Consider the equilibrium If the pressure applied over the system increases by two fold at constant temperature then: (A) Concentration of reactants and products increases. (B) Equilibrium will shift in forward direction. (C) Equilibrium constant increases since concentration of products increases. (D) Equilibrium constant remains unchanged as concentration of reactants and products remain same. Choose the correct answer from the options given below:

Statement (A) is correct: Increasing pressure by compressing the volume increases active mass/concentration (c = n/V) for both reactants and products instantly. Statement (B) is correct: The reaction has Delta ng = 2 - 4 = -2. Increasing pressure shifts equilibrium towards the direction of fewer gaseous moles, which is the forward path. Statement (C) is incorrect: Equilibrium constant (K) is exclusively temperature-dependent and does not alter with pressure changes. Statement (D) is correct: Confirms that equilibrium constant remains unchanged. Always remember: pressure changes shift positions but NEVER alter the value of the equilibrium constant Kc or Kp. Only temperature changes can change K. Class 11 Chemistry: Equilibrium

Consider the equilibrium If the pressure applied over the system increases by two fold at constant temperature then: (A) Concentration of reactants and products increases. (B) Equilibrium will shift i

Statement (A) is correct: Increasing pressure by compressing the volume increases active mass/concentration (c = n/V) for both reactants and products instantly. Statement (B) is correct: The reaction has Delta ng = 2 - 4 = -2. Increasing pressure shifts equilibrium towards the direction of fewer gaseous moles, which is the forward path. Statement (C) is incorrect: Equilibrium constant (K) is exclusively temperature-dependent and does not alter with pressure changes. Statement (D) is correct: Con

Consider the equilibrium If the pressure applied over the system increases by two fold at constant temperature then: (A) Concentration of reactants and products increases. (B) Equilibrium will shift in forward direction. (C) Equilibrium constant increases since concentration of products increases. (D) Equilibrium constant remains unchanged as concentration of reactants and products remain same. Choose the correct answer from the options given below:

Statement (A) is correct: Increasing pressure by compressing the volume increases active mass/concentration (c = n/V) for both reactants and products instantly. Statement (B) is correct: The reaction has Delta ng = 2 - 4 = -2. Increasing pressure shifts equilibrium towards the direction of fewer gaseous moles, which is the forward path. Statement (C) is incorrect: Equilibrium constant (K) is exclusively temperature-dependent and does not alter with pressure changes. Statement (D) is correct: Confirms that equilibrium constant remains unchanged. Always remember: pressure changes shift positions but NEVER alter the value of the equilibrium constant Kc or Kp. Only temperature changes can change K. Class 11 Chemistry: Equilibrium

Consider the equilibrium If the pressure applied over the system increases by two fold at constant temperature then: (A) Concentration of reactants and products increases. (B) Equilibrium will shift i

Statement (A) is correct: Increasing pressure by compressing the volume increases active mass/concentration (c = n/V) for both reactants and products instantly. Statement (B) is correct: The reaction has Delta ng = 2 - 4 = -2. Increasing pressure shifts equilibrium towards the direction of fewer gaseous moles, which is the forward path. Statement (C) is incorrect: Equilibrium constant (K) is exclusively temperature-dependent and does not alter with pressure changes. Statement (D) is correct: Con

Consider the equilibrium If the pressure applied over the system increases by two fold at constant temperature then…

Year	2026	2025	2024	Total
Questions	10	17	8	35

Q84 jee_main_2024_29_jan_morning Kp and Kc Relationship

For the reaction

N₂O₄(g) leftharpoons 2NO₂(g)

Kₚ = 0.492

atm at

300K

.

Kc

for the reaction at same temperature is

______ \times 10⁻²

.

(Given: R = 0.082 L atm mol⁻¹ K⁻¹)

Numerical Answer. Answer: 2 to 2

Solution

Related Formula

Kₚ = Kc \cdot (RT)Δ ng

Core Logic

For the given gaseous equilibrium reaction:

N₂O₄(g) leftharpoons 2NO₂(g)

First, find the change in the number of moles of gas ( $Δ ng$ $\Delta n_g$):

Δ ng = nₚ - nᵣ = 2 - 1 = 1

Step 1: Calculation

Substitute the given values into the $Kₚ - Kc$ $K_p - K_c$ relationship: $Kₚ = 0.492$ $K_p = 0.492$ $R = 0.082$ $R = 0.082$ $T = 300 K$ $T = 300\text{ K}$

0.492 = Kc \cdot (0.082 \times 300)¹

Kc = (0.492)/(0.082 \times 300)

Kc = (0.492)/(24.6)

$Kc = 0.02$ $K_c = 0.02$

Converting to the requested format ( $x \times 10⁻²$ $x \times 10^{-2}$):

Kc = 2 \times 10⁻²

So, the value is 2.

Chapter Mix

Class 11 Chemistry: Equilibrium

Q85 jee_main_2024_30_january_evening Buffer Solutions

The pH of an aqueous solution containing 1M benzoic acid

(pKₐ = 4.20)

and 1M sodium benzoate is 4.5. The volume of benzoic acid solution in 300 mL of this buffer solution is ______ mL.

Numerical Answer. Answer: 100 to 100

Solution

Related Formula

Henderson-Hasselbalch Equation for Acidic Buffers:

pH = pKₐ + ( [Salt][Acid] )

Core Logic

Let the volume of 1M Benzoic acid be $Vₐ$ $V_a$ mL and the volume of 1M Sodium benzoate be $Vₛ$ $V_s$ mL. Total volume = $Vₛ + Vₐ = 300 mL$ $V_s + V_a = 300\,\text{mL}$.

Millimoles of acid = $1 \times Vₐ = Vₐ$ $1 \times V_a = V_a$ Millimoles of salt = $1 \times Vₛ = Vₛ$ $1 \times V_s = V_s$

Applying Henderson's Equation:

4.5 = 4.2 + ((Vₛ)/(Vₐ))

Step 1: Calculate Volume Ratio

((Vₛ)/(Vₐ)) = 4.5 - 4.2 = 0.3

Since $2 \approx 0.3$ $\log 2 \approx 0.3$, we have:

(Vₛ)/(Vₐ) = 2

$Vₛ = 2 Vₐ$ $V_s = 2 V_a$

Step 2: Substitute and Solve

We know $Vₛ + Vₐ = 300$ $V_s + V_a = 300$ Substituting $Vₛ = 2 Vₐ$ $V_s = 2 V_a$:

2 Vₐ + Vₐ = 300

$3 Vₐ = 300$ $3 V_a = 300$

Vₐ = 100 mL

Chapter Mix

Class 11 Chemistry: Equilibrium

Q82 jee_main_2024_30_jan_morning Solubility Product

The pH at which

Mg(OH)₂

[Kₛₚ=1\times 10⁻¹¹]

begins to precipitate from a solution containing

0.10 M Mg²⁺

ions is

Numerical Answer. Answer: 9 to 9

Solution

Related Formula

Kₛₚ = [Mg²⁺][OH^-]²

pOH = - [OH^-]

$pH + pOH = 14$ $pH + pOH = 14$

Core Logic

Precipitation begins just when the ionic product equals the solubility product ( $Qₛₚ = Kₛₚ$ $Q_{sp} = K_{sp}$).

Step 1: Calculating required [OH-]

[Mg²⁺][OH^-]² = 10⁻¹¹

Given $[Mg²⁺] = 0.10 M$ $[Mg^{2+}] = 0.10 \text{ M}$

0.10 \times [OH^-]² = 10⁻¹¹

[OH^-]² = 10⁻¹⁰

[OH^-] = 10⁻⁵ M

Step 2: Finding pH

pOH = - (10⁻⁵) = 5

$pH = 14 - pOH$ $pH = 14 - pOH$ $pH = 14 - 5 = 9$ $pH = 14 - 5 = 9$

Chapter Mix

Class 11 Chemistry: Equilibrium

Q71 jee_main_2024_31_jan_evening Equilibrium Constants (Kp and Kc)

A(g) leftharpoons B(g) + (C)/(2)(g)

The correct relationship between

KP

,

α

and equilibrium pressure

P

is

A. $KP = α(1)/(2)P(1)/(2)(2 + α)(1)/(2)$
B. $KP = α(3)/(2)P(1)/(2)(2 + α)(1)/(2)(1 - α)$
C. $KP = α(1)/(2)P(3)/(2)(2 + α)(3)/(2)$
D. $KP = α(1)/(2)P(1)/(2)(2 + α)(3)/(2)$

Solution

Related Formula

KP = PB \cdot (PC)(1)/(2)PA

where $Pᵢ$ $P_i$ is the partial pressure of component $i$ $i$.

Step 1: Setting up the ICE Table

For the reaction $A(g) leftharpoons B(g) + (1)/(2) C(g)$

Let initial moles of $A = 1$ $A = 1$. At equilibrium: Moles of $A = 1 - α$ $A = 1 - \alpha$ Moles of $B = α$ $B = \alpha$ Moles of $C = (α)/(2)$ $C = \frac{\alpha}{2}$

Total moles at equilibrium = $(1 - α) + α + (α)/(2) = 1 + (α)/(2) = (2 + α)/(2)$

Step 2: Calculating Partial Pressures

Using mole fraction $\times$ $\times$ Total Pressure ( $P$ $P$): $PA = (1 - α)/(1 + (α)/(2)) \cdot P$ $PB = (α)/(1 + (α)/(2)) \cdot P$ $PC = ((α)/(2))/(1 + (α)/(2)) \cdot P$

Step 3: Calculating Kp

KP = PB \cdot (PC)(1)/(2)PA

KP = ( (α)/(1 + α/2) P ) \cdot ( (α/2)/(1 + α/2) P )1/2(1 - α)/(1 + α/2) P

KP = α \cdot (α/2)1/2 \cdot P3/2(1 + α/2)3/2 \cdot (1 + α/2)/((1 - α) P)

KP = α3/2 \cdot P1/2\sqrt(2) \cdot (1 + α/2)1/2 \cdot (1 - α)

Since $1 + α/2 = (2+α)/(2)$ $1 + \alpha/2 = \frac{2+\alpha}{2}$, the $\sqrt(2)$ $\sqrt{2}$ in denominator cancels out perfectly leaving:

KP = α(3)/(2) P(1)/(2)(2 + α)(1)/(2)(1 - α)

Chapter Mix

Class 11 Chemistry: Equilibrium

Q62 jee_main_2024_31_jan_morning Equilibrium Constant

For the given reaction, choose the correct expression of

KC

from the following :-

Fe(aq)³⁺ + SCN(aq)⁻ leftharpoons (FeSCN)(aq)²⁺

A. $KC = [FeSCN²⁺][Fe³⁺][SCN⁻]$
B. $KC = [Fe³⁺][SCN⁻][FeSCN²⁺]$
C. $KC = [FeSCN²⁺][Fe³⁺]²[SCN⁻]²$
D. $KC = [FeSCN²⁺]²[Fe³⁺][SCN⁻]$

Solution

Related Formula

KC = [Products][Reactants]

Core Logic

KC = Products ion conc.Reactants ion conc.

KC = [FeSCN²⁺][Fe³⁺][SCN⁻]

Chapter Mix

Class 11 Chemistry: Equilibrium

Solution & Explanation

Core Logic

Pattern Recognition

Chapter Mix

Reference Study Guides

More Equilibrium Previous-Year Questions — Page 7

Solution

Related Formula

Core Logic

Step 1: Calculation

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Volume Ratio

Step 2: Substitute and Solve

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculating required [OH-]

Step 2: Finding pH

Chapter Mix

Solution

Related Formula

Step 1: Setting up the ICE Table

Step 2: Calculating Partial Pressures

Step 3: Calculating Kp

Chapter Mix

Solution

Related Formula

Core Logic

Chapter Mix

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