🧪 Chemistry · Physical 🔥 Day 1

0/9 topics mastered

Physical Chemistry — Chemistry

🧪

Some Basic Concepts

Mole concept · laws of combination · stoichiometry · significant figures

1. Matter, Elements & Compounds

Matter is anything with mass that occupies space. An element cannot be broken into simpler substances by chemical means; a compound is two or more elements chemically combined in a fixed ratio (H₂O is always 2 H : 1 O); a mixture combines substances in variable ratio, separable physically.

2. Laws of Chemical Combination

Conservation of Mass (Lavoisier): mass of reactants = mass of products.
Definite Proportions (Proust): a given compound always has the same elements in the same mass ratio.
Multiple Proportions (Dalton): for two elements forming several compounds, the masses of one combining with a fixed mass of the other are in a simple whole-number ratio (CO and CO₂ → 16 g : 32 g = 1 : 2).
Gay-Lussac's Law of gaseous volumes and Avogadro's Law (equal volumes of gases at the same T and P contain equal numbers of molecules).

3. The Mole Concept

Atoms are too small to count, so chemists use the mole — exactly $6.022\times10^{23}$ entities (the Avogadro constant $N_A$), just as "a dozen" always means 12. The mole is defined so the mass of one mole in grams (the molar mass) numerically equals the atomic/molecular mass in $u$. This links the atomic world to the weighable world:

$$ n = \frac{\text{given mass}}{\text{molar mass}} = \frac{N}{N_A} = \frac{V}{22.4\ \text{L}}\ (\text{gas at STP}) $$

One mole of any gas occupies $22.4\ \text{L}$ at STP — a direct consequence of Avogadro's law.

4. Empirical & Molecular Formula · Stoichiometry

The empirical formula is the simplest whole-number atom ratio; the molecular formula $= \text{empirical} \times n$, where $n = \dfrac{\text{molecular mass}}{\text{empirical formula mass}}$. In a reaction, the limiting reagent is found by comparing (moles ÷ coefficient) for each reactant — the smallest value runs out first and caps the product.

5. Significant Figures (measurement)

Non-zero digits and zeros between them are significant; leading zeros never are; trailing zeros count only after a decimal point. For addition/subtraction the result keeps the fewest decimal places; for multiplication/division it keeps the fewest significant figures.

💡 Highest-yield idea: nearly every numerical in Physical Chemistry — solutions, equilibrium, electrochemistry, kinetics — reduces to $n = \dfrac{\text{mass}}{\text{molar mass}}$. Master the mole and the rest gets far easier.

⚠ Common mistake: assuming the reactant of smaller mass is the limiting reagent. Always compare moles ÷ stoichiometric coefficient, not raw mass.

🧪

Some Basic Concepts — Formulas

Mole, formula and concentration relations

Moles

$$ n = \frac{\text{mass}}{\text{molar mass}} = \frac{N}{N_A} = \frac{V}{22.4} $$

Gas volume term valid at STP

mol

Avogadro Constant

$$ N_A = 6.022\times10^{23}\ \text{mol}^{-1} $$

Entities per mole

mol⁻¹

Molecular Formula factor

$$ n = \frac{\text{molecular mass}}{\text{empirical formula mass}} $$

Molecular formula = empirical × $n$

—

Molarity

$$ M = \frac{\text{moles of solute}}{\text{volume of solution (L)}} $$

Temperature-dependent (volume)

mol/L

Atomic Mass Unit

$$ 1\ u = 1.66\times10^{-24}\ \text{g} $$

$\tfrac{1}{12}$ mass of a ¹²C atom

🧪

Some Basic Concepts — Solved Examples

NEET · JEE Main level

Calculate the number of moles in 4.4 g of CO₂ (molar mass 44 g/mol), and the number of molecules in it.

$n = \dfrac{4.4}{44} = 0.1$ mol
$N = n\,N_A = 0.1 \times 6.022\times10^{23} = 6.022\times10^{22}$ molecules

✓ 0.1 mol · $6.022\times10^{22}$ molecules

A compound is 40% C, 6.7% H, 53.3% O by mass and has molecular mass 180. Find its molecular formula.

Assume 100 g → moles: C = 40/12 = 3.33, H = 6.7/1 = 6.7, O = 53.3/16 = 3.33
Divide by smallest (3.33): C = 1, H ≈ 2, O = 1 → empirical formula CH₂O (mass 30)
$n = \dfrac{180}{30} = 6$ → molecular formula $= \text{CH}_2\text{O} \times 6 = \text{C}_6\text{H}_{12}\text{O}_6$

✓ C₆H₁₂O₆ (glucose)

5.6 L of a gas at STP weighs 11 g. Find its molar mass and a possible formula (C and O only).

Moles $= \dfrac{5.6}{22.4} = 0.25$ mol
Molar mass $= \dfrac{11}{0.25} = 44$ g/mol
CO₂ has molar mass $12 + 32 = 44$ → matches

✓ Molar mass 44 g/mol; gas is likely CO₂

🧪

Some Basic Concepts — Quick Quiz

tap an option to reveal the answer

Question 1

CO and CO₂: for a fixed mass of carbon, the oxygen masses are in a 1:2 ratio. Which law does this illustrate?

Two elements forming several compounds in simple whole-number mass ratios is the law of multiple proportions (Dalton).

Question 2

A compound has empirical formula CH₂ and molecular mass 42. Its molecular formula is:

Empirical mass of CH₂ = 14; $n = \dfrac{42}{14} = 3$ → molecular formula C₃H₆.

Question 3

How many significant figures are in 0.0025?

Leading zeros only locate the decimal point and are never significant, so only 2 and 5 count → 2 significant figures.

⚛

Atomic Structure

Bohr model · Quantum numbers · Orbitals · Electronic configuration · Spectra

Early Atomic Models

Model	Scientist	Key Feature	Limitation
Plum-pudding	Thomson (1904)	Electrons embedded in positive sphere	Couldn't explain α-scattering
Nuclear model	Rutherford (1911)	Tiny dense nucleus; electrons orbit	Couldn't explain atomic stability or line spectra
Planetary (quantised)	Bohr (1913)	Fixed circular orbits; energy quantised	Only valid for H-like atoms; ignored wave nature

Bohr's Model (Hydrogen-like atoms)

Radius of nth orbit: $r_{n} = 0.529 \times \dfrac{n^{2}}{Z} Å$ (a₀ = 0.529 Å = Bohr radius)
Velocity of electron: $v_{n} = 2.18\times10^{6} \times \dfrac{Z}{n} m/s$
Energy of nth level: $E_{n} = -13.6 \times \dfrac{Z^{2}}{n^{2}} eV$
Frequency of emitted photon: $h\nu = E_i - E_f = 13.6Z^{2}(\dfrac{1}{n_f^{2}} - \dfrac{1}{n_i^{2}}) eV$
Wave number of spectral line: $\nũ = RZ^{2}(\dfrac{1}{n_f^{2}} - \dfrac{1}{n_i^{2}})$; $R = 1.097\times10^{7} m^{-1}$ (Rydberg constant)

Hydrogen Spectral Series

Series	$n_f$	$n_i$	Region	First line ($n_i = n_f+1$)
Lyman	1	2,3,4…	UV	n=2→1 (121.6 nm)
Balmer	2	3,4,5…	Visible	n=3→2 (656.3 nm, red)
Paschen	3	4,5,6…	Near IR	n=4→3
Brackett	4	5,6,7…	IR	n=5→4
Pfund	5	6,7,8…	Far IR	n=6→5

Wave-Particle Duality and Uncertainty

de Broglie wavelength: $\lambda = \dfrac{h}{mv} = \dfrac{h}{p}$; for electron accelerated through V volts: $\lambda = \dfrac{12.27}{\sqrt V} Å$
Heisenberg's uncertainty principle: Δx·Δp ≥ h/4π; ΔE·Δt ≥ h/4π
Consequence: electrons cannot have defined orbits — described by wave functions (orbitals)
Photoelectric effect: $KE_{max} = h\nu - \phi$ (φ = work function); threshold ν₀ = φ/h

Quantum Numbers

QN	Symbol	Values	Describes
Principal	$n$	1, 2, 3, …	Shell, energy level, size
Azimuthal	$l$	$0$ to $n-1$	Sub-shell, shape (s,p,d,f)
Magnetic	$m_l$	$-l$ to $+l$	Orientation in space
Spin	$m_s$	$+\tfrac{1}{2}$ or $-\tfrac{1}{2}$	Spin of electron

l = 0 (s), 1 (p), 2 (d), 3 (f); number of orbitals in subshell = $2l+1$
Maximum electrons per subshell = $2(2l+1)$: s=2, p=6, d=10, f=14
Total orbitals in nth shell = $n^2$; maximum electrons = $2n^2$

Orbital explorer: n, l and mₗ — Step through every orbital up to n = 3: 1s; 2s and the three 2p; 3s, the three 3p and the five 3d. n fixes the shell and the size; l fixes the shape (0 = s, 1 = p, 2 = d) and the number of orbitals in the subshell, 2l + 1. The real orbitals in textbooks combine mₗ states: pz and dz² are mₗ = 0, but px and py are each built from mₗ = +1 and −1 together, so px is not “mₗ = +1”; dxz and dyz come from ±1, and dxy and dx²−y² from ±2. Radial nodes = n − l − 1 and angular nodes = l, so every orbital in shell n has n − 1 nodes. Each surface encloses 90% of the probability, and colour shows the sign of ψ, not charge. Orbitals with radial nodes (2s, 3s, 3p) are cut open to show them.

The seven 4f orbitals — l = 3 gives 2l + 1 = 7 orbitals. Each has three angular nodes (planes or cones) and, in the 4f subshell, no radial node (n − l − 1 = 0). The set shown is the usual general set: fz³, fxz², fyz², fxyz, fz(x²−y²), fx(x²−3y²) and fy(3x²−y²). fxyz and fz(x²−y²) have eight lobes; fz³ has two lobes and two rings; the other four have six lobes. Colour shows the sign of ψ, which alternates from lobe to lobe.

Electronic Configuration Rules

Aufbau principle: Fill orbitals in order of increasing energy — 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p…
Pauli exclusion: No two electrons in an atom can have identical set of all four quantum numbers — max 2 e⁻ per orbital, with opposite spins
Hund's rule: Electrons occupy degenerate orbitals singly before pairing; singly occupied orbitals have parallel spins
Exceptions: Cr = [Ar] 3d⁵ 4s¹; Cu = [Ar] 3d¹⁰ 4s¹ (extra stability of half-filled and fully-filled d)

💡 JEE Tip: Number of spectral lines when electron falls from nth level to ground state = n(n−1)/2. For n=4: 4×3/2 = 6 lines. In a single transition from n=4 to n=1, only one photon is emitted — but multiple transitions collectively give 6 lines.

⚠ Common Error: Bohr's model gives $E \propto \dfrac{Z^2}{n^2}$ — always negative (bound state). Ground state of H: $E_1 = -13.6$ eV. Ionisation energy = 13.6 eV (energy to remove electron, making $E = 0$). Don't confuse energy of orbit with ionisation energy.

📐

Atomic Structure — Formula Sheet

Bohr's Model

$$ r_{n} = 0.529 \times \frac{n^{2}}{Z} Å $$

Radius of nth orbit; $a_{0} = 0.529 Å$ (Bohr radius for H, n=1, Z=1)

$$ E_{n} = -13.6 \times \frac{Z^{2}}{n^{2}} eV $$

Energy of nth orbit; negative = bound; ionisation energy = $13.6 Z^2$ eV from ground state

$$ v_{n} = 2.18\times10^{6} \times \frac{Z}{n} m/s $$

Speed of electron in nth orbit; increases toward nucleus

$$ \Delta E = 13.6 Z^{2} (\frac{1}{n_f^{2}} - \frac{1}{n_i^{2}}) eV $$

Energy of emitted/absorbed photon; $n_i > n_f$ for emission

Spectral Lines

$$ \frac{1}{\lambda} = RZ^{2}(\frac{1}{n_f^{2}} - \frac{1}{n_i^{2}}) $$

Rydberg formula; $R = 1.097\times10^{7} m^{-1}$; Lyman: n_f=1, Balmer: n_f=2, Paschen: n_f=3

$$ \text{No. of spectral lines} = \frac{n(n-1)}{2} $$

Total lines when electron falls from nth level; each pair $(n_i, n_f)$ gives one line

Wave-Particle Duality & Uncertainty

$$ \lambda = \frac{h}{p} = \frac{h}{mv} $$

de Broglie wavelength; $h = 6.626\times10^{-34} J\cdot s$ (Planck's constant)

$\lambda = \dfrac{12.27}{\sqrt V}$ Å (electron)

de Broglie wavelength of electron accelerated through potential V volts

Δx · Δp ≥ h/4π

Heisenberg's uncertainty principle; also ΔE·Δt ≥ h/4π

$$ KE_{max} = h\nu - \phi = h\nu - h\nu_{0} $$

Photoelectric effect; $\phi$ = work function; $\nu_0$ = threshold frequency = $\dfrac{\phi}{h}$

Quantum Numbers & Orbitals

$$ \text{Orbitals in subshell} = 2l+1 \quad|\quad e^- \text{ max} = 2(2l+1) $$

l=0 (s): 1 orbital, 2e⁻; l=1 (p): 3 orbitals, 6e⁻; l=2 (d): 5 orbitals, 10e⁻

Total orbitals in nth shell = n² | $Max e^{-} = 2n^{2}$

n=1: 1 orbital, 2e⁻; n=2: 4 orbitals, 8e⁻; n=3: 9 orbitals, 18e⁻

🧮

Atomic Structure — Solved Examples

Calculate for He⁺ ion (Z=2): (a) the radius of the 2nd orbit, (b) the energy of the 3rd orbit, (c) the ionisation energy from the ground state, and (d) the wavelength of radiation emitted when electron falls from n=3 to n=2.

(a) r₂ (He⁺, Z=2): $r = 0.529 \times \dfrac{n^{2}}{Z} = 0.529 \times \dfrac{4}{2} =$ 1.058 Å
(b) E₃ (He⁺): $E = -13.6 \times \dfrac{Z^{2}}{n^{2}} = -13.6 \times \dfrac{4}{9} = -\dfrac{6.04}{1} =$ −6.04 eV
(c) Ionisation energy from ground state: $IE = |E_{1}| = 13.6 \times \dfrac{Z^{2}}{1^{2}} = 13.6 \times 4 =$ 54.4 eV
(d) Wavelength (n=3→2): $\Delta E = 13.6 \times 4 \times (\dfrac{1}{4} - \dfrac{1}{9}) = 54.4 \times \frac{5}{36} = 7.56 eV$
$\lambda = \dfrac{hc}{\Delta} E = \dfrac{6.626\times10^{-34} \times 3\times10^{8}}{7.56 \times 1.6\times10^{-19}} = 1.988\dfrac{\times10^{-25}}{1.21}\times10^{-18} =$ 164 nm (UV region)

✓ Note: He⁺ Balmer series (n→2) falls in UV, unlike H where it's visible — because Z²=4 scales all energies up by 4×.

(a) Find the de Broglie wavelength of an electron accelerated through 100 V. (b) A metal has work function $\phi = 2.0$ eV. Find the maximum KE of photoelectrons when light of wavelength 300 nm falls on it, and the stopping potential.

(a) de Broglie λ: $\lambda = \dfrac{12.27}{\sqrt V} = \dfrac{12.27}{\sqrt{100}} = \dfrac{12.27}{10}$ = 1.227 Å
(b) Energy of photon (λ=300 nm): $E = \dfrac{hc}{\lambda} = \dfrac{6.626\times10^{-34} \times 3\times10^{8}}{300\times10^{-9}} = 6.626\times10^{-19} J = 6.626\dfrac{\times10^{-19}}{1.6}\times10^{-19} =$ 4.14 eV
KE_max: $= h\nu - \phi = 4.14 - 2.0 =$ 2.14 eV
Stopping potential $V_0$: $eV_0 = KE_{max} \to V_0 = 2.14$ V (stopping potential = $KE_{max}$ in eV numerically)

✓ Answers: $\lambda = 1.227$ Å; $KE_{max} = 2.14$ eV; $V_0 = 2.14$ V. Key: doubling light intensity doubles electron count but doesn't change $KE_{max}$ — only frequency matters for energy.

(a) An electron in an atom has quantum numbers $n=4, l=2, m_l=-1, m_s=+\tfrac{1}{2}$. Identify the subshell, orbital, and state whether this is valid. (b) Write the ground state electronic configuration of Cr (Z=24) and Cu (Z=29), explaining the exceptions. (c) How many electrons in Fe (Z=26) have $l=2$?

(a) Identification: $n=4, l=2$ → 4d subshell; $m_l=-1$ → one of the five 4d orbitals (valid: $-2$ to $+2$); $m_s=+\tfrac{1}{2}$ → spin up. Valid — 4d orbital, spin up.
(b) Cr (Z=24) expected: [Ar] 3d⁴ 4s². Actual: [Ar] 3d⁵ 4s¹ — half-filled 3d⁵ is extra stable (symmetric distribution of electrons); one electron from 4s migrates to 3d.
Cu (Z=29) expected: [Ar] 3d⁹ 4s². Actual: [Ar] 3d¹⁰ 4s¹ — fully-filled 3d¹⁰ is extra stable; one electron from 4s moves to complete 3d.
(c) Fe (Z=26): Configuration = [Ar] 3d⁶ 4s². Electrons with $l=2$ are all d-electrons → only in 3d subshell = 6 electrons.

✓ Remember: l=0 (s), l=1 (p), l=2 (d), l=3 (f). In Fe, only 3d has l=2. The 4s electrons have l=0. Counting by l is a common JEE question type.

📊

Atomic Structure — Visual Concepts

Click any diagram to zoom

Hydrogen Energy Levels and Spectral Series

Hydrogen energy levels (not to scale). Lyman series: transitions to n=1 (UV). Balmer series: to n=2 (visible, first line at 656 nm is red). Paschen series: to n=3 (infrared). Energy spacing decreases for higher n — levels get closer to ionisation limit.

Atomic Orbital Shapes — s, p, d

s orbitals are spherically symmetric (1 orientation). p orbitals are dumbbell-shaped along x, y, z axes (3 orientations). Of the five d orbitals, four (dxy, dyz, dxz, dx²−y²) are cloverleaf-shaped; dz² is a dumbbell along z with a ring (torus) round its middle. The + and − signs indicate phase of the wave function, not charge. Each surface encloses 90% of the probability of finding the electron; it is not an orbit. In 3D, 2s and 3p are cut open to show their radial nodes: 2s at r = 2a₀ and 3p at 6a₀ (a₀ = 52.9 pm).

Aufbau Filling Order — (n+l) Rule

Aufbau filling order determined by (n+l) rule. 4s fills before 3d because 4s has n+l = 4+0 = 4, while 3d has n+l = 3+2 = 5. Likewise 6s (n+l = 6) fills before 4f (n+l = 7): the order runs 5p ⑪ → 6s ⑫ → 4f ⑬ → 5d → 6p. When (n+l) is equal, the orbital with lower n fills first. Note exceptions: Cr and Cu deviate from this for extra stability.

⚡

Atomic Structure — Quick Quiz

Question 1

The wavelength of the first line of the Lyman series for hydrogen is 121.6 nm. The wavelength of the second line of the Balmer series for He⁺ (Z=2) is:

Second Balmer line of He⁺ corresponds to $n_i=4 \to n_f=2$. Using the Rydberg formula $\dfrac{1}{\lambda} = RZ^2(\dfrac{1}{n_f^2} - \dfrac{1}{n_i^2})$ with $Z=2$: $\dfrac{1}{\lambda} = R\times 4\times(\dfrac{1}{4} - \dfrac{1}{16}) = 4R\times \dfrac{3}{16} = \dfrac{3R}{4}$. For the first Lyman line of H: $\dfrac{1}{\lambda} = R(1 - \dfrac{1}{4}) = \dfrac{3R}{4}$ — identical wavenumber, so both have $\lambda = 121.6$ nm. The Z² factor for He⁺ exactly cancels the smaller $(\dfrac{1}{4}-\dfrac{1}{16}) = \dfrac{3}{16}$ versus the H $(1-\dfrac{1}{4}) = \dfrac{3}{4}$. Answer: A ✓ — this is a known coincidence in JEE problems.

Question 2

Which of the following sets of quantum numbers is NOT possible?

For $n=2$, $l$ can only be 0 or 1 ($l$ ranges from 0 to $n-1 = 1$). $l=2$ with $n=2$ is not allowed. Answer: B ✓. Check others: A ($n=3, l=2$ valid since $l\le 2$), C ($n=4, l=0$ valid), D ($n=3, l=1$ valid, $m_l=+1$ valid).

Question 3

An electron jumps from the 4th to the 2nd orbit of hydrogen. The number of spectral lines that can be observed in this process is:

From n=4, electron can reach any lower level: 4→3, 4→2, 4→1, 3→2, 3→1, 2→1 — but we're told it falls from n=4 to n=2 overall. Intermediate transitions possible: 4→3→2, 4→2 directly, 4→3, 3→2. Lines between levels 1 and 4 (considering all possible intermediate transitions): n(n−1)/2 where n is the span. Levels involved: 2,3,4 → that's 3 levels → 3(3−1)/2 = 3 lines. Answer: B ✓. (4→2, 4→3, 3→2)

🔥

Chemical Thermodynamics

Laws of thermodynamics · Enthalpy · Entropy · Gibbs free energy · Spontaneity

Basic Concepts

System & Surroundings

System: part under study. Surroundings: everything else. Universe = system + surroundings.

Types: Open (exchange mass+energy), Closed (energy only), Isolated (neither).

State functions: H, S, G, U, T, P, V — depend only on current state, not path. q and w are path functions.

First Law

ΔU = q + w (IUPAC sign: w = −PₑₓₜΔV for expansion)

For ideal gas: ΔU = nCᵥΔT (depends only on T).

Enthalpy: H = U + PV; ΔH = ΔU + ΔngRT for reactions involving gases (Δng = moles of gaseous products − reactants).

Enthalpy Changes

Term	Symbol	Definition
Standard enthalpy of formation	ΔH°f	For 1 mol compound from elements in standard states; ΔH°f(elements) = 0
Combustion	ΔH°c	Complete combustion of 1 mol in excess O₂ (always negative)
Atomisation	ΔH°at	1 mol gaseous atoms from element in standard state (always +ve)
Bond dissociation	BE	Energy to break 1 mol bonds homolytically in gas phase (always +ve)
Lattice enthalpy	ΔH°L	1 mol ionic solid → gaseous ions (always +ve; Born-Haber cycle)
Solution	ΔH°soln	ΔH°soln = ΔH°L + ΔH°hydration

Hess's Law & Bond Enthalpy

Hess's Law

ΔH of a reaction is independent of the path — only depends on initial and final states.

ΔH°rxn = Σ ΔH°f(products) − Σ ΔH°f(reactants)

Flip a reaction → change sign of ΔH. Multiply by n → multiply ΔH by n.

Bond Enthalpy Method

ΔH°rxn = Σ BE(bonds broken) − Σ BE(bonds formed)

Bonds broken = reactants; bonds formed = products.

Breaking bonds requires energy (+); forming bonds releases energy (−). Used when ΔH°f data is unavailable.

Second Law — Entropy & Spontaneity

Entropy (S): measure of disorder/randomness. $\Delta S = \dfrac{q_{rev}}{T}$. For universe: $\Delta S_{univ} = \Delta S_{sys} + \Delta S_{surr} \ge 0$ (always for spontaneous processes).

Entropy trends

S(gas) >> S(liquid) > S(solid)
Dissolution usually increases S
More complex molecules have higher S
Higher temperature → higher S
Δng > 0 → ΔS > 0 for gas-phase reactions

Gibbs Free Energy

G = H − TS; ΔG = ΔH − TΔS (at constant T, P)

ΔG < 0: spontaneous; ΔG = 0: equilibrium; ΔG > 0: non-spontaneous.

ΔG° = −RT ln K = −nFE°cell

Standard free energy of formation: ΔG°rxn = Σ ΔG°f(products) − Σ ΔG°f(reactants)

ΔH	ΔS	ΔG = ΔH−TΔS	Spontaneity
−	+	Always −	Spontaneous at all T
+	−	Always +	Non-spontaneous at all T
−	−	− at low T	Spontaneous at low T only
+	+	− at high T	Spontaneous at high T only

💡Third Law: S = 0 for a perfect crystal at 0 K. Absolute entropy values are always positive. Cp − Cv = R (for ideal gas). Kirchhoff's law: ΔH changes with T via ΔCp.

📐

Thermodynamics — Key Formulas

First Law

ΔU = q + w | w = −PₑₓₜΔV | ΔH = ΔU + ΔngRT

Hess's Law

ΔH°rxn = Σ ΔH°f(products) − Σ ΔH°f(reactants)

Bond Enthalpy

ΔH°rxn = Σ BE(bonds broken) − Σ BE(bonds formed)

Gibbs Free Energy

ΔG = ΔH − TΔS | ΔG° = −RT ln K | ΔG° = −nFE°cell

Entropy Change

$$ \Delta S = \frac{q_{rev}}{T} \quad|\quad \Delta S_{univ} \ge 0 \text{ (spontaneous)} \quad|\quad \Delta S^\circ_{rxn} = \Sigma S^\circ(\text{products}) - \Sigma S^\circ(\text{reactants}) $$

Transition Temperature

At equilibrium $\Delta G = 0$: $T_{transition} = \Delta \dfrac{H}{\Delta} S$ (when $\Delta H$ and $\Delta S$ same sign)

🧮

Thermodynamics — Solved Examples

Calculate ΔH°rxn for CH₄(g)+2O₂(g)→CO₂(g)+2H₂O(l) using ΔH°f: CH₄=−74.8, CO₂=−393.5, H₂O(l)=−285.8 kJ/mol

ΔH°rxn = [ΔH°f(CO₂) + 2×ΔH°f(H₂O)] − [ΔH°f(CH₄) + 2×ΔH°f(O₂)]
$$ = [-393.5 + 2\times(-285.8)] - [-74.8 + 0] = [-393.5 - 571.6] - [-74.8] $$
= −965.1 + 74.8 = −890.3 kJ/mol

For a reaction ΔH = −100 kJ/mol and ΔS = −200 J/mol·K. At what temperature does spontaneity reverse?

ΔG = ΔH − TΔS. Reaction is spontaneous when ΔG < 0.
$$ \Delta G = -100\times10^{3} - T\times(-200) = -100000 + 200T $$
ΔG = 0 at T = 100000/200 = 500 K
For T < 500 K: ΔG < 0 (spontaneous). For T > 500 K: ΔG > 0 (non-spontaneous). This is the ΔH−,ΔS− case — spontaneous only at low T.

Calculate ΔH for C–H bond energy given: C(g)→C(s) ΔH=−718 kJ, H₂→2H ΔH=+436 kJ, CH₄ ΔH°f=−74.8 kJ

Target: C(s)+2H₂(g)→CH₄(g) ΔH=−74.8 kJ … use atomisation
C(s)→C(g): +718; 2H₂→4H: $+436\times2=+872$; C(g)+4H(g)→CH₄: ΔH=?
By Hess: −74.8 = 718 + 872 + ΔH(formation from atoms) → ΔH = −1664.8 kJ for 4 C–H bonds
Average C–H BE = 1664.8/4 = ~416 kJ/mol

For N₂(g)+3H₂(g)⇌2NH₃(g), ΔG°=−33 kJ at 298 K. Find K.

$$ \Delta G^{\circ} = -RT \ln K \to \ln K = -\Delta \frac{G^{\circ}}{RT} = -\frac{-33000}{8.314\times298} = \frac{33000}{2477.6} = 13.32 $$
$K = e^{13}.32 \approx$ 6×10⁵
Large K >> 1 confirms the reaction strongly favours products at 298 K.

📊

Thermodynamics — Key Diagrams

Click any diagram to zoom

ΔG vs Temperature — Spontaneity Regions

ΔG = ΔH − TΔS. Green line (ΔH−,ΔS+) always negative — spontaneous everywhere. Red line always positive. Orange and blue lines cross the x-axis at T* = ΔH/ΔS — the transition temperature.

Born-Haber Cycle for NaCl

Born-Haber cycle: ΔH°f(NaCl) = sublimation + ½ dissociation + IE₁ − EA + (−Lattice E). The lattice energy dominates and makes the overall process exothermic despite the endothermic steps.

⚡

Thermodynamics — Quick Quiz

Question 1

For a reaction ΔH = +50 kJ and ΔS = +150 J/K. The reaction is spontaneous:

ΔH+, ΔS+ → spontaneous at high T. T* = ΔH/ΔS = 50000/150 = 333 K. Above 333 K, TΔS > ΔH so ΔG < 0.

Question 2

The relationship between ΔH and ΔU for the reaction N₂(g)+3H₂(g)→2NH₃(g) is:

Δng = 2−(1+3) = −2. ΔH = ΔU + ΔngRT = ΔU + (−2)RT = ΔU − 2RT.

Question 3

Which process has ΔS < 0?

ΔS < 0 when disorder decreases. A: 2 mol gas → 1 mol gas (Δng=−1) → ΔS < 0. B, C, D all increase disorder (gas produced, dissolution, or vaporisation).

⚖

Chemical & Ionic Equilibrium

Kc, Kp, Le Chatelier · pH · Buffer · Ksp · Solubility

Chemical Equilibrium

At equilibrium: rate of forward = rate of reverse reaction. Concentrations are constant (not necessarily equal).

Equilibrium Constants

For aA+bB ⇌ cC+dD:

Kc = [C]ᶜ[D]^d / [A]ᵃ[B]^b (concentrations)

Kp $= (pC)ᶜ(pD)^d / (pA)ᵃ(pB)^b$ (partial pressures)

$Kp = Kc(RT)^\Delta ng$ where Δng = (c+d)−(a+b) for gases only

Pure solids and liquids excluded from K expression.

Reaction Quotient Q

Q has same form as K but uses non-equilibrium concentrations.

Q < K: reaction proceeds forward (→)

Q > K: reaction proceeds reverse (←)

Q = K: system is at equilibrium

Le Chatelier's Principle

System at equilibrium subjected to a stress shifts to partially relieve that stress.

Add reactant/remove product: shifts forward
Increase P: shifts to fewer moles of gas
Increase T: shifts in endothermic direction (K changes)
Catalyst: no shift in equilibrium (reaches equilibrium faster)
Inert gas at constant V: no shift

Relationship with ΔG°

ΔG° = −RT ln K

K > 1 → ΔG° < 0 (products favoured)

K < 1 → ΔG° > 0 (reactants favoured)

K = 1 → ΔG° = 0

Ionic Equilibrium — Acids & Bases

pH Scale & Water

$K_w = [\mathrm{H}^+][\mathrm{OH}^-] = 10^{-14}$ at 25°C; pH + pOH = 14

$\text{pH} = -\log[\mathrm{H}^+]$; $\text{pOH} = -\log[\mathrm{OH}^-]$

Acid: pH < 7; Base: pH > 7; Neutral: pH = 7 (at 25°C)

Weak Acid/Base

For weak acid HA ⇌ H⁺+A⁻: $K_a = [\mathrm{H}^+]\dfrac{\mathrm{A}^-}{\mathrm{HA}}$

$[\mathrm{H}^+] = \sqrt{K_a \cdot C}$ (if $\alpha \ll 1$)

$\text{pH} = \tfrac{1}{2}(\text{p}K_a - \log C)$

Degree of dissociation: $\alpha = \sqrt{K_a/C}$

$K_a\times K_b = K_w$ for conjugate pair

Buffer Solutions

Henderson-Hasselbalch: $\text{pH} = \text{p}K_a + \log(\dfrac{\mathrm{A}^-}{\mathrm{HA}})$

Maximum buffer capacity when $[\mathrm{A}^-]=[\mathrm{HA}]$, i.e. $\text{pH} = \text{p}K_a$.

Acidic buffer: weak acid + its salt. Basic buffer: weak base + its salt.

Solubility Product Ksp

For $A_m B_n$(s) ⇌ $m\mathrm{A}^{n+} + n\mathrm{B}^{m-}$:

$K_{sp} = [\mathrm{A}^{n+}]^m [\mathrm{B}^{m-}]^n$

Solubility s: $K_{sp} = m^m \cdot n^n \cdot s^{m+n}$

Common ion effect: adding common ion decreases solubility (shifts equilibrium left).

💡Salt hydrolysis: Salt of strong acid + weak base → acidic solution (pH < 7). Salt of weak acid + strong base → basic (pH > 7). Salt of weak acid + weak base: $\text{pH} = 7 + \tfrac{1}{2}(\text{p}K_a - \text{p}K_b)$. Salt of strong acid + strong base → neutral (pH = 7).

📐

Equilibrium — Key Formulas

Kc and Kp

$$ K_c = [C]^c[D]\frac{^d}{[A]^a}[B]^b \quad|\quad K_p = K_c(RT)^{\Delta n_g} $$

pH and pKa

$$ \text{pH} = -\log[\mathrm{H}^+] \quad|\quad \text{pOH} = -\log[\mathrm{OH}^-] \quad|\quad \text{pH+pOH} = 14\ (25°C) $$
$[\mathrm{H}^+] = \sqrt{K_a\cdot C}$ | $\text{pH} = \tfrac{1}{2}(\text{p}K_a - \log C)$ for weak acid

Henderson-Hasselbalch

$$ \text{pH} = \text{p}K_a + \log(\frac{\mathrm{A}^-}{\mathrm{HA}}) \quad|\quad \text{pOH} = \text{p}K_b + \log(\frac{\mathrm{BH}^+}{\mathrm{B}}) $$

Solubility Product

$$ K_{sp} = [\mathrm{A}^{n+}]^m[\mathrm{B}^{m-}]^n \quad|\quad \text{For AB: } K_{sp} = s^2 \quad|\quad \text{For AB}_2\text{: } K_{sp} = 4s^3 $$

Degree of Dissociation

$\alpha = \sqrt{K_a/C}$ for weak acid (valid when α << 1) | $K_a\times K_b = K_w = 10^{-14}$

🧮

Equilibrium — Solved Examples

For N₂+3H₂⇌2NH₃, Kc=0.5 at 400°C. Find Kp if RT=50 L·atm/mol.

Δng = 2−(1+3) = −2
$K_p = K_c(RT)^{\Delta n_g} = 0.5\times(50)^{-2} = \dfrac{0.5}{2500} =$ 2×10⁻⁴

Calculate pH of 0.1 M CH₃COOH ($K_a = 1.8\times 10^{-5}$).

$[\mathrm{H}^+] = \sqrt{K_a\times C} = \sqrt{1.8\times10^{-5}\times 0.1} = \sqrt{1.8\times10^{-6}} = 1.34\times10^{-3}$ M
$\text{pH} = -\log(1.34\times10^{-3}) = 3 - \log(1.34) \approx 3 - 0.127 =$ 2.87
Check $\alpha = 1.34\dfrac{\times10^{-3}}{0.1} = 0.0134$ (1.34% $\ll$ 100% → approximation valid ✓)

Find the solubility of BaSO₄ in 0.1 M Na₂SO₄. $K_{sp}$(BaSO₄) = 1×10⁻¹⁰.

BaSO₄ ⇌ Ba²⁺ + SO₄²⁻. Common ion: [SO₄²⁻] from Na₂SO₄ = 0.1 M.
$K_{sp} = [\mathrm{Ba}^{2+}][\mathrm{SO}_4^{2-}] = s\times(0.1+s) \approx s\times 0.1$ (since $s \ll 0.1$)
$s = \dfrac{K_{sp}}{0.1} = 1\dfrac{\times10^{-10}}{0.1} =$ 10⁻⁹ M
Compare pure water: $s = \sqrt{10^{-10}} = 10^{-5}$ M. Common ion reduces solubility 10000-fold.

A buffer is made with 0.2 M acetic acid and 0.3 M sodium acetate ($\text{p}K_a = 4.74$). Find pH and the ratio after adding 0.05 mol HCl per litre.

Initial pH: $\text{pH} = \text{p}K_a + \log(\dfrac{0.3}{0.2}) = 4.74 + \log(1.5) = 4.74 + 0.176 =$ 4.92
After HCl: HCl reacts with CH₃COO⁻: [A⁻] = 0.3 − 0.05 = 0.25 M; [HA] = 0.2 + 0.05 = 0.25 M
$\text{pH} = 4.74 + \log(\dfrac{0.25}{0.25}) = 4.74 + 0 =$ 4.74 (equals $\text{p}K_a$, maximum buffer capacity)

📊

Equilibrium — Key Diagrams

Click any diagram to zoom

Reaction Quotient Q vs K — Direction of Shift

Q < K means the system has fewer products than equilibrium — it shifts right. Q > K means too many products — shifts left. At Q = K, the system is at equilibrium.

pH of Common Solutions

pH scale 0–14 at 25°C. Acids have pH < 7 (H⁺ > OH⁻); bases pH > 7 (OH⁻ > H⁺). Neutral at pH = 7 only at 25°C ($K_w = 10^{-14}$); at higher T, neutral pH drops below 7.

⚡

Equilibrium — Quick Quiz

Question 1

For PCl₅⇌PCl₃+Cl₂, if pressure is increased at constant T, equilibrium shifts:

Increased pressure shifts towards fewer moles of gas. Products have 1+1=2 moles; reactant has 1 mole. Shift backward (to fewer gas moles = PCl₅ side).

Question 2

What is the pH of a buffer made with equal moles of NH₃ and NH₄Cl? (pKb of NH₃ = 4.74)

pKa of NH₄⁺ = 14 − pKb = 14 − 4.74 = 9.26. Using Henderson-Hasselbalch for basic buffer: pOH = pKb + log([NH₄⁺]/[NH₃]) = 4.74 + log(1) = 4.74. pH = 14 − 4.74 = 9.26.

Question 3

Adding a catalyst to a system at equilibrium:

A catalyst lowers activation energy equally for both forward and reverse reactions — the system was already at equilibrium. K is unchanged, equilibrium position unchanged. The catalyst only helps reach equilibrium faster if the system is not yet there.

🔌

Electrochemistry

EMF · Nernst equation · Electrolysis · Faraday's laws · Conductance

Electrochemical Cells

Galvanic vs Electrolytic Cell

Galvanic: spontaneous redox → electrical energy. Anode (oxidation, −ve), Cathode (reduction, +ve). E°cell = E°cathode − E°anode.

Electrolytic: electrical energy → non-spontaneous redox. External EMF applied. Anode = +ve, Cathode = −ve.

Standard Electrode Potential

Measured against SHE (Standard Hydrogen Electrode, E° = 0 V by convention).

More positive E° → stronger oxidising agent (reduced at cathode). More negative E° → stronger reducing agent.

E°cell = E°cathode(reduction) − E°anode(reduction)

Spontaneous cell: E°cell > 0 (ΔG° = −nFE°cell < 0).

Nernst Equation

E = E° − (RT/nF)ln Q = E° − (0.0592/n)log Q (at 25°C)

At Equilibrium

E = 0 when Q = K (cell has fully discharged).

$$ ∴ E^{\circ} = (\frac{0.0592}{n})\log K = \frac{RT}{nF} \times \ln K $$

Combined: ΔG° = −nFE° = −RT ln K

Concentration Cell

Both electrodes same metal but different concentrations: E° = 0.

E = (0.0592/n)log([higher]/[lower])

Reaction proceeds until concentrations equalise.

The Electrochemical Series

Arranging electrodes by their standard reduction potential (E°) gives the electrochemical series. A more positive E° means a stronger oxidising agent (readily reduced); a more negative E° means a stronger reducing agent (readily oxidised).

Top (most negative E°, e.g. Li, K, Ca, Na): strong reducing agents; most reactive metals; displace metals below them.
Bottom (most positive E°, e.g. Ag, Pt, Au, F₂): strong oxidising agents; least reactive.
A metal higher in the series displaces a metal lower down from its salt solution (e.g. Zn displaces Cu²⁺). Metals above hydrogen liberate H₂ from dilute acids.
Cell EMF $= E^{\circ}_{cathode} - E^{\circ}_{anode}$; a positive EMF means the reaction is spontaneous.

Faraday's Laws of Electrolysis

First Law

Mass deposited ∝ charge passed: m = ZQ = ZIt

Z = electrochemical equivalent (g/C) = M/nF where M = molar mass, n = electrons.

Second Law

Same charge deposits masses in ratio of their equivalent weights (M/n).

m = (M/nF)×Q | 1 Faraday = 96500 C ≈ 1 mol electrons.

Conductance

Term	Formula	Units
Resistance $R$	$R = \rho\cdot \dfrac{l}{A}$	Ω
Conductance $G$	$G = \dfrac{1}{R}$	S (siemens)
Specific conductance $\kappa$	$\kappa = \dfrac{1}{\rho} = G\cdot(\dfrac{l}{A})$	S·m⁻¹
Molar conductance $\Lambda_m$	$\Lambda_m = \dfrac{\kappa}{C}$ (C in mol/m³)	S·m²·mol⁻¹
Kohlrausch's law	$\Lambda^\circ_m = \Sigma\, \lambda^\circ(\text{ions})$ (at infinite dilution)	additive

💡Strong electrolyte: Λm increases with dilution (approaches Λ°m). Weak electrolyte: Λm increases sharply at high dilution. Kohlrausch's law used to find Λ°m for weak electrolytes indirectly. Degree of dissociation $\alpha = \Lambda \dfrac{m}{\Lambda^{\circ}}m$.

📐

Electrochemistry — Key Formulas

Cell EMF

E°cell = E°cathode − E°anode | ΔG° = −nFE°cell | ΔG° = −RT ln K

Nernst Equation (25°C)

$$ E = E^\circ - (\frac{0.0592}{n})\log Q \quad|\quad E^\circ = (\frac{0.0592}{n})\log K $$

Faraday's Laws

$$ m = ZIt = (\frac{M}{nF})\cdot It \quad|\quad 1\text{ F} = 96500\text{ C} = 1\text{ mol e}^- $$

Molar Conductance

$$ \Lambda_m = \kappa\times \frac{1000}{C(\text{M})} \quad|\quad \Lambda^\circ_m = \Sigma\, \lambda^\circ(\text{cation}) + \Sigma\, \lambda^\circ(\text{anion}) \quad|\quad \alpha = \frac{\Lambda_m}{\Lambda^\circ_m} $$

🧮

Electrochemistry — Solved Examples

Calculate $E^\circ_{cell}$ for Zn|Zn²⁺||Cu²⁺|Cu. $E^\circ$(Zn²⁺/Zn) = −0.76 V, $E^\circ$(Cu²⁺/Cu) = +0.34 V.

Zn is oxidised (anode), Cu²⁺ is reduced (cathode). $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.34 - (-0.76) =$ +1.10 V
$E^\circ_{cell} > 0 \to \Delta G^\circ < 0 \to$ spontaneous ✓

How many grams of copper deposit when 2 A flows for 30 min through CuSO₄ solution? (M(Cu)=63.5, n=2)

Charge Q = I×t = 2×(30×60) = 3600 C
$m = \frac{M}{nF}\times Q = (\dfrac{63.5}{2}\times96500)\times3600 = \dfrac{63.5\times3600}{193000} = \dfrac{228600}{193000} \approx$ 1.18 g

For Daniel cell at 25°C, [Zn²⁺]=0.1 M, [Cu²⁺]=0.01 M. $E^\circ_{cell}$=1.10 V. Find E.

Zn + Cu²⁺ → Zn²⁺ + Cu. $n=2$. $Q = \dfrac{\mathrm{Zn}^{2+}}{\mathrm{Cu}^{2+}} = \dfrac{0.1}{0.01} = 10$
$E = E^{\circ} - (\dfrac{0.0592}{n})\log Q = 1.10 - (\dfrac{0.0592}{2})\log 10 = 1.10 - 0.0296\times1 =$ 1.0704 V

📊

Electrochemistry — Key Diagrams

Click any diagram to zoom

Daniel Cell — Galvanic Cell Schematic

Daniel cell: Zn oxidises at the anode (−), electrons flow through external circuit to Cu cathode (+) where Cu²⁺ is reduced. Salt bridge maintains charge balance by allowing ion migration between half-cells.

⚡

Electrochemistry — Quick Quiz

Question 1

In a galvanic cell, oxidation occurs at the:

In a galvanic cell, anode = oxidation (metal dissolves, releases electrons). Cathode = reduction (ions gain electrons, deposit). Anode is negative (source of electrons). Mnemonic: AN OX (ANode OXidation) and RED CAT (REDuction CATthode).

Question 2

96500 C of charge is passed through AgNO₃ solution (M(Ag)=108, n=1). Mass of Ag deposited:

$m = \frac{M}{nF}\times Q = (\dfrac{108}{1}\times96500)\times96500 = \dfrac{108}{1} \times 1 =$ 108 g. 96500 C = 1 Faraday = 1 mol of electrons = 1 mol Ag (n=1) = 108 g. This is the definition of Faraday.

Question 3

If E°cell for a reaction is positive, which statement is correct?

ΔG° = −nFE°cell. If E°cell > 0, then ΔG° < 0 (spontaneous). Since ΔG° = −RT ln K, ΔG° < 0 means ln K > 0, so K > 1 (products favoured at equilibrium).

⏱

Chemical Kinetics

Rate laws · Order · Half-life · Arrhenius equation · Collision theory

Rate of Reaction

For aA+bB→cC+dD: Rate = −(1/a)d[A]/dt = −(1/b)d[B]/dt = +(1/c)d[C]/dt = +(1/d)d[D]/dt

Rate Law & Order

Rate = k[A]^m[B]^n (determined experimentally, not from stoichiometry)

Order = m+n (overall); m = order w.r.t A; n w.r.t B.

Units of k: $(\mathrm{mol/L})^{1-n}\cdot s^{-1}$ where n = overall order.

Pseudo-order Reactions

One reactant in large excess → its concentration is effectively constant.

Hydrolysis of ester in dilute acid: Rate = k'[ester] — pseudo first-order (water in excess).

k' = k[H₂O] — pseudo rate constant.

Integrated Rate Laws

Order	Rate law	Integrated form	Half-life t½	Units of k
Zero	$\text{Rate} = k$	$[A] = [A]_0 - kt$	$t_{1/2} = [A]\dfrac{_0}{2k}$	mol·L⁻¹·s⁻¹
First	$\text{Rate} = k[A]$	$\ln[A] = \ln[A]_0 - kt$ or $[A]=[A]_0 e^{-kt}$	$t_{1/2} = \ln 2/k = \dfrac{0.693}{k}$	s⁻¹
Second	$\text{Rate} = k[A]^2$	$\dfrac{1}{A} = \dfrac{1}{[A]_0} + kt$	$t_{1/2} = \dfrac{1}{k[A]_0}$	L·mol⁻¹·s⁻¹

💡Key property: First-order half-life is constant (independent of [A]₀). Zero-order t½ depends on [A]₀. Second-order t½ also depends on [A]₀. Radioactive decay is always first-order.

Arrhenius Equation & Activation Energy

Arrhenius Equation

$k = A\cdot e^{-E_a/RT}$

ln k = ln A − Ea/RT

$$ \ln(\frac{k_{2}}{k_{1}}) = \frac{Ea}{R}(\frac{1}{T_{1}} - \frac{1}{T_{2}}) $$

A = frequency factor (pre-exponential); Ea = activation energy (J/mol)

Effect of Temperature

Rule of thumb: rate doubles for every 10°C rise (temperature coefficient $\mu \approx 2$).

Higher $E_a$ → rate more sensitive to temperature.

Catalyst lowers $E_a$ → increases $k$ (doesn't change $\Delta H$ or equilibrium).

Collision Theory

Rate = Z_AB·f·p where Z_AB = collision frequency, $f = e^{-E_a/RT}$ (fraction with enough energy), p = steric factor.

Not all collisions are effective — need correct orientation AND energy ≥ Ea.

Mechanisms & Rate

Rate-determining step (RDS) = slowest step → determines overall rate law.

Intermediates cancel out; only species present before RDS appear in rate law.

Molecularity = number of species in elementary step (1, 2, or 3). Order = from experiment.

Collision theory in 3D: orientation and energy — In CH₃Br + OH⁻ → CH₃OH + Br⁻ a collision reacts only if OH⁻ meets the carbon from the side opposite Br (proper orientation) and brings at least the activation energy E_a along the line of centres. Hitting the Br end, or arriving too gently, OH⁻ just bounces off. In a reacting collision the O···C and C···Br bonds are half-made at the halfway point, with the three H atoms flat; then Br⁻ leaves and the CH₃ group is turned inside out. Molecules tumble, so collisions come from every direction in 3D. In the model drawn, only those within 60° of the side opposite Br are well aimed: that cone is (1 − cos 60°)/2 = 0.25 of all directions (not a third, as a flat drawing suggests), so the steric factor p = 0.25 < 1, and 12 of the 48 collisions drawn are in the cone. Only a fraction $e^{-E_a/RT}$ of collisions have enough energy: 0.22 at the drawing’s E_a = 1.5RT, giving 3 reactions out of 48 (2.7 expected). So rate = $Z\,p\,e^{-E_a/RT}$, where Z is the collision frequency. For a real barrier of 50 kJ/mol at 298 K, $e^{-E_a/RT} = 1.7 \times 10^{-9}$. The cone is a model; real values of p are measured.

📐

Chemical Kinetics — Key Formulas

First-Order Integrated Rate Law

ln[A] = ln[A]₀ − kt | t½ = 0.693/k | $[A] = [A]_0\cdot e^{-kt}$

Zero & Second Order

Zero: [A]=[A]₀−kt; t½=[A]₀/2k
Second: 1/[A]=1/[A]₀+kt; t½=1/(k[A]₀)

Arrhenius Equation

$k = A\cdot e^{-E_a/RT}$ | $\ln(\dfrac{k_{2}}{k_{1}}) = \frac{E_a}{R}(\dfrac{1}{T_{1}} - \dfrac{1}{T_{2}})$

Units of Rate Constant k

k units = $(\mathrm{mol/L})^{1-n}\cdot s^{-1}$ where n = order
n=0: mol·L⁻¹·s⁻¹ | n=1: s⁻¹ | n=2: L·mol⁻¹·s⁻¹

🧮

Chemical Kinetics — Solved Examples

A first-order reaction has $k = 0.0693$ min⁻¹. Find $t_{1/2}$ and time for 75% completion.

$t_{1/2} = \dfrac{0.693}{k} = \dfrac{0.693}{0.0693} =$ 10 min
75% complete → [A] = 25% of [A]₀ = [A]₀/4 — reached after 2 half-lives
$t = 2\times t_{1/2} =$ 20 min

For a reaction, rate = $k[A]^2[B]$. [A]=0.1 M, [B]=0.2 M, rate = 2×10⁻⁶ mol/L/s. Find k.

$k = \dfrac{\text{rate}}{[A]^2[B]} = 2\dfrac{\times10^{-6}}{(0.1)^2\times 0.2} = 2\dfrac{\times10^{-6}}{0.01\times 0.2} = 2\dfrac{\times10^{-6}}{2\times10^{-3}} =$ $10^{-3}$ L²·mol⁻²·s⁻¹
Order = 2+1 = 3 (third order). Units check: $(\mathrm{mol/L})^{1-3}\cdot s^{-1} = L^2\cdot mol^{-2}\cdot s^{-1}$ ✓

Ea for a reaction is 50 kJ/mol. By what factor does k increase going from 300 K to 310 K?

$$ \ln(\frac{k_{2}}{k_{1}}) = \frac{Ea}{R}(\frac{1}{T_{1}}-\frac{1}{T_{2}}) = (\frac{50000}{8.314})(\frac{1}{300}-\frac{1}{310}) $$
$$ = 6013\frac{\times(310-300)}{300\times310} = 6013\frac{\times10}{93000} = \frac{60130}{93000} = 0.6466 $$
$\dfrac{k_{2}}{k_{1}} = e^0.6466 \approx$ 1.91 (~double, consistent with the rule of thumb)

📊

Chemical Kinetics — Key Diagrams

Click any diagram to zoom

Energy Profile — Activation Energy & Effect of Catalyst

A catalyst provides an alternate pathway with lower Ea. Both pathways start and end at the same energy levels — ΔH and K are unchanged. The lower Ea means more molecules have sufficient energy → faster rate (larger k).

⚡

Chemical Kinetics — Quick Quiz

Question 1

A first-order reaction is 50% complete in 20 min. In how many minutes will it be 87.5% complete?

$t_{1/2} = 20$ min. 87.5% complete → 12.5% remains = $(\dfrac{1}{2})^3$ of original → 3 half-lives. $t = 3\times 20 =$ 60 min.

Question 2

The unit of rate constant for a second-order reaction is:

k units = $(\mathrm{mol/L})^{1-n}\cdot s^{-1}$. For n=2: $(\mathrm{mol/L})^{-1}\cdot s^{-1}$ = L·mol⁻¹·s⁻¹.

Question 3

Which graph is LINEAR for a first-order reaction?

First-order: ln[A] = ln[A]₀ − kt. This is y = mx + c form, so ln[A] vs t is linear with slope = −k. [A] vs t is exponential (curved). 1/[A] vs t is linear for second-order.

💧

Solutions

Concentration · Raoult's law · Colligative properties · Henry's law · Abnormal molar masses

Concentration Terms

Term	Formula	Units
Molarity (M)	moles of solute / volume of solution (L)	mol L⁻¹
Molality (m)	moles of solute / mass of solvent (kg)	mol kg⁻¹
Mole fraction ($\chi$)	$n_A / (n_A + n_B)$; $\chi_A + \chi_B = 1$	dimensionless
Mass fraction (w)	mass of solute / total mass	dimensionless
ppm	mass of solute / total mass $\times 10^6$	mg kg⁻¹

💡Molality is temperature-independent (uses mass of solvent); molarity changes with temperature (uses volume of solution). JEE often asks to interconvert — key relation: $M = (1000 \times m \times \rho) / (1000 + m \times M_2)$ where $\rho$ = density (g/mL), $M_2$ = molar mass of solute.

Raoult's Law and Ideal Solutions

Raoult's Law

For a volatile solute in a volatile solvent:

$p_A = \chi_A \cdot p_A^\circ$ (partial pressure = mole fraction × vapour pressure of pure component)

Total pressure: $P = p_A + p_B = \chi_A p_A^\circ + \chi_B p_B^\circ$

For a non-volatile solute: $p_{solution} = \chi_{solvent} \cdot p^\circ_{solvent}$

Relative Lowering of Vapour Pressure (RLVP)

$(p^\circ - p) / p^\circ = \chi_{solute} = n_2 / (n_1 + n_2)$

For dilute solutions: $\approx \dfrac{n_{2}}{n_{1}} = w_{2}M_{1} / (w_{1}M_{2})$

RLVP is a colligative property — depends only on number of solute particles, not their nature.

Ideal vs Non-Ideal Solutions

Ideal: $\Delta H_{mix} = 0$, $\Delta V_{mix} = 0$; obeys Raoult's law at all concentrations. A-B interactions = A-A = B-B.

Positive deviation: A-B < A-A/B-B; p > Raoult prediction. Examples: ethanol-water, acetone-CS₂.

Negative deviation: A-B > A-A/B-B; p < Raoult prediction. Examples: HCl-water, acetone-chloroform.

Henry's Law (Gases in Liquid)

$p = K_H \cdot \chi$ where $K_H$ = Henry's law constant (large $K_H$ → low solubility).

Solubility of gas ↑ with ↑ pressure and ↓ temperature.

Applications: carbonated drinks (CO₂ under pressure), scuba diving (N₂ narcosis), oxygen in blood.

Colligative Properties

Depend only on the number of solute particles in solution, not their identity.

Property	Formula	Key point
Relative lowering of VP	$\dfrac{p^{\circ}-p}{p^{\circ}} = \chi_{2}$	χ₂ = mole fraction of solute
Elevation of boiling point	$\Delta T_b = K_b \cdot m$	$K_b$ = ebullioscopic constant (solvent property)
Depression of freezing point	$\Delta T_f = K_f \cdot m$	$K_f$ = cryoscopic constant; larger than $K_b$
Osmotic pressure	$\pi = CRT = \frac{n}{V}RT$	Van't Hoff equation; C in mol/L

💡Water: $K_f = 1.86$ K·kg·mol⁻¹, $K_b = 0.52$ K·kg·mol⁻¹. $K_f > K_b$ always (freezing point depression is more sensitive). Osmotic pressure is most sensitive — used for high molar mass polymers and proteins.

Abnormal Molar Masses and Van't Hoff Factor

i = observed colligative property / expected colligative property = actual moles / formula moles

Case	i value	Example
No association/dissociation	i = 1	Glucose, urea in water
Dissociation (electrolytes)	i > 1	NaCl → Na⁺ + Cl⁻ → i = 2 (ideal); KCl, MgCl₂
Association	i < 1	Acetic acid in benzene: 2CH₃COOH ⇌ (CH₃COOH)₂ → i = 0.5

Modified formula: ΔT_f = i · K_f · m; $\pi = i \cdot CRT$

For α degree of dissociation into n ions: $i = 1 + \alpha(n - 1)$

🔢

Solutions — Key Formulas

Raoult's Law

$$ p_A = \chi_A \cdot p_A^\circ \quad|\quad P_{total} = \chi_A p_A^\circ + \chi_B p_B^\circ $$
$\dfrac{p^\circ - p}{p^\circ} = \chi_{solute}$ (RLVP for non-volatile solute)

Colligative Properties

$$ \Delta T_b = K_b \cdot m \quad|\quad \Delta T_f = K_f \cdot m $$
$\pi = CRT$ (C in mol/L, R = 0.0821 L·atm/mol·K)
Water: $K_f = 1.86$, $K_b = 0.52$ K·kg·mol⁻¹

Van't Hoff Factor

$i$ = actual moles / formula moles
$$ \Delta T_f = i\cdot K_f\cdot m \quad|\quad \pi = i\cdot CRT $$
$i = 1 + \alpha(n - 1)$ ($\alpha$ = degree of dissociation, $n$ = ions formed)

Henry's Law

$p = K_H \cdot \chi$ ($K_H$ = Henry's constant; larger $K_H$ → less soluble gas)
Solubility of gas ↑ with ↑ pressure, ↓ temperature

Molarity ↔ Molality Conversion

$$ M = (1000\cdot m\cdot\rho) / (1000 + m\cdot M_{2}) $$
$$ m = (1000\cdot M) / (1000\rho - M\cdot M_{2}) $$
ρ = density (g/mL), M₂ = molar mass of solute (g/mol)

🧮

Solutions — Solved Examples

Find the boiling point of a solution of 18 g glucose (M=180) in 100 g water. K_b(water) = 0.52 K kg mol⁻¹

Moles of glucose = 18/180 = 0.1 mol. Mass of solvent = 0.1 kg.
m = 0.1/0.1 = 1 mol kg⁻¹
$$ \Delta T_b = K_b \times m = 0.52 \times 1 = 0.52 K $$
Boiling point = 100 + 0.52 = 100.52°C

The vapour pressure of pure water at 25°C is 23.8 mmHg. What is the VP of a solution of 36 g glucose in 180 g water?

n(glucose) = 36/180 = 0.2 mol; n(water) = 180/18 = 10 mol
χ(water) = 10/(10+0.2) = 10/10.2 = 0.9804
p = χ(water) × p° = 0.9804 × 23.8 = 23.33 mmHg
RLVP = (23.8−23.33)/23.8 = 0.0196 ≈ χ(glucose) = 0.2/10.2 = 0.0196 ✓

A 0.5 molal aqueous NaCl solution has ΔT_f = 1.72°C. Find the Van't Hoff factor. K_f = 1.86.

Expected ΔT_f (if i=1) = 1.86 × 0.5 = 0.93°C
i = observed ΔT_f / expected ΔT_f = 1.72 / 0.93 = 1.85
For NaCl → Na⁺ + Cl⁻: ideal i = 2. Observed i = 1.85 → not fully dissociated due to ion-pairing.
Using i = 1 + α(n−1): 1.85 = 1 + α(2−1) → α = 0.85 (85% dissociation)

An aqueous solution of a polymer has osmotic pressure 1.8×10⁻³ atm at 27°C. Find the molar mass of the polymer if 2 g is dissolved in 500 mL of solution.

$$ \pi = CRT \to C = \frac{\pi}{RT} = (1.8\times10^{-3}) / (0.0821 \times 300) = 1.8\times10^{-3} / 24.63 = 7.31\times10^{-5} mol/L $$
Moles in 500 mL = 7.31×10⁻⁵ × 0.5 = 3.655×10⁻⁵ mol
M = mass/moles = 2 / 3.655×10⁻⁵ = 54,720 g/mol ≈ 54.7 kg/mol
This illustrates why osmotic pressure (not ΔT_b or ΔT_f) is used for high-molar-mass substances — even tiny concentrations give measurable π.

📊

Solutions — Key Diagrams

Click any diagram to zoom

Vapour Pressure vs Composition — Ideal, Positive and Negative Deviation

Left: ideal solution — partial pressures are linear in mole fraction (Raoult's law); total pressure is constant. Centre: positive deviation — A-B interactions weaker than A-A/B-B; total pressure exceeds Raoult prediction; minimum boiling azeotrope (e.g. ethanol-water 95.6%). Right: negative deviation — A-B stronger; pressure below Raoult; maximum boiling azeotrope (e.g. HCl-water, HNO₃-water).

Freezing Point Depression and Boiling Point Elevation

The solution (red) has a lower vapour pressure curve than the pure solvent (blue). It crosses atmospheric pressure at a higher temperature (boiling point elevation ΔT_b) and crosses the solid-phase line at a lower temperature (freezing point depression ΔT_f). Both are proportional to molality via K_b and K_f respectively.

⚡

Solutions — Quick Quiz

Question 1

Which colligative property is most suitable for determining the molar mass of a large polymer?

For high molar mass solutes, concentration is very low. Osmotic pressure (π=CRT) gives measurable values even at very low concentrations, while ΔT_b and ΔT_f are too small to measure accurately. Osmotic pressure is used for polymers, proteins, and colloids.

Question 2

The van't Hoff factor for a 0.1 m aqueous solution of K₂SO₄ that is 90% dissociated is:

K₂SO₄ → 2K⁺ + SO₄²⁻ gives n=3 ions. α=0.90. i = 1 + α(n−1) = 1 + 0.90×(3−1) = 1 + 1.8 = 2.8.

Question 3

Ethanol-water mixture shows positive deviation from Raoult's law because:

Positive deviation: A-B interactions < A-A and B-B interactions → molecules escape more easily → vapour pressure exceeds Raoult prediction → ΔH_mix > 0, ΔV_mix > 0. Ethanol-water forms a minimum boiling azeotrope at 95.6% ethanol, not maximum. Answer: weaker A-B interactions.

💨

States of Matter

Gas laws · Ideal gas · Kinetic theory · Real gases · van der Waals

Gas Laws

Law	Relation	Constant
Boyle's Law	PV = constant → P₁V₁ = P₂V₂	T, n fixed
Charles's Law	V/T = constant → V₁/T₁ = V₂/T₂	P, n fixed
Gay-Lussac's Law	P/T = constant → P₁/T₁ = P₂/T₂	V, n fixed
Avogadro's Law	V ∝ n at constant T, P	T, P fixed
Combined Gas Law	PV/T = constant → $\dfrac{P_{1}V_{1}}{T_{1}} = \dfrac{P_{2}V_{2}}{T_{2}}$	n fixed

💡Always use temperature in Kelvin (T = °C + 273.15). At STP: T=273 K, P=1 atm, 1 mol of ideal gas occupies 22.4 L. At NTP (new): T=293 K, P=1 bar, 1 mol = 24.04 L.

Ideal Gas Equation

PV = nRT

R = 8.314 J mol⁻¹ K⁻¹ = 0.0821 L·atm mol⁻¹ K⁻¹ = 2 cal mol⁻¹ K⁻¹

Molar mass from ideal gas: $M = \dfrac{mRT}{PV} = \rho \dfrac{RT}{P}$ (where ρ = density, m = mass)

Dalton's Law of Partial Pressures: P_total = p₁ + p₂ + p₃ + … ; pᵢ = χᵢ · P_total

Kinetic Molecular Theory

Assumptions

Gas molecules are point masses (negligible volume)
No intermolecular forces between molecules
Collisions are perfectly elastic
Average KE ∝ absolute temperature: KE = (3/2)RT per mole

Molecular Speeds

u_rms = $\sqrt{\dfrac{3RT}{M}}$ (root mean square)

ū = $\sqrt{\dfrac{8RT}{\pi M}}$ (mean/average)

u_mp = $\sqrt{\dfrac{2RT}{M}}$ (most probable)

Ratio: $u_{mp} : \bar{u} : u_{rms} = \sqrt2 : \sqrt{\dfrac{8}{\pi}} : \sqrt3 = 1 : 1.128 : 1.225$

All $\propto \sqrt T$ and $\propto \dfrac{1}{\sqrt M}$

Graham's Law of Diffusion

Rate of diffusion $\propto \dfrac{1}{\sqrt M}$ (at same T and P)

$$ \frac{r_{1}}{r_{2}} = \sqrt{\frac{M_{2}}{M_{1}}} = \sqrt{\frac{d_{2}}{d_{1}}} $$

Used to compare diffusion rates of gases and to identify unknown gases by rate comparison.

Mean Free Path (λ)

$$ \lambda = RT / (\sqrt2 \cdot \pi d^{2} \cdot N_{a} \cdot P) $$

λ ↑ as P ↓ (fewer collisions) and T ↑

λ ↓ as molecular size (d) ↑

Collision frequency $Z \propto \dfrac{P^2}{\sqrt T}$

Kinetic theory in 3D: Graham’s law and the mean free path — H₂ (M = 2) and O₂ (M = 32) at the same temperature have the same average kinetic energy, (3/2)RT = 3.72 kJ/mol at 298 K, so the lighter molecule moves faster: u_rms = 1.93 km/s for H₂ and 0.48 km/s for O₂, in the ratio $\sqrt{32/2} = 4$. With equal numbers of each in one box (the same partial pressure), molecules find a pinhole in proportion to their speed, so H₂ effuses four times as fast as O₂: rate $\propto \dfrac{1}{\sqrt{M}}$ (Graham’s law). In the drawn run 14 H₂ and 4 O₂ get through to the pumped side; a long run of the same model, 20,000 molecules of each, gives a ratio of 3.97 (with molar masses 2.016 and 31.998 the law gives 3.98). Real molecules have a size, so they collide: one molecule followed through the gas travels in straight runs between collisions, and the average run is the mean free path $\lambda = \dfrac{kT}{\sqrt{2}\,\pi d^{2} P}$. The $\sqrt{2}$ is there because the other molecules move too. For N₂ at 298 K and 1 atm (d = 0.37 nm) λ = 67 nm, about 180 molecular diameters; λ grows as T rises and as P falls. Numbers of molecules, sizes and speeds in the drawing are not to scale.

Real Gases and Van der Waals Equation

Ideal gas deviates at high P and low T. Two corrections:

$$ (P + \frac{an^{2}}{V^{2}})(V - nb) = nRT $$

Constant	Corrects for	Effect
a (pressure correction)	Intermolecular attractions	Observed P < ideal; add a/V² to correct
b (volume correction)	Finite volume of molecules	Observed V > free space; subtract nb

Compressibility factor Z = PV/nRT

Z value	Behaviour	Dominant effect
Z = 1	Ideal gas	—
Z < 1 (low P)	More compressible than ideal	Intermolecular attractions (a dominates)
Z > 1 (high P)	Less compressible than ideal	Finite volume of molecules (b dominates)

Boyle temperature T_B = a/(Rb) — temperature at which a real gas behaves ideally over a wide pressure range (Z ≈ 1).

⚠️H₂ and He have very weak attractions (a ≈ 0) so Z > 1 even at low pressures — b dominates throughout. All other gases show Z < 1 at low P before rising above 1 at high P.

Real gases in 3D: what a and b correct — 27 molecules of CO₂ at 1 mol per litre, drawn to scale: each 0.32 nm across, on average 1.18 nm apart. An ideal gas has point molecules and no forces, so 1 mol in 1 L at 300 K would exert P = nRT/V = 24.6 atm. b corrects for size: no other molecule’s centre can come within one diameter of a molecule, so the free volume is V − nb, where nb = 0.043 L is four times the molecules’ own volume; on its own this raises P to 25.7 atm. a corrects for attractions: inside the gas the pulls cancel on average, but a molecule about to hit the wall has neighbours on one side only, so it is pulled back, slows (its dots, equal times apart, bunch up) and hits the wall more softly; on its own this lowers P by an²/V² = 3.59 atm, to 21.0 atm. Together, $(P + \dfrac{an^{2}}{V^{2}})(V - nb) = nRT$ gives P = 22.1 atm and Z = PV/nRT = 0.90: at this density attractions win and Z < 1. For CO₂, a = 3.592 L² atm mol⁻² and b = 0.04267 L mol⁻¹ (CRC Handbook). Speeds are not to scale.

🔢

States of Matter — Key Formulas

Ideal Gas Equation & Dalton's Law

PV = nRT | R = 0.0821 L·atm mol⁻¹K⁻¹ = 8.314 J mol⁻¹K⁻¹
M = ρRT/P | P_total = Σpᵢ | pᵢ = χᵢ·P_total

Molecular Speeds

$u_{rms} = \sqrt{\dfrac{3RT}{M}}$ | $\bar u = \sqrt{\dfrac{8RT}{\pi M}}$ | $u_{mp} = \sqrt{\dfrac{2RT}{M}}$
Ratio u_mp : ū : u_rms = 1 : 1.128 : 1.225

Graham's Law of Diffusion

$\dfrac{r_1}{r_2} = \sqrt{\dfrac{M_2}{M_1}} = \sqrt{\dfrac{d_2}{d_1}}$ (same T, P)

Van der Waals Equation

(P + an²/V²)(V − nb) = nRT
a: intermolecular attractions | b: finite molecular volume
Boyle temperature T_B = a/(Rb)

Compressibility Factor

Z = PV/nRT | Z=1 ideal; Z<1 attractions; Z>1 repulsions
H₂ and He: Z > 1 at all pressures (a ≈ 0)

🧮

States of Matter — Solved Examples

A gas occupies 5 L at 27°C and 1 atm. What volume does it occupy at 127°C and 2 atm?

P₁V₁/T₁ = P₂V₂/T₂; $T_{1} = 300 K, T_{2} = 400 K$
$V_{2} = P_{1}V_{1}T_{2} / (T_{1}P_{2}) = (1 \times 5 \times 400) / (300 \times 2) = \dfrac{2000}{600} =$ 10/3 ≈ 3.33 L

Calculate the rms speed of N₂ molecules at 27°C. (M = 28 g/mol)

T = 300 K, M = 0.028 kg/mol, R = 8.314 J/mol·K
$$ u_{rms} = \sqrt{\frac{3RT}{M}} = \sqrt{3 \times 8.314 \times 300 / 0.028} $$
$= \sqrt{7482.6 / 0.028} = \sqrt{267,236} =$ 517 m/s

A mixture contains 4 g H₂ and 32 g O₂ at total pressure 3 atm. Find partial pressures.

n(H₂) = 4/2 = 2 mol; n(O₂) = 32/32 = 1 mol; n_total = 3 mol
χ(H₂) = 2/3; χ(O₂) = 1/3
p(H₂) = (2/3) × 3 = 2 atm; p(O₂) = (1/3) × 3 = 1 atm

At what temperature will the rms speed of SO₂ equal the rms speed of O₂ at 25°C?

$u_{rms} \propto \sqrt{\dfrac{T}{M}}$. Set equal: $\sqrt{\dfrac{T_{SO_2}}{64}} = \sqrt{\dfrac{298}{32}}$
T_SO₂/64 = 298/32
T_SO₂ = 298 × 64/32 = 298 × 2 = 596 K = 323°C
SO₂ (M=64) needs twice the temperature of O₂ (M=32) for the same rms speed, since M_SO₂ = 2 × M_O₂.

📊

States of Matter — Key Diagrams

Click any diagram to zoom

Maxwell-Boltzmann Speed Distribution

Maxwell-Boltzmann speed distribution for a gas at two temperatures. At higher T (red): the curve is broader, flatter, and the peak shifts to higher speed. The three characteristic speeds always satisfy u_mp < ū < u_rms. All three scale as $\sqrt T$ and $\propto \dfrac{1}{\sqrt M}$.

Compressibility Factor Z vs Pressure for Real Gases

Z vs P for real gases. H₂ and He (green): Z > 1 at all pressures because intermolecular attractions are negligible (a ≈ 0) and molecular volume effect (b) dominates. N₂ (blue): Z dips below 1 at moderate P (attractions) then rises above 1 at high P (volume). CO₂ (red): dips deeper because it is more easily liquefied (stronger intermolecular forces, larger a).

⚡

States of Matter — Quick Quiz

Question 1

The ratio of rms speeds of O₂ and H₂ at the same temperature is:

$u_{rms} \propto \dfrac{1}{\sqrt M}$. $\dfrac{u(O_2)}{u(H_2)} = \sqrt{\dfrac{M_{H_2}}{M_{O_2}}} = \sqrt{\dfrac{2}{32}} = \sqrt{\dfrac{1}{16}}$ = 1/4. So ratio O₂:H₂ = 1:4. H₂ molecules move 4× faster than O₂ at the same temperature.

Question 2

For a real gas, Z < 1 at moderate pressures indicates:

Z = PV/nRT < 1 means PV < nRT — the gas is more compressible than ideal. This happens because intermolecular attractions pull molecules together, reducing the effective pressure. The correction factor 'a' in van der Waals equation accounts for this.

Question 3

A gas diffuses 4 times faster than SO₂ (M=64). The molar mass of the gas is:

Graham's law: $\dfrac{r_{1}}{r_{2}} = \sqrt{\dfrac{M_{2}}{M_{1}}}. \dfrac{4}{1} = \sqrt{\dfrac{64}{M_{1}}}. 16 = \dfrac{64}{M_{1}}. M_{1} = \dfrac{64}{16} =$ 4 g/mol. This is He (helium) — the lightest noble gas after H₂.

🧫

Surface Chemistry

Adsorption · Catalysis · Colloids · Emulsions

Surface Chemistry

Surface chemistry deals with phenomena occurring at interfaces — solid-gas, solid-liquid, and liquid-liquid — including adsorption, catalysis, colloids, and emulsions.

Adsorption

Adsorption vs Absorption

Adsorption: accumulation of a substance (adsorbate) on the surface of another (adsorbent). Surface phenomenon.

Absorption: uniform distribution throughout the bulk of the absorbing material.

Sorption: both adsorption and absorption occurring simultaneously.

Physisorption vs Chemisorption

Property	Physisorption	Chemisorption
Force	van der Waals	Chemical bonds
ΔH	20–40 kJ/mol	40–400 kJ/mol
Reversibility	Reversible	Irreversible
Specificity	Non-specific	Highly specific
Layers	Multilayer	Monolayer only
Temp effect	Decreases with T	Initially increases, then decreases

Factors Affecting Adsorption

Nature of adsorbent: porous materials (charcoal, silica gel) have high surface area → greater adsorption
Surface area: more surface area → more adsorption (finely divided > coarse)
Nature of adsorbate: easily liquefiable gases (high $T_{c}$) are better adsorbed
Temperature: physisorption decreases; chemisorption peaks then decreases
Pressure: adsorption increases with pressure until saturation

Adsorption Isotherms

Freundlich isotherm: x/m = k·$P^{1/n}$ (empirical; n > 1)

log(x/m) = log k + (1/n) log P — straight line graph; 1/n = slope

Langmuir isotherm: P/(x/m) = 1/(ab) + P/a — monolayer, homogeneous surface

At low P: x/m ∝ P (linear); At high P: x/m → constant (saturation)

Catalysis

Homogeneous Catalysis

Catalyst and reactants in the same phase.

Example: SO₂ oxidation — 2SO₂ + O₂ → 2SO₃ (catalyst: NO in gas phase)

Mechanism: catalyst forms intermediate compound with reactant; intermediate decomposes to give product and regenerate catalyst

Ea is lowered → rate increases without changing thermodynamics.

Heterogeneous Catalysis

Catalyst and reactants in different phases. Reaction occurs on catalyst surface.

Steps: (1) Diffusion to surface → (2) Adsorption → (3) Reaction → (4) Desorption → (5) Diffusion away

Examples: Haber process (Fe catalyst); Contact process (V₂O₅); Hydrogenation of oils (Ni)

Promoters: enhance catalyst activity (e.g., Mo in Haber process)

Poisons: reduce activity (e.g., CO poisons Fe in Haber process)

Enzyme Catalysis

Biological catalysts; highly specific (lock-and-key model).

Properties: colloidal in nature; specific; greatly accelerate reaction; optimum temperature and pH; destroyed by high T or pH extremes

Michaelis-Menten kinetics: rate = $V_{max}$[S] / ($K_{m}$ + [S])

At [S] << $K_{m}$: first order; At [S] >> $K_{m}$: zero order (saturation)

Adsorption and heterogeneous catalysis in 3D — Physisorption: at low temperature N₂ is held on a solid by weak van der Waals forces, the molecules whole and loosely packed in several layers (three in the drawing); it is not specific and is reversible, and it falls as the temperature rises (on heating 17 of the 22 molecules leave, and one layer is left). Chemisorption is chemical bonding to the surface: a single layer (monolayer), and specific to the gas and the solid. Heterogeneous catalysis on Fe, in the Haber process N₂ + 3H₂ → 2NH₃: (1) N₂ and H₂ diffuse to the surface; (2) they are chemisorbed, splitting into N and H atoms bonded one per site (16 of 25 sites in the drawing), while Ar, which does not bond to Fe, bounces off; (3) the atoms react on the surface, N + 3H → NH₃, four times, with 4 N and 12 H atoms before and after; (4) NH₃ desorbs and (5) diffuses away, leaving the catalyst unchanged and its sites free. Sizes are not to scale.

Colloids

Colloidal State

Particle size: 1–1000 nm (intermediate between true solution and suspension)

Type	Particle size	Visibility
True solution	<1 nm	Not visible
Colloidal sol	1–1000 nm	Ultramicroscope
Suspension	>1000 nm	Naked eye

Classification of Colloids

By interaction with medium:

Lyophilic: attracted to solvent (gelatin, starch, gum) — stable, reversible
Lyophobic: repelled by solvent (metal sols, sulfur sol) — unstable, irreversible

Dispersed phase / Dispersion medium combinations:

Sol: solid in liquid (Fe(OH)₃ sol)
Gel: liquid in solid (cheese, butter)
Emulsion: liquid in liquid (milk, mayonnaise)
Aerosol: solid/liquid in gas (fog, smoke)
Foam: gas in liquid (shaving cream)

Properties of Colloids

Tyndall effect: scattering of light by colloidal particles — cone of light visible (true solutions don't show this)
Brownian motion: zigzag random motion of colloidal particles; caused by unequal bombardment by dispersion medium molecules
Electrophoresis: migration of colloidal particles toward oppositely charged electrode under electric field
Coagulation: aggregation and settling of colloidal particles. Hardy-Schulze rule: higher the valence of coagulating ion → greater the coagulating power

Preparation and Purification

Dispersion methods: Bredig's arc (metals), peptization (adding electrolyte to precipitate)

Condensation methods: double decomposition (As₂S₃ from H₃AsO₃ + H₂S), reduction (Au/Ag sol), oxidation

Purification:

Dialysis: removal of small ions/molecules through semipermeable membrane
Electrodialysis: same but speeded by electric field
Ultrafiltration: use of ultrafilter paper (very fine pores)

Emulsions

Types of Emulsions

O/W (oil in water): oil droplets dispersed in water — milk, cold cream. Emulsifier: soap, protein

W/O (water in oil): water droplets dispersed in oil — butter, cream. Emulsifier: heavy metal soaps

Emulsifiers: stabilise emulsions by adsorbing at interface and reducing surface tension

Demulsification: breaking emulsion by heating, centrifugation, adding demulsifier, or electric field

Surface Chemistry — Key Formulas

Freundlich Isotherm

$$ \frac{x}{m} = k \cdot P^{1/n} $$

log(x/m) = log k + (1/n) log P | x = mass adsorbed, m = mass adsorbent, P = pressure | At high P: $\dfrac{x}{m} \propto P (\dfrac{1}{n} = 1)$; At low P: x/m = const (1/n = 0)

Langmuir Isotherm

P/(x/m) = 1/(ab) + P/a

where a = saturation capacity, b = Langmuir constant | Plot: P/(x/m) vs P → straight line; slope = 1/a; intercept = 1/(ab)

Hardy-Schulze Rule

Coagulating power $\propto \text{(valence of ion)}^n$

For negative sol (Fe(OH)₃): PO₄³⁻ > SO₄²⁻ > Cl⁻ | For positive sol (As₂S₃): Al³⁺ > Ba²⁺ > Na⁺ | Flocculation value (mmol/L) = concentration needed to coagulate

Tyndall Cone

Rayleigh scattering: I ∝ 1/λ⁴

Blue light (shorter λ) scattered more than red light | Applies to colloidal solutions (not true solutions) | Used to distinguish colloid from true solution

Michaelis-Menten (Enzyme)

v = $V_{max}$[S] / ($K_{m}$ + [S])

$K_{m}$ = Michaelis constant = [S] when v = $V_{max}$/2 | Low [S]: v ≈ ($V_{max}/K_{m}$)[S] — first order | High [S]: v ≈ $V_{max}$ — zero order

Solved Examples

Example 1: At a given temperature, a gas is adsorbed on charcoal. When pressure is doubled, the amount adsorbed increases by a factor of $2^{0.4}$. Find 1/n in the Freundlich isotherm.

Solution:
Freundlich: $\dfrac{x}{m} = k\cdot P^{1/n}$
When P → 2P: $new \frac{x}{m} = k\cdot(2P)^{1/n} = 2^{1/n} \cdot \frac{x}{m}$
Given factor = $2^{0.4}$
$∴ 2^{1/n} = 2^{0.4} \to$ 1/n = 0.4
This means n = 2.5. The adsorption increases sub-linearly with pressure.

Example 2: Fe(OH)₃ sol is coagulated by adding electrolytes. Arrange in order of coagulating power: NaCl, Na₂SO₄, Na₃PO₄.

Solution:
Fe(OH)₃ sol is positively charged (Fe³⁺ ions on surface).
Coagulation is done by the oppositely charged (negative) ions.
By Hardy-Schulze rule: coagulating power ∝ valence of ion
PO₄³⁻ (valence 3) > SO₄²⁻ (valence 2) > Cl⁻ (valence 1)
∴ Coagulating power: Na₃PO₄ > Na₂SO₄ > NaCl

Example 3: A catalyst lowers the activation energy of a reaction from 80 kJ/mol to 50 kJ/mol at 300 K. By what factor does the rate increase? (R = 8.314 J/mol·K)

Solution:
Arrhenius equation: k = A·$e^{-Ea/RT}$
Rate factor = $\dfrac{k_{cat}}{k_{uncat}}$ = $e^{-(Ea,cat - Ea,uncat)/RT}$
ΔEa = 80 − 50 = 30 kJ/mol = 30,000 J/mol
$$ \Delta \frac{Ea}{RT} = 30000 / (8.314 \times 300) = 12.03 $$
Rate factor = $e^{12.03}$ ≈ 1.67 × 10⁵
Rate increases by about 167,000 times — dramatic effect of catalysis.

Example 4: Explain why lyophilic colloids are more stable than lyophobic colloids.

Solution:
Lyophilic colloids (e.g., gelatin, starch):
- Have high affinity for the dispersion medium
- Surrounded by a solvation (hydration) shell that prevents aggregation
- Charge is not essential for stability
- Reversible — can be reconstituted after drying
Lyophobic colloids (e.g., metal sols):
- Low affinity for dispersion medium
- Stability depends entirely on charge (electrostatic repulsion between particles)
- Adding electrolytes neutralises charge → coagulation occurs
- Irreversible — cannot be reconstituted easily
Addition of lyophilic colloid to lyophobic protects it (protective colloid action — Gold Number concept).

Diagrams

Freundlich isotherm: curve in P vs x/m plot; straight line in log-log plot with slope 1/n

Key properties of colloidal systems: Tyndall effect, Brownian motion, electrophoresis, coagulation

Quick Quiz — Surface Chemistry

Question 1

In a Freundlich isotherm plot of log(x/m) vs log P, the slope equals:

The linearised Freundlich equation is log(x/m) = log k + (1/n)·log P. This is y = c + mx form, so slope = 1/n. The y-intercept = log k gives the Freundlich constant.

Question 2

Which property distinguishes a colloidal sol from a true solution?

The Tyndall effect (scattering of light to form a visible cone) is shown by colloidal sols but not by true solutions. True solutions do not scatter light because their particles (ions/molecules) are too small (<1 nm). This is the key distinguishing test.

Question 3

For coagulation of a positively charged Fe(OH)₃ sol, which electrolyte is most effective?

Fe(OH)₃ sol is positively charged. By Hardy-Schulze rule, coagulation is caused by the oppositely charged (negative) ion, and effectiveness increases with valence. PO₄³⁻ (valence 3) > SO₄²⁻ (valence 2) > Cl⁻ = NO₃⁻ (valence 1). So Na₃PO₄ is most effective.

QN	Symbol	Values	Describes
Principal	\(n\)	1, 2, 3, …	Shell, energy level, size
Azimuthal	\(l\)	\(0\) to \(n-1\)	Sub-shell, shape (s,p,d,f)
Magnetic	\(m_l\)	\(-l\) to \(+l\)	Orientation in space
Spin	\(m_s\)	\(+\tfrac{1}{2}\) or \(-\tfrac{1}{2}\)	Spin of electron

Term	Formula	Units
Resistance \(R\)	\(R = \rho\cdot \dfrac{l}{A}\)	Ω
Conductance \(G\)	\(G = \dfrac{1}{R}\)	S (siemens)
Specific conductance \(\kappa\)	\(\kappa = \dfrac{1}{\rho} = G\cdot(\dfrac{l}{A})\)	S·m⁻¹
Molar conductance \(\Lambda_m\)	\(\Lambda_m = \dfrac{\kappa}{C}\) (C in mol/m³)	S·m²·mol⁻¹
Kohlrausch's law	\(\Lambda^\circ_m = \Sigma\, \lambda^\circ(\text{ions})\) (at infinite dilution)	additive

Order	Rate law	Integrated form	Half-life t½	Units of k
Zero	\(\text{Rate} = k\)	\([A] = [A]_0 - kt\)	\(t_{1/2} = [A]\dfrac{_0}{2k}\)	mol·L⁻¹·s⁻¹
First	\(\text{Rate} = k[A]\)	\(\ln[A] = \ln[A]_0 - kt\) or \([A]=[A]_0 e^{-kt}\)	\(t_{1/2} = \ln 2/k = \dfrac{0.693}{k}\)	s⁻¹
Second	\(\text{Rate} = k[A]^2\)	\(\dfrac{1}{A} = \dfrac{1}{[A]_0} + kt\)	\(t_{1/2} = \dfrac{1}{k[A]_0}\)	L·mol⁻¹·s⁻¹

Property	Formula	Key point
Relative lowering of VP	\(\dfrac{p^{\circ}-p}{p^{\circ}} = \chi_{2}\)	χ₂ = mole fraction of solute
Elevation of boiling point	\(\Delta T_b = K_b \cdot m\)	\(K_b\) = ebullioscopic constant (solvent property)
Depression of freezing point	\(\Delta T_f = K_f \cdot m\)	\(K_f\) = cryoscopic constant; larger than \(K_b\)
Osmotic pressure	\(\pi = CRT = \frac{n}{V}RT\)	Van't Hoff equation; C in mol/L