JEE Main & Advanced

Equilibrium

Equilibrium for JEE Main & Advanced

1
Module 1

Chemical Equilibrium

Law of Mass Action, Kc, Kp, Reaction QuotientTopic 1

Reversible Reaction: Reaction that can proceed both forward and backward. Indicated by ⇌.

Chemical Equilibrium: State where forward rate = backward rate; concentrations remain constant (dynamic, not static).

Law of Mass Action (Guldberg-Waage): Rate of reaction $\propto$ product of active masses (concentrations) of reactants.

For: $aA + bB \rightleftharpoons cC + dD$ $$K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}$$

$K_c$ (concentration): Used when concentrations expressed in mol/L. Units depend on $\Delta n = (c+d)-(a+b)$.

$K_p$ (partial pressure): Used for gaseous reactions. Relation: $$K_p = K_c(RT)^{\Delta n_g}$$ $\Delta n_g$ = (gaseous product moles) - (gaseous reactant moles).

  • If $\Delta n = 0$: $K_p = K_c$
  • If $\Delta n > 0$: $K_p > K_c$
  • If $\Delta n < 0$: $K_p < K_c$

$K_x$ (mole fraction): $K_p = K_x(P_{\text{total}})^{\Delta n}$.

Characteristics of Equilibrium Constant:

  1. Constant at given temperature
  2. Independent of initial concentrations
  3. Independent of presence of catalyst
  4. Changes with temperature
  5. Reverse reaction: $K_{\text{reverse}} = 1/K_{\text{forward}}$
  6. Adding reactions: $K_{\text{net}} = K_1 \times K_2$

Magnitude of $K$:

  • $K > 10^3$: forward strongly favored (products dominate)
  • $K < 10^{-3}$: backward favored (reactants dominate)
  • $10^{-3} < K < 10^3$: appreciable concentrations of both

Reaction Quotient ($Q$): Same expression as $K$, but with current (not equilibrium) concentrations.

  • $Q < K$: forward shifts (toward products)
  • $Q > K$: backward shifts (toward reactants)
  • $Q = K$: at equilibrium

Relation with Free Energy: $$\Delta G^\ominus = -RT\ln K = -2.303RT\log K$$ $$\Delta G = \Delta G^\ominus + RT\ln Q$$

Degree of dissociation ($\alpha$): Fraction of molecules dissociating.

For: $PCl_5 \rightleftharpoons PCl_3 + Cl_2$, starting with $n$ moles in volume $V$: $$K_c = \frac{n\alpha^2}{V(1-\alpha)} \quad \text{(if } \alpha \ll 1 \text{: } K_c \approx \alpha^2 \cdot n/V)$$

Worked Examples
1

For $N_2 + 3H_2 \rightleftharpoons 2NH_3$ at $500$ K, $K_p = 1.44 \times 10^{-5}$. Find $K_c$. ($R = 0.0821$ L·atm/mol·K)

Show solution

$\Delta n_g = 2 - 4 = -2$. $$K_p = K_c(RT)^{\Delta n} \implies K_c = K_p(RT)^{-\Delta n} = K_p (RT)^2$$ $$K_c = 1.44 \times 10^{-5} \times (0.0821 \times 500)^2 = 1.44 \times 10^{-5} \times 1685 \approx 0.0243$$

Final Answer: $K_c \approx 0.0243$.

2

$1$ mol PCl₅ in $1$ L container; $\alpha = 0.2$. Find $K_c$.

Show solution

$PCl_5 \rightleftharpoons PCl_3 + Cl_2$. Initial: $1, 0, 0$. At eq: $1-\alpha = 0.8$, $\alpha = 0.2$, $\alpha = 0.2$. $$K_c = \frac{(0.2)(0.2)}{0.8} = \frac{0.04}{0.8} = 0.05 \text{ M}$$

Final Answer: $K_c = 0.05$.

✎ Self-Check — 5 questions0 / 5
Q1.

At equilibrium:

Q2.

$K_p = K_c$ when:

Q3.

If $Q < K$, reaction proceeds:

Q4.

The unit of $K_c$ for $N_2 + 3H_2 \to 2NH_3$:

Q5.

Catalyst changes:

Le Chatelier's Principle and Factors Affecting EquilibriumTopic 2

Le Chatelier's Principle (1884): If a system in equilibrium is disturbed, it shifts in a direction that minimizes the disturbance.

Factors:

ChangeDirection of Shift
Increase [reactant]Forward
Increase [product]Backward
Increase volume (gases)Toward more moles of gas
Decrease volume (gases)Toward fewer moles of gas
Increase $P$ (gases)Toward fewer moles
Decrease $P$ (gases)Toward more moles
Increase $T$ (exothermic forward)Backward
Increase $T$ (endothermic forward)Forward

Effect of inert gas:

  • At constant volume: No effect on equilibrium ($K$ same)
  • At constant pressure: Effectively increases volume → shifts toward more gas moles

Effect of temperature:

  • Exothermic: $K$ decreases with increasing $T$
  • Endothermic: $K$ increases with increasing $T$

Van't Hoff Equation: $$\ln\left(\frac{K_2}{K_1}\right) = \frac{\Delta H^\ominus}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$$

Industrial Applications:

ProcessReactionConditions
Haber's process (NH₃)$N_2 + 3H_2 \rightleftharpoons 2NH_3 + \text{heat}$High $P$, moderate $T$ ($450°$C), Fe catalyst
Contact process (SO₃)$2SO_2 + O_2 \rightleftharpoons 2SO_3 + \text{heat}$Moderate $T$ ($720$ K), V₂O₅ catalyst
Ostwald (HNO₃)$4NH_3 + 5O_2 \to 4NO + 6H_2O$Pt catalyst

Catalyst:

  • Speeds up forward and reverse reactions equally
  • Does NOT shift equilibrium
  • Only reduces time to reach equilibrium

Heterogeneous Equilibrium: Pure solids and pure liquids don't appear in $K$ expression (their activities = 1). e.g., $CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)$: $K_p = P_{CO_2}$.

Worked Examples
1

For $N_2 + 3H_2 \rightleftharpoons 2NH_3 + Q$, predict effect of: (a) increasing pressure, (b) increasing temperature, (c) adding more N₂.

Show solution

(a) Increase $P$: shifts toward fewer gas moles. Left side: 4 mol, right: 2 mol. → forward (more NH₃). (b) Increase $T$: exothermic forward, so shifts backward (less NH₃). Lower $K$. (c) Add N₂: increases reactant; shifts forward (more NH₃).

Final Answer: (a) Forward, (b) Backward, (c) Forward.

2

$K_1 = 4 \times 10^{-2}$ at $300$ K, $K_2$ at $400$ K. $\Delta H = +20$ kJ/mol. Find $K_2$ using van't Hoff.

Show solution

$$\ln(K_2/K_1) = (20000/8.314)(1/300 - 1/400) = 2406 \times 0.000833 = 2.0$$ $K_2 = K_1 \cdot e^{2.0} = 4 \times 10^{-2} \times 7.39 \approx 0.295$.

Final Answer: $K_2 \approx 0.30$.

✎ Self-Check — 5 questions0 / 5
Q1.

Le Chatelier's principle states:

Q2.

For exothermic reaction, increasing $T$:

Q3.

Effect of catalyst on equilibrium constant:

Q4.

In Haber process, high pressure favors:

Q5.

Effect of inert gas at constant pressure:

2
Module 2

Ionic Equilibrium

Acid-Base Concepts, pH, BuffersTopic 1

Theories of Acids and Bases:

TheoryAcidBase
ArrheniusReleases H⁺ in waterReleases OH⁻ in water
Bronsted-LowryProton donorProton acceptor
LewisElectron pair acceptorElectron pair donor

Conjugate Pairs: Acid → its conjugate base after losing H⁺. e.g., HCl/Cl⁻, H₂O/OH⁻, NH₃/NH₄⁺.

Strong vs Weak:

  • Strong acid (HCl, HBr, HI, HNO₃, H₂SO₄, HClO₄): fully dissociates
  • Weak acid (CH₃COOH, HCN, HF): partial dissociation; Ka small
  • Strong base (NaOH, KOH, LiOH, group I/II hydroxides): fully dissociates
  • Weak base (NH₃, RNH₂): partial; Kb small

Auto-ionization of Water: $H_2O \rightleftharpoons H^+ + OH^-$, $K_w = [H^+][OH^-] = 10^{-14}$ at $25°$C.

pH Scale: $$pH = -\log[H^+], \quad pOH = -\log[OH^-], \quad pH + pOH = 14 \text{ at } 25°C$$

  • Acidic: pH < 7
  • Neutral: pH = 7
  • Basic: pH > 7

Ostwald's Dilution Law (for weak electrolytes): $K_a = c\alpha^2/(1-\alpha)$. If $\alpha \ll 1$: $K_a \approx c\alpha^2$, so $\alpha = \sqrt{K_a/c}$.

For weak acid: $[H^+] = \sqrt{K_aC}$, $pH = -(1/2)(\log K_a + \log C) = (1/2)(pK_a - \log C)$.

For weak base: $pOH = (1/2)(pK_b - \log C)$.

Buffer Solution: Resists change in pH on addition of small amounts of acid/base.

TypeCompositionExample
Acidic bufferWeak acid + salt of its conjugate baseCH₃COOH + CH₃COONa
Basic bufferWeak base + salt of its conjugate acidNH₄OH + NH₄Cl

Henderson-Hasselbalch Equation: $$pH = pK_a + \log\frac{[\text{salt}]}{[\text{acid}]} \text{ (acidic buffer)}$$ $$pOH = pK_b + \log\frac{[\text{salt}]}{[\text{base}]} \text{ (basic buffer)}$$

Buffer Capacity: Maximum when $[\text{salt}] = [\text{acid}]$ (or base), giving pH = pKa.

Worked Examples
1

$0.01$ M acetic acid has $K_a = 1.8 \times 10^{-5}$. Find pH.

Show solution

$[H^+] = \sqrt{K_a C} = \sqrt{1.8 \times 10^{-5} \times 0.01} = \sqrt{1.8 \times 10^{-7}} \approx 4.24 \times 10^{-4}$ M. $pH = -\log(4.24 \times 10^{-4}) \approx 3.37$.

Final Answer: $pH \approx 3.37$.

2

Buffer: $0.1$ M CH₃COOH + $0.2$ M CH₃COONa. $pK_a = 4.74$. Find pH.

Show solution

$$pH = pK_a + \log\frac{[\text{salt}]}{[\text{acid}]} = 4.74 + \log(0.2/0.1) = 4.74 + 0.30 = 5.04$$

Final Answer: $pH = 5.04$.

✎ Self-Check — 5 questions0 / 5
Q1.

pH of $0.01$ M HCl:

Q2.

Conjugate base of NH₃:

Q3.

Buffer of weak acid and its salt is called:

Q4.

pH range of acidic buffer:

Q5.

A strong acid completely dissociates because:

Solubility Product, Common Ion Effect, HydrolysisTopic 2

Solubility Product ($K_{sp}$): Ionic product of saturated solution of sparingly soluble salt.

For: $A_xB_y(s) \rightleftharpoons xA^{p+}(aq) + yB^{q-}(aq)$ $$K_{sp} = [A^{p+}]^x[B^{q-}]^y$$

Relation with solubility $s$ (mol/L):

Salt type$K_{sp}$Example
AB (e.g., AgCl)$K_{sp} = s^2$$s = \sqrt{K_{sp}}$
AB₂ (e.g., CaF₂)$K_{sp} = 4s^3$$s = (K_{sp}/4)^{1/3}$
A₂B (e.g., Ag₂CrO₄)$K_{sp} = 4s^3$$s = (K_{sp}/4)^{1/3}$
A₂B₃ (e.g., Fe(OH)₃ → Fe³⁺ + 3OH⁻)$K_{sp} = 27s^4$(for $AB_3$)

Ionic Product (Q): Calculated like $K_{sp}$ but with current concentrations.

  • $Q < K_{sp}$: unsaturated; more solid dissolves
  • $Q = K_{sp}$: saturated (equilibrium)
  • $Q > K_{sp}$: supersaturated; precipitation occurs

Common Ion Effect: Solubility of a salt is reduced by the presence of a common ion. e.g., AgCl solubility decreases in NaCl solution.

Application: Salt precipitation, qualitative analysis (group separation in salt analysis).

Salt Hydrolysis: Ions of dissolved salt react with water, changing pH.

Salt typeExamplepH of solution
Strong acid + Strong baseNaClNeutral (7)
Strong acid + Weak baseNH₄ClAcidic (< 7)
Weak acid + Strong baseCH₃COONaBasic (> 7)
Weak acid + Weak baseCH₃COONH₄Depends on $K_a$ and $K_b$

For salt of weak acid + strong base (e.g., CH₃COONa): $$K_h = K_w/K_a, \quad pH = 7 + (1/2)(pK_a + \log C)$$

For salt of weak base + strong acid (e.g., NH₄Cl): $$K_h = K_w/K_b, \quad pH = 7 - (1/2)(pK_b + \log C)$$

For salt of weak acid + weak base: $pH = 7 + (1/2)(pK_a - pK_b)$ (independent of concentration to first order).

Degree of hydrolysis ($h$): Fraction of salt that hydrolyzes. $h = \sqrt{K_h/C}$.

Worked Examples
1

Solubility of AgCl is $1.3 \times 10^{-5}$ M. Find $K_{sp}$.

Show solution

$$K_{sp} = s^2 = (1.3 \times 10^{-5})^2 = 1.69 \times 10^{-10}$$

Final Answer: $K_{sp} = 1.69 \times 10^{-10}$.

2

$0.1$ M CH₃COONa solution; $pK_a$ of acetic acid = $4.74$. Find pH.

Show solution

$$pH = 7 + (1/2)(pK_a + \log C) = 7 + (1/2)(4.74 + \log 0.1)$$ $$= 7 + (1/2)(4.74 - 1) = 7 + 1.87 = 8.87$$

Final Answer: $pH = 8.87$ (basic).

✎ Self-Check — 5 questions0 / 5
Q1.

Solubility product of $CaF_2$ in terms of solubility $s$:

Q2.

Common ion effect:

Q3.

Salt of strong acid and weak base in water:

Q4.

If $Q > K_{sp}$:

Q5.

The hydrolysis constant of CH₃COONa:

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