Chemical Equilibrium MCQ Quiz in தமிழ் - Objective Question with Answer for Chemical Equilibrium - இலவச PDF ஐப் பதிவிறக்கவும்

Concept:

Equilibrium Constant (K): The equilibrium constant (K) is a dimensionless value that describes the ratio of the concentration of products to reactants at equilibrium for a reversible chemical reaction. The expression for the equilibrium constant depends on the balanced chemical equation for the reaction.

For the generic reaction: aA + bB ⇌ cC + dD

The equilibrium constant (Kc) expression is given by: 

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

where [A], [B], [C], and [D] are the molar concentrations of the reactants and products at equilibrium.

Explanation:

For, X(g) \(\rightleftharpoons \) Y(g)

Given K_c = 1.1:

\(K_c= \frac{[Y]}{[X]}=1.1\)

Therefore:

[Y] = 1.1 [X]

Let's consider different cases:

If ([X] = 1): [Y] = \(1.1 \times 1 = 1.1\), Therefore, both ([X]) and ([Y]) are greater than 1.

If ([X] > 1): [Y] = 1.1 [X] Since ([X]) is greater than 1, [Y] will also be greater than 1. For example, if [X] = 2, then [Y] = 2.2.

If ([X] < 1): Consider an initial concentration of ( [X] = 0.5 ). Then using: [Y] = \(1.1 \times 0.5 = 0.55\) In this case, [Y] would be less than 1. However, we are analyzing the molar concentrations that justify equilibrium constraints.

Conclusion:

Based on the equilibrium constant K_c = 1.1, both gases X and Y have molar concentrations greater than 1.

Concept:
The relationship between the equilibrium constants K_c (concentration) and K_p (partial pressure) for a gaseous reaction is given by the equation: K_p = K_c(RT)^Δn, where Δn is the change in the number of moles of gas (products minus reactants), R is the gas constant, and T is the temperature in Kelvin.

Explanation:
Consider the reaction:

2Ag ⇌ Bg + Cg

The change in the number of moles of gas (Δn) is calculated as follows:

Δn = (moles of products) - (moles of reactants)

Δn = (1 + 1) - 2

Δn = 2 - 2 = 0

Using the relationship K_p = K_c(RT)^Δn:

K_p = K_c(RT)⁰

K_p = K_c × 1

K_p = K_c

Conclusion:
The correct relationship between K_p and K_c for the reaction 2Ag ⇌ Bg + Cg is: K_p = K_c

Concept:

The stability of a chemical compound can be inferred using the equilibrium constant (K) of its decomposition reaction. A smaller equilibrium constant indicates that the reactants are favored over the products, meaning the compound is more stable. Conversely, a larger equilibrium constant implies the compound decomposes easily, suggesting it is less stable.

Explanation:

We are given four decompositions of nitrogen oxides with their equilibrium constants:

1. \(2 \mathrm{NO_2(g) \leftrightarrow N_2(g) + 2O_2(g)}; K = 6.7 \times 10^{16} \, \text{mol}^{-1} \, \text{litre}\)

2. \(2 \mathrm{NO(g) \leftrightarrow N_2(g) + O_2(g)}; K = 2.2 \times 10^{30}\)

3. \(2 \mathrm{N_2O(g) \leftrightarrow 2N_2(g) + 5O_2(g)}; K = 1.2 \times 10^{24} \, \text{mol}^{5} \, \text{litre}^{-5}\)

4. \(2 \mathrm{N_2O(g) \leftrightarrow 2N_2(g) + O_2(g)}; K = 3.5 \times 10^{33} \, \text{mol}^{-1} \, \text{litre}\)

The equilibrium constant is inversely proportional to the stability because equilibrium constant is ratio of concentration of product to concentration of reactant. Greater equilibrium constant means more product formation and less stability of reactant. 

Conclusion:

Therefore, the most stable oxide of nitrogen among the given options is: \(2 \mathrm{NO_2(g) \leftrightarrow N_2(g) + 2O_2(g)}; K = 6.7 \times 10^{16} \, \text{mol}^{-1} \, \text{litre}\)

Concept:

Equilibrium Constant (K): The equilibrium constant (K) is a dimensionless value that describes the ratio of the concentration of products to reactants at equilibrium for a reversible chemical reaction. The expression for the equilibrium constant depends on the balanced chemical equation for the reaction.

Equilibrium constant (Kp) is related to the pressures of the gases in a reaction at equilibrium. When a reaction is reversed, the new Kp is the reciprocal of the original Kp. When a reaction equation is multiplied by a factor, the new Kp is the original Kp raised to the power of that factor.

For a general reaction aA +bB → cC+dD 

\(K_p = {[P_C]^c[P_D]^d\over{[P_A]^a[P_B]^b}}\)

For cC+dD → aA +bB

\(K_p' = {[P_A]^a[P_B]^b\over{[P_C^c[P_D]^d}}={1\over{K_p}}\)

Explanation

Given:

SO₂(g) + 1/2 O₂(g) ⇌ SO₃(g) with K_p = 1.7 × 10¹²

K_p for:

2SO₃(g) ⇌ 2SO₂(g) + O₂(g)

Step 1: Reverse the given reaction

Reversing the given reaction:

SO₃(g) ⇌ SO₂(g) + 1/2 O₂(g)

K_p for this reaction = \({1\over{(1.7 × 10^{12})}}\)

Step 2: Multiply the reversed reaction by 2

2(SO₃(g) ⇌ SO₂(g) + 1/2 O₂(g))

This gives us:

2SO₃(g) ⇌ 2SO₂(g) + O₂(g)

Step 3: Adjust Kp for the multiplied reaction

When scaling a reaction by a factor of 2, the new Kp is the square of the original Kp:

\(K_p' = {1\over{(1.7 × 10^{12})^2}}\)

\(K_p' ≈ (5.882 × 10^{-13})^2\)

Conclusion

The equilibrium constant, K_p, for the reaction 2SO₃(g) ⇌ 2SO₂(g) + O₂(g) at 300 K is approximately 5.8 × 10^–26

Concept:

Nernst Distribution Law: It describes how a solute partitions between two immiscible solvents. At equilibrium, the ratio of the concentration of a solute in one solvent to its concentration in another solvent is constant.

Distribution Coefficient:

The distribution coefficient, K, quantifies how a solute divides between two solvents. It is calculated using the formula:

Where:

• K = distribution coefficient
• C₁ = concentration of the solute in solvent 1
• C₂ = concentration of the solute in solvent 2

Explanation:

Step 1: Initial Conditions

Let the initial concentrations of the solute in the upper and lower layers be [A]_u and [A]_l , respectively, with a distribution coefficient \( K_D = \frac{[A]_u}{[A]_l}\) .

Step 2: Effect of Volume Reduction

When the volume of the system is reduced to 1/4th of its original volume, the concentration of the solute in each layer will increase by a factor of 4:

• New concentration in the upper layer: \([A]_u' = 4[A]_u\)

• New concentration in the lower layer: \([A]_l' = 4[A]_l\)

Step 3: Rate of Reaction

If the rate of reaction is proportional to the concentration of the solute, the new rates will be:

• New rate of reaction in the upper layer: \(R_u' = k \times 4[A]_u = 4R_u\)

• New rate of reaction in the lower layer: \(R_l' = k \times 4[A]_l = 4R_l\)

Step 4: Relative Rates Using Nernst Distribution

By Nernst distribution law, the ratio of concentrations remains constant:

\\(\frac{[A]_u'}{[A]_l'} = \frac{4[A]_u}{4[A]_l} = \frac{[A]_u}{[A]_l} = K_D\)

Hence, the relative rates will remain constant:

\(\frac{R_u'}{R_l'} = \frac{4R_u}{4R_l} = \frac{R_u}{R_l}\)

Conclusion

The rate of reaction (relative to the original rate) would be: In fixed ratio with that in the lower layer

Concept:

Equilibrium Constant (K): The equilibrium constant (K) is a dimensionless value that describes the ratio of the concentration of products to reactants at equilibrium for a reversible chemical reaction. The expression for the equilibrium constant depends on the balanced chemical equation for the reaction.

For the generic reaction: aA + bB ⇌ cC + dD

The equilibrium constant (K_c) expression is given by: \(K_c = {[C]^c[D]^d\over{[A]^a[B]^b}}\)

where [A], [B], [C], and [D] are the molar concentrations of the reactants and products at equilibrium.

Relations Between Different Equilibrium Constants:

There are different forms of equilibrium constants depending on the phase of the reactants and products:

K_c vs. K_p:

K_c is the equilibrium constant in terms of concentration, while K_p is the equilibrium constant in terms of partial pressures of gases. For a reaction involving gases, the relationship between K_c and K_p is given by:

K_p = K_c(RT)^Δn

where R is the gas constant, T is the temperature in Kelvin, and Δn is the change in the number of moles of gas (moles of products - moles of reactants).

K_forward vs. K_reverse

The equilibrium constant for the forward reaction (K_forward) and the reverse reaction (K_reverse) are reciprocals:

\(K_{reverse} = {1\over{K_{forward}}}\)

Overall Reaction Equilibrium Constant

If a reaction can be expressed as the sum of two or more reactions, the overall equilibrium constant is the product of the equilibrium constants for the individual steps:

K_overall = K₁ × K₂ × ... × K_n

Relation with Reaction Quotient (Q)

The reaction quotient (Q) is calculated using the same expression as K but for non-equilibrium conditions. Comparing Q with K helps predict the direction the reaction will proceed to reach equilibrium:

• Q = K: The system is at equilibrium.

• Q < K: The reaction will proceed forward (towards products) to reach equilibrium.

• Q > K: The reaction will proceed in reverse (towards reactants) to reach equilibrium.

Explanation:

Given the equilibrium constant for the dissociation reaction:

[Ag(CN)₂]^– ⇌ Ag⁺ + 2CN^–

K_c = 4.0 × 10^–19

We can use the initial concentrations to determine the extent of the reaction.

Assume complex formation reaction is complete:

Ag⁺ + 2CN^– ⇌ [Ag(CN)₂]^–

The equilibrium constant for this reverse reaction can be approximated as K_reverse = 1 / K_c

K_reverse ≈ 2.5 × 10¹⁸

Due to the large K_reverse, almost all Ag⁺ ions will form the complex with CN^–.

If we start with 0.03 M AgNO₃ (thus 0.03 M Ag⁺) and 0.1 M KCN (thus 0.1 M CN^–), almost all Ag⁺ will form [Ag(CN)₂]^–.

[Ag(CN)₂]^– formed ≈ 0.03 M

Remaining [CN^–] ≈ 0.1 - 2 × 0.03 = 0.04 M

Using the equilibrium constant for dissociation:

K_c = [Ag⁺][CN^–]² / [[Ag(CN)₂]^–]

Substitute the known values into the expression:

4.0 × 10^–19 = [Ag⁺] × (0.04)² / 0.03

Solve for [Ag⁺]:

[Ag⁺] = 4.0 × 10^–19 × 0.03 / (0.04)²

[Ag⁺] = 4.0 × 10^–19 × 0.03 / 0.0016

[Ag⁺] = 7.5 × 10^–18 M

Conclusion:

The concentration of Ag⁺ ions in the solution is 7.5 × 10^–18 M.

Concept:

The equilibrium constants for a perfect gaseous mixture are \(K_{P}\) and \(K_{C}\).

\(\mathbf{K_{P}}\): When gas phase equilibrium concentrations are stated in terms of atmospheric pressure, the equilibrium constant is designated by Kp.

\(\mathbf{K_{C}}\): When equilibrium concentrations are expressed in molarity, the equilibrium constant is expressed by Kc.

The relationship between \(K_{P}\) and \(K_{C}\) is:

\(K_{P}\;=\;K_{C}(RT)^{\Delta n_{g}} \)

Where \(\Delta n_{g}\) is the number of products - number of moles of reactants

Explanation:

The given reaction is:

\(4NH_{3}(g)\;+\;7O_{2}(g)\;\rightleftharpoons \;6H_{2}O(g)\;+\;4NO_{2}(g)\)

The equilibrium constant of the reaction is expressed as:

The product and reactant both are in the gaseous phase, so the concentration of both comes under the equilibrium equation:

\(K_{P}\;=\;\dfrac{[NO_{2}]^{4}\;[H_{2}O]^{6}}{[NH_{3}]^{4}\;[O_{2}]^{7}}\)

Calculating the value of \(\Delta n_{g}\):

\(\Delta n_{g}\;=\;10-11\)

\(\Delta n_{g}=-1\)

The  \(K_{P}\) and \(K_{C}\) relationship is:

\(K_{P}\;=\;K_{C}(RT)^{\Delta n_{g}}\)

OR

\(\mathbf{K_{P}\;=\;\dfrac{K_{C}}{RT}}\)

Conclusion:

For the reaction, \(\mathbf{4NH_{3}(g)\;+\;7O_{2}(g)\;\rightleftharpoons \;6H_{2}O(g)\;+\;4NO_{2}(g)}\) the \(\mathbf{K_{P}}\) value is \(\mathbf{K_{P}}\;\mathbf{=\;\dfrac{K_{C}}{RT}}\)

Concept:

Degree of Dissociation - It is the fraction of a mole of reactant that undergoes dissociation in a chemical reaction is called the degree of dissociation.

It is represented by α.

Factor affecting degree of dissociation -

1. Nature of electrolye
2.  Temperature
3. Nature of solvent

Calculation:

XY(g) ⇋ X(g) + Y(g)

Given,α = 33% or o.33

Total no. of moles = x -0.33x +0.33x +0.33x = x + 0.33x = 1.33x

\(K_p = \frac{P_A × P_B}{P_{AB}} \;\;\;\;------(1)\)

and we know,

 \(P_{AB} =x_{AB}× P_t\)

And P_t = P

∴ P_AB = \(\frac{0.67c}{1.33c}× P\)     ------(2)

\(P_A = \frac{0.33c}{1.33c}× P \)        -------(3)

\(P_B = \frac{0.33c}{1.33c}× P\)        -----(4)

Put the value of equation (2), (3) and (4) in (1)

\(K_p = \frac{\frac{0.33c}{1.33c}× P × \frac{0.33c}{1.33c}× P}{\frac{0.67c}{1.33c}× P}\)

\(K_p = 0.122 P\)

\(P = \frac{K_p}{0.122}\) = 8.2 × K_p ≈  8Kp

or

P = 8Kp

Conclusion:

Therefore,  the correct relation between P and K p is -

P = 8Kp

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:

Relationship between Kc and Kp

• The relationship between the equilibrium constant in terms of concentration (Kc) and the equilibrium constant in terms of pressure (Kp) is given by the equation:

  Kp = Kc(RT)^Δn

• Here, R is the gas constant, T is the temperature in Kelvin, and Δn is the change in the number of moles of gas.
• Δn is calculated as the difference between the number of moles of gaseous products and the number of moles of gaseous reactants.

EXPLANATION:

• For the given reaction:

  NH₄Cl (s) ↔ NH₃ (g) + HCl (g)

• Identify the number of moles of gases on both sides of the reaction:
  • Reactants: NH₄Cl (s) (0 moles of gas)
  • Products: NH₃ (g) (1 mole of gas) + HCl (g) (1 mole of gas)
• Calculate Δn:
  • Δn = (moles of gaseous products) - (moles of gaseous reactants)
  • Δn = (1 + 1) - 0
  • Δn = 2

Therefore, the value of Δn for the reaction NH₄Cl (s) ↔ NH₃ (g) + HCl (g) is 2.