Equilibrium – MHT CET Chemistry Study Notes
1. Simple Explanation
Chemical equilibrium is the state in a reversible reaction when the forward and reverse reaction rates become equal.
- Forward reaction (reactants → products)
- Reverse reaction (products → reactants)
When the rates are equal, the concentrations of all species remain constant (though not necessarily equal). The system is dynamic: molecules are still reacting, but the net change is zero.
2. Key Points
| Topic | Details |
|---|---|
| Reversible Reaction | aA + bB ⇌ cC + dD |
| Rate Equality | (r_f = r_r) → (k_f[A]^a[B]^b = k_r[C]^c[D]^d) |
| Equilibrium Constant (K) | (K = \frac{[C]^c[D]^d}{[A]^a[B]^b}) (units depend on reaction) |
| Types of Equilibria | • Chemical (dynamic) equilibrium • Solubility equilibrium (Ksp) • Acid–base equilibrium (Ka, Kb) • Redox equilibrium (Eh) |
| Le Chatelier’s Principle | If a stress (change in concentration, pressure, temperature, catalyst) is applied, the system shifts to counteract it. |
| Effect of Temperature | Exothermic: Increasing T shifts left (decreases K). Endothermic: Increasing T shifts right (increases K). |
| Effect of Pressure | For gas reactions, change in pressure shifts equilibrium toward the side with fewer moles of gas. |
| Effect of Concentration | Adding reactant shifts right; adding product shifts left. |
| Effect of Solvent | Changes solubility equilibrium; can alter Ksp. |
| Catalysts | Speed the approach to equilibrium but do not change K. |
3. Worked Example
Problem
For the dissolution of silver chloride in water:
[ \ce{AgCl(s) ⇌ Ag^+(aq) + Cl^-(aq)} ]
Given the solubility product (K_{sp} = 1.8 \times 10^{-10}).
(a) Calculate the equilibrium concentrations of (\ce{Ag^+}) and (\ce{Cl^-}).
(b) If 0.10 M (\ce{NaCl}) is added, what is the new solubility of (\ce{AgCl})?
(a) Solubility in pure water
Let the molar solubility of AgCl be (s) mol L⁻¹.
At equilibrium:
[ [\ce{Ag^+}] = [\ce{Cl^-}] = s ]
[ K_{sp} = [\ce{Ag^+}][\ce{Cl^-}] = s^2 ]
[ s = \sqrt{K_{sp}} = \sqrt{1.8 \times 10^{-10}} = 1.34 \times 10^{-5}\ \text{M} ]
Thus,
[
[\ce{Ag^+}] = [\ce{Cl^-}] = 1.34 \times 10^{-5}\ \text{M}
]
(b) Common‑ion effect (adding NaCl)
Adding 0.10 M (\ce{NaCl}) supplies 0.10 M (\ce{Cl^-}).
Let the new solubility be (s').
[ K_{sp} = [\ce{Ag^+}][\ce{Cl^-}] = s' \times (0.10 + s') ]
Since (s') is much smaller than 0.10, we approximate:
[ K_{sp} \approx s' \times 0.10 ]
[ s' = \frac{1.8 \times 10^{-10}}{0.10} = 1.8 \times 10^{-9}\ \text{M} ]
So the solubility drops from (1.34\times10^{-5}) M to (1.8\times10^{-9}) M – a huge decrease due to the common‑ion effect.
4. Quick Revision Summary
- Equilibrium: Forward and reverse rates are equal → constant concentrations.
- Equilibrium constant (K): Ratio of product concentrations to reactant concentrations (raised to stoichiometric powers).
- Le Chatelier’s Principle: Stress → shift that opposes the stress.
- Temperature: Exothermic → lower K with ↑T; Endothermic → higher K with ↑T.
- Pressure: Shift toward fewer moles of gas.
- Concentration: Adding a reactant shifts right; adding a product shifts left.
- Common‑ion effect: Adding an ion from a soluble salt reduces solubility of the sparingly soluble compound.
- Catalysts: Speed equilibrium attainment but do not alter K.
Use these points to solve equilibrium problems, compute (K) values, and predict the direction of shift under various conditions. Good luck with MHT CET!