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1 Some Basic Concepts of Chemistry 2 Structure of Atom 3 Classification of Elements and Periodicity in Properties 4 Chemical Bonding and Molecular Structure 5 Thermodynamics 6 Equilibrium 7 Redox Reactions 8 Organic Chemistry – Some Basic Principles and Techniques 9 Hydrocarbons

6 Equilibrium

6.1 Equilibrium in Physical Processes 6.2 Equilibrium in Chemical Processes - Dynamic Equilibrium 6.3 Law of Chemical Equilibrium and Equilibrium Constant 6.4 Homogeneous Equilibria 6.5 Heterogeneous Equilibria 6.6 Applications of Equilibrium Constants 6.7 Relationship between Equilibrium Constant K, Reaction Quotient Q and Gibbs Energy G 6.8 Factors Affecting Equilibria 6.9 Ionic Equilibrium in Solutions 6.10 Acids, Bases and Salts 6.11 Ionization of Acids and Bases 6.12 Buffer Solutions 6.13 Solubility Equilibria of Sparingly Soluble Salts

Applications of Equilibrium Constants

NCERT Reference: Chapter 6 – Equilibrium – Pages 174–176

Quick Notes

  • K (equilibrium constant) indicates how far a reaction proceeds.
  • Extent of Reaction:
    • Very large K (K ≫ 1): Reaction proceeds nearly to completion.
    • Very small K (K ≪ 1): Hardly any reaction occurs.
  • Direction of Reaction:
    • We can predict the direction a reaction will proceed in using the reaction quotient Q:
      • Q < K → Forward reaction favored
      • Q > K → Reverse reaction favored
      • Q = K → At equilibrium
  • Equilibrium Concentrations: ICE tables can be used to calculate unknown equilibrium concentrations.

Full Notes

Applications of Equilibrium Constants

Before looking at how to use equilibrium constants in practice, let’s first recap a few important characteristics of them:

Key Features of Equilibrium Constants

  1. Applies at Equilibrium Only
    • The expression for an equilibrium constant is valid only when the system has reached equilibrium, meaning the concentrations of reactants and products have stabilized.
  2. Independent of Initial Concentrations
    • The value of the equilibrium constant depends only on the reaction and temperature, not on how much of each substance you started with.
  3. Temperature Dependent
    • For any specific reaction, the value of the equilibrium constant changes if the temperature changes.
  4. Reversing a Reaction Inverts K
    • If you reverse the direction of a reaction, the new equilibrium constant is simply the reciprocal (1/K) of the original.
  5. Multiplying the Equation Affects K
    • If you multiply the entire balanced reaction by a number, raise K to that power.
      (e.g., if you double the reaction, new K = K2)

We can use equilibrium constants to:

6.6.1 Predicting the Extent of a Reaction

The magnitude of K tells us how "complete" a reaction is – but it doesn't tell us how fast it reaches equilibrium.

This helps us judge whether a reaction will go close to completion (high concentration of products) or hardly proceed (high concentration of reactants).

NCERT 11 Chemistry chart categorising reaction extent by K values showing K much greater than 10^3 product-favoured, K near 1 comparable amounts, and K much less than 10^-3 reactant-favoured.

6.6.2 Predicting the Direction of the Reaction

What is the Reaction Quotient (Q)?

The reaction quotient (Q) is a snapshot of a reaction’s progress. It is calculated by using concentration values at a specific point in time, which might not be equilibrium values (unlike for Kc and Kp).

General formula for a reaction:

NCERT 11 Chemistry general reaction aA plus bB in equilibrium with cC plus dD used to define Q. NCERT 11 Chemistry expression for reaction quotient Q as products over reactants raised to stoichiometric powers.

Comparing Q and K

The reaction quotient (Q) can be compared to the equilibrium constant (K) to predict the direction in which a reversible reaction will proceed.

It’s important to compare like with like – for example, Qc must be compared with Kc, and Qp with Kp. Mixing types (e.g., comparing Qc with Kp) is not valid.

Comparison Interpretation Shift Direction
Q < K Too many reactants, not enough products Shift right (products)
Q > K Too many products, not enough reactants Shift left (reactants)
Q = K System is already at equilibrium No shift
Worked Example

For the reaction: N2(g) + 3H2(g) ⇌ 2NH3(g) the equilibrium constant K = 0.50 at a given temperature. At one point during the reaction, the concentrations are: [N2] = 0.20 mol dm−3, [H2] = 0.60 mol dm−3, [NH3] = 0.10 mol dm−3.

Using this information: calculate the reaction quotient Q and determine the direction in which the reaction will shift to reach equilibrium.

  1. Write the expression for Q
    Q = \[NH3\]2 / \(\[N2\]\[H2\]3\)
  2. Substitute the concentrations
    Q = (0.10)2 / \(\,0.20 × 0.603\,\)
    Q = 0.01 / \(0.20 × 0.216\) = 0.01 / 0.0432 ≈ 0.231
  3. Compare Q with K
    Q ≈ 0.231, K = 0.50Q < K
  4. Decide the shift
    Since Q < K, the reaction will proceed in the forward direction to form more NH3 until equilibrium is reached.

Answer: Q ≈ 0.231. Because Q < K, the system shifts right (forward), favouring formation of more NH3.

6.6.3 Calculating Equilibrium Concentrations

You need to be able to determine equilibrium concentrations, to do this we can use ICE tables.

  1. Write the balanced equation.
  2. Fill in Initial concentrations.
  3. Use ‘x’ to denote Change.
  4. Express Equilibrium concentrations.
  5. Substitute into K expression and solve.
Worked Example

For the reaction: H2(g) + I2(g) ⇌ 2HI(g) At 298 K, the equilibrium constant is K = 50.0. At equilibrium: [HI] = 0.80 mol dm−3 [I2] = 0.10 mol dm−3 Calculate the equilibrium concentration of H2.

  1. Step 1: Use an ICE table
    Let the change in [HI] be +2x, so [HI] = 0.80 → x = 0.40.
    The change in [H2] and [I2] is −x = −0.40.
    Species Initial Change Equilibrium
    H2 a −0.40 a − 0.40
    I2 −0.40 0.10
    HI 0 +0.80 0.80
  2. Step 2: Apply the equilibrium expression
    K = [HI]2 / ([H2][I2])
    50.0 = (0.80)2 / [(a − 0.40)(0.10)]
    50.0 = 0.64 / [0.10(a − 0.40)]
    0.10(a − 0.40) = 0.64 / 50 = 0.0128
    a − 0.40 = 0.128
    a = 0.528 mol dm−3
  3. Step 3: Final answer
    [H2] at equilibrium = a − 0.40 = 0.128 mol dm−3

Summary