Given below are two statements: Statement I : A catalyst cannot alter the equilibrium constant (Kc) of the reaction, temperature remaining constant. Statement II : A homogenous catalyst can change the equilibrium composition of a system, temperature remaining constant. In the light of the above statements, choose the correct answer from the options given below:

where kf and kb are forward and backward rate constants. Statement I is true: A catalyst increases both forward (kf) and backward (kb) rate constants to the same extent by lowering the activation energy barrier. Therefore, Kc = kf / kb remains unchanged at constant temperature. Statement II is false: A catalyst (whether homogeneous or heterogeneous) helps the reaction attain equilibrium faster but cannot alter the equilibrium composition or the position of equilibrium. Statement I is true but Statement II is false. Catalyst effect = Speeds up reaching equilibrium. Catalyst NEVER changes: Equilibrium Constant (Kc) or Equilibrium Concentrations/Composition. Class 11 Chemistry: Chemical Equilibrium

Given below are two statements: Statement I : A catalyst cannot alter the equilibrium constant (Kc) of the reaction, temperature remaining constant. Statement II : A homogenous catalyst can change the

where kf and kb are forward and backward rate constants. Statement I is true: A catalyst increases both forward (kf) and backward (kb) rate constants to the same extent by lowering the activation energy barrier. Therefore, Kc = kf / kb remains unchanged at constant temperature. Statement II is false: A catalyst (whether homogeneous or heterogeneous) helps the reaction attain equilibrium faster but cannot alter the equilibrium composition or the position of equilibrium. Statement I is true but St

Given below are two statements: Statement I : A catalyst cannot alter the equilibrium constant (Kc) of the reaction, temperature remaining constant. Statement II : A homogenous catalyst can change the equilibrium composition of a system, temperature remaining constant. In the light of the above statements, choose the correct answer from the options given below:

where kf and kb are forward and backward rate constants. Statement I is true: A catalyst increases both forward (kf) and backward (kb) rate constants to the same extent by lowering the activation energy barrier. Therefore, Kc = kf / kb remains unchanged at constant temperature. Statement II is false: A catalyst (whether homogeneous or heterogeneous) helps the reaction attain equilibrium faster but cannot alter the equilibrium composition or the position of equilibrium. Statement I is true but Statement II is false. Catalyst effect = Speeds up reaching equilibrium. Catalyst NEVER changes: Equilibrium Constant (Kc) or Equilibrium Concentrations/Composition. Class 11 Chemistry: Chemical Equilibrium

Given below are two statements: Statement I : A catalyst cannot alter the equilibrium constant (Kc) of the reaction, temperature remaining constant. Statement II : A homogenous catalyst can change the

where kf and kb are forward and backward rate constants. Statement I is true: A catalyst increases both forward (kf) and backward (kb) rate constants to the same extent by lowering the activation energy barrier. Therefore, Kc = kf / kb remains unchanged at constant temperature. Statement II is false: A catalyst (whether homogeneous or heterogeneous) helps the reaction attain equilibrium faster but cannot alter the equilibrium composition or the position of equilibrium. Statement I is true but St

Given below are two statements: Statement I : A catalyst cannot alter the equilibrium constant (Kc) of the reaction…

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

Q30 jee_main_2025_28_jan_evening Solubility Product

Arrange the following in increasing order of solubility product:

Ca(OH)₂, AgBr, PbS, HgS

A. $PbS < HgS < Ca(OH)₂ < AgBr$ $PbS < HgS < Ca(OH)_{2} < AgBr$
B. $HgS < PbS < AgBr < Ca(OH)₂$ $HgS < PbS < AgBr < Ca(OH)_{2}$
C. $Ca(OH)₂ < AgBr < HgS < PbS$ $Ca(OH)_{2} < AgBr < HgS < PbS$
D. $HgS < AgBr < PbS < Ca(OH)₂$ $HgS < AgBr < PbS < Ca(OH)_{2}$

Solution

Related Formula

The solubility product constant ( $Kₛₚ$ $K_{sp}$) reflects the equilibrium position of a sparingly soluble salt in water.

Core Logic

Based on standard literature $Kₛₚ$ $K_{sp}$ values at $298 K$ $298\text{ K}$:

$HgS$ $HgS$: $\approx 4 \times 10⁻⁵³$ $\approx 4 \times 10^{-53}$ (extremely insoluble, Group IIB cation analysis)
$PbS$ $PbS$: $\approx 8 \times 10⁻²⁸$ $\approx 8 \times 10^{-28}$ (highly insoluble, Group IIA cation analysis)
$AgBr$ $AgBr$: $\approx 5 \times 10⁻¹³$ $\approx 5 \times 10^{-13}$ (sparingly soluble halide salt)
$Ca(OH)₂$ $Ca(OH)_2$: $\approx 5.5 \times 10⁻⁶$ $\approx 5.5 \times 10^{-6}$ (moderately soluble base)

Step 1: Arrangement

Comparing these $Kₛₚ$ $K_{sp}$ orders:

4 \times 10⁻⁵³ < 8 \times 10⁻²⁸ < 5 \times 10⁻¹³ < 5.5 \times 10⁻⁶

Hence, the correct increasing sequence is: $HgS < PbS < AgBr < Ca(OH)₂$ $HgS < PbS < AgBr < Ca(OH)_2$.

Pattern Recognition

Sulphides of heavy transition metals like $Hg²⁺$ $Hg^{2+}$ and $Pb²⁺$ $Pb^{2+}$ have exceptionally small $Kₛₚ$ $K_{sp}$ values compared to halides or hydroxides. Among sulphides, $HgS$ $HgS$ is famously known to have one of the lowest solubility products found in inorganic qualitative analysis.

Chapter Mix

Class 11 Chemistry: Equilibrium

Q jee_main_2025_29_jan_morning Degree of Dissociation and Equilibrium Constant

At temperature

T

, compound

AB2(g)

dissociates as:

AB2(g) leftharpoons AB(g) + (1)/(2)B2(g)

having a degree of dissociation

x

(

x ll 1

). The correct expression for

x

in terms of

Kₚ

and total pressure

p

is:

x = ((2Kₚ²)/(p))1/3

A. $[3] 2Kₚp$ $\sqrt[3]{\frac{2K_{p}}{p}}$
B. $[4] 2Kₚp$ $\sqrt[4]{\frac{2K_{p}}{p}}$
C. $[3] 2Kₚ²p$ $\sqrt[3]{\frac{2K_{p}^{2}}{p}}$
D. $Kₚ$ $\sqrt{\mathrm{K_p}}$

Solution

Related Formula

Kₚ = pAB \cdot pB₂1/2pAB₂

Core Logic

Consider the equilibrium reaction setup:

State	$AB2(g)$ $\mathrm{AB}_{2(g)}$	$leftharpoons$ $\rightleftharpoons$	$AB(g)$ $\mathrm{AB}_{(g)}$	$+$ $+$	$(1)/(2)B2(g)$ $\frac{1}{2}\mathrm{B}_{2(g)}$
Initial moles:	$1$ $1$		$0$ $0$		$0$ $0$
Equilibrium moles:	$1 - x$ $1 - x$		$x$ $x$		$(x)/(2)$ $\frac{x}{2}$

Total equilibrium moles:

ntotal = 1 - x + x + (x)/(2) = 1 + (x)/(2)

Since $x ll 1$ $x \ll 1$, total moles $ntotal \approx 1$ $n_{\text{total}} \approx 1$ and $(1 - x) \approx 1$ $(1 - x) \approx 1$.

Partial pressures:

pAB₂ \approx p

pAB \approx x p

pB₂ \approx (x)/(2) p

Substituting into $Kₚ$ $K_p$:

Kₚ = (x p) \cdot ((x p)/(2))1/2p = x \cdot ((x p)/(2))1/2 = x3/2 p1/2\sqrt(2)

Squaring both sides and solving for $x$ $x$:

Kₚ² = (x³ p)/(2) x³ = (2Kₚ²)/(p)

x = 3\sqrt((2Kₚ²)/(p))

Pattern Recognition

For $Δ ng = 0.5$ $\Delta n_g = 0.5$ involving degree of dissociation $x ll 1$ $x \ll 1$, tracking total pressure approximations ensures an immediate analytical solution without full polynomial expansion.

Chapter Mix

Class 11 Chemistry: Chemical Equilibrium

Q90 jee_main_2024_01_february_morning Hydrolysis of Salts

Kₐ

for

CH₃COOH

is

1.8 \times 10⁻⁵

and

Kb

for

NH₄OH

is

1.8 \times 10⁻⁵

. The pH of ammonium acetate solution will be

Numerical Answer. Answer: 7 to 7

Solution

Related Formula

For a salt of weak acid and weak base (like ammonium acetate):

pH = (1)/(2) (pKw + pKₐ - pKb)

Core Logic

Ammonium acetate ( $CH₃COONH₄$ $CH_3COONH_4$) is a salt formed from a weak acid ( $CH₃COOH$ $CH_3COOH$) and a weak base ( $NH₄OH$ $NH_4OH$). Given: $Kₐ = 1.8 \times 10⁻⁵$ $K_a = 1.8 \times 10^{-5}$ $Kb = 1.8 \times 10⁻⁵$ $K_b = 1.8 \times 10^{-5}$

Since $Kₐ = Kb$ $K_a = K_b$, taking the negative logarithm gives $pKₐ = pKb$ $\mathrm{p}K_a = \mathrm{p}K_b$.

Step 1: Calculate pH

pH = pKw + pKₐ - pKb2

Substitute $pKₐ = pKb$ $\mathrm{p}K_a = \mathrm{p}K_b$:

pH = pKw2

At standard temperature (298 K), $pKw = 14$ $\mathrm{p}K_w = 14$.

pH = (14)/(2) = 7

Pattern Recognition

If $Kₐ = Kb$ $K_a = K_b$ for a weak acid-weak base salt, the hydrolysis of cation and anion perfectly balance out, making the resulting solution exactly neutral (pH = 7) regardless of the concentration of the salt.

Chapter Mix

Class 11 Chemistry: Equilibrium

Q jee_main_2024_29_january_evening Equilibrium Constant Calculation

The following concentrations were observed at

500 K

for the formation of

NH₃

from

N₂

and

H₂

. At equilibrium:

[N₂] = 2 \times 10⁻² M

,

[H₂] = 3 \times 10⁻² M

, and

[NH₃] = 1.5 \times 10⁻² M

. Equilibrium constant for the reaction is ________.

Numerical Answer. Answer: 417 to 417

Solution

Related Formula

N₂(g) + 3H₂(g) leftharpoons 2NH₃(g)

Kc = [NH₃]²[N₂][H₂]³

Core Logic

Substituting the given equilibrium concentrations into the equilibrium constant expression:

Kc = (1.5 \times 10⁻²)²(2 \times 10⁻²) \times (3 \times 10⁻²)³

Evaluating the values step-by-step:

Kc = 2.25 \times 10⁻⁴(2 \times 10⁻²) \times (27 \times 10⁻⁶)

Step 1: Final Arithmetic Integration

Kc = 2.25 \times 10⁻⁴54 \times 10⁻⁸ = (2.25)/(54) \times 10⁴ = 0.041666 \times 10⁴ \approx 416.67

Rounding to the nearest integer yields 417.

Pattern Recognition

Pay close attention to the cubic exponent in the denominator derived from the hydrogen stoichiometric coefficient ( $3$ $3$). Small calculation errors here can significantly alter the result.

Chapter Mix

Class 11 Chemistry: Chemical Equilibrium

Solution & Explanation

Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Reference Study Guides

More Chemical Equilibrium Previous-Year Questions — Page 6

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Arrangement

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate pH

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Final Arithmetic Integration

Pattern Recognition

Chapter Mix

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