Equilibrium Law's And Equilibrium Constant MCQ Quiz in தமிழ் - Objective Question with Answer for Equilibrium Law's And Equilibrium Constant - 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] is concentration of gas A, B, C and D respectively at equilibrium.

Explanation:

Given the initial moles and volume of the flask, we can calculate the initial concentrations:

• Initial moles of PCl₃(g): 1.0 mole

• Initial moles of Cl₂(g): 2.0 moles

• Volume of the flask: 3.0 L

• Equilibrium moles of PCl₃(g): 0.70 mole

The initial concentrations are:

• \([\mathrm{PCl_3}]_0 = \frac{1.0 \text{ mol}}{3.0 \text{ L}} = 0.333 \text{ M}\)

• \([\mathrm{Cl_2}]_0 = \frac{2.0 \text{ mol}}{3.0 \text{ L}} = 0.667 \text{ M}\)

Let the change in moles of PCl₃(g) and Cl₂(g) that react be denoted by x. At equilibrium:

• \(\text{Moles of } \mathrm{PCl_3(g)} = 1.0 - x = 0.70 \text{ mole}\)

Thus, x = 1.0 - 0.70 = 0.30 mole. The equilibrium moles are:

• \(\text{Moles of } \mathrm{Cl_2(g)} = 2.0 - x = 1.70 \text{ mole}\)

• \(\text{Moles of } \mathrm{PCl_5(g)} = x = 0.30 \text{ mole}\)

The equilibrium concentrations are:

• \([\mathrm{PCl_3}]_{\text{eq}} = \frac{0.70 \text{ mol}}{3.0 \text{ L}} = 0.233 \text{ M}\)

• \([\mathrm{Cl_2}]_{\text{eq}} = \frac{1.70 \text{ mol}}{3.0 \text{ L}} = 0.567 \text{ M}\)

• \([\mathrm{PCl_5}]_{\text{eq}} = \frac{0.30 \text{ mol}}{3.0 \text{ L}} = 0.100 \text{ M}\)

For \(\mathrm{PCl_3(g) + Cl_2(g)\rightleftharpoons PCl_5(g)}\)

The equilibrium constant K_c is given by:

\(K_c = \frac{[\mathrm{PCl_5}]}{[\mathrm{PCl_3}][\mathrm{Cl_2}]}\)

Substituting the equilibrium concentrations:

\(K_c = \frac{0.100}{0.233 \times 0.567} \approx \frac{0.100}{0.132} \approx 0.756 \text{ L mol}^{-1}\)

Conclusion:

The value of equilibrium constant K_c for the reaction \(\mathrm{PCl_3(g) + Cl_2(g)\rightleftharpoons PCl_5(g)}\) at 400 K is approximately \(0.756\ L mol^{-1}\)

Concept:

Equilibrium constant ( \(\mathbf{K_{P}} \))

Any equilibrium reaction can have one or more gas reactants or products, in which case the equilibrium constant can be represented in terms of partial pressure. \(K_{P}\) stands for equilibrium constant expressed in terms of partial pressure.

A balanced equation's equilibrium constant, \(K_{P}\), is determined by dividing the partial pressure of products by the partial pressure of reactants. The partial pressure is then raised by a factor that is equal to the substance's coefficient.

\(aA(g)\;+\;bB\;(g)\;\rightleftharpoons \;cC(g)\;+\;dD(g)\)

\(K_{P}\;=\;\dfrac{[P_{C}^{c}[P_{D}^{d}]}{[P_{A}^{a}[P_{B}^{b}]} \)

Explanation:

Initial pressure  = 1 atm

Fall in pressure at equilibrium = 29% 

 = \(\dfrac{29}{100}\)

= 0.29 %

 Again, 

                         \(S_{8}(g)\;\rightleftharpoons \;4S_{2}(g) \)

Initial Pressure  1 atm         0

At equilibrium    1- 0.29       \(4\times 0.29 \)

                        = 0.71 atm   1.16 atm

The equilibrium equation of the equation at constant pressure is given by:

\(K_{P}\;=\;\dfrac{p_{[S_{2}]^4{}}}{p_{[S_{8}]}} \)

\(K_{P}\;=\;\dfrac{[1.16\;atm]^{4}}{[0.71\;atm]} \)

\(K_{P}\;=\;2.55\;atm^{3} \)

Conclusion:

The \(\mathbf{K_{P}} \) for the reaction when \(\mathbf{S_{8}}\) converted to \(\mathbf{S_{2}}\) is \(\mathbf{2.55\;atm^{3}} \).

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.

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:

Equilibrium Partial Pressures and Equilibrium Constant Kp

• Kp is the equilibrium constant calculated using the partial pressures of the gases involved in the reaction.
• For the reaction: NH4COONH2(s) ⇌ 2NH3(g) + CO2(g)
• The total pressure at equilibrium is given as 6 atm.

EXPLANATION:

\(NH_4COONH_2(s) \rightleftharpoons \underset{2p}{2NH_3(g)} + \underset{p}{CO_2(g)}\)

\(2p + p = 6\)

\(p = 2\) atm

\(P_{NH_3} = 4\) atm; \(P_{CO_2} =2\) atm

\(K_p = [P_{NH_3}]^2 [P_{CO_2}] = 4^2 \times 2 = 32\)

Therefore, the equilibrium constant K_p is 32 atm.

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:

Given: 2A + B ⇌ C + 4D

• A and B are mixed in the ratio 5:3.

• They react in the ratio 2:1.

• At equilibrium, the concentration of B and C are equal.

Let the initial concentrations be:

• [A]₀ = 5x

• [B]₀ = 3x

Let the change in the concentration of B at equilibrium be -y. Then:

• Change in A: -2y

• Change in C: +y

• Change in D: +4y

Equilibrium concentrations:

• [A]_eq = 5x - 2y

• [B]_eq = 3x - y

• [C]_eq = y

• [D]_eq = 4y

Given that [B]_eq and [C]_eq are equal:

3x - y = y

3x = 2y

\(y = \frac{3x}{2}\)

Substituting y back into the equilibrium concentrations:

• \([A]_{\text{eq}} = 5x - 2\left(\frac{3x}{2}\right) = 2x\)

• \([B]_{\text{eq}} = 3x - \frac{3x}{2} = \frac{3x}{2}\)

• \([C]_{\text{eq}} = \frac{3x}{2}\)

• \([D]_{\text{eq}} = 4\left(\frac{3x}{2}\right) = 6x\)

Expression for the equilibrium constant K_c :

\(K_c = \frac{[C][D]^4}{[A]^2[B]}\)

Substitute the equilibrium concentrations:

\(K_c = \frac{\left(\frac{3x}{2}\right)\left(6x\right)^4}{(2x)^2\left(\frac{3x}{2}\right)}\)

Simplify:

\(K_c = \frac{3x \cdot 1296x^4}{2 \cdot 4x^2 \cdot \frac{3x}{2}}\)

\(K_c = \frac{1296x^4}{4x^2}\)

\(K_c = \frac{1296}{4} x^{4-2}\)

\( K_c = 324x^2\)

Conclusion:

The calculated equilibrium constant K_c is 324, but none of the given options match this value. Therefore, the correct answer is: None of these

Concept:

Partial Pressure

P_i = x_i × P_total

Where:

• P_i = Partial pressure of gas i

• x_i = Mole fraction of gas i

• P_total = Total pressure of the gas mixture

At temperature T, a compound AB2(g) dissociates according to the reaction:

2AB₂(g) ⇌ 2AB(g) + B₂(g)

The degree of dissociation x is small compared to unity. The goal is to express the equilibrium constant Kp in terms of x and the total pressure p .

Explanation:

For, 2AB₂(g) ⇌ 2AB(g) + B₂(g)

Consider the initial pressure of AB₂ as p .

• At equilibrium, let the degree of dissociation of AB₂ be x .

• The initial moles of AB₂ are 2 (since the coefficient is 2 in the balanced reaction).

• At equilibrium, the moles dissociated: \(2AB_2 \rightarrow (1 - x)\) . So, the remaining moles of AB₂ are (1 - x) .

• Also, the moles of AB are x .

• The moles of B₂ are x/2  (only 1 mole of B₂ is formed per 2 moles AB₂).

The total moles at equilibrium:

\((1 - x) + x + \frac{x}{2} = \frac{2+x}{2}\)

The partial pressures of each species (assuming ideal gas behavior and Dalton's Law), considering the initial pressure p and small x , will be:

\(P_{AB_2} = \frac{2(1 - x)}{2 + x} \cdot p \)

\(P_{AB} = \frac{2x}{2 + x} \cdot p \)

\( P_{B_2} = \frac{x}{2 + x} \cdot p \) (since x is small relative to unity)

Now, the expression for the equilibrium constant Kp is:

\(K_p = \frac{(P_{AB})^2 \cdot P_{B_2}}{(P_{AB_2})^2}\)

Substituting the partial pressures:

\(K_p =\frac{ [\frac{2x}{2 + x} \cdot p ]^2[\frac{x}{2 + x} \cdot p]}{[\frac{2(1 - x)}{2 + x} \cdot p]^2}\)

Simplifying, we get:

\(K_p = \frac{x^3p}{(2+x)(1-x)^2}\)

Since x is very small then, 2+x = 2 and 1-x = 1

\(K_p = \frac{ x^3p}{2}\)

Conclusion:

The correct expression for Kp in terms of x and p is: \(K_p = \frac{ px^3}{2}\)

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

where [A], [B], [C] and [D] is concentration of gas A, B, C and D respectively at equilibrium.

Explanation:

Given the initial quantities and volume of the vessel, we calculate the equilibrium concentrations:

• Volume of the vessel: 500 cm³ = 0.5 L

• Moles of \(\mathrm{COCl_2}\) at equilibrium: 0.3 mol

• Moles of \(\mathrm{CO}\) at equilibrium: 0.1 mol

• Moles of \(\mathrm{Cl_2}\) at equilibrium: 0.1 mol

The equilibrium concentrations are:

• \([\mathrm{COCl_2}]_{\text{eq}} = \frac{0.3 \text{ mol}}{0.5 \text{ L}} = 0.6 \text{ M}\)

• \([\mathrm{CO}]_{\text{eq}} = \frac{0.1 \text{ mol}}{0.5 \text{ L}} = 0.2 \text{ M}\)

• \([\mathrm{Cl_2}]_{\text{eq}} = \frac{0.1 \text{ mol}}{0.5 \text{ L}} = 0.2 \text{ M}\)

For, CO + Cl2 \(\rightleftharpoons\) COCl2

The equilibrium constant Kc is given by:

\(K_c = \frac{[\mathrm{COCl_2}]}{[\mathrm{CO}][\mathrm{Cl_2}]}\)

Substituting the equilibrium concentrations:

\(K_c = \frac{0.6}{0.2 \times 0.2} = \frac{0.6}{0.04} = 15\)

Conclusion:

The value of equilibrium constant Kc for the reaction CO(g) + Cl₂(g) \(\rightleftharpoons\)COCl₂(g) is 15.