Chemistry Physical - How to Solve: Equilibrium (Kc, Kp, Le Chatelier’s Principle, Degree of Dissociation) – NEET UG Guide

How to Solve: Equilibrium (Kc, Kp, Le Chatelier’s Principle, Degree of Dissociation) – NEET UG Guide

Introduction

"Mastering equilibrium can get you 8–10 marks in NEET Chemistry—enough to push you into the top 1%! From drug reactions in your body to industrial ammonia synthesis, equilibrium decides whether a reaction happens… or just stops. Today, you’ll learn the exact steps to solve any Kc, Kp, Le Chatelier, or dissociation problem—no guessing, no panic."

WHAT YOU NEED TO KNOW FIRST

Before diving in, ensure you understand:
1. Balanced chemical equations – Coefficients become exponents in equilibrium expressions.
2. Mole concept & concentration – Moles per liter (mol/L) = molarity (M).
3. Partial pressure – For gases, pressure is proportional to moles (Dalton’s Law).

KEY TERMS & FORMULAS

1. Equilibrium Constant (Kc)

Formula: Kc = [Products] / [Reactants] - [ ] = Molar concentration at equilibrium (mol/L) - Exponents = Coefficients from balanced equation - Solids & pure liquids are NOT included in Kc.

MEMORISE THIS: Kc is temperature-dependent but pressure/concentration-independent.

2. Equilibrium Constant (Kp)

Formula: Kp = (P_products) / (P_reactants) - P = Partial pressure of gases at equilibrium (in atm or bar) - Exponents = Coefficients from balanced equation - Only gases are included (solids/liquids ignored).

Relation between Kc and Kp: Kp = Kc (RT)^Δn - R = 0.0821 L·atm·mol⁻¹·K⁻¹ (gas constant) - T = Temperature (Kelvin) - Δn = (Moles of gaseous products) – (Moles of gaseous reactants)

MEMORISE THIS: If Δn = 0, Kp = Kc.

3. Le Chatelier’s Principle

Definition: If a system at equilibrium is disturbed, it shifts to counteract the change and restore equilibrium.

Factors & Effects: | Change | Shift Direction | Effect on K | |------------------|---------------------|-----------------| | ↑ Concentration | Away from added species | No change | | ↑ Pressure | Toward fewer gas moles | No change | | ↑ Temperature | Toward endothermic reaction | K changes | | Catalyst | No shift | No change |

MEMORISE THIS: Only temperature changes K!

4. Degree of Dissociation (α)

Definition: Fraction of reactant that dissociates into products. Formula: α = (Amount dissociated) / (Initial amount)

Relation to Kc (for weak electrolytes): For A ⇌ nB + mC, Kc = [C]^m [B]^n / [A] = (α^n × α^m × C^(n+m-1)) / (1 – α) - C = Initial concentration of reactant - α = Degree of dissociation

MEMORISE THIS: For small α (α < 0.1), 1 – α ≈ 1, so Kc ≈ α²C.

STEP-BY-STEP METHOD

Step 1: Write the Balanced Equation

• Include states of matter (s, l, g, aq).
• Example: N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

Step 2: Identify What’s Given & What’s Asked

• Given: Initial concentrations, Kc/Kp, volume, temperature, etc.
• Asked: Equilibrium concentrations, Kc/Kp, shift direction, α, etc.

Step 3: Set Up an ICE Table (Initial-Change-Equilibrium)

Key Rule: Change row follows stoichiometric coefficients.

Step 4: Write the Equilibrium Expression

• For Kc: Kc = [NH₃]² / ([N₂][H₂]³)
• For Kp: Kp = (P_NH₃)² / (P_N₂ × P_H₂³)

Step 5: Plug in Equilibrium Values & Solve for x

• Substitute equilibrium concentrations from ICE table.
• Example: Kc = (2x)² / [(1.0 – x)(3.0 – 3x)³] = 0.5
• Solve for x (may require approximation if K is small).

Step 6: Find Required Quantities

• Equilibrium concentrations: Substitute x back into ICE table.
• Degree of dissociation (α): α = x / initial concentration
• Shift direction: Apply Le Chatelier’s Principle.

Step 7: Check Units & Significant Figures

• Kc: No units (concentrations cancel out).
• Kp: Units depend on Δn (e.g., atm⁻² if Δn = -2).
• α: Unitless (fraction or percentage).

WORKED EXAMPLES

Example 1 – Basic (Kc Calculation)

Problem: For H₂(g) + I₂(g) ⇌ 2HI(g), at equilibrium, [H₂] = 0.2 M, [I₂] = 0.2 M, [HI] = 0.6 M. Find Kc.

Solution:
1. Balanced equation: H₂ + I₂ ⇌ 2HI
2. Kc expression: Kc = [HI]² / ([H₂][I₂])
3. Substitute values: Kc = (0.6)² / (0.2 × 0.2) = 0.36 / 0.04 = 9

What we did and why: - Wrote Kc expression directly from balanced equation. - Plugged in equilibrium concentrations (no ICE table needed here). - Kc = 9 means products are favored at equilibrium.

Example 2 – Medium (ICE Table + Kp)

Problem: For PCl₅(g) ⇌ PCl₃(g) + Cl₂(g), Kp = 0.82 atm at 500 K. If initial PCl₅ pressure is 1 atm, find equilibrium partial pressures.

Solution:
1. Balanced equation: PCl₅ ⇌ PCl₃ + Cl₂
2. ICE Table: | Species | Initial (atm) | Change (atm) | Equilibrium (atm) | |---------|--------------|--------------|-------------------| | PCl₅ | 1.0 | -x | 1.0 – x | | PCl₃ | 0 | +x | x | | Cl₂ | 0 | +x | x |

1. Kp expression: Kp = (P_PCl₃ × P_Cl₂) / P_PCl₅ = (x × x) / (1.0 – x) = 0.82
2. Solve for x: x² = 0.82(1.0 – x) x² + 0.82x – 0.82 = 0 x = 0.58 atm (using quadratic formula)
3. Equilibrium pressures:
4. P_PCl₅ = 1.0 – 0.58 = 0.42 atm
5. P_PCl₃ = P_Cl₂ = 0.58 atm

What we did and why: - Used ICE table to track pressure changes. - Solved quadratic equation (no approximation since Kp is large). - Δn = +1, so Kp ≠ Kc (but Kp was given, so no conversion needed).

Example 3 – Exam-Style (Le Chatelier + Degree of Dissociation)

Problem: For 2SO₃(g) ⇌ 2SO₂(g) + O₂(g), ΔH = +198 kJ/mol. If: - Volume is doubled, - Temperature is increased, - A catalyst is added, Predict the effect on: a) Equilibrium position b) Kc c) Degree of dissociation (α) of SO₃.

Solution:
1. Balanced equation: 2SO₃ ⇌ 2SO₂ + O₂ (ΔH = +198 kJ/mol)
2. Effect of volume doubling (↓ pressure): - Shift: Toward more gas moles (right, since 3 moles > 2 moles). - Kc: No change (only temperature affects K). - α: Increases (more dissociation to counteract pressure drop).
3. Effect of temperature increase: - Shift: Toward endothermic reaction (right, ΔH = +ve). - Kc: Increases (favors products at higher T). - α: Increases (more dissociation).
4. Effect of catalyst: - Shift: No change (catalyst speeds up both forward & reverse reactions equally). - Kc: No change. - α: No change (only reaches equilibrium faster).

What we did and why: - Applied Le Chatelier’s Principle separately for each change. - Temperature is the only factor that changes Kc. - Catalysts never affect equilibrium position or K.

COMMON MISTAKES

EXAM TRAPS

1-MINUTE RECAP (Night Before Exam)

"Listen up—this is your 60-second equilibrium survival guide for NEET:

1. Kc vs. Kp: Kc uses concentrations, Kp uses partial pressures. Kp = Kc (RT)^Δn—memorize this!
2. ICE Table: Write Initial, Change, Equilibrium for every problem. Change row follows coefficients.
3. Le Chatelier: Only temperature changes K. Pressure/volume shifts toward fewer gas moles. Catalysts? No effect.
4. Degree of dissociation (α): For A ⇌ nB, Kc = α^n × C^(n-1) (if α is small, Kc ≈ α²C).
5. Common traps: Solids/liquids don’t go in K expressions. Coefficients become exponents. Check units for Kp!

You’ve got this. Now go crush that equilibrium question!