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

Q71 jee_main_2026_21_jan_evening Ionic Equilibrium and Polyprotic Acids

The first and second ionization constants of

H₂X

are

2.5 \times 10⁻⁸

and

1.0 \times 10⁻¹³

respectively. The concentration of

X²⁻

in

0.1 M H₂X

solution is

_ \times 10⁻¹⁵ M

. (Nearest Integer)

Numerical Answer. Answer: 100 to 100

Solution

Core Logic

For a weak diprotic acid $H₂X$ $\text{H}_2\text{X}$, the concentration of $X²⁻$ $\text{X}^{2-}$ in a solution of concentration $C$ $C$ is approximately equal to the second dissociation constant $Ka₂$ $K_{a_2}$ under standard weak acid approximation conditions ( $[X²⁻] \approx Ka₂ = 1.0 \times 10⁻¹³ M$ ).

Converting to the required format: $1.0 \times 10⁻¹³ = 100 \times 10⁻¹⁵ M$ .

Step 1: Final Calculation

Thus, the integer value is 100.

Pattern Recognition

Sees: polyprotic acid dissociation equilibrium calculation for secondary anion concentration. Trap: Assuming complete dissociation for the second step.

Chapter Mix

Class 11 Chemistry: Equilibrium

Q63 jee_main_2026_22_january_evening Buffer Solution pH Calculation

Which of the following mixture gives a buffer solution with pH = 9.25 ? Given:

pKb(NH₄OH) = 4.75

A. $0.2M NH₄OH (0.4 L) + 0.1M HCl (1 L)$
B. $0.2M NH₄OH (0.5 L) + 0.1M HCl (0.5 L)$
C. $0.5M NH₄OH (0.2 L) + 0.2M HCl (0.5 L)$
D. $0.4M NH₄OH (1 L) + 0.1M HCl (1 L)$

Solution

Related Formula

pOH = 14 - pH = 14 - 9.25 = 4.75

pOH = pKb + [Salt][Base]

Core Logic

Step 1: Determine condition for $pOH = 4.75$ $\text{pOH} = 4.75$:

4.75 = 4.75 + [Salt][Base] [Salt][Base] = 0 [Salt] = [Base]

Step 2: Test Option (2):

Initial millimoles of NH₄OH = 0.2 \times 500 = 100 mmol

Initial millimoles of HCl = 0.1 \times 500 = 50 mmol

Millimoles of NH₄Cl formed = 50 mmol

Millimoles of NH₄OH remaining = 100 - 50 = 50 mmol

Since $[Salt] = [Base]$ $[\text{Salt}] = [\text{Base}]$, $pOH = pKb = 4.75$ $\text{pOH} = pK_b = 4.75$ and $pH = 9.25$ $\text{pH} = 9.25$.

Pattern Recognition

Sees: Basic buffer when $pH = 14 - pKb$ $\text{pH} = 14 - pK_b$. Shortcut: Requires millimoles of remaining weak base to equal millimoles of formed conjugate salt (i.e. acid amount must be exactly half of base amount).

Chapter Mix

Class 11 Chemistry: Equilibrium

Q70 jee_main_2026_23_january_morning Relation between Kp and Kc

Consider the general reaction given below at 400 K

xA(g) leftharpoons yB(g)

The values of

Kₚ

and

Kc

are studied under the same condition of temperature but variation in x and y. (i)

Kₚ = 85.87

and

Kc = 2.586

appropriate units (ii)

Kₚ = 0.862

and

Kc = 28.62

appropriate units The value of x and y in (i) and (ii) respectively are:

A. $(i) 3, 1 ; (ii) 3, 1$ $\text{(i) 3, 1 ; (ii) 3, 1}$
B. $(i) 4, 1 ; (ii) 4, 1$ $\text{(i) 4, 1 ; (ii) 4, 1}$
C. $(i) 1, 3 ; (ii) 2, 1$ $\text{(i) 1, 3 ; (ii) 2, 1}$
D. $(i) 1, 2 ; (ii) 2, 1$ $\text{(i) 1, 2 ; (ii) 2, 1}$

Solution

Related Formula

Kₚ = Kc (RT)Δ ng

Core Logic

For the reaction $xA(g) leftharpoons yB(g)$ $xA(g) \rightleftharpoons yB(g)$, $Δ ng = y - x$ $\Delta n_g = y - x$. We use $R = 0.0821 L atm K⁻¹ mol⁻¹$ and $T = 400 K$ $T = 400 \text{ K}$.

$RT = 0.0821 \times 400 = 32.84$ $RT = 0.0821 \times 400 = 32.84$.

Step 1: Case (i) Analysis

In case (i), $Kₚ = 85.87$ $K_p = 85.87$ and $Kc = 2.586$ $K_c = 2.586$. Since $Kₚ > Kc$ $K_p > K_c$, $Δ ng > 0$ $\Delta n_g > 0$, implying $y - x > 0$ $y - x > 0$.

85.87 = 2.586 \times (32.84)y-x

33.20 \approx (32.84)y-x

Solving gives $y - x = 1$ $y - x = 1$. Looking at options for (i), "1, 2" (where $x=1, y=2$ $x=1, y=2$) fits perfectly because $2 - 1 = 1$ $2 - 1 = 1$.

Step 2: Case (ii) Analysis

In case (ii), $Kₚ = 0.862$ $K_p = 0.862$ and $Kc = 28.62$ $K_c = 28.62$. Since $Kₚ < Kc$ $K_p < K_c$, $Δ ng < 0$ $\Delta n_g < 0$, implying $y - x < 0$ $y - x < 0$. Looking at options for (ii), "2, 1" (where $x=2, y=1$ $x=2, y=1$) gives $y - x = -1$ $y - x = -1$.

Kₚ = Kc \times (32.84)⁻¹ 0.862 = 28.62 / 32.84 \approx 0.87

This matches correctly.

Pattern Recognition

Just check the signs! $Kₚ > Kc \Rightarrow y > x$ $K_p > K_c \Rightarrow y > x$. $Kₚ < Kc \Rightarrow y < x$ $K_p < K_c \Rightarrow y < x$. Only option (4) satisfies both conditions simultaneously without even needing full calculation.

Chapter Mix

Class 11 Chemistry: Equilibrium

Q72 jee_main_2026_23_january_morning Degree of Dissociation and Kp

For the following gas phase equilibrium reaction at constant temperature,

NH₃(g) leftharpoons (1)/(2)N₂(g) + (3)/(2)H₂(g)

If the total pressure is

\sqrt(3)

atm and the pressure equilibrium constant (

Kₚ

) is 9 atm, then the degree of dissociation is given as

(x \times 10⁻²)-1/2

. The value of x is ____ (Nearest integer)

Numerical Answer. Answer: 125 to 125

Solution

Related Formula

Kₚ = (PN₂)1/2 (PH₂)3/2(PNH₃)

Core Logic

Let the degree of dissociation of $NH₃$ $NH_3$ be $α$ $\alpha$. Initial moles: $NH₃ = 1$ $NH_3 = 1$ Equilibrium moles: $NH₃ = 1 - α$ $NH_3 = 1 - \alpha$, $N₂ = (α)/(2)$ $N_2 = \frac{\alpha}{2}$, $H₂ = (3α)/(2)$ $H_2 = \frac{3\alpha}{2}$ Total moles = $1 - α + (α)/(2) + (3α)/(2) = 1 + α$

Step 1: Setting up Partial Pressures

Using mole fractions and total pressure $PT = \sqrt(3)$ $P_T = \sqrt{3}$: $PNH₃ = (1-α)/(1+α) PT$ $PN₂ = (α/2)/(1+α) PT$ $PH₂ = (3α/2)/(1+α) PT$

Step 2: Solving for alpha

Kₚ = ((α/2)/(1+α) PT)1/2 \cdot ((3α/2)/(1+α) PT)3/2 (1-α)/(1+α) PT

Kₚ = (α/2)1/2 (3α/2)3/2 1-α \cdot (PT²)/(PT (1+α)) \cdot (1+α)

Wait, simplify the $PT$ $P_T$ term:

Kₚ = (α/2)1/2 (3α/2)3/2 1-α \cdot ((PT)/(1+α))

Substitute $Kₚ = 9$ $K_p = 9$ and $PT = \sqrt(3)$ $P_T = \sqrt{3}$:

9 = (α/2)1/2 (3α/2)3/2 (1-α) \cdot \sqrt(3)1+α

9 = \sqrt(27) \cdot (α/2)² (1-α)(1+α) \cdot \sqrt(3)

9 = ( 9 \cdot (α²/4) )/( 1 - α² )

1 = ( α²/4 )/( 1 - α² )

1 - α² = (α²)/(4)

1 = (5α²)/(4)

α² = (4)/(5) = 0.8

α = (0.8)1/2

Step 3: Matching the given format

We are given $α = (x \times 10⁻²)-1/2$ $\alpha = (x \times 10^{-2})^{-1/2}$. Equating both forms:

(0.8)1/2 = (x \times 10⁻²)-1/2

Invert both sides:

x \times 10⁻² = (1)/(0.8)

x \times 10⁻² = 1.25

$x = 125$ $x = 125$

Pattern Recognition

When doing fractional stoichiometric power $Kₚ$ $K_p$ calculations, grouping the $α$ $\alpha$ terms early simplifies algebra massively. $33/2 \times 31/2 = 9$ $3^{3/2} \times 3^{1/2} = 9$, which perfectly cancels the $Kₚ$ $K_p$ value.

Chapter Mix

Class 11 Chemistry: Equilibrium

Q71 jee_main_2026_23_january_evening Chemical Equilibrium and Le Chatelier's Principle

X₂(g) + Y₂(g) leftharpoons 2Z(g)

X₂(g)

and

Y₂(g)

are added to a

1 L

flask and it is found that the system attains the above equilibrium at T(K) with the number of moles of

X₂(g)

,

Y₂(g)

and Z(g) being 3, 3 and 9 mol respectively (equilibrium moles). Under this conditions of equilibrium, 10 mol of Z(g) is added to the flask and the temperature is maintained at T(K). Then the number of moles of Z(g) in the flask when the new equilibrium is established is ____. (Nearest integer).

Numerical Answer. Answer: 15 to 15

Solution

Related Formula

Kc = ([Z]²)/([X₂][Y₂])

Core Logic

For the reaction $X₂(g) + Y₂(g) leftharpoons 2Z(g)$ : First, calculate the equilibrium constant $Kc$ $K_c$ using the initial equilibrium concentrations (moles in 1 L volume). $[X₂] = 3 M$ $[X_2] = 3 \text{ M}$ $[Y₂] = 3 M$ $[Y_2] = 3 \text{ M}$ $[Z] = 9 M$ $[Z] = 9 \text{ M}$

Kc = ((9)²)/((3)(3)) = (81)/(9) = 9

Step 1: Shift due to addition of Z

Now, 10 moles of Z(g) are added. The new initial moles are: $X₂ = 3$ $X_2 = 3$ $Y₂ = 3$ $Y_2 = 3$ $Z = 9 + 10 = 19$ $Z = 9 + 10 = 19$

The reaction quotient $Qc > Kc$ $Q_c > K_c$, so the reaction shifts backward. Let the change in $X₂$ $X_2$ and $Y₂$ $Y_2$ be $+x$ $+x$ moles, then Z changes by $-2x$ $-2x$ moles. At new equilibrium: $X₂ = 3 + x$ $X_2 = 3 + x$ $Y₂ = 3 + x$ $Y_2 = 3 + x$ $Z = 19 - 2x$ $Z = 19 - 2x$

Step 2: Calculate New Equilibrium

Kc = ((19 - 2x)²)/((3 + x)(3 + x)) = 9

Take the square root of both sides (concentrations must be positive):

(19 - 2x)/(3 + x) = 3

19 - 2x = 3(3 + x)

19 - 2x = 9 + 3x

10 = 5x x = 2

The number of moles of Z(g) at the new equilibrium is:

Moles of Z = 19 - 2x = 19 - 2(2) = 19 - 4 = 15

Pattern Recognition

If a reaction's $Kc$ $K_c$ evaluates to a perfect square and $Δ ng = 0$ $\Delta n_g = 0$, adding products sets up a neat quadratic that can be directly resolved by taking the root of both sides, heavily speeding up the algebraic work.

Chapter Mix

Class 11 Chemistry: Equilibrium

Solution & Explanation

Core Logic

Pattern Recognition

Chapter Mix

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