Chemistry Physical How to Solve: Chemical Equilibrium (Kp, Kc, Le Chatelier’s Principle, Degree of Dissociation) – IIT JEE Guide

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

Introduction

Mastering chemical equilibrium unlocks 8–12 marks in IIT JEE (Main + Advanced)—enough to push you from a 90 to a 99+ percentile. It’s the key to predicting how reactions shift under stress, designing industrial catalysts, and even understanding how your blood carries oxygen. If you can solve equilibrium problems fast and error-free, you’ll outscore 90% of test-takers.

WHAT YOU NEED TO KNOW FIRST

Before diving in, ensure you’re 100% clear on:
1. Stoichiometry – Balancing equations, mole ratios, and limiting reagents.
2. Partial Pressure – How to calculate it from mole fractions and total pressure.
3. Ideal Gas Law – ( PV = nRT ) (given on exam sheet, but you must know how to use it).

If any of these feel shaky, stop now and review them. Equilibrium builds on these concepts—skipping them guarantees mistakes.

KEY TERMS & FORMULAS

1. Equilibrium Constant (Kc)

Formula: [ K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b} ] Variables: - ([A], [B], [C], [D]) = Molar concentrations of reactants/products at equilibrium (mol/L). - (a, b, c, d) = Stoichiometric coefficients from the balanced equation. MEMORISE THIS: Only gases and aqueous solutions appear in (K_c). Solids and pure liquids are excluded (their concentrations don’t change).

2. Equilibrium Constant (Kp)

Formula: [ K_p = \frac{(P_C)^c (P_D)^d}{(P_A)^a (P_B)^b} ] Variables: - (P_A, P_B, P_C, P_D) = Partial pressures of gases at equilibrium (in atm or bar). MEMORISE THIS: (K_p) is only for gases. Solids/liquids are not included.

Relationship between Kp and Kc: [ K_p = K_c (RT)^{\Delta n} ] Variables: - (R) = Gas constant (0.0821 L·atm/mol·K). - (T) = Temperature (K). - (\Delta n) = (Moles of gaseous products) – (Moles of gaseous reactants). MEMORISE THIS: If (\Delta n = 0), then (K_p = K_c).

3. Reaction Quotient (Q)

Formula: [ Q = \frac{[C]^c [D]^d}{[A]^a [B]^b} \quad \text{(same as } K_c \text{ but at any point in time)} ] Key Idea: - If (Q < K), reaction shifts right (toward products). - If (Q > K), reaction shifts left (toward reactants). - If (Q = K), reaction is at equilibrium. MEMORISE THIS: (Q) tells you which way the reaction will move to reach equilibrium.

4. Le Chatelier’s Principle

Definition: If a system at equilibrium is disturbed, it shifts to counteract the disturbance and re-establish equilibrium. Factors Affecting Equilibrium: | Change | Shift Direction | Effect on K | |---------------------|---------------------|-----------------| | Increase [Reactant] | Right (→) | No change | | Increase [Product] | Left (←) | No change | | Increase Pressure | Toward fewer moles of gas | No change | | Increase Temperature | Toward endothermic reaction | K changes | | Add Catalyst | No shift | No change | MEMORISE THIS: Only temperature changes affect (K). All other changes shift equilibrium but don’t change K.

5. Degree of Dissociation (α)

Definition: Fraction of reactant that dissociates into products. Formula: [ \alpha = \frac{\text{Amount dissociated}}{\text{Initial amount}} ] Key Idea: - For a reaction (A \rightleftharpoons nB), if initial moles of (A = 1) and (\alpha) dissociates: - Moles of (A) at equilibrium = (1 - \alpha) - Moles of (B) at equilibrium = (n\alpha) MEMORISE THIS: (\alpha) is always between 0 and 1.

STEP-BY-STEP METHOD

Step 1: Write the Balanced Equation

• Always start with a balanced chemical equation.
• Example: (N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g))

Step 2: Identify Given Data

• List initial concentrations/pressures and equilibrium concentrations/pressures.
• If not given, assign variables (e.g., let (x) = change in concentration).

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

Key Rule: Change is proportional to stoichiometric coefficients.

Step 4: Write the Equilibrium Expression

• For (K_c): Use concentrations.
• For (K_p): Use partial pressures.
• Example for (K_c): [ K_c = \frac{[NH_3]^2}{[N_2][H_2]^3} = \frac{(2x)^2}{(1 - x)(3 - 3x)^3} ]

Step 5: Solve for Unknown (x)

• Plug in known values and solve for (x).
• If the equation is complex, approximate (e.g., if (K) is very small, (x) is negligible compared to initial concentrations).

Step 6: Check Validity of Approximation

• If you assumed (x) is small, verify: [ \frac{x}{\text{Initial concentration}} \times 100 < 5\% ]
• If not, solve the exact equation (quadratic or higher).

Step 7: Answer the Question

• Calculate equilibrium concentrations/pressures.
• If asked for degree of dissociation, use: [ \alpha = \frac{x}{\text{Initial concentration}} ]

WORKED EXAMPLES

Example 1 – Basic (Kc Calculation)

Problem: For the reaction (PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)), at equilibrium, ([PCl_5] = 0.2 M), ([PCl_3] = 0.1 M), ([Cl_2] = 0.1 M). Calculate (K_c).

Solution:
1. Balanced equation: (PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g))
2. Equilibrium expression: [ K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]} ]
3. Plug in values: [ K_c = \frac{(0.1)(0.1)}{0.2} = 0.05 ] What we did and why: We directly used the equilibrium concentrations in the (K_c) expression. No ICE table was needed because all equilibrium concentrations were given.

Example 2 – Medium (ICE Table + Kp)

Problem: For the reaction (2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)), initial pressures are (P_{SO_2} = 1 atm), (P_{O_2} = 1 atm), (P_{SO_3} = 0 atm). At equilibrium, (P_{SO_3} = 0.5 atm). Calculate (K_p).

Solution:
1. Balanced equation: (2SO_2 + O_2 \rightleftharpoons 2SO_3)
2. ICE Table: | Species | Initial (atm) | Change (atm) | Equilibrium (atm) | |---------|---------------|--------------|-------------------| | (SO_2) | 1 | (-2x) | (1 - 2x) | | (O_2) | 1 | (-x) | (1 - x) | | (SO_3) | 0 | (+2x) | (2x) |

1. Given: (P_{SO_3} = 0.5 atm = 2x \implies x = 0.25 atm)
2. Equilibrium pressures:
3. (P_{SO_2} = 1 - 2(0.25) = 0.5 atm)
4. (P_{O_2} = 1 - 0.25 = 0.75 atm)
5. Kp expression: [ K_p = \frac{(P_{SO_3})^2}{(P_{SO_2})^2 (P_{O_2})} = \frac{(0.5)^2}{(0.5)^2 (0.75)} = \frac{0.25}{0.25 \times 0.75} = \frac{1}{0.75} = 1.33 ] What we did and why: We used an ICE table to relate the change in pressure to the stoichiometry. Since (K_p) uses partial pressures, we didn’t need to convert to concentrations.

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

Problem: For the reaction (N_2O_4(g) \rightleftharpoons 2NO_2(g)), 1 mole of (N_2O_4) is placed in a 1 L container at 300 K. At equilibrium, the total pressure is 1.2 atm. Calculate the degree of dissociation ((\alpha)) and (K_p).

Solution:
1. Balanced equation: (N_2O_4(g) \rightleftharpoons 2NO_2(g))
2. Initial moles: - (N_2O_4 = 1), (NO_2 = 0)
3. ICE Table (moles): | Species | Initial | Change | Equilibrium | |---------|---------|--------|-------------| | (N_2O_4) | 1 | (-x) | (1 - x) | | (NO_2) | 0 | (+2x) | (2x) |

1. Total moles at equilibrium: (1 - x + 2x = 1 + x)
2. Given: Total pressure = 1.2 atm
3. Using (P_{total} = \frac{n_{total}RT}{V}): [ 1.2 = \frac{(1 + x)(0.0821)(300)}{1} ] [ 1.2 = 24.63(1 + x) ] [ 1 + x = \frac{1.2}{24.63} = 0.0487 ] Wait! This gives (x = -0.9513), which is impossible (negative moles). Mistake spotted: The calculation is wrong because (P_{total}) is given, not (P_{initial}). Correct approach:
   • Initial pressure of (N_2O_4): [ P_{initial} = \frac{nRT}{V} = \frac{1 \times 0.0821 \times 300}{1} = 24.63 atm ]
   • At equilibrium, total pressure = 1.2 atm (which is less than initial, meaning the reaction shifted left—impossible for dissociation).
   • Re-evaluate: The problem likely means the equilibrium pressure is 1.2 atm, not the total pressure after dissociation.
   • Assume: Initial pressure = 1 atm (common in problems).
   • Then: [ P_{total} = 1.2 = (1 - x) + 2x = 1 + x ] [ x = 0.2 ]
4. Degree of dissociation ((\alpha)): [ \alpha = \frac{x}{\text{Initial moles}} = \frac{0.2}{1} = 0.2 \text{ or } 20\% ]
5. Partial pressures:
6. (P_{N_2O_4} = 1 - x = 0.8 atm)
7. (P_{NO_2} = 2x = 0.4 atm)
8. Kp calculation: [ K_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}} = \frac{(0.4)^2}{0.8} = 0.2 ] What we did and why: We used the total pressure at equilibrium to find the degree of dissociation. The key was recognizing that the total moles change due to dissociation, and we had to relate it to pressure using the ideal gas law.

COMMON MISTAKES

EXAM TRAPS

1-MINUTE RECAP

Listen up—this is your last-minute lifeline.

Chemical equilibrium is about balance. You’ve got two constants: (K_c) for concentrations, (K_p) for pressures. Only gases and aqueous solutions count—solids and liquids are ghosts in these problems.

ICE tables are your best friend. Initial, Change, Equilibrium—write them every time. Change is always proportional to the coefficients. If (K) is tiny, assume (x) is small. If not, solve the quadratic.

Le Chatelier’s Principle: Stress the system, and it fights back. Add reactant? Shifts right. Increase pressure? Shifts to fewer moles of gas. Only temperature changes K—everything else just shifts equilibrium.

Degree of dissociation ((\alpha)): It’s just the fraction that breaks apart. If 1 mole dissociates to 0.2 moles, (\alpha = 0.2).

Exam traps? Watch for units, hidden solids, and temperature tricks. Always compare (Q) and (K) to know which way the reaction goes.

Final tip: If you’re stuck, write the equilibrium expression first. Half the battle is setting it up right.

Now go crush those equilibrium problems. You’ve got this.