Equilibrium Law's And Equilibrium Constant MCQ Quiz - Objective Question with Answer for Equilibrium Law's And Equilibrium Constant - Download Free PDF

Concept:

1. Degree of Dissociation (α)

The degree of dissociation (α) represents the fraction of the initial amount of a compound that dissociates into its ions or smaller molecules at equilibrium.

2. Equilibrium Constant (K_eq)

The equilibrium constant (K_eq) is given by:

\(K_{eq} = \frac{[Products]}{[Reactants]}\)

3. Relationship Between K_eq and α

Dissociation of a Weak Acid

For a weak acid HA dissociating:

HA ↔ H⁺ + A⁻

The equilibrium constant K_a is:

\(K_a = \frac{[H+][A−]}{[HA]} = \frac{C^2 α^2}{C(1 - α)}=\frac{C α^2}{1 - α}\)

Explanation:

Consider the decomposition reaction:

X A ⇌ C + D

Denote the initial concentration of A as [A]₀.

• Initial moles of A: [A]₀

• Initial moles of C: 0

• Initial moles of D: 0

At equilibrium, a fraction α of A has decomposed:

• Moles of A decomposed: α [A]₀

• Moles of A remaining: [A]₀ (1 - α)

• Moles of C formed: \(α [A]_0\over{X}\)

• Moles of D formed: \(α [A]_0\over{X}\)

Total moles before the reaction: [A]₀

Total moles after the reaction:

\([A]_0 (1 - α) + ({2α [A]_0\over{X}})\) = [A]₀

For the moles to remain constant:

\(1 - α + {2α\over{X}} = 1\)

Simplifying:

\(-α + {2α\over{X}} = 0\)

X = 2

Conclusion:

The value of X is 2.

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:

Dissociation of Gaseous Compound and Degree of Dissociation

• For the reaction: AB2(g) ⇌ AB(g) + 1/2 B2(g)
• Let the initial pressure of AB2 be P, and the degree of dissociation be x.
• The equilibrium pressures will be:
  • AB2: (1 - x)P
  • AB: xP
  • B2: (x/2)P
• For small values of x (x << 1), approximations can be made: (1 + x/2) ≈ 1 and (1 - x) ≈ 1.

Explanation:-

\(\mathrm{AB}_{2(\mathrm{g})} \rightleftharpoons \mathrm{AB}_{(\mathrm{g})}+\frac{1}{2} \mathrm{B}_{2(\mathrm{g})} \)

\(\mathrm{t}_{\mathrm{eq}} \frac{(1-\mathrm{x})}{1+\frac{\mathrm{x}}{2}} \mathrm{P} \frac{\mathrm{xP}}{1+\frac{\mathrm{x}}{2}} \frac{\left(\frac{\mathrm{x}}{2}\right) \mathrm{P}}{1+\frac{\mathrm{x}}{2}} \)

\(\Rightarrow \mathrm{x} <<1 \Rightarrow 1+\frac{\mathrm{x}}{2} \simeq 1\) and 1 - x ≃ 1

\(\Rightarrow \mathrm{k}_{\mathrm{P}}=\frac{(\mathrm{xp}) \cdot\left(\frac{\mathrm{xp}}{2}\right)^{\frac{1}{2}}}{\mathrm{P}}\)

\(\Rightarrow \mathrm{k}_{\mathrm{P}}^{2}=\mathrm{x}^{2} . \frac{\mathrm{xP}}{2}\)

\(\mathrm{x}=\sqrt[3]{\frac{2 \mathrm{k}_{\mathrm{P}}^{2}}{\mathrm{P}}} \)

Therefor the correct answer is x ≈ √(2K_p/P √3) (option 3).

CONCEPT:

Solubility Product (Ksp)

• The solubility product (Ksp) is an equilibrium constant that applies to the dissolution of a sparingly soluble compound.
• A higher Ksp value indicates higher solubility of the compound in water. Conversely, a lower Ksp value indicates lower solubility.

EXPLANATION:

• Given the Ksp values:
  • HgS: Ksp = 4 × 10⁻⁵³
  • PbS: Ksp = 8 × 10⁻²⁸
  • AgBr: Ksp = 5 × 10⁻¹³
  • Ca(OH)₂: Ksp = 5.5 × 10⁻⁶
• Arranging the compounds in increasing order of their Ksp values (i.e., increasing order of solubility):
  • HgS: Ksp = 4 × 10⁻⁵³
  • PbS: Ksp = 8 × 10⁻²⁸
  • AgBr: Ksp = 5 × 10⁻¹³
  • Ca(OH)₂: Ksp = 5.5 × 10⁻⁶

Therefore, the correct order of increasing solubility product is HgS < PbS < AgBr < Ca(OH)₂

CONCEPT:

The degree of dissociation (x) of a gas in equilibrium can be related to its equilibrium constant (Kp) using the reaction equation:

• Consider the reaction:

  X2Y(g) ⇌ X2(g) + 1/2 Y2(g)

• The degree of dissociation (x) can be related to the changes in moles and partial pressures at equilibrium.

EXPLANATION:

\(\mathrm{X}_{2} \mathrm{Y}_{(\mathrm{g})} \rightleftharpoons \mathrm{X}_{2(\mathrm{~g})}+\frac{1}{2} \mathrm{Y}_{2(\mathrm{~g})}\)

\(1 \text {-x mole } \quad x \text { mole } \quad \frac{x}{2} \text { mole }\)

∴ \(\mathrm{P}_{\mathrm{X}_{2} \mathrm{Y}}=\frac{1-\mathrm{x}}{1+\frac{\mathrm{x}}{2}} \times \mathrm{P}\)

\(\mathrm{P}_{\mathrm{X}_{2}}=\frac{\mathrm{x}}{1+\frac{\mathrm{x}}{2}} \times \mathrm{P}\)

\(\mathrm{P}_{\mathrm{Y}_{2}}=\frac{\mathrm{x} / 2}{1+\frac{\mathrm{x}}{2}} \times \mathrm{P}\)

∴ \(\mathrm{K}_{\mathrm{p}}=\left(\frac{\mathrm{x}}{1+\frac{\mathrm{x}}{2}} \mathrm{P}\right)\left(\frac{\mathrm{x}}{2\left(1+\frac{\mathrm{x}}{2}\right)} \mathrm{P}\right)^{\frac{1}{2}} /\left(\frac{1-\mathrm{x}}{1+\frac{\mathrm{x}}{2}}\right) \times \mathrm{P}\)

∴ \(\mathrm{K}_{\mathrm{p}}=\left(\frac{\mathrm{x}}{1-\mathrm{x}}\right)\left(\frac{\mathrm{x}}{2\left(1+\frac{\mathrm{x}}{2}\right)}\right)^{\frac{1}{2}} \times \mathrm{p}^{\frac{1}{2}}\)

∵ x to be very very small

∴ \(\mathrm{K}_{\mathrm{p}}=\frac{\mathrm{x}^{3 / 2}}{(2)^{\frac{1}{2}}} \times \mathrm{P}^{\frac{1}{2}}\)

∴ \(\mathrm{x}^{\frac{3}{2}}=\frac{\mathrm{K}_{\mathrm{p}} \times 2^{\frac{1}{2}}}{\mathrm{P}^{\frac{1}{2}}}\)

∴ \(\mathrm{x}^{3}=\frac{\mathrm{K}_{\mathrm{p}}^{2} \times 2}{\mathrm{P}}\)

\(\mathrm{x}=\left(\frac{\mathrm{K}_{\mathrm{p}}^{2} \times 2}{\mathrm{P}}\right)^{\frac{1}{3}}\)

The correct answer is option (2).

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:

• 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:

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.

The correct answer is 

Concept:-

• Equilibrium Constant (K): The equilibrium constant, often denoted by K, is a value that expresses the balance between reactants and products of a reaction at equilibrium. In general terms, it's equal to the ratio of the concentrations of the products to the concentrations of the reactants, each raised to the power of their stoichiometric coefficient.
• Direction of Reaction & Equilibrium Constant: The equilibrium constant can help to predict the direction of a reaction to reach equilibrium if non-equilibrium concentrations of reactants and products are provided. A comparison between the reaction quotient (Q), calculated the same way as K but from initial concentrations, and K determines the direction of the reaction.
• Dependence of K on Initial Concentrations: The value of the equilibrium constant for a given reaction at a given temperature is a constant value, independent of the initial concentrations of reactants and products. However, it is worth noting that the individual equilibrium concentrations of reactants and products depend on initial concentrations.

Explanation:-

K_equ = kf/kb

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

The equilibrium constant can indeed be used for calculating the equilibrium concentrations given initial concentrations and other equilibrium constants, so this statement is correct.

The direction of a reaction can be predicted by comparing the reaction quotient (Q) to the equilibrium constant.

• If Q > K, the reaction will proceed in the direction of the reactants.
• If Q < K, the reaction will proceed in the direction of the products.
• If Q = K, the reaction is at equilibrium.
• So, this statement is also correct.​

​

The value of the equilibrium constant is independent of initial concentrations of reactants and products but is a function of temperature. Therefore, this statement is incorrect.

So, the correct answer is 4) More than one of the above, as both statements 1 and 2 are correct.

Concept:

• Effect of pressure on equilibrium: When a system at equilibrium is subjected to a change in the pressure of gaseous reactants and if pressure is increased, then equilibrium shifts towards the fewer moles of the gaseous reactants.
• Since pressure and volume are mutually inversely proportional to each other in the case of gaseous substrates.

Explanation:

The given equilibrium reaction is shown below:

\(\rm N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)\)

 The pressure at the equilibrium position is increased.

• When pressure is increased, then equilibrium will move in such a direction that it opposes the change according to LeChatlier's principle.
• The total number of moles of gaseous substrates on the left side of the equilibrium is 4 mol.
• The total number of moles of gaseous substrates on the left side of the equilibrium is 2 mol.
• An increase in pressure favours the decrease in the volume of gases at a constant temperature according to Boyle's law.
• Hence, an increase in pressure at this equilibrium, allows the equilibrium to move towards a lesser number of moles.

Hence, the formation of ammonia NH₃ gas will be there.

Explanation:-

2NO(g) ⇌ N₂(g) + O₂(g)

• [N₂] = 3.0 × 10^–3 M
• [O₂] = 4.2 × 10^–3 M
• [NO] = 2.8 × 10^–3 M
• Initial concentration of NO = 0.1 mol L^–1

Step 1: Calculate K_c

The equilibrium constant K_c is given by:

\(K_c = \frac{[N_2][O_2]}{[NO]^2}\)

Substitute the given concentrations:

\(K_c = \frac{(3.0 \times 10^{-3})(4.2 \times 10^{-3})}{(2.8 \times 10^{-3})^2}\)

Calculate K_c :

\(K_c = \frac{(3.0 \times 10^{-3}) \times (4.2 \times 10^{-3})}{(2.8 \times 10^{-3})^2}\)

\(K_c = \frac{12.6 \times 10^{-6}}{7.84 \times 10^{-6}}\)

Kc = 1.607

Step 2: Determine the degree of dissociation ( α  )

\( K_c = \frac{0.05 \alpha \times 0.05 \alpha}{(0.1 - 0.1 \alpha)^2} \\ K_c = \frac{0.05 \alpha \times 0.05 \alpha}{0.01 (1 - \alpha)^2} \\ 1.607 = \frac{(0.05)^2 \alpha^2}{0.01 (1 - \alpha)^2} \\ \frac{\alpha^2}{(1 - \alpha)^2}= \frac{1.607 \times (0.1)^2}{(0.05)^2} \\ \frac{\alpha}{1 - \alpha} = \frac{1.27 \times 0.1}{0.05} \\ \frac{\alpha}{1 - \alpha} = 2.54 \\ \alpha = 2.54 - 2.54 \alpha \\ 3.54 \alpha = 2.54 \\ \alpha = \frac{2.54}{3.54} = 0.717 \)

Conclusion:-

The correct degree of dissociation ( α ) of NO(g) at equilibrium is0.717.

Correct answer: 1)

Concept:

• According to the law of mass action, it is the proposition that the rate of the chemical reaction is directly proportional to the product of the activities or concentration of the reactants.
• Equilibrium is when the rate of the forward reaction equals the rate of the reverse reaction.
• All reactant and product concentrations are constant at equilibrium.

• For reactions that are not at equilibrium, we can write a similar expression called the reaction quotient Q, which is equal to K_c​ at equilibrium. K_c​ and Q can be used to determine if a reaction is at equilibrium, to calculate concentrations at equilibrium, and to estimate whether a reaction favors products or reactants at equilibrium.

Explanation:

Given, the equilibrium concentration of N2, O2, and NO is found to be 4 × 10−3, 3 × 10−3 and 3 × 10−3 M respectively.

Thus,

C_N2 = 4 × 10−3 M

C_O2 = 3 × 10−3  M

C_NO = 3 × 10−3 M

On applying the law of chemical equilibrium for the reaction

N2 + O2 \(\rightleftharpoons\) 2NO

\(K_{c}=\frac{[NO]^{2}}{[N_{2}][O_{2}]}\)

On putting the value of equilibrium concentration in the above expression, we get

\(K_{c}=\frac{(3\times 10^{-3}M)^{2}}{(4\times 10^{-3}M)(3\times 10^{-3}M)}\)

=0.750

Conclusion:

• Hence, the equilibrium constant (Kc) for the given reaction is 0.750.

Explanation:

Given Reaction are

\({H_{2\left( g \right)}} + \frac{1}{2}{S_2}\left( s \right)\rightleftharpoons{H_2}{S_{\left( g \right)}},{K_1}\)

\({H_{2\left( g \right)}} + {B_{{r_2}\left( g \right)}} \rightleftharpoons2H{B_r}_{(g)},{K_2}\)

we need to final K for the following Reaction

\({B_{r2\left( g \right)}} + {H_2}{S_g}\rightleftharpoons 2H{B_r}_{(g)} + \frac{1}{2}{S_2}\left( s \right)\)

We know

\({K_1} = \frac{{\left[ {{H_2}S} \right]}}{{\left[ {{H_2}} \right]{{\left[ {{S_2}} \right]}^{\frac{1}{2}}}}}\)

\({K_2} = \frac{{{{\left[ {H{B_r}} \right]}^2}}}{{\left[ {{H_2}} \right]\left[ {{B_r}_2} \right]}}\)

Dividing K₂ by K₁,

\(\frac{{{K_2}}}{{{K_1}}} = \frac{{{{\left[ {H{B_r}} \right]}^2}}}{{\left[ {{H_2}} \right]\left[ {{B_r}_2} \right]}} \times \frac{{\left[ {{H_2}} \right]{{\left[ {{S_2}} \right]}^{\frac{1}{2}}}}}{{\left[ {{H_2}S} \right]}}\)

\(\frac{{{K_2}}}{{{K_1}}} = \frac{{{{\left[ {H{B_r}} \right]}^2}{{\left[ {{S_2}} \right]}^{\frac{1}{2}}}}}{{\left[ {{H_2}S} \right]\left[ {{B_r}_2} \right]}} = K\)

Concept:

Relationship between K_p and K_c - For a reaction, aA + bB  \(\leftrightharpoons \) xX + yY

• K_p - It is the equilibrium constant when concentration is expressed in terms of partial pressure(generally when reactant and products are gaseous).
  • \(K_p = \frac{p^x_X\;.\;p^y_Y}{p^a_A\;.\;p^b_B}\)
• K_c - It is the equilibrium constant when concentration is expressed in terms of molarity.
  • \(K_c = \frac{C^x_X\;.\;C^y_Y}{C^a_A\;.\;C^b_B}\)

Relationship between Kp and Kc:

\(K_p = K_c {(RT)}^{Δ n} \)

Where 

• Δn is the change in number of moles of gaseous substance.
• R is gas constant.
• T is temperature

Calculation:

The chemical reaction is -

C10H8(s) \(\leftrightharpoons \) C10H8(v) 

The vapour pressure above the balls was found to be 0.10 mm Hg.

∴ K_p =  0.10 mm Hg.

Relation between K_p and K_c is -

\(K_p = K_c {(Rt)}^{Δ n}\)

or \(K_c = \frac {K_p}{{(RT)}^{Δ n}} \)

Where 

• Δn is the change in number of moles.
• R is gas constant.
• T is temperature

Given,  Kp =  0.10 mm Hg or \(\frac{0.1}{760}\) atm

R = 0.0821 L(atm) mol^-1K^-1.

T = 273 + 27 = 300 K

Δn = 1

 \(K_c = \frac {0.1}{760 \times 0.0821 \times 300} \) 

= 5.36 × 10-6

Conclusion:

Therefore, the value of Kc(sublimation) = 5.36 × 10-6

Concept:

Relation between K_c and Q_c - 

1. The comparison between Kc and Qc determines the direction of the reaction to take place.
2. If Kc > Qc the reaction proceed to form products or to the right.
3. Kc = Qc this is condition of equillibrium
4. Kc < Qc the reaction proceed in the reverse direction or to the left.

Explanation:

For the reaction, 2A ⇋ B + C, reaction quaotient(Q_c) is given by -

\(Q_c = \frac{[B][C]}{[A]^2}\)

Given, the reaction mixture has [A] = [B] = [C] = 3 × 10-3 M.

\(Q_c = \frac{(3\times10^{-3})(3\times10^{-3})}{(3\times10^{-3})^2}\) = 1

Given, K_c = 2 × 10-3 or  0.002 

As (Qc) >  (Kc )

So, the reaction will proceed in the reverse manner or to the left.

Conclusion:

For a reaction 2A ⇋ B + C, Kc is 2 × 10-3. At a given time, the reaction mixture has [A] = [B] = [C] = 3 × 10-3 M,  the reaction will proceed in the reverse manner or to the left.