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

Q49 jee_main_2025_04_april_morning pH of Weak Acid and Dilution

The pH of a

0.01mathrm~M

weak acid HX (

Kₐ = 4 \times 10⁻¹⁰

) is found to be 5. Now the acid solution is diluted with excess of water so that the pH of the solution changes to 6. The new concentration of the diluted weak acid is given as

x \times 10⁻⁴mathrm~M

. The value of x is _______ (nearest integer).

Numerical Answer. Answer: 25 to 25

Solution

Related Formula

HX(aq) leftharpoons H⁺(aq) + X⁻(aq)

Kₐ = ([H^+][X^-])/([HX]) = ((Cα)²)/(C(1-α)) \approx Cα²

Core Logic

Official Answer Path Analysis: When the solution is diluted until $pH = 6$ $\mathrm{pH} = 6$, the hydronium ion concentration becomes:

[H^+] = 10⁻⁶~M = Cnewαnew

Applying the equilibrium constant expression without approximations for very high dilutions:

Kₐ = (Cα²)/(1-α) = ([H^+]α)/(1-α) = 4 \times 10⁻¹⁰

10⁻⁶ \cdot α1-α = 4 \times 10⁻¹⁰ 10⁴ α = 4(1-α)

2500α = 1 - α 2501α = 1 α \approx (1)/(2500)

Now, substitute $α$ $\alpha$ back to isolate the absolute concentration parameter $Cnew$ $C_{\text{new}}$:

Cnewα = 10⁻⁶ Cnew \cdot ((1)/(2500)) = 10⁻⁶

Cnew = 2500 \times 10⁻⁶ = 25 \times 10⁻⁴~M

Therefore, comparing with $x \times 10⁻⁴~M$ $x \times 10^{-4}\mathrm{~M}$ yields $x = 25$ $x = 25$.

Pattern Recognition

When dealing with extreme dilution states where $α$ $\alpha$ becomes large, you must avoid the standard $(1-α) \approx 1$ $(1-\alpha) \approx 1$ simplification step to ensure mathematically accurate answers.

Chapter Mix

Class 11 Chemistry: Ionic Equilibrium

Q46 jee_main_2025_07_april_evening Buffer Solutions

One litre buffer solution was prepared by adding

0.10 mol

each of

NH₃

and

NH₄Cl

in deionised water. The change in pH on addition of

0.05 mol

of

HCl

to the above solution is

\times 10⁻²

(Nearest integer) [cite: 433, 434] Given:

pKb

of

NH₃ = 4.745

and

₁₀3 = 0.477

Numerical Answer. Answer: 47.5 to 48.5

Solution

Related Formula

pOH = pKb + [Salt][Base]

pH = 14 - pOH

Core Logic

Initially, the basic buffer solution contains:

[Salt] = [NH4^+] = 0.10 mol, [Base] = [NH3] = 0.10 mol

pOHinitial = 4.745 + (0.10)/(0.10) = 4.745

When $0.05 mol$ $0.05\text{ mol}$ of strong acid $HCl$ $\text{HCl}$ is introduced, it reacts stoichiometrically with the weak base $NH₃$ $\text{NH}_3$: [cite: 1049, 1050]

$arrayrcccc & NH3 & + & H^+ & arrow & NH4^+ Initial (mol): & 0.10 & & 0.05 & & 0.10 Final (mol): & 0.05 & & 0 & & 0.15 array$

Step 1: Computing Post-Acid pOH and pH

Recalculating via Henderson's equation:

pOHfinal = 4.745 + (0.15)/(0.05) = 4.745 + 3

The total shift value follows as:

Δ pOH = pOHfinal - pOHinitial = 3 = 0.477

Since $pH = 14 - pOH$ $\text{pH} = 14 - \text{pOH}$:

Δ pH = -Δ pOH = -0.477

Expressing the structural magnitude in scientific notation format:

|Δ pH| = 0.477 = 47.7 \times 10⁻² \approx 48 \times 10⁻²

Pattern Recognition

Buffer shifting rule: Adding an acid consumes base and builds salt. The base drops from $0.1$ $0.1$ to $0.05$ $0.05$ (halved), while salt grows from $0.1$ $0.1$ to $0.15$ $0.15$ (tripled). The ratio flips to 3, introducing a clean $3$ $\log 3$ change factor into the solution.

Chapter Mix

Class 11 Chemistry: Ionic Equilibrium

Q34 jee_main_2025_24_jan_evening Chemical Equilibrium and Law of Mass Action

For the reaction,

H₂(g) + I₂(g) leftharpoons 2HI(g)

Attainment of equilibrium is predicted correctly by the concentration profiles plotted over time in option:

A. \text{Graph Option (1)}
B. \text{Graph Option (2)}
C. \text{Graph Option (3)}
D. \text{Graph Option (4)}

Solution

Core Logic

Let's track the concentration changes for the reversible reaction starting with reactants $H₂$ $\mathrm{H}_{2}$ and $I₂$ $\mathrm{I}_{2}$:

As the forward reaction proceeds, the concentrations of reactants ( $H₂$ $\mathrm{H}_2$ and $I₂$ $\mathrm{I}_2$) decrease over time.
Simultaneously, the concentration of the product ( $HI$ $\mathrm{HI}$) increases from zero.
Once dynamic equilibrium is attained, the rates of the forward and reverse reactions become equal. Consequently, the concentrations of all reactants and products become constant over time, appearing as horizontal lines on a concentration vs. time plot.

Graph (2) correctly depicts the concentration of reactants smoothly decreasing and the product concentration increasing until they all plateau horizontally at dynamic equilibrium.

Pattern Recognition

On a concentration vs. time graph, look for lines that become perfectly horizontal after a certain point. This horizontal plateau signifies that equilibrium has been established.

Chapter Mix

Class 11 Chemistry: Equilibrium

Q jee_main_2025_24_jan_morning Chemical Equilibrium and Kp calculation

37.8 ~g ~N₂ O₅

was taken in a

1 ~L

reaction vessel and allowed to undergo the following reaction at

500 ~K

:

2 N _ 2 O _ 5 (g) leftharpoons 2 N _ 2 O _ 4 (g) + O _ 2 (g)

The total pressure at equilibrium was found to be 18.65 bar. Then,

Kp = _ _ _ _ \times 10⁻²

[nearest integer] Assume

N₂O₅

to behave ideally under these conditions Given:

R = 0.082

bar

L mol⁻¹ K⁻¹

Numerical Answer. Answer: 962 to 962

Solution

Related Formula

P = (nRT)/(V) and Kₚ = (PN₂O₄)² \cdot PO₂(PN₂O₅)²

Core Logic

First, find the initial moles of $N₂O₅$ $\mathrm{N}_2\mathrm{O}_5$ using its molar mass ( $108 g/mol$ $108\text{ g/mol}$):

n₀ = (37.8)/(108) = 0.35 moles

Using the ideal gas equation, compute the initial pressure ( $Pᵢ$ $P_i$):

Pᵢ = (0.35 \times 0.082 \times 500)/(1) = 14.35 bar

Setting up the equilibrium partial pressures table: $arraylccccc & 2N₂O5(g) & leftharpoons & 2N₂O4(g) & + & O2(g) Initially: & 14.35 & & 0 & & 0 At equilibrium: & 14.35 - 2P & & 2P & & P array$

The total pressure at equilibrium is given as:

Ptotal = (14.35 - 2P) + 2P + P = 14.35 + P = 18.65 bar

P = 18.65 - 14.35 = 4.3 bar

Now, calculate the equilibrium partial pressures for each component:

$PN₂O₅ = 14.35 - 2(4.3) = 5.75 bar$
$PN₂O₄ = 2(4.3) = 8.6 bar$
$PO₂ = 4.3 bar$ $P_{\mathrm{O}_2} = 4.3\text{ bar}$

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

Kₚ = ((8.6)² \times 4.3)/((5.75)²) = (73.96 \times 4.3)/(33.0625) \approx 9.619

Expressing the result in the requested format ( $x \times 10⁻²$ $x \times 10^{-2}$):

Kₚ = 961.9 \times 10⁻²

Rounding to the nearest integer yields 962.

Pattern Recognition

Always calculate the initial pressure first using the ideal gas law ( $PV=nRT$ $PV=nRT$). This provides a clear baseline for tracking equilibrium partial pressures.

Chapter Mix

Class 11 Chemistry: Equilibrium

Q32 jee_main_2025_24_jan_morning Solubility Product Constant

Ksp for

Cr(OH)₃

is

1.6\times10⁻³⁰.

What is the molar solubility of this salt in water ?

A. $[4] 1.6\times10⁻³⁰27$
B. $1.8\times10⁻³⁰27$ $\frac{1.8\times10^{-30}}{27}$
C. $[5]1.8\times10⁻³⁰$ $\sqrt[5]{1.8\times10^{-30}}$
D. $[2]1.6\times10⁻³⁰$ $\sqrt[2]{1.6\times10^{-30}}$

Solution

Related Formula

Kₛₚ = x^x y^y sx+y

Core Logic

The dissolution equilibrium for chromium hydroxide is written as:

Cr(OH)3(s) leftharpoons Cr³⁺(aq) + 3OH⁻(aq)

If the molar solubility is denoted by $s$ $s$:

[Cr³⁺] = s and [OH⁻] = 3s

Substituting values into the expressions:

Kₛₚ = (s) \cdot (3s)³ = 27s⁴

Given $Kₛₚ = 1.6 \times 10⁻³⁰$ $$K_{sp} = 1.6 \times 10^{-30}$$:

27s⁴ = 1.6 \times 10⁻³⁰

s = ( 1.6 \times 10⁻³⁰27 )1/4

Pattern Recognition

For a binary-quaternary salt of type $AB₃$ $AB_3$, the relationship simplifies strictly to $Kₛₚ = 27s⁴$ $K_{sp} = 27s^4$.

Chapter Mix

Class 11 Chemistry: Equilibrium

Solution & Explanation

Core Logic

Pattern Recognition

Chapter Mix

Reference Study Guides

More Equilibrium Previous-Year Questions — Page 5

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Computing Post-Acid pOH and pH

Pattern Recognition

Chapter Mix

Solution

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

More Equilibrium Questions — jee_main_2025_29_jan_evening

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

Exam Analysis Tools

Chapter Weightage & Trends

Topic-wise Practice Hub

JEE Main Exam Papers

Chapter Weightage Board

Formula Recognition

Exam Questions

Select an Exam

Welcome back