📝 Chapter Notes & Revision

Equilibrium

🏫 MP BoardClass 11Chemistry

📐 Formula & Cheat Sheet (English)

Class 11 Chemistry: Chapter 6 - Equilibrium (साम्यावस्था)

Quick Revision Notes & Formula Sheet


Part 1: Chemical Equilibrium (रासायनिक साम्यावस्था)

1. Key Definitions & Concepts

  • Reversible Reaction (उत्क्रमणीय अभिक्रिया): A reaction that proceeds in both forward and backward directions simultaneously.
  • Chemical Equilibrium (रासायनिक साम्यावस्था): A dynamic state in a reversible reaction where the rate of forward reaction ($R_f$) equals the rate of backward reaction ($R_b$). At this state, concentrations of reactants and products remain constant over time.
  • Law of Mass Action (द्रव्यमान अनुपाती क्रिया का नियम): Given by Guldberg and Waage. At a constant temperature, the rate of a chemical reaction is directly proportional to the product of the active masses (molar concentrations) of the reacting substances.

2. Equilibrium Constant Formulas

For a general reversible reaction:
aA + bB ⇌ cC + dD

Equilibrium Constant in terms of Concentration ($K_c$):

$$K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b}$$ (Note: Active mass of pure solids and pure liquids is taken as 1)

Equilibrium Constant in terms of Partial Pressure ($K_p$):

$$K_p = \frac{(P_C)^c \cdot (P_D)^d}{(P_A)^a \cdot (P_B)^b}$$

Relationship between $K_p$ and $K_c$:

$$K_p = K_c (RT)^{\Delta n_g}$$

  • $R$ = Gas constant ($0.0821 \text{ L atm K}^{-1} \text{mol}^{-1}$ or $8.314 \text{ J K}^{-1} \text{mol}^{-1}$)
  • $T$ = Temperature in Kelvin
  • $\Delta n_g$ = (Moles of gaseous products) - (Moles of gaseous reactants) $= (c + d) - (a + b)$

Cases of $\Delta n_g$:

  1. If $\Delta n_g = 0 \implies K_p = K_c$ (e.g., $H_2 + I_2 \rightleftharpoons 2HI$)
  2. If $\Delta n_g > 0 \implies K_p > K_c$ (e.g., $PCl_5 \rightleftharpoons PCl_3 + Cl_2$)
  3. If $\Delta n_g < 0 \implies K_p < K_c$ (e.g., $N_2 + 3H_2 \rightleftharpoons 2NH_3$)

3. Characteristics of Equilibrium Constant ($K$)

  • Reversing a Reaction: If $A \rightleftharpoons B$ has constant $K$, then $B \rightleftharpoons A$ has $K' = \frac{1}{K}$.
  • Multiplying Coefficients: If reaction is multiplied by a factor $n$, new constant $K'' = K^n$.
  • Adding Reactions: If Reaction 3 = Reaction 1 + Reaction 2, then $K_3 = K_1 \times K_2$.

4. Reaction Quotient ($Q$) & Predicting Reaction Direction

For aA + bB ⇌ cC + dD at any state (not necessarily equilibrium): $$Q_c = \frac{[C]^c [D]^d}{[A]^a [B]^b}$$

ConditionDirection of Reaction
$Q_c < K_c$Reaction moves in Forward direction (उत्तरोत्तर / आगे)
$Q_c = K_c$Reaction is at Equilibrium (साम्यावस्था)
$Q_c > K_c$Reaction moves in Backward direction (पश्चात / पीछे)

5. Thermodynamics of Equilibrium

  • Standard Free Energy Change ($\Delta G^\circ$): $$\Delta G^\circ = -2.303 R T \log_{10} K$$
  • Free Energy Change ($\Delta G$): $$\Delta G = \Delta G^\circ + 2.303 R T \log_{10} Q$$ (At equilibrium, $\Delta G = 0$)

6. Le Chatelier's Principle (ला-शातेलिए का नियम)

"If a system at equilibrium is subjected to a change in concentration, temperature, or pressure, the system shifts in a direction that tends to counteract the effect of the change."

Factor / Factor ShiftChange AppliedEffect on Equilibrium Shift
ConcentrationIncrease Reactant conc.Shifts Forward
Increase Product conc.Shifts Backward
PressurePressure IncreaseShifts towards fewer gaseous moles ($\Delta n_g$)
Pressure DecreaseShifts towards more gaseous moles
VolumeVolume IncreaseShifts towards more gaseous moles
TemperatureTemp IncreaseFavors Endothermic reaction ($\Delta H > 0$)
Temp DecreaseFavors Exothermic reaction ($\Delta H < 0$)
CatalystAddition of CatalystNo shift (Only speeds up rates equally)
Inert GasAdded at Constant VolumeNo effect
Added at Constant PressureShifts towards more gaseous moles

Part 2: Ionic Equilibrium (आयनिक साम्यावस्था)

1. Acid-Base Theories

  1. Arrhenius Theory:
    • Acid: Gives $H^+$ ions in water.
    • Base: Gives $OH^-$ ions in water.
  2. Brönsted-Lowry Theory:
    • Acid: Proton ($H^+$) donor.
    • Base: Proton ($H^+$) acceptor.
    • Conjugate Acid-Base Pair: Differ by a single proton ($H^+$). $$\text{Acid} \rightleftharpoons \text{Conjugate Base} + H^+$$
  3. Lewis Theory:
    • Acid: Electron pair acceptor (e.g., $BF_3, AlCl_3, H^+$).
    • Base: Electron pair donor (e.g., $NH_3, H_2O, F^-$).

2. Ostwald's Dilution Law (for Weak Electrolytes)

For a weak acid $HA \rightleftharpoons H^+ + A^-$: $$K_a = \frac{C \alpha^2}{1 - \alpha}$$

If degree of dissociation $\alpha \ll 1$ (i.e., $\alpha < 5%$): $$\alpha = \sqrt{\frac{K_a}{C}} = \sqrt{K_a \cdot V}$$ $$[H^+] = C \cdot \alpha = \sqrt{K_a \cdot C}$$

  • $\alpha$ = Degree of dissociation (वियोजन की मात्रा)
  • $C$ = Concentration in $\text{mol/L}$
  • $V$ = Dilution (Volume containing 1 mole of electrolyte)

3. Ionic Product of Water ($K_w$) & pH Concept

  • Ionic Product of Water ($K_w$): $$K_w = [H^+][OH^-] = 1.0 \times 10^{-14} \quad \text{at } 298 \text{ K } (25^\circ\text{C})$$
  • pH Scale Formulation: $$pH = -\log_{10}[H^+] \implies [H^+] = 10^{-pH}$$ $$pOH = -\log_{10}[OH^-] \implies [OH^-] = 10^{-pOH}$$ $$pK_w = pH + pOH = 14 \quad (\text{at } 25^\circ\text{C})$$

Nature of Solutions at $25^\circ\text{C}$:

  • Neutral: $[H^+] = 10^{-7}\text{ M} \implies pH = 7$
  • Acidic: $[H^+] > 10^{-7}\text{ M} \implies pH < 7$
  • Basic: $[H^+] < 10^{-7}\text{ M} \implies pH > 7$

4. Common Ion Effect (सम-आयन प्रभाव)

The suppression of dissociation of a weak electrolyte by the addition of a strong electrolyte containing a common ion.
Example: Dissociation of weak acid $CH_3COOH$ decreases upon adding $CH_3COONa$ (common ion: $CH_3COO^-$).


5. Buffer Solutions (बफर विलयन)

Solutions that resist changes in pH upon addition of small amounts of acid or base.

Types & Henderson-Hasselbalch Equations:

  1. Acidic Buffer (अम्लीय बफर): Weak Acid + Salt of Weak Acid with Strong Base (e.g., $CH_3COOH + CH_3COONa$) $$pH = pK_a + \log_{10}\left( \frac{[\text{Salt}]}{[\text{Acid}]} \right)$$ where $pK_a = -\log_{10} K_a$

  2. Basic Buffer (क्षारीय बफर): Weak Base + Salt of Weak Base with Strong Acid (e.g., $NH_4OH + NH_4Cl$) $$pOH = pK_b + \log_{10}\left( \frac{[\text{Salt}]}{[\text{Base}]} \right)$$ $$pH = 14 - pOH$$ where $pK_b = -\log_{10} K_b$


6. Solubility Product ($K_{sp}$) (विलेयता गुणनफल)

For a sparingly soluble salt $A_x B_y \rightleftharpoons x A^{y+} + y B^{x-}$: $$K_{sp} = [A^{y+}]^x [B^{x-}]^y$$

If $S$ is the solubility in $\text{mol/L}$: $$K_{sp} = (x^x \cdot y^y) \cdot S^{(x+y)}$$

Examples:

  • 1:1 Salt (e.g., $AgCl$): $K_{sp} = S^2 \implies S = \sqrt{K_{sp}}$
  • 1:2 or 2:1 Salt (e.g., $CaCl_2, Ag_2CrO_4$): $K_{sp} = 4S^3 \implies S = \left(\frac{K_{sp}}{4}\right)^{1/3}$
  • 1:3 Salt (e.g., $AlCl_3$): $K_{sp} = 27S^4$

Criteria for Precipitation (अवक्षेपण की शर्त):

  • $Q_{sp} < K_{sp}$: Unsaturated solution (No precipitation)
  • $Q_{sp} = K_{sp}$: Saturated solution (Equilibrium state)
  • $Q_{sp} > K_{sp}$: Supersaturated solution (Precipitation occurs / अवक्षेप बनेगा)

Key Exam Tips for MP Board

  1. Units of $K_c$ and $K_p$:
    • Unit of $K_c = (\text{mol L}^{-1})^{\Delta n_g}$
    • Unit of $K_p = (\text{atm})^{\Delta n_g}$ or $(\text{bar})^{\Delta n_g}$
  2. Conjugate Pair Questions: Remember that a Strong Acid has a Weak Conjugate Base, and a Weak Acid has a Strong Conjugate Base.
  3. Le Chatelier Applications: Be ready for direct conceptual questions on Haber's Process ($N_2 + 3H_2 \rightleftharpoons 2NH_3, \Delta H < 0$) and Contact Process ($2SO_2 + O_2 \rightleftharpoons 2SO_3, \Delta H < 0$).