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

More Chemical 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

Q31 jee_main_2025_29_jan_evening Le Chatelier's Principle

Consider the equilibrium

CO(g) + 3H₂(g) leftharpoons CH₄(g) + H₂O(g)

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:

A. (A) and (B) only
B. (A), (B) and (D) only
C. (B) and (C) only
D. (A), (B) and (C) only

Solution

Core Logic

Statement (A) is correct: Increasing pressure by compressing the volume increases active mass/concentration ( $c = n/V$ $c = n/V$) for both reactants and products instantly. Statement (B) is correct: The reaction has $Delta ng = 2 - 4 = -2$ $Delta n_g = 2 - 4 = -2$. Increasing pressure shifts equilibrium towards the direction of fewer gaseous moles, which is the forward path. Statement (C) is incorrect: Equilibrium constant ( $K$ $K$) is exclusively temperature-dependent and does not alter with pressure changes. Statement (D) is correct: Confirms that equilibrium constant remains unchanged.

Pattern Recognition

Always remember: pressure changes shift positions but NEVER alter the value of the equilibrium constant $Kc$ $K_c$ or $Kₚ$ $K_p$. Only temperature changes can change $K$ $K$.

Chapter Mix

Class 11 Chemistry: Equilibrium

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

Q28 jee_main_2025_03_april_morning Factors Affecting Equilibrium and Catalyst

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:

A. Statement I is false but Statement II is true
B. Both Statement I and Statement II are true
C. Both Statement I and Statement II are false
D. Statement I is true but Statement II is false

Solution

Related Formula

Kc = (kf)/(kb)

where $kf$ $k_f$ and $kb$ $k_b$ are forward and backward rate constants.

Core Logic

Statement I is true: A catalyst increases both forward ( $kf$ $k_f$) and backward ( $kb$ $k_b$) rate constants to the same extent by lowering the activation energy barrier. Therefore, $Kc = kf / kb$ $K_c = k_f / k_b$ 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.

Step 1: Final Conclusion

Statement I is true but Statement II is false.

Pattern Recognition

Catalyst effect = Speeds up reaching equilibrium. Catalyst NEVER changes: Equilibrium Constant ( $Kc$ $K_c$) or Equilibrium Concentrations/Composition.

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 4

Solution

Related Formula

Execution

Pattern Recognition

Chapter Mix

Solution

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Step 1: Expressing Concentration

Step 2: Logarithmic Rearrangement

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Final Conclusion

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: