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

More Equilibrium Previous-Year Questions — Page 4

Q49 jee_main_2025_08_april_evening Equilibrium Constant

The equilibrium constant (

Kₚ

) for the thermal decomposition of water vapor:

H₂O(g) leftharpoons H₂(g) + (1)/(2)O₂(g) (Δ G^° = 92.34 kJ mol⁻¹)

is evaluated as

8.0 \times 10⁻³

2300 K

under a total pressure of

1 bar

. Under these specific conditions, the degree of dissociation (

α

) of water is _________

\times 10⁻²

(as the nearest integer value). [Assume

α ll 1

Numerical Answer. Answer: 5 to 5

Solution

Related Formula

Gas phase dissociation equilibrium setup:

H₂O(g) leftharpoons H₂(g) + (1)/(2)O₂(g)

Partial pressure equilibrium expression:

Kₚ = PH₂ \cdot (PO₂)1/2PH₂O

Execution

Step 1: Set up the mole distribution table at equilibrium assuming 1 initial mole:

$H₂O = 1 - α$ $\text{H}_2\text{O} = 1 - \alpha$
$H₂ = α$ $\text{H}_2 = \alpha$
$O₂ = (α)/(2)$ $\text{O}_2 = \frac{\alpha}{2}$

Step 2: Calculate the total moles ( $nT$ $n_T$) at equilibrium:

nT = (1 - α) + α + (α)/(2) = 1 + (α)/(2)

Given $α ll 1$ $\alpha \ll 1$, we can approximate $nT \approx 1$ $n_T \approx 1$.

Step 3: Express the partial pressures using total pressure $P = 1 bar$ $P = 1 \text{ bar}$:

PH₂O = (1-α)/(1) \cdot P \approx 1 \cdot 1 = 1

PH₂ = α \cdot P = α

PO₂ = (α)/(2) \cdot P = (α)/(2)

Step 4: Substitute these partial pressures into the $Kₚ$ $K_p$ expression:

Kₚ = α \cdot ((α)/(2))1/21 = α3/2\sqrt(2)

Step 5: Equate to the given value of $Kₚ = 8.0 \times 10⁻³$ $K_p = 8.0 \times 10^{-3}$ and solve for $α$ $\alpha$:

8.0 \times 10⁻³ = α3/2\sqrt(2) α3/2 = 8\sqrt(2) \times 10⁻³

Cube both sides to clear fractional exponents:

α³ = (8\sqrt(2) \times 10⁻³)² = 128 \times 10⁻⁶

α = 3\sqrt(128) \times 10⁻² \approx 5.03 \times 10⁻²

Matching the target template $α = 5.03 \times 10⁻²$ $\alpha = 5.03 \times 10^{-2}$, the integer value is 5.

Pattern Recognition

When $α ll 1$ $\alpha \ll 1$, the total mole expression simplifies to 1, and the denominator $(1-α)$ $(1-\alpha)$ drops out. This simplifies the expression to $Kₚ \propto α1 + Δ ng$ $K_p \propto \alpha^{1 + \Delta n_g}$, allowing you to quickly isolate $α$ $\alpha$ via standard powers.

Chapter Mix

Class 11 Chemistry: Chemical Equilibrium Class 11 Chemistry: Chemical Thermodynamics

Q30 jee_main_2025_28_jan_morning Degree of Dissociation and pH

A weak acid

HA

has degree of dissociation

x

. Which option gives the correct expression of

pH - pKₐ

A. $(1 + 2x)$ $\log (1 + 2x)$
B. $((1 - x)/(x))$ $\log \left(\frac{1 - x}{x}\right)$
C. $0$ $0$
D. $( x1 - x)$

Solution

Related Formula

For a weak acid solution:

HA leftharpoons H^+ + A^-

Kₐ = [H^+][A^-][HA]

Step 1: Expressing Concentration

Let the initial concentration be $a$ $a$. At equilibrium:

[HA] = a(1-x), [H^+] = ax, [A^-] = ax

Substituting into the equilibrium expression:

Kₐ = ((ax)(x))/(1-x) = [H^+] ((x)/(1-x))

Step 2: Logarithmic Rearrangement

Taking negative logarithms on both sides:

- Kₐ = - [H^+] - ((x)/(1-x))

pKₐ = pH - ((x)/(1-x))

pH - pKₐ = ((x)/(1-x))

Pattern Recognition

Sees: $pH - pKₐ$ $\mathrm{pH} - \mathrm{pK}_a$ for weak acid equilibrium. Shortcut: This is equivalent to the Henderson-Hasselbalch framework: $pH = pKₐ + [Salt][Acid] = pKₐ + (x)/(1-x)$ .

Chapter Mix

Class 11 Chemistry: Ionic Equilibrium

Q jee_main_2025_03_april_morning Le Chatelier's Principle - Addition of Inert Gas

In the following system,

PCl₅(g) leftharpoons PCl₃(g) + Cl₂(g)

at equilibrium, upon addition of xenon gas at constant

T

p

, the concentration of:

A. $PCl₅$ $\text{PCl}_{5}$ will increase
B. $Cl₂$ $\text{Cl}_{2}$ will decrease
C. $PCl₅$ $\text{PCl}_{5}$, $PCl₃$ $\text{PCl}_{3}$ & $Cl₂$ $\text{Cl}_{2}$ remain constant
D. $PCl₃$ $\text{PCl}_{3}$ will increase

Solution

Core Logic

When an inert gas like Xenon is added at constant temperature and pressure, the total volume of the container must expand significantly to maintain constant pressure.

According to Le Chatelier's principle, the equilibrium shifts towards the side with a higher number of gas moles (the forward direction here, since $Δ ng = 1 > 0$ $\Delta n_g = 1 > 0$). This increases the absolute number of moles of $PCl₃$ $\text{PCl}_3$ and $Cl₂$ $\text{Cl}_2$.

Step 1: Concentration Analysis

Concentration is calculated as moles divided by total volume ( $[X] = (n)/(V)$ $[\text{X}] = \frac{n}{V}$). Even though the forward shift generates a few more moles of products, the fractional volume expansion factor is larger. Thus, the actual molar concentrations of all species ( $PCl₅$ $\text{PCl}_5$, $PCl₃$ $\text{PCl}_3$, and $Cl₂$ $\text{Cl}_2$) ultimately decrease.

Pattern Recognition

Trap Alert: Distinguish clearly between "moles" and "concentration". The forward shift increases product moles, but the volume expansion dilutes all chemical species, dropping total concentration values.

Chapter Mix

Class 11 Chemistry: Chemical Equilibrium

Q50 jee_main_2025_04_april_evening Solubility Product and pH

x mg of

Mg(OH)₂

(molar mass

= 58

) is required to be dissolved in

1.0~L

of water to produce a pH of

10.0

298~K

. The value of x is ________ mg. (Nearest integer) (Given:

Mg(OH)₂

is assumed to dissociate completely in

H₂O

)

Numerical Answer. Answer: 2.5 to 3.5

Solution

Related Formula

pH + pOH = 14 pOH = 14 - pH

[OH^-] = 10-pOH

Moles of Mg(OH)₂ = ([OH^-])/(2)

Core Logic

Convert the given pH into standard hydroxide concentration:

pOH = 14 - 10.0 = 4.0 [OH^-] = 10⁻⁴ ~mol \cdot L⁻¹

For $1.0 ~L$ $1.0 \mathrm{~L}$ of solution, the number of moles of $OH^-$ $OH^-$ required is $10⁻⁴$ $10^{-4}$ moles.

Since each formula unit of $Mg(OH)₂$ $Mg(OH)_2$ releases 2 moles of $OH^-$ $OH^-$ ions:

moles of Mg(OH)₂ = 10⁻⁴2 = 5 \times 10⁻⁵ moles

Convert this molar value into absolute mass units:

mass = 5 \times 10⁻⁵ \times 58 ~g = 2.9 \times 10⁻³ ~g = 2.9 ~mg

Rounding to the nearest integer gives 3.

Pattern Recognition

Always remember that $Mg(OH)₂$ $Mg(OH)_2$ is a diacidic base. Forgetting to divide the hydroxide concentration by 2 is a frequent pitfall that leads to a value double the correct answer.

Chapter Mix

Class 11 Chemistry: Ionic Equilibrium

Solution & Explanation

Core Logic

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More Equilibrium Previous-Year Questions — Page 4

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Related Formula

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