Equilibrium – Study Notes (D.Pharm Admission – Chemistry)
1. Simple Explanation
- Equilibrium is a state in which the forward and reverse rates of a reaction are equal, so the concentrations of reactants and products no longer change with time.
- It is dynamic: molecules are still reacting, but the macroscopic composition stays constant.
- In an equilibrium mixture the system is minimally disturbed by external changes.
2. Key Points
| Concept | What it Means | Key Formula / Symbol |
|---|---|---|
| Dynamic Equilibrium | Reaction proceeds in both directions at the same rate. | ( r_{\text{forward}} = r_{\text{reverse}} ) |
| Equilibrium Concentration | Concentrations of species that satisfy the equilibrium constant. | ( [A]{\text{eq}}, [B]{\text{eq}}, \dots ) |
| Equilibrium Constant (K) | Ratio of product concentrations to reactant concentrations (raised to stoichiometric powers). | ( K = \frac{[C]^c[D]^d}{[A]^a[B]^b} ) |
| Kc vs. Kp | (K_c) uses molar concentrations (mol L⁻¹). (K_p) uses partial pressures (atm). | ( K_p = K_c (RT)^{\Delta n} ) |
| Le Chatelier’s Principle | System shifts to oppose the change applied. | Increase concentration → shift to consume it; increase temperature → shift to endothermic side, etc. |
| Reaction Quotient (Q) | Same expression as K but with actual concentrations/pressures. | ( Q = \frac{[C]^c[D]^d}{[A]^a[B]^b} ) |
| Comparison of Q and K | • (Q < K): forward reaction predominates. • (Q > K): reverse reaction predominates. • (Q = K): equilibrium. |
|
| Solubility Product (Ksp) | Equilibrium constant for dissolution of sparingly soluble salts. | ( K_{sp} = [M^+][X^-] ) |
| Acid–Base Equilibria | Ka (acid dissociation), Kb (base dissociation). | ( K_a = \frac{[H^+][A^-]}{[HA]} ) |
3. Worked Example
Reaction
[
\mathrm{N_2(g) + 3H_2(g) ;\rightleftharpoons; 2NH_3(g)} \quad (1)
]
Initial amounts (in a 1.00 L flask)
| Species | moles |
|---|---|
| N₂ | 0.200 |
| H₂ | 0.600 |
| NH₃ | 0.000 |
Goal: Find (K_c) and the equilibrium concentrations.
Step 1: ICE Table
| Species | Initial (mol) | Change (mol) | Equilibrium (mol) |
|---|---|---|---|
| N₂ | 0.200 | (-x) | (0.200 - x) |
| H₂ | 0.600 | (-3x) | (0.600 - 3x) |
| NH₃ | 0.000 | (+2x) | (2x) |
Step 2: Write (K_c) Expression
[ K_c = \frac{[NH_3]^2}{[N_2][H_2]^3} ]
Convert moles → molarity (volume = 1 L):
[ K_c = \frac{(2x)^2}{(0.200 - x)(0.600 - 3x)^3} ]
Step 3: Determine (x) from Experimental Data
Assume after reaching equilibrium the measured concentration of NH₃ is (0.120) M.
[ 2x = 0.120 ;\Rightarrow; x = 0.060 ]
Now compute equilibrium concentrations:
[ \begin{aligned} [N_2]{\text{eq}} &= 0.200 - 0.060 = 0.140;\text{M} \ [H_2]{\text{eq}} &= 0.600 - 3(0.060) = 0.420;\text{M} \ [NH_3]_{\text{eq}} &= 0.120;\text{M} \end{aligned} ]
Step 4: Calculate (K_c)
[ \begin{aligned} K_c &= \frac{(0.120)^2}{(0.140)(0.420)^3} \ &= \frac{0.0144}{0.140 \times 0.074088} \ &= \frac{0.0144}{0.010372} \ &\approx 1.39 \times 10^2 \end{aligned} ]
Result:
(K_c \approx 1.4 \times 10^2)
4. Quick Revision Summary
- Equilibrium: forward = reverse rate → constant concentrations.
- Equilibrium constant: (K_c) (concentration) or (K_p) (pressure).
(K_p = K_c (RT)^{\Delta n}). - Le Chatelier’s principle: shift opposite to