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

CONCEPT:

Le Chatelier's Principle and the Effect of Inert Gases

• Le Chatelier's principle states that if a system at equilibrium is disturbed by changing the conditions (such as pressure, temperature, or concentration), the system will adjust itself to counteract the change and restore equilibrium.
• In the case of adding an inert gas (like xenon) to a system at constant temperature and pressure, the addition of the gas does not affect the partial pressures of the reactants or products. However, it increases the total volume of the system, which effectively reduces the partial pressures of the reactants and products.
• At constant pressure, adding an inert gas will shift the equilibrium towards the side with a greater number of moles of gas, as this compensates for the decrease in partial pressure of the reacting species. This is because the system will try to increase the number of moles to increase the pressure again.

EXPLANATION:

• The reaction given is:

  PCl₅ (g) ⇌ PCl₃ (g) + Cl₂ (g)

• On adding xenon (an inert gas) at constant temperature and pressure:
  • The volume of the system increases, which causes the partial pressures of the components to decrease.
  • The equilibrium will shift towards the side with more moles of gas to counteract this change and restore equilibrium. In this case, the right-hand side of the reaction has 2 moles of gas (PCl₃ and Cl₂) compared to the left-hand side, which has only 1 mole (PCl₅).
  • This results in an increase in the concentration of PCl₃ and Cl₂ and a decrease in the concentration of PCl₅.

Therefore, the correct answer is: PCl₃ will increase.

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:

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:

Molarity (M)

• Molarity is defined as the number of moles of solute per litre of solution.
• It is calculated using the formula:

  M = (moles of solute) / (volume of solution in L)

• Moles of solute = Given molarity × initial volume (in L)
• If some solute is lost, subtract the lost moles from initial moles before recalculating the final molarity using the new volume.

EXPLANATION:

• Initial Data:
  • Initial volume = 1000 mL = 1 L
  • Initial molarity = 0.5 M
  • Moles of HCl initially = 0.5 × 1 = 0.5 mol
• Loss of HCl:
  • Molecular weight of HCl ≈ 36.5 g/mol
  • Moles lost = 2.25 g / 36.5 g/mol ≈ 0.0616 mol
• Moles left = 0.5 – 0.0616 = 0.4384 mol
• Final volume = 750 mL = 0.75 L
• Final Molarity = 0.4384 mol / 0.75 L = 0.5845 M
• Rounded to two decimal places: 0.58 M ≈ 0.60 M (as per given range)

Therefore, the molarity of the resulting solution is approximately 0.60 M.

Calculation:

We know that, K_P = K_c(RT)^∆ng

K_P and K_c are equilibrium constant.

R is gas constant

T is temperature

∆n_g is differences of moles.

∴ If ∆ng ≠ 0 then K_p ≠ K_c

Now, 2C (s) + O₂ (g) ⇌ 2CO (g)

∆n_g = +1

∴ K_P = K_c (RT)¹

Concept:

Van't Hoff equation -The Van't Hoff equation gives the relationship between the standard Gibbs free energy change and the equilibrium constant.

It is represented by the equation -

-ΔG° = RT log_eK_p

where, R = gas constant 

T = temperature 

K_p = equilibrium constant

or 

\(\frac{dlnK}{dT} =\frac {\Delta G}{RT^2}\)

Thus, the equilibrium constant K is temperature-dependent.

Explanation:

The given reaction is -

N2 (g) + 3H2 (g) \(\rightleftharpoons\) 2NH3 (g)

Given condition - the total pressure at which the equilibrium is established, is increased without changing the temperature or at a constant temperature.

According to Van't Hoff equation, the equilibrium constant K is temperature-dependent. 

If the temperature is constant, the reaction is neither exothermic nor endothermic.

Therefore, K will remain same as temperature is constant.

Conclusion:

Hence, if the total pressure at which the equilibrium is established, is increased without changing the temperature the equilibrium constant K for the reaction 

N2 (g) + 3H2 (g) \(\rightleftharpoons\) 2NH3 (g), remains same.

 Hence, the correct answer is option 1.

Concept:

• pH (Power of Hydrogen): The degree of acidity or alkalinity of a substance is expressed in pH value.
• pH is a scale (0 to 14) used to specify the acidity or basicity of an aqueous solution.
• Lower pH values of solutions represent more acidic in nature,
• While higher pH values of solutions represent more basic or alkaline.

Normality: It is the concentration of a solution expressed as no. of gram equivalents per litre solution.

\({\rm{Normality}} = \frac{{{\rm{Gram\;equivalents}}}}{{{\rm{volume\;of\;solution\;in\;liter}}}}\)

Where, \({\rm{No}}.{\rm{\;of\;gram\;equivalents}} = \frac{{{\rm{Given\;weight}}}}{{{\rm{Equivalent\;weight}}}}\)

\({\rm{And\;equivalent\;weight}} = \frac{{{\rm{Molecular\;weight}}}}{{{\rm{Valency}}}}\)

Calculation:

Given that, 

Volume of N/5 of NaOH = 40 ml

Normality NaOH = N/5

The volume of N/5 of HCl= 60 ml

Normality HCl= N/5

NaOH + HCL → NaCl + H2O

N1V1 - N2V2 = N3(V1 + V2)

N3 = (12 - 8)/ 100 = 4 X 10-2

Hydrogen ion conc. = 4 X 10-2

 pH = -log[H+].

∴ pH = -log [4 X 10-2] = 1.397 = 1.4

• The scale runs from 0 to 14, with acids having a pH less than 7, 7 being neutral, and bases having a pH higher than 7.
• Acids and bases react with each other in what is called a neutralization reaction.

CONCEPT:

Le Chatelier's Principle

• Le Chatelier’s principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change.
• Common disturbances include changes in concentration, pressure, and temperature.
• In the context of a chemical reaction, the system will adjust in such a way as to partially oppose the disturbance.

APPLICATION TO THE REACTION:

• The given reaction is:

  \(Fe_2O_3(s) + 3CO(g) \rightleftharpoons 2Fe(s) + 3CO_2(g)\)

1. Addition of Fe₂O₃:

• Since Fe₂O₃ is a solid, its addition generally does not affect the equilibrium position significantly.

1. Addition of CO₂:
   • Adding a product will shift the equilibrium to the left, favoring the formation of reactants.
2. Removal of CO:
   • Removing a reactant will shift the equilibrium to the left, favoring the formation of reactants.
3. Removal of CO₂:
   • Removing a product will shift the equilibrium to the right, favoring the formation of products.

CONCLUSION:

The correct answer is (addition of Fe₂O₃

Concept:
Equilibrium Constant

For the dissociation reaction of oxygen:

\(\text{O}_2 \leftrightarrow 2\text{O}\)

The equilibrium constant K is defined as:

\(K = \frac{[\text{O}]^2}{[\text{O}_2]}\)

At high temperatures, the dissociation of O2  to oxygen atoms is more significant. However, at 1000 K, the equilibrium concentration of atomic oxygen (O) is generally quite low, indicating a small value for K .

Analysis of Given Options

1. 3.3 × 10-20
   • This is a very low value, indicating very little dissociation of {O}_2 to {O} .
2. 2.5 × 10-10
   • A small value representing minimal dissociation, but larger than option 1.
3. 5.2 × 10-7
   • This value is higher and might be reasonable for some equilibrium reactions, but not typical for the dissociation of {O}_2 at 1000 K.
4. 1.4 × 10-37
   • Extremely low, suggesting virtually no dissociation, which is unrealistic for a measurable equilibrium constant at 1000 K.

At 1000 K, oxygen molecules ( O2 ) do not dissociate significantly into oxygen atoms ( {O} ) due to the strong O=O double bond. The equilibrium constant for such a reaction is usually very small, but still within a measurable range.

Conclusion

The most reasonable equilibrium constant for the dissociation reaction of oxygen at 1000 K would be 2.5 × 10^-10

Concept:

• Reversible reactions involve the condition where the rate of forward reaction becomes equal to the rate of backward reaction.
• It is termed an equilibrium condition.

Explanation:

• Consider a hypothetical reversible reaction: \(\rm aA(g) +bB(g) \rightleftharpoons cC(g)+dD(g)\)
• The expression for the rate constant of the forward reaction is shown below:
• \(\rm K_f=\dfrac{C^c \times D^d}{A^a \times B^b}\)------(1)
• The expression for the rate constant of the backward reaction is shown below:
• \(\rm K_b=\dfrac{A^a \times B^b}{C^c \times D^d}\)------(2)
• Thus, from (1) and (2) the relation between \(\rm K_f\,and\,K_b\) is shown below:
• \(\rm K_f=\frac{1}{K_b}\)

Concept:

Vapour pressure and its relation with boiling point - 

• The pressure exerted by the vapours on the liquid surface when equilibrium is achieved between the liquid and vapour is called the vapour pressure.
• The temperature at which vapour pressure of liquid becomes equal to the atmospheric pressure is called the boiling point of the liquid.
• The vapour pressure is inversely proportional to the boiling point.
• If the intermolecular forces are weak then the liquid will have low boiling point and high vapour pressure.

Explanation:

We know that the greater is the boiling boiling point the lower will be the vapour pressure as both are inversely related to each other.

The given Compounds are having boiling point in order - Ether < acetone < water.

Therefore, the order of vapour pressure will be exactly reverse to boiling point.

And hence the order of vapour pressure is - Water < acetone < ether.

Conclusion:

The correct order of vapour pressure of water, acetone and ether at 30°C is Water < acetone < ether when water has maximum boiling point and ether has minimum boiling point.

Hence, the correct answer is option 2.

Concept:

Relation between standard Gibbs free energy (∆GΘ)and equilibrium constant(K) - 

The change is Gibbs free energy G is represented by ΔG.

If K is the equilibrium constant then the relation between standard Gibbs free energy and K is given by the formula - 

∆GΘ  = - RT lnK

where, R is the gas constant and T is the temperature.

Explanation:

For the reaction  A \(\rightleftharpoons\) B, the equilibrium constant K is given as -

K =\(\frac{[product]}{[reactant]}\) =  \(\frac{[B]}{[A]}\)

At half completion of the reaction, the concentration of the reactant and product are equal.

Therefore, [A] = [B]

Put it in the above equation, we get the value of K.

K = \(\frac{[B]}{[A]}\) = 1

We know that ∆GΘ  = - RTlnK

 ∆GΘ  = - RT ln1

As ln 1 = 0

 ∆GΘ  = - RT × 0

 ∆GΘ  = 0

Conclusion:

Therefore for the stage of half completion of the reaction A \(\rightleftharpoons\) B, the value of ∆GΘ  is equal to zero or ∆GΘ = 0.

Hence, the correct answer is option 1.

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.

The correct answer is Acetone

Concept:-

• Vapor Pressure: It is the pressure exerted by a vapor in thermodynamic equilibrium with its condensed phases at a specific temperature. In a closed container, the particles of a volatile liquid will continue to evaporate until a state of dynamic equilibrium is reached, where the vapor pressure above the liquid is constant.
• Boiling Point: Boiling point of a substance is the temperature at which the vapor pressure of the liquid equals the environmental pressure surrounding the liquid. If the vapor pressure of the liquid is high at a given temperature, it means that the liquid can more readily form vapor at that temperature, so it has a lower boiling point.
• Correlation between Vapor Pressure and Boiling Point: Vapor pressure and boiling point have an inverse relationship. This means that substances with higher vapor pressures have lower boiling points and vice versa. The reason is that a high vapor pressure indicates a high propensity of the substance to transition from the liquid phase to the gaseous phase, so it doesn't need as much heat to boil.

Explanation:-

The substance with the lowest boiling point will be the one with the highest vapor pressure at 293 K.

Comparing the vapor pressures:

• The vapor pressure of water at 293 K: 2.34 kPa
• The vapor pressure of acetone at 293 K: 12.36 kPa
• The vapor pressure of ethanol at 293 K: 5.85 kPa

The substance with the highest vapor pressure at 293 K is acetone. Therefore, according to the relationship between the boiling point and vapor pressure, acetone has the lowest boiling point among the substances.

Conclusion:-

So, the answer is 2) Acetone.