {{\rm{product}}} \right]}^{\rm{m}}}}}{{{{\left[ {{\rm{reactant}}} \right]}^{\rm{n}}}}}\)

Consider the given first equation:

\({A_2}\left( g \right)\; + \;{B_2}\;\left( g \right)\begin{array}{*{20}{c}} {{K_1}}\\ \rightleftharpoons \end{array}\;2AB\left( g \right)\)

\({K_1} = \frac{{{{\left[ {AB} \right]}^2}}}{{\left[ {{A_2}} \right]\left[ {{B_2}} \right]}}\)

Consider the given second equation:

\(6AB\left( g \right)\begin{array}{*{20}{c}} {{K_2}}\\ \rightleftharpoons \end{array}\;3{A_2}\left( g \right) + 3{B_2}\left( g \right)\)

\({K_2} = \frac{{{{\left[ {{A_2}} \right]}^3}{{\left[ {{B_2}} \right]}^3}}}{{{{\left[ {AB} \right]}^6}}}\)

\(= \frac{1}{{{{\left( {\frac{{{{\left[ {AB} \right]}^2}}}{{\left[ {{A_2}} \right]\left[ {{B_2}} \right]}}} \right)}^3}}}\)

\(= \frac{1}{{K_1^3}}\)

 → K2 = K1-3

 Concept:

Lewis concept of acids and bases is given as follows:

The interaction between Lewis acid and base- 

• Lewis acid has a vacant unoccupied orbital called LUMO.
• Lewis bases have fulfilled occupied orbital called HOMO.
• Electron density from HOMO is donated to LUMO of Lewis acid.
• A simple HOMO and LUMO interaction is shown below:

This donation of electron density forms a covalent coordinate between the acid and the base,

Explanation:

BCl₃ -

• BCl₃ molecule has an empty p orbital because its octet is not fulfilled.
• It can accept an electron pair inti this orbital

Thus it electron-deficient and a Lewis acid

FeCl3:

• Fe is a transition metal having empty d orbitals.
• It can easily accept a lone pair in the empty 4d orbital and can act as Lewis acid.

​

Hence, BCl3, FeCl3 is a pair of Lewis acids.

Additional Information

• AlCl3: AlCl3 also has empty p orbitals and thus is an electrophile.

• Proton acceptors are Bronsted-Lowry bases, while Proton donors are Bronsted-Lowry acids

Thus, HCl, HNO3 and H₂CO₃ NH4+ are Bronsted acids.

The correct answer is high pressure and low temperature.

Key Points

• This equation is an example of Equilibrium.
• Equilibrium is a condition that occurs when a chemical reaction is reversible, and the forward and reverse reactions occur simultaneously, at the same rate.
• Le Chatlier's Principle- "if a chemical system at equilibrium experiences a change in concentration, temperature or total pressure, the equilibrium will shift in order to minimize that change" 
• Any decrease in nitrogen or hydrogen pulls the reaction towards the left side.
• Any decrease in ammonia or temperature pulls the reaction towards the right side.