Equilibrium
Equilibrium for JEE Main & Advanced
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:
- Constant at given temperature
- Independent of initial concentrations
- Independent of presence of catalyst
- Changes with temperature
- Reverse reaction: $K_{\text{reverse}} = 1/K_{\text{forward}}$
- 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 $V$: $$K_c = \frac{n\alpha^2}{V(1-\alpha)} \quad \text{(if } \alpha \ll 1 \text{: } K_c \approx \alpha^2 \cdot n/V)$$
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: $K_c \approx 0.0243$.
$1$ 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: $K_c = 0.05$.
At equilibrium:
$K_p = K_c$ when:
If $Q < K$, reaction proceeds:
The unit of $K_c$ for $N_2 + 3H_2 \to 2NH_3$:
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:
| Change | Direction 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:
| Process | Reaction | Conditions |
|---|---|---|
| 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., $CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)$: $K_p = P_{CO_2}$.
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.
$K_1 = 4 \times 10^{-2}$ 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: $K_2 \approx 0.30$.
Le Chatelier's principle states:
For exothermic reaction, increasing $T$:
Effect of catalyst on equilibrium constant:
In Haber process, high pressure favors:
Effect of inert gas at constant pressure:
Ionic Equilibrium
Acid-Base Concepts, pH, BuffersTopic 1
Theories of Acids and Bases:
| Theory | Acid | Base |
|---|---|---|
| Arrhenius | Releases H⁺ in water | Releases OH⁻ in water |
| Bronsted-Lowry | Proton donor | Proton acceptor |
| Lewis | Electron pair acceptor | Electron 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.
| Type | Composition | Example |
|---|---|---|
| Acidic buffer | Weak acid + salt of its conjugate base | CH₃COOH + CH₃COONa |
| Basic buffer | Weak base + salt of its conjugate acid | NH₄OH + 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.
$0.01$ 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: $pH \approx 3.37$.
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: $pH = 5.04$.
pH of $0.01$ M HCl:
Conjugate base of NH₃:
Buffer of weak acid and its salt is called:
pH range of acidic buffer:
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 type | Example | pH of solution |
|---|---|---|
| Strong acid + Strong base | NaCl | Neutral (7) |
| Strong acid + Weak base | NH₄Cl | Acidic (< 7) |
| Weak acid + Strong base | CH₃COONa | Basic (> 7) |
| Weak acid + Weak base | CH₃COONH₄ | Depends 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. $h = \sqrt{K_h/C}$.
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: $K_{sp} = 1.69 \times 10^{-10}$.
$0.1$ 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).
Solubility product of $CaF_2$ in terms of solubility $s$:
Common ion effect:
Salt of strong acid and weak base in water:
If $Q > K_{sp}$:
The hydrolysis constant of CH₃COONa:
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