Colligative properties, Gibbs free energy, equilibrium constants, Bohr model, and quantum theory derivations for NEET.
1. Colligative Properties — ΔTb, ΔTf, π
💡 Why this matters in real life
Why does adding salt to water make it boil at a higher temperature? Why is antifreeze added to car radiators in winter? Why do intravenous drips use saline at a specific concentration? These everyday phenomena are explained by colligative properties — properties that depend only on the number of solute particles, not their identity. There are four key colligative properties: relative lowering of vapor pressure, boiling point elevation, freezing point depression, and osmotic pressure.
Fig 1.1: Osmosis — solvent molecules (blue) flow through the semipermeable membrane from pure solvent to solution, equalizing concentrations.
🧠 The Big Idea — In Plain Words
Colligative properties arise because solute particles reduce the number of solvent molecules at the surface, lowering the solvent's tendency to escape into the vapor phase or freeze. All four properties can be derived from Raoult's law: Pₐ = P°ₐ·Xₐ. The key is that the relative lowering of vapor pressure ΔP/P°ₐ = X_B (mole fraction of solute). For dilute solutions, this leads to ΔT_b = K_b·m, ΔT_f = K_f·m, and π = iCRT.
🔍 Step-by-Step: Boiling Point Elevation
1
Start from Raoult's law for a non-volatile solute. Vapor pressure of solution: P = P°·X_A = P°(1 − X_B). The lowering: ΔP = P° − P = P°·X_B. Relative lowering: ΔP/P° = X_B
2
For dilute solutions, X_B ≈ n_B/n_A. Since n_A = W_A/M_A and n_B = W_B/M_B: X_B = (W_B/M_B) / (W_A/M_A) = (W_B·M_A)/(M_B·W_A)
3
The boiling point elevation ΔT_b is proportional to the lowering of vapor pressure. From the Clausius-Clapeyron equation: ΔT_b = (RT_b²/ΔH_vap)·X_B. Since X_B ∝ molality (m), we get: ΔT_b = K_b·m where K_b = RT_b²M_A/(1000·ΔH_vap)
🔍 Step-by-Step: Freezing Point Depression
1
At the freezing point, solid and liquid solvent coexist in equilibrium. The presence of solute lowers the freezing point. The derivation is analogous to boiling point elevation: ΔT_f = K_f·m
2
K_f depends on the solvent properties.K_f = RT_f²M_A/(1000·ΔH_fus). For water: K_f = 1.86 K·kg/mol. The negative sign indicates freezing point decreases: T_f(solution) = T_f(pure) − ΔT_f
🔍 Step-by-Step: Osmotic Pressure
1
Osmotic pressure (π) is the pressure needed to prevent solvent flow across a semipermeable membrane. At equilibrium, the chemical potential of solvent is equal on both sides.
2
For dilute solutions, van't Hoff derived:π = CRT. This is analogous to the ideal gas law! C is the molar concentration of solute. For electrolytes: π = iCRT where i = van't Hoff factor (number of particles per formula unit).
ΔT_b = K_b·m ΔT_f = K_f·m π = iCRT ΔP/P° = X_B
🔍 Common Mistake
Common Mistake — Molality vs Molarity
ΔT_b and ΔT_f use molality (m) — moles of solute per kg of solvent. Osmotic pressure uses molarity (C) — moles per liter of solution. Also: for electrolytes like NaCl (i = 2) or CaCl₂ (i = 3), the effect is multiplied by i. Always check whether the solute dissociates or associates.
🎯 NEET Pattern
🎯 NEET Pattern — Colligative Properties
• "K_f for water is 1.86 K·kg/mol. Find FP of 0.1 m NaCl solution." → ΔT_f = i·K_f·m = 2 × 1.86 × 0.1 = 0.372 K. T_f = −0.372°C.
• "Which has highest boiling point: 0.1 M glucose, 0.1 M NaCl, or 0.1 M CaCl₂?" → CaCl₂ (i = 3, greatest number of particles).
• "Osmotic pressure of 0.01 M glucose at 27°C?" → π = CRT = 0.01 × 0.0821 × 300 = 0.246 atm.
✍️ Worked Example
Example 1: Boiling Point Elevation
K_b for water = 0.512 K·kg/mol. A solution contains 18 g glucose (M = 180 g/mol) in 500 g water. Find ΔT_b.
Solution: Molality = (18/180) / (500/1000) = 0.1/0.5 = 0.2 m. ΔT_b = 0.512 × 0.2 = 0.1024°C. Boiling point = 100.102°C. Key insight: Glucose doesn't dissociate (i = 1), so ΔT_b is small.
2. Gibbs Free Energy — ΔG = ΔH − TΔS
💡 Why this matters in real life
Why do some chemical reactions happen spontaneously while others don't? Why does ice melt above 0°C? Why do batteries produce electricity? The answer lies in Gibbs free energy. It combines enthalpy (heat content) and entropy (disorder) to predict whether a process is spontaneous. Every living system operates under the principles of Gibbs free energy — your body uses ATP precisely because its hydrolysis has a favorable ΔG.
Fig 2.1: ΔG combines ΔH (enthalpy) and TΔS (entropy). When ΔG is negative, the process is spontaneous.
🧠 The Big Idea — In Plain Words
Gibbs free energy (G) tells us whether a reaction will happen on its own. The change ΔG = ΔH − TΔS determines spontaneity. If ΔG is negative, the reaction is spontaneous (thermodynamically favorable). If ΔG is zero, the system is at equilibrium. If ΔG is positive, the reaction is non-spontaneous and needs external energy. The equation ΔG° = −RT ln K connects thermodynamics with equilibrium — a large K means ΔG° is strongly negative.
🔍 Step-by-Step: Deriving ΔG = ΔH − TΔS
1
Define Gibbs free energy.G = H − TS. H is enthalpy, T is temperature (K), S is entropy. For a change at constant temperature: ΔG = ΔH − TΔS
2
Spontaneity condition. At constant T and P, a process is spontaneous when ΔG < 0. Three cases: (a) ΔH negative, ΔS positive → ΔG always negative (always spontaneous). (b) ΔH positive, ΔS negative → ΔG always positive (never spontaneous). (c) Both same sign → temperature determines spontaneity.
3
At equilibrium, ΔG = 0. Setting ΔG = 0: 0 = ΔH − TΔS → T = ΔH/ΔS. This is the temperature at which the system transitions between spontaneous and non-spontaneous behavior.
🔍 Step-by-Step: Deriving ΔG° = −RT ln K
1
Start with the relationship between ΔG and reaction quotient Q. For a reaction aA + bB ⇌ cC + dD: ΔG = ΔG° + RT ln Q. At equilibrium, ΔG = 0 and Q = K.
2
Substitute equilibrium conditions.0 = ΔG° + RT ln K → ΔG° = −RT ln K. In base 10: ΔG° = −2.303RT log K. At 298 K: ΔG° = −5.71 log K kJ/mol.
ΔG = ΔH − TΔS ΔG° = −RT ln K Spontaneous when ΔG < 0
🔍 Common Mistake
Common Mistake — ΔG vs ΔG°
ΔG° (standard Gibbs free energy change) refers to standard conditions (1 M, 1 atm, 298 K). ΔG (actual) depends on current concentrations. A reaction with positive ΔG° can still proceed if concentrations are far from standard (ΔG becomes negative due to the RT ln Q term).
🎯 NEET Pattern
🎯 NEET Pattern — Gibbs Free Energy
• "For a reaction, ΔH = −100 kJ, ΔS = −200 J/K. Is it spontaneous at 300 K?" → ΔG = −100 − 0.3×(−200) = −100 + 60 = −40 kJ. Yes, spontaneous.
• "ΔG° = 5.71 kJ/mol at 298 K. Find K." → 5.71 = −5.71 log K → log K = −1 → K = 0.1.
• "At what T does ΔG = 0 if ΔH = 40 kJ and ΔS = 100 J/K?" → T = 40000/100 = 400 K.
✍️ Worked Example
Example 2: Gibbs Free Energy
For the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = −92.4 kJ and ΔS = −198 J/K at 298 K. Is it spontaneous?
Solution: ΔG = −92400 − 298×(−198) = −92400 + 59004 = −33396 J = −33.4 kJ. ΔG < 0, so the reaction is spontaneous at 298 K. Key insight: At higher temperature, −TΔS becomes large positive, making ΔG positive — that's why NH₃ synthesis is run at moderate temperatures.
3. Kp = Kc(RT)^Δn
💡 Why this matters in real life
Chemical equilibrium constants can be expressed in terms of concentrations (Kc) or partial pressures (Kp). The relationship between them is essential because NEET questions often give one and ask for the other. Understanding this relationship is crucial for predicting how changing pressure or volume affects equilibrium in gaseous reactions.
🧠 The Big Idea — In Plain Words
For a gaseous reaction aA + bB ⇌ cC + dD, the equilibrium constants are Kc = [C]ᶜ[D]ᵈ/[A]ᵃ[B]ᵇ and Kp = P_Cᶜ·P_Dᵈ/P_Aᵃ·P_Bᵇ. Since PV = nRT, we have P = (n/V)RT = CRT. Substituting this relationship gives Kp = Kc(RT)^Δn, where Δn = (c + d) − (a + b) is the change in the number of moles of gas.
🔍 Step-by-Step: Deriving Kp = Kc(RT)^Δn
1
Write the expression for Kp. For aA + bB ⇌ cC + dD: Kp = P_Cᶜ·P_Dᵈ / P_Aᵃ·P_Bᵇ. Each P is the partial pressure at equilibrium.
2
Use the ideal gas relation P = CRT. Where C = n/V = concentration. Substitute: Kp = ([C]RT)ᶜ·([D]RT)ᵈ / ([A]RT)ᵃ·([B]RT)ᵇ
3
Factor out RT terms.Kp = [C]ᶜ[D]ᵈ/[A]ᵃ[B]ᵇ × (RT)^{(c+d)−(a+b)} = Kc × (RT)^{Δn}. Where Δn = moles of gaseous products − moles of gaseous reactants.
When using Kp = Kc(RT)^Δn, R must be 0.0821 L·atm/(mol·K) (not 8.314 J/(mol·K)). This is because Kc uses mol/L and Kp uses atm. If Δn = 0, Kp = Kc. Also: only include gaseous species — solids and liquids don't appear in either Kp or Kc.
🎯 NEET Pattern
🎯 NEET Pattern — Kp and Kc
• "For N₂ + 3H₂ ⇌ 2NH₃, Kc = 4.0 at 400 K. Find Kp. (R = 0.0821)" → Δn = 2 − 4 = −2. Kp = 4.0 × (0.0821×400)^(−2) = 4.0 / (32.84)².
• "For H₂ + I₂ ⇌ 2HI, Δn = 0, so Kp = Kc." — a common direct answer.
• "For which reaction does Kp > Kc?" → When Δn > 0 (more product moles than reactant moles).
✍️ Worked Example
Example 3: Kp and Kc
For 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), Kc = 280 at 1000 K. Find Kp. (R = 0.0821 L·atm/mol·K)
Solution: Δn = 2 − (2+1) = −1. Kp = 280 × (0.0821 × 1000)^(−1) = 280/(82.1) = 3.41. Key insight: Kp is smaller than Kc because the reaction reduces the number of gas molecules.
4. Bohr Model — Radius & Energy of Orbits
💡 Why this matters in real life
The Bohr model, despite being superseded by quantum mechanics, remains the simplest way to understand atomic spectra. It explains why hydrogen emits specific colors of light (like the red 656 nm line in the Balmer series). NEET frequently asks about the radius, energy, and velocity of electrons in Bohr orbits.
Fig 4.1: Bohr orbits for hydrogen (n = 1, 2, 3). Electron transition from n = 3 to n = 2 emits a photon of red light (656 nm).
🧠 The Big Idea — In Plain Words
Bohr proposed that electrons move in fixed circular orbits around the nucleus. The key idea is quantization of angular momentum: mvr = nh/2π. This leads to discrete energy levels for the electron. The radius increases as n², and the energy becomes less negative (higher) as n increases. When an electron jumps from a higher to a lower orbit, it emits a photon of energy hν = ΔE.
🔍 Step-by-Step: Deriving the Radius of nth Orbit
1
Coulomb force provides the centripetal force. For an electron orbiting a nucleus of charge Ze: mv²/r = (1/4πε₀)·(Ze·e)/r². This gives: mv² = Ze²/(4πε₀r)
2
Bohr's quantization condition. Angular momentum is quantized: mvr = nh/2π. So: v = nh/(2πmr)
3
Substitute v into the force equation.m(nh/2πmr)² = Ze²/(4πε₀r). Solve for r: r_n = (n²h²ε₀)/(πmZe²)
4
For hydrogen (Z = 1). Substituting constants: r_n = 0.529 × n² Å. The first Bohr radius r₁ = 0.529 Å. This is the most probable distance of the electron from the nucleus.
🔍 Step-by-Step: Deriving the Energy of nth Orbit
1
Total energy = KE + PE. Kinetic energy: KE = ½mv². Potential energy: PE = −Ze²/(4πε₀r). From force equation: mv² = Ze²/(4πε₀r), so KE = Ze²/(8πε₀r)
2
Total energy E = KE + PE.E = Ze²/(8πε₀r) − Ze²/(4πε₀r) = −Ze²/(8πε₀r). The negative sign means the electron is bound to the nucleus.
3
Substitute r_n.E_n = −(mZ²e⁴)/(8ε₀²h²)·(1/n²). For hydrogen: E_n = −13.6/n² eV. Ground state (n = 1): −13.6 eV. First excited (n = 2): −3.4 eV.
r_n = 0.529·n²/Z Å E_n = −13.6·Z²/n² eV ΔE = hν = hc/λ
🔍 Common Mistake
Common Mistake — Energy Sign Convention
Energy of an electron in an orbit is negative. This is because energy is measured relative to the free electron (E = 0 when n = ∞). The more negative the energy, the more stable the electron. Also: when an electron absorbs energy, it jumps to a higher (less negative) n — this is excitation.
🎯 NEET Pattern
🎯 NEET Pattern — Bohr Model
• "Radius of 2nd orbit of He⁺?" → r₂ = 0.529 × 4/2 = 1.058 Å.
• "Energy of 1st orbit of Li²⁺?" → E₁ = −13.6 × 9 = −122.4 eV.
• "Wavelength emitted when electron drops from n = 3 to n = 2 in H?" → 1/λ = R(1/4 − 1/9) = R×5/36 → λ = 656 nm (red, Balmer series).
✍️ Worked Example
Example 4: Bohr Model
Calculate the radius of the 3rd Bohr orbit of hydrogen and the velocity of the electron in it.
Solution: r₃ = 0.529 × 9 = 4.761 Å. v = nh/(2πmr) = 3 × 6.626×10⁻³⁴ / (2π × 9.1×10⁻³¹ × 4.761×10⁻¹⁰) = 7.29×10⁵ m/s. Key insight: Velocity decreases as n increases (v ∝ 1/n).
5. de Broglie Wavelength & Uncertainty Principle
💡 Why this matters in real life
Electron microscopes use the wave nature of electrons to see objects as small as atoms — much smaller than visible light can resolve. The de Broglie wavelength explains why particles like electrons exhibit wave-like behavior. The Heisenberg uncertainty principle sets a fundamental limit on measurement accuracy, which is why we can't pin down exactly where an electron is at any moment.
🧠 The Big Idea — In Plain Words
de Broglie proposed that every moving particle has a wavelength: λ = h/p = h/mv. For macroscopic objects, this wavelength is tiny and undetectable. For electrons (mass 9.1×10⁻³¹ kg), the wavelength is comparable to atomic spacings, so electron diffraction is observable. Heisenberg's uncertainty principle states: Δx·Δp ≥ h/4π. You cannot simultaneously know both position and momentum with perfect accuracy.
🔍 Step-by-Step: De Broglie Wavelength Derivation from Bohr Model
1
Bohr's quantization condition mvr = nh/2π can be reinterpreted. Rewrite as: 2πr = nh/mv = nλ. This means the circumference of the orbit must contain an integer number of de Broglie wavelengths.
2
This is the standing wave condition. An electron wave in an orbit forms a standing wave only if the circumference is an integer multiple of the wavelength: 2πr = nλ → λ = 2πr/n. This naturally leads to the quantization of angular momentum.
The uncertainty principle is a fundamental quantum mechanical result.Δx·Δp ≥ h/4π. Δx is the uncertainty in position, Δp is the uncertainty in momentum. The product cannot be less than h/4π ≈ 5.27×10⁻³⁵ J·s.
2
Physical implication. If we know position very precisely (Δx small), then Δp must be large, meaning we can't know momentum (and thus velocity) accurately. Conversely, if we know momentum precisely, position becomes uncertain. This is not a limitation of instruments — it's a fundamental property of nature.
λ = h/mv Δx·Δp ≥ h/4π h = 6.626×10⁻³⁴ J·s
🔍 Common Mistake
Common Mistake — de Broglie for Macroscopic Objects
Students apply de Broglie's equation to macroscopic objects without checking the magnitude. For a cricket ball (m = 0.16 kg) moving at 30 m/s: λ = 6.63×10⁻³⁴/(0.16×30) ≈ 1.38×10⁻³⁴ m — far too small to detect. The wave nature is only observable for microscopic particles.
🎯 NEET Pattern
🎯 NEET Pattern — de Broglie & Uncertainty
• "de Broglie wavelength of an electron accelerated through 100 V?" → λ = h/√(2meV) = 12.27/√V Å = 1.227 Å.
• "If uncertainty in position is 1 Å, find minimum uncertainty in momentum." → Δp ≥ h/(4πΔx) = 5.27×10⁻²⁵ kg·m/s.
• "Which has longer λ: electron or proton at same speed?" → Electron (smaller mass → larger wavelength).
✍️ Worked Example
Example 5: de Broglie Wavelength
An electron has a de Broglie wavelength of 1 Å. Find its velocity. (m_e = 9.1×10⁻³¹ kg, h = 6.626×10⁻³⁴ J·s)
Solution: v = h/(mλ) = 6.626×10⁻³⁴/(9.1×10⁻³¹ × 10⁻¹⁰) = 7.28×10⁶ m/s. Key insight: This velocity is about 2.4% of the speed of light — relativistic effects are small but not negligible at this speed.
6. Quantum Numbers & Rydberg Formula
💡 Why this matters in real life
Atomic spectra are like fingerprints for elements — each element has a unique set of spectral lines. This is used in astronomy to determine the composition of stars, in forensic science to identify substances, and in neon signs where different gases produce different colors. Quantum numbers describe the exact state of every electron in an atom.
🧠 The Big Idea — In Plain Words
Four quantum numbers completely describe an electron in an atom: n (principal — energy level), l (azimuthal — shape of orbital), m_l (magnetic — orientation of orbital), and m_s (spin — +½ or −½). The Pauli exclusion principle states that no two electrons can have the same set of four quantum numbers. The Rydberg formula 1/λ = R(1/n₁² − 1/n₂²) predicts the wavelengths of spectral lines for hydrogen-like atoms.
🔍 Step-by-Step: Deriving the Rydberg Formula
1
Energy of an electron in the nth orbit of hydrogen.E_n = −13.6/n² eV = −(2π²me⁴)/(h²)·(1/n²) in SI units.
2
Energy of photon emitted during transition. When electron drops from n₂ to n₁ (n₂ > n₁): ΔE = E_{n₂} − E_{n₁} = hν = hc/λ. This gives: ΔE = −13.6(1/n₂² − 1/n₁²) = 13.6(1/n₁² − 1/n₂²)
3
Express in terms of wavelength.1/λ = ΔE/(hc) = (13.6 eV)/(hc)·(1/n₁² − 1/n₂²) = R(1/n₁² − 1/n₂²). The Rydberg constant R = 1.097×10⁷ m⁻¹.
4
Spectral series. For hydrogen: Lyman (n₁ = 1, UV), Balmer (n₁ = 2, visible), Paschen (n₁ = 3, IR), Brackett (n₁ = 4, IR), Pfund (n₁ = 5, IR). NEET frequently asks Balmer series questions because it's in the visible range.
1/λ = R(1/n₁² − 1/n₂²) R = 1.097×10⁷ m⁻¹ n₂ > n₁
🔍 Common Mistake
Common Mistake — n₁ vs n₂
Students often reverse n₁ and n₂. n₁ is the lower energy level (where the electron ends up), and n₂ is the higher energy level (where it started). Also: for hydrogen-like ions (He⁺, Li²⁺), the Rydberg formula becomes 1/λ = RZ²(1/n₁² − 1/n₂²).
🎯 NEET Pattern
🎯 NEET Pattern — Atomic Spectra
• "Which series of hydrogen spectrum lies in the visible region?" → Balmer series (n₁ = 2).
• "Shortest wavelength of Lyman series?" → n₂ = ∞: 1/λ = R(1/1² − 0) = R → λ = 1/R ≈ 911.6 Å.
• "Number of spectral lines when electron drops from n = 5 to n = 1?" → Number = n(n−1)/2 = 5×4/2 = 10 lines.
✍️ Worked Example
Example 6: Rydberg Formula
Calculate the wavelength of the first line of the Balmer series for hydrogen. (R = 1.097×10⁷ m⁻¹)
Solution: First Balmer line: n₁ = 2, n₂ = 3. 1/λ = 1.097×10⁷(1/4 − 1/9) = 1.097×10⁷ × 5/36 = 1.524×10⁶ m⁻¹. λ = 656 nm (red light). Key insight: This is the famous H-α line, used to identify hydrogen in stars and nebulae.