Equilibrium in Chemistry
Study notes for the ITI Admission Test – Chemistry – Equilibrium
1. Simple Explanation
An equilibrium is a steady state in a chemical reaction where the forward and reverse reaction rates are equal.
- Forward reaction (reactants → products)
- Reverse reaction (products → reactants)
At equilibrium:
- Concentrations of all species stay constant (though the reaction still occurs).
- The equilibrium constant (K) defines the ratio of product concentrations to reactant concentrations, each raised to the power of their stoichiometric coefficients.
2. Key Points
| Concept | What It Means | Why It Matters |
|---|---|---|
| Chemical Equilibrium | The state where (k_f = k_r) (forward = reverse rate). | Determines how much of each species is present in a reaction vessel. |
| Le Chatelier’s Principle | A system at equilibrium will oppose any change imposed on it. | Predicts direction of shift when conditions change (concentration, pressure, temperature). |
| Equilibrium Constant (K) | (K = \frac{[C]^c[D]^d}{[A]^a[B]^b}) for (aA + bB ⇌ cC + dD). | Quantifies the position of equilibrium; (K>1) → products favored, (K<1) → reactants favored. |
| Reaction Quotient (Q) | Same expression as K but with current concentrations. | If (Q<K) → forward reaction; if (Q>K) → reverse reaction. |
| Factors Affecting Equilibrium | • Concentration of reactants/products • Pressure (for gaseous reactions) • Temperature (endothermic vs exothermic) • Presence of catalysts (no effect on K) |
They change the value of K or the reaction direction. |
| Solubility Equilibrium | (K_{sp} = [M^+][X^-]) for (MX ⇌ M^+ + X^-). | Governs how much solid dissolves in a solution. |
| Common‑Ion Effect | Adding an ion that is part of the equilibrium shifts the balance toward the solid. | Useful for precipitation and purification. |
| Industrial Examples | • Haber process (NH₃ synthesis) • Ostwald process (NO production) • Cracking of hydrocarbons |
Equilibrium considerations dictate reactor design, pressure, temperature, and catalyst choice. |
3. Worked Example
Problem
Calculate the equilibrium concentrations of (NH_3) and (HCl) when 0.100 mol of each is mixed in a 1.00 L vessel. The equilibrium constant for the reaction
[ NH_3(g) + HCl(g) ⇌ NH_4Cl(s) ]
is (K = 1.0 \times 10^{10}) at 298 K.
Assume the solid is insoluble and does not affect the gas‑phase concentrations.
Solution
-
Write the equilibrium expression
Since (NH_4Cl) is a solid, it does not appear in the K expression.
[ K = \frac{1}{[NH_3][HCl]} = 1.0 \times 10^{10} ]
(The equilibrium constant for this reaction is often given as (K = 1/[NH_3][HCl]).) -
Set up the ICE table
Species Initial (M) Change (M) Equilibrium (M) (NH_3) 0.100 (-x) (0.100 - x) (HCl) 0.100 (-x) (0.100 - x) (NH_4Cl(s)) 0 (+x) (x) -
Express K in terms of (x)
[ K = \frac{1}{(0.100 - x)(0.100 - x)} = 1.0 \times 10^{10} ]
-
Solve for (x)
[ (0.100 - x)^2 = \frac{1}{1.0 \times 10^{10}} = 1.0 \times 10^{-10} ] [ 0.100 - x = \sqrt{1.0 \times 10^{-10}} = 1.0 \times 10^{-5} ] [ x = 0.100 - 1.0 \times 10^{-5} \approx 0.09999 ]
-
Find equilibrium concentrations
[ [NH_3]{eq} = 0.100 - x \approx 1.0 \times 10^{-5},\text{M} ] [ [HCl]{eq} = 0.100 - x \approx 1.0 \times 10^{-5},\text{M} ]
Answer
At equilibrium, both (NH_3) and (HCl) are essentially consumed to form the solid salt:
[ [NH_3]{eq} \approx [HCl]{eq} \approx 1.0 \times 10^{-5}\ \text{M} ]
4. Quick Revision Summary
- Equilibrium: (k_f = k_r); concentrations constant.
- Le Chatelier’s Principle
- Increase reactants → forward shift.
- Increase products → reverse shift.