Equilibrium Law's And Equilibrium Constant MCQ Quiz in मल्याळम - Objective Question with Answer for Equilibrium Law's And Equilibrium Constant - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

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 in terms of partial pressure of gases (Kp) expression is given by: 

\(K_p= \frac{[p_C]^c[p_D]^d}{[p_A]^a[p_B]^b}\)

where p_A, p_B, p_C, and p_D is partial pressure of gaseous A, B, C and D respectively at equilibrium.

Ammonium carbamate dissociates into ammonia and carbon dioxide as per the following equilibrium reaction:

NH₂COONH₄(s) ⇌ 2NH₃(g) + CO₂(g)

In a closed vessel at equilibrium, the total pressure is the sum of the partial pressures of ammonia and carbon dioxide. When additional ammonia is added such that its partial pressure equals the original total pressure, the new total pressure needs to be determined.

Explanation

Initially, at equilibrium, let: Initial Total Pressure = P

• NH₃ : CO₂ = 2:1(according to reaction)

• Thus, pNH₃ = \(\frac{2}{3}P\) & pCO₂ = \(\frac{1}{3}P\)

• For given reaction,

• \(K_p = (p_{NH_3})^2(p_{CO_2})= (\frac{2}{3}P)^2(\frac{1}{3}P)= \frac{4}{27}P^3\)

After adding ammonia such that its partial pressure equals the original total pressure (P):

• The new partial pressure of NH₃(g) is: P' = P (original total pressure).

• \(K_p = (p_{NH_3})^2(p_{CO_2})= P^2(p_{CO_2})\)
• Thus, \(P^2(p_{CO_2})= \frac{4}{27}P^3\)
• \(p_{CO_2}= \frac{4}{27}P\)

Thus, the new total pressure (P_new) is:

• \(P'=P+\frac{4}{27}P= \frac{31}{27}P\)

​\(\text{Ratio of new total pressure to original total pressure}= \frac{31}{27}\)

Conclusion:

Therefore, the ratio of the new total pressure to the original total pressure is: \(\frac{31}{27}\)

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:

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.