In the following system, PCl5(g) leftharpoons PCl3(g) + Cl2(g) at equilibrium, upon addition of xenon gas at constant T…

Q: 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:

Addition of inert gas at constant pressure increases total volume V, shifting equilibrium toward the side with more gaseous moles (Delta ng > 0). For PCl5(g) leftharpoons PCl3(g) + Cl2(g), Delta ng = 1 + 1 - 1 = +1 > 0. Addition of an inert gas (xenon) at constant temperature and total pressure increases system volume. To counteract this change, equilibrium shifts in the forward direction (towards greater number of gaseous moles), resulting in an increase in the moles/amount of PCl3 and Cl2 form

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

In the following system, PCl₅(g) leftharpoons PCl₃(g) + Cl₂(g)$\mathrm{PCl}_5(\mathrm{g}) \rightleftharpoons \mathrm{PCl}_3(\mathrm{g}) + \mathrm{Cl}_2(\mathrm{g})$ at equilibrium, upon addition of xenon gas at constant T$T$ & p$p$, the concentration of:

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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In the following system, $PCl₅(g) leftharpoons PCl₃(g) + Cl₂(g)$ at equilibrium, upon addition of xenon gas at constant $T$ $T$ & $p$ $p$, the concentration of: