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General Chemistry

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General Chemistry - Complete Interactive Lesson

Part 1: Atomic Structure & Periodic Trends

General Chemistry for the MCAT

Part 1 of 7 — Atomic Structure & Periodic Trends

General chemistry accounts for roughly 30% of the Chemical and Physical Foundations section of the MCAT. Atomic structure is the foundation everything else builds on — bonding, reactivity, acid-base behavior, and redox all trace back to electron configuration.

Quantum Numbers

Every electron in an atom is described by four quantum numbers. The MCAT tests your ability to identify invalid combinations.

Quantum Number	Symbol	What It Describes	Allowed Values
Principal	$n$	Energy level / shell	1, 2, 3, …
Angular momentum	$l$	Subshell shape	0 to $n-1$
Magnetic	$m_l$	Orbital orientation	$-l$ to $+l$
Spin	$m_s$	Electron spin	$+\frac{1}{2}$ or

Quick reference — subshell shapes:

$l = 0$ → s orbital (spherical)
$l = 1$ → p orbital (dumbbell)
$l = 2$ → d orbital (cloverleaf)
$l = 3$ → f orbital

Three Key Rules

Pauli Exclusion Principle: No two electrons in the same atom can have identical sets of all four quantum numbers. Each orbital holds at most 2 electrons with opposite spins.

Hund's Rule: When filling degenerate (equal-energy) orbitals, electrons occupy them singly before pairing up. This minimizes electron–electron repulsion.

Aufbau Principle: Electrons fill lower-energy orbitals first. The general order is: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, …

Electron Configuration & MCAT Exceptions

Standard Filling

Write configurations using noble gas shorthand. For example:

Na (Z=11): $[\text{Ne}]\,3s^1$
Fe (Z=26): $[\text{Ar}]\,3d^6\,4s^2$

Quantum Numbers & Electron Configuration 🎯

Periodic Trends — Reason, Don't Memorize

All periodic trends reduce to one concept: effective nuclear charge ( $Z_{eff}$ ).

$Z_{eff} = Z - S$

Periodic Trends 🎯

Worked Example: Connecting Electron Config to Properties

Problem: A metal M forms a 2+ ion with configuration $[\text{Ar}]\,3d^5$ . Identify M and predict whether $M^{2+}$ is paramagnetic or diamagnetic.

Step 1 — Identify M:
$M^{2 +}$ is . To get the neutral atom, add back 2 electrons. For transition metals, they go back to 4s: Counting electrons: = 18, plus = 5, plus = 2 → total 25 electrons → .

Key Takeaways — Part 1

Quantum numbers: $l$ ranges from 0 to $n-1$ ; $m_l$ from $-l$ to . Invalid or = most common MCAT trap.

Part 2: Chemical Bonding

General Chemistry for the MCAT

Part 2 of 7 — Bonding & Molecular Geometry

Bonding determines shape, shape determines polarity, and polarity determines intermolecular forces — which in turn control boiling points, solubility, and biological behavior. The MCAT tests this entire chain of reasoning.

Types of Chemical Bonds

Electronegativity Difference → Bond Type

ΔEN	Bond Type	Example
0 – 0.4	Nonpolar covalent	$\text{H}_2$ , $\text{Cl}_2$ ,

Part 3: Stoichiometry & Solutions

General Chemistry for the MCAT

Part 3 of 7 — Stoichiometry, Solutions & Concentration

Stoichiometry is the arithmetic of chemistry. On the MCAT you will encounter stoichiometry problems embedded in biochemistry passages (enzyme reactions, metabolic pathways) and lab-technique passages. Connecting moles to biological quantities is a high-yield skill.

Stoichiometry: The Mole Map

The Mole is the chemist's counting unit: 1 mol = $6.022 \times 10^{23}$ particles (Avogadro's number).

$\text{mol} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} = \frac{\text{volume of gas (L)}}{22.4 \text{ L/mol at STP}}$

Part 4: Acids, Bases & Buffers

General Chemistry for the MCAT

Part 4 of 7 — Acids, Bases, pH & Buffers (ULTRA HIGH YIELD)

Acid-base chemistry appears in nearly every MCAT section — chemistry passages, biochemistry passages, and physiology contexts (blood pH, enzyme activity, kidney function). This is one of the most tested areas on the exam.

Acid-Base Theories

Theory	Acid Definition	Base Definition	Scope
Arrhenius	Produces $\text{H}^+$ in water	Produces $\text{OH}^-$ in water

Part 5: Thermodynamics & Equilibrium

General Chemistry for the MCAT

Part 5 of 7 — Thermodynamics & Equilibrium

Thermodynamics tells us whether a reaction is favorable; kinetics tells us how fast. The MCAT extensively tests your ability to connect $\Delta G$ , $\Delta H$ , $\Delta S$ , and $K$ to real chemical and biological scenarios (ATP hydrolysis, protein folding, metabolic reactions).

Enthalpy, Entropy, and Free Energy

Enthalpy ()

Part 6: Chemical Kinetics

General Chemistry for the MCAT

Part 6 of 7 — Chemical Kinetics

Kinetics answers the question how fast? Thermodynamics only tells us if a reaction is favorable ( $\Delta G$ ); kinetics tells us the rate and what factors control it. On the MCAT, kinetics questions often appear in enzyme kinetics passages (Michaelis-Menten is direct kinetics) and analytical chemistry passages.

Rate Laws & Reaction Orders

The Rate Law

$\text{Rate} = k[A]^m[B]^n$

Part 7: Electrochemistry & Redox

General Chemistry for the MCAT

Part 7 of 7 — Electrochemistry & Redox

Electrochemistry bridges general chemistry, biochemistry, and physiology. The MCAT tests galvanic cells, electrolytic cells, the Nernst equation, and — critically — biological redox: electron transport chain, $NADH/FADH_{2}$ as electron carriers, and oxidation state assignments in metabolic intermediates.

Oxidation States & Half-Reactions

Assigning Oxidation States — Rules (in priority order)

Free element = 0 (e.g., $\text{O}_2$ , )

Cl (Z=17):

[\text{Ne}]\,3s^2\,3p^5

High-Yield Exceptions (Memorize These Two)

Element	Expected	Actual	Reason
Cr (Z=24)	$[\text{Ar}]\,3d^4\,4s^2$	$[\text{Ar}]\,3d^5\,4s^1$	Half-filled $d$ is extra stable
Cu (Z=29)	$[\text{Ar}]\,3d^9\,4s^2$	$[\text{Ar}]\,3d^{10}\,4s^1$

Transition Metal Cations: Remove 4s First

Although 4s fills before 3d, electrons are removed from 4s first when forming cations:

$\text{Fe}^{2+}: [\text{Ar}]\,3d^6 \qquad \text{Fe}^{3+}: [\text{Ar}]\,3d^5$

This is because once 3d is occupied, 3d electrons become lower in energy than 4s.

Diamagnetic vs. Paramagnetic

Paramagnetic: has one or more unpaired electrons → weakly attracted to magnetic fields
Diamagnetic: all electrons paired → weakly repelled by magnetic fields

Example: $\text{Cu}^{2+}$ is $[\text{Ar}]\,3d^9$ — one unpaired $d$ electron → paramagnetic.

$Z$ = atomic number (number of protons)
$S$ = shielding constant (approximate number of core electrons shielding valence electrons from the nucleus)

Across a period (left → right): $Z$ increases but shielding stays nearly constant → $Z_{eff}$ increases → valence electrons are pulled in tighter.

Down a group (top → bottom): New shells are added → valence electrons are farther from nucleus and more shielded → $Z_{eff}$ is lower for valence electrons.

Property	Across Period (→)	Down Group (↓)	Driven By
Atomic radius	Decreases	Increases	Higher $Z_{eff}$ pulls electrons in; more shells add distance
Ionization energy $(IE_{1})$	Increases	Decreases	Harder to remove from tighter-held valence shell
Electronegativity	Increases	Decreases	Same as IE — ability to attract bonding electrons
Electron affinity	Generally more negative	Generally less negative	More favorable to add $e^{-}$ with high $Z_{eff}$
Metallic character	Decreases	Increases	Inverse of IE — metals lose electrons easily

$IE_{1}$ Exceptions (MUST Know for MCAT)

Ionization energy generally rises across a period, but there are two important dips:

Group IIA → IIIA: $\text{Mg} \to \text{Al}$
Al's highest-energy electron is in 3p (higher energy, easier to remove) vs. Mg's 3s.
Group VA → VIA: $\text{P} \to \text{S}$
P has a half-filled $3p$ (extra stable, each orbital singly occupied). S has one paired $3p$ electron that experiences extra repulsion → easier to ionize.

Ionization Energy Jump Logic (MCAT Favorite)

A large jump between successive ionization energies reveals the valence electron count:

Big jump between $IE_{2}$ and $IE_{3}$ → 2 valence electrons → Group IIA
Big jump between $IE_{1}$ and $IE_{2}$ → 1 valence electron → Group IA

M = [\text{Ar}]\,3d^5\,4s^2

M = Mn (manganese)

Step 2 — Paramagnetic or diamagnetic?
$[\text{Ar}]\,3d^5$ means 5 electrons in 5 separate $d$ orbitals (Hund's rule) → 5 unpaired electrons → paramagnetic.

MCAT Connection: Iron in hemoglobin is $\text{Fe}^{2+}$ ( $[\text{Ar}]\,3d^6$ ) with 4 unpaired electrons. This paramagnetic property is exploited in MRI contrast agents.

Electron removal from transition metals: always remove 4s before 3d when forming cations.

Cr and Cu exceptions: half-filled and fully filled

d

subshells are extra stable.

All periodic trends trace to $Z_{eff}$ : increasing

Z_{eff}

→ smaller radius, higher IE, higher EN.

IE exceptions: Al < Mg and S < P due to subshell energy and pairing effects.

Ionization energy jump: locates valence electron count — important for group identification.

Use formal charge to identify the best Lewis structure (lowest formal charges, negative charge on most electronegative atom):

$\text{FC} = V - L - \frac{1}{2}B$

where $V$ = valence electrons, $L$ = lone-pair electrons, $B$ = bonding electrons

Example — $\text{CO}_2$ :
Central C: $V=4$ , $L=0$ , $B=8$ (two double bonds) → $\text{FC} = 4 - 0 - 4 = 0$ ✓

When multiple valid Lewis structures exist (e.g., $\text{SO}_3$ , $\text{NO}_3^-$ , benzene), the molecule is best described as a resonance hybrid — all bonds are intermediate in character, not alternating.

All bonds in $\text{NO}_3^-$ are equivalent (bond order = $1\tfrac{1}{3}$ )
Resonance structures share the same skeleton but differ in electron distribution

VSEPR & Molecular Geometry

Strategy: Count electron domains (bonds + lone pairs) on the central atom → get electron-domain geometry → remove lone pairs → get molecular geometry.

Electron Domains	Lone Pairs	Molecular Geometry	Bond Angles
2	0	Linear	180°
3	0	Trigonal planar	120°
3	1	Bent	<120°
4	0	Tetrahedral	109.5°
4	1	Trigonal pyramidal	~107°
4	2	Bent	~104.5°
5	0	Trigonal bipyramidal	90°/120°
6	0	Octahedral	90°
6	2	Square planar	90°

Lone pairs compress angles because lone pair–bonding pair repulsion > bonding pair–bonding pair repulsion.

Polarity

A molecule is polar if:

It has polar bonds AND
The bond dipoles do NOT cancel symmetrically

Nonpolar despite polar bonds: $\text{CO}_2$ (linear), $\text{BF}_3$ (trigonal planar), $\text{CCl}_4$ (tetrahedral) — bond dipoles cancel.

Polar: $\text{H}_2\text{O}$ (bent), $\text{NH}_3$ (trigonal pyramidal), $\text{HCl}$ (linear, only one dipole).

Bonding & Molecular Geometry 🎯

Intermolecular Forces (IMFs)

IMFs determine physical properties: boiling point, melting point, viscosity, surface tension, vapor pressure.

Strength ranking (weakest → strongest):

$\text{London Dispersion} < \text{Dipole-Dipole} < \text{Hydrogen Bonding} < \text{Ion-Dipole}$

London Dispersion Forces (LDF)

Present in all molecules (even nonpolar)
Arise from temporary fluctuating dipoles
Increase with molecular size (molar mass) and surface area
Linear molecules have more surface contact than branched → higher BP

Dipole-Dipole Forces

In polar molecules (permanent dipoles)
Larger dipole moment → stronger force

Hydrogen Bonding

Requires: H bonded directly to N, O, or F

Examples: $\text{H}_2\text{O}$ , $\text{NH}_3$ , $\text{HF}$ , alcohols, carboxylic acids, DNA base pairs

Why water is anomalous: Each water molecule can form up to 4 H-bonds (2 donor, 2 acceptor) → unusually high BP (100°C vs. expected ~−80°C).

MCAT Application — Boiling Point Comparisons

Comparison	Winner	Reason
$\text{CH}_3\text{OH}$ vs $\text{CH}_3\text{CH}_3$

Intermolecular Forces 🎯

Key Takeaways — Part 2

Bond polarity depends on electronegativity difference; molecular polarity depends on geometry.
VSEPR: count all electron domains, determine geometry, then remove lone pairs for molecular shape.
Lone pairs compress bond angles more than bonding pairs.
$\text{CO}_2$ , $\text{BF}_3$ , $\text{CCl}_4$ = nonpolar despite polar bonds (symmetric cancellation).
IMF strength: LDF < dipole-dipole < H-bonding < ion-dipole.
Hydrogen bonding requires H directly bonded to N, O, or F.
Larger molecular surface area → stronger LDF → higher boiling point (branching lowers BP).

molar mass (g/mol)mass (g)=

22.4 L/mol at STPvolume of gas (L)

Four-Step Stoichiometry Workflow

Balance the chemical equation.
Convert all given quantities to moles.
Apply mole ratio from balanced equation.
Convert moles to requested units (grams, liters, molarity, particles).

The limiting reagent is the reactant that runs out first and determines the maximum yield.

Short method: Divide each reactant's moles by its stoichiometric coefficient. The smallest ratio identifies the limiting reagent.

Example:
$2\text{H}_2 + \text{O}_2 \to 2\text{H}_2\text{O}$

If you have 3 mol $\text{H}_2$ and 2 mol $\text{O}_2$ :

$\text{H}_2$ : $3 / 2 = 1.5$
$\text{O}_2$ : $2 / 1 = 2.0$

Smallest = 1.5 → $\text{H}_2$ is limiting. Moles $\text{H}_2\text{O}$ = $1.5 \times 2 = 3$ mol.

$\%\text{yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\%$

$\%\text{ element} = \frac{\text{molar mass of element in formula}}{\text{molar mass of compound}} \times 100\%$

Solutions & Concentration

Key Concentration Units

Measure	Formula	Temperature Dependent?
Molarity (M)	$\dfrac{\text{mol solute}}{\text{L solution}}$	Yes (volume changes with T)
Molality (m)	$\dfrac{\text{mol solute}}{\text{kg solvent}}$	No
Mole fraction (χ)	$\dfrac{n_A}{n_A + n_B + \ldots}$

Dilution

When you dilute a solution, moles of solute are conserved:

$M_1 V_1 = M_2 V_2$

Example: Preparing 250 mL of 0.50 M $\text{HCl}$ from 12 M $\text{HCl}$ : $V_1 = \frac{M_2 V_2}{M_1} = \frac{0.50 \times 0.250}{12} = 0.0104 \text{ L} = \mathbf{10.4 \text{ mL}}$

Solubility Rules (MCAT High-Yield)

Always soluble: all $Na^{+}$ , $K^{+}$ , $NH_{4}^{+}$ , , salts

Usually soluble: halides ( $Cl^{-}$ , $Br^{-}$ , $I^{-}$ ) AgX, ,

Usually insoluble: carbonates $(CO_{3}^{2-})$ , phosphates $(PO_{4}^{3-})$ , hydroxides Group IA +

Usually insoluble: sulfates $(SO_{4}^{2-})$ except $MgSO_{4}$ , (slightly), (insoluble)

Stoichiometry & Solutions 🎯

Colligative Properties

Colligative properties depend only on the number of dissolved particles, not their identity.

$i = \text{van't Hoff factor (particles per formula unit in solution)}$

Solute	$i$	Reason
Glucose (nonelectrolyte)	1	No dissociation
NaCl	2	$\text{Na}^+ + \text{Cl}^-$
$\text{CaCl}_2$	3	$\text{Ca}^{2+} + 2\text{Cl}^-$
$\text{AlCl}_3$	4	$\text{Al}^{3+} + 3\text{Cl}^-$

Freezing Point Depression & Boiling Point Elevation

$\Delta T_f = K_f \cdot m \cdot i \qquad (\text{solution freezes } \Delta T_f \text{ lower than pure solvent})$

For water: $K_f = 1.86\;°\text{C}\cdot\text{kg/mol}$ , $K_b = 0.512\;°\text{C}\cdot\text{kg/mol}$

Osmotic Pressure

$\Pi = iMRT$

where $M$ = molarity, $R = 0.0821\;\text{L}\cdot\text{atm/(mol}\cdot\text{K)}$ , $T$ = temperature in K

MCAT Connection: Osmosis is critical in biology (cells shrink in hypertonic solution, swell in hypotonic). Dissolving more particles = higher osmolarity = more osmotic pressure.

Vapor Pressure Lowering (Raoult's Law)

$P_{\text{solution}} = \chi_{\text{solvent}} \cdot P°_{\text{solvent}}$

Adding a nonvolatile solute always lowers vapor pressure.

Colligative Properties 🎯

Key Takeaways — Part 3

Stoichiometry workflow: balance → moles → mole ratio → convert to final units.
Limiting reagent: divide reactant moles by coefficient; smallest ratio wins.
Molarity vs molality: M changes with temperature (volume changes); m does not.
Dilution: $M_1V_1 = M_2V_2$ — moles of solute are conserved.
Colligative properties depend on particle count ( $i$ ): more particles = greater effect.
Osmosis: water moves toward higher solute concentration (lower water potential).
Key solubility rules: all $NO_{3}^{-}$ soluble; AgCl insoluble; $BaSO_{4}$ insoluble.

Conjugate pairs: When a Brønsted acid donates $H^{+}$ , it forms its conjugate base:

$\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-$

Stronger acid → weaker conjugate base
Weaker acid → stronger conjugate base

Lewis acids/bases (MCAT favorites):

$\text{BF}_3$ , $\text{AlCl}_3$ , metal cations, $\text{H}^+$ are Lewis acids (accept $e^{-}$ pair)
$\text{NH}_3$ , $\text{OH}^-$ , $\text{Cl}^-$ , water are Lewis bases (donate pair)

6 Strong Acids (memorize — complete dissociation): HCl, HBr, HI, $\text{HNO}_3$ , $\text{HClO}_4$ , $\text{H}_2\text{SO}_4$ (first proton)

Strong Bases (complete dissociation): LiOH, NaOH, KOH, $\text{Ba(OH)}_2$ , $\text{Ca(OH)}_2$

Everything else is weak (partial dissociation, governed by $K_a$ or $K_b$ ).

pH Calculations

Fundamental Relationships

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

$\text{pH} + \text{pOH} = 14 \quad (\text{at 25°C})$

$K_w = [\text{H}^+][\text{OH}^-] = 1.0 \times 10^{-14} \quad (\text{at 25°C})$

$K_a \times K_b = K_w \quad (\text{for conjugate pair})$

Strong Acid/Base pH (Direct)

For 0.010 M HCl (strong acid): $[\text{H}^+] = 0.010$ M → $\text{pH} = 2.0$

For 0.010 M NaOH (strong base): $[\text{OH}^-] = 0.010$ M → $\text{pOH} = 2.0$ → $\text{pH} = 12.0$

Weak Acid pH (Approximate Formula)

For weak acid HA with $K_a$ at concentration $C$ (valid when $C \gg K_a$ ):

$[\text{H}^+] \approx \sqrt{K_a \cdot C}$

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

Example: 0.10 M acetic acid ( $K_a = 1.8 \times 10^{-5}$ ): $[H^{+}] = \sqrt{1.8 \times 10^{- 5} \times 0.10} = \sqrt{}$

Henderson-Hasselbalch Equation (Buffer pH)

$\text{pH} = \text{p}K_a + \log\frac{[\text{A}^-]}{[\text{HA}]}$

At half-equivalence point: $[\text{A}^-] = [\text{HA}]$ → $\text{pH} = \text{p}K_a$ (most tested point!)

pH Calculations & Acid-Base Theory 🎯

Titration Curves & Buffer Behavior

Strong Acid + Strong Base

Initial pH determined by strong acid concentration.
At equivalence point: pH = 7.0 (salt of strong acid/strong base is neutral).
Sharp pH jump at equivalence point.

Weak Acid + Strong Base (Most Common MCAT Type)

Key points on the curve:

Point	pH Relationship	Significance
Initial	pH calculated from $K_a$ , $[\text{HA}]$	Higher than strong acid of same concentration
Half-equivalence	pH = p $K_a$	Equal amounts HA and $A^{-}$ — maximum buffer capacity
Equivalence	pH > 7	$A^{-}$ hydrolyzes water: $\text{A}^- + \text{H}_2\text{O} \rightleftharpoons \text{HA} + \text{OH}^-$
Past equivalence	pH determined by excess base

Buffer Capacity

A buffer resists pH change by:

Adding acid $(H^{+})$ : consumed by conjugate base → $\text{A}^- + \text{H}^+ \to \text{HA}$

Maximum capacity is at pH = p $K_a$ (equal concentrations of HA and $A^{-}$ ).

Physiological Buffer: Bicarbonate

$\text{CO}_2 + \text{H}_2\text{O} \rightleftharpoons \text{H}_2\text{CO}_3 \rightleftharpoons \text{HCO}_3^- + \text{H}^+$

Normal blood pH = 7.40. $\text{p}K_a$ of $\text{H}_2\text{CO}_3$ ≈ 6.1.

Henderson-Hasselbalch: $7.40 = 6.1 + \log([\text{HCO}_3^-]/[\text{H}_2\text{CO}_3])$

Ratio ≈ 20:1 ( $HCO_{3}^{-}$ : $H_{2}CO_{3}$ ), maintained by lungs and kidneys.

Titrations & Buffers 🎯

Key Takeaways — Part 4

6 strong acids and 4–5 strong bases: complete dissociation, pH calculated directly.
Weak acid: $[\text{H}^+] \approx \sqrt{K_a C}$ — valid when $C \gg K_a$ .
Henderson-Hasselbalch: pH = p $K_a + \log([A^-]/[HA])$ — memorize and practice.
Half-equivalence point: pH = p $K_a$ — most frequent MCAT titration question.
Equivalence point: weak acid + strong base → pH > 7 (conjugate base hydrolysis).
Buffer capacity is maximum at pH = p $K_a$ .
Blood buffer: bicarbonate system, pH 7.40, controlled by lungs $(CO_{2})$ and kidneys $(HCO_{3}^{-})$ .

Enthalpy measures heat flow at constant pressure.

$\Delta H < 0$ : exothermic (heat released to surroundings)
$\Delta H > 0$ : endothermic (heat absorbed from surroundings)

Hess's Law: $\Delta H_{rxn}$ can be calculated by algebraically combining reaction enthalpies:

$\Delta H_{rxn} = \sum \Delta H_f°(\text{products}) - \sum \Delta H_f°(\text{reactants})$

Standard enthalpy of formation of any element in its standard state = 0 (e.g., $O_2(g)$ , $C(\text{graphite})$ ).

Entropy ( $\Delta S$ )

Entropy measures disorder or dispersal of energy.

$\Delta S > 0$ : increase in disorder (favored)
$\Delta S < 0$ : decrease in disorder (unfavored)

Predicting sign of $\Delta S$ :

Solid → liquid → gas: $\Delta S > 0$
Dissolving most salts: $\Delta S > 0$
More moles of gas products than reactants: $\Delta S > 0$
Protein folding, crystallization: $\Delta S < 0$

Gibbs Free Energy ( $\Delta G$ )

$\Delta G = \Delta H - T\Delta S$

$\Delta G < 0$ : spontaneous (thermodynamically favorable)
$\Delta G > 0$ : non-spontaneous
$\Delta G = 0$ : system at equilibrium

Temperature crossover: For reactions where $\Delta H$ and $\Delta S$ have the same sign, spontaneity depends on temperature. Set $\Delta G = 0$ to find the crossover temperature:

$T = \frac{\Delta H}{\Delta S}$

Spontaneity Analysis — The Four Cases

$\Delta H$	$\Delta S$	$\Delta G = \Delta H - T\Delta S$	Spontaneous?
$-$	$+$	Always negative	Always (at all T)
$+$	$-$	Always positive	Never (at any T)
$-$

Biological example: ATP hydrolysis ( $\text{ATP} + \text{H}_2\text{O} \to \text{ADP} + P_i$ ) has kJ/mol under standard biochemical conditions — spontaneous, drives unfavorable reactions when coupled.

Connecting $\Delta G°$ to Equilibrium

$\Delta G° = -RT\ln K$

$\ln K = -\frac{\Delta G°}{RT}$

$\Delta G°$	$K$	Meaning
$< 0$	$> 1$	Products favored at equilibrium

Reaction at Non-Standard Conditions

$\Delta G = \Delta G° + RT\ln Q$

When $Q < K$ : $\Delta G < 0$ (forward reaction spontaneous)
When $Q > K$ : $\Delta G > 0$ (reverse reaction spontaneous) When : (equilibrium)

Thermodynamics: ΔG, ΔH, ΔS 🎯

Chemical Equilibrium

Equilibrium Constant

For $a\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D}$ :

$K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}$

Important rules:

Pure solids and pure liquids are NOT included in $K$ expressions
$K_p$ uses partial pressures; $K_c$ uses molar concentrations
where = moles gas products − moles gas reactants

Le Chatelier's Principle

When a system at equilibrium is disturbed, it shifts to partially counteract the disturbance.

Disturbance	Direction of Shift
Add reactant	Forward (→)
Remove reactant	Reverse (←)
Add product	Reverse (←)
Remove product	Forward (→)
Increase pressure (gas)	Toward fewer moles of gas
Decrease pressure (gas)	Toward more moles of gas
Increase temperature	Toward endothermic direction
Decrease temperature	Toward exothermic direction
Add catalyst	No shift (reaches equilibrium faster)

Reaction Quotient Q

$Q = \frac{[\text{products (current)}]}{[\text{reactants (current)}]}$

$Q < K$ : too many reactants → shifts forward
$Q > K$ : too many products → shifts reverse
$Q = K$ : at equilibrium

Equilibrium & Le Chatelier 🎯

Key Takeaways — Part 5

$\Delta G = \Delta H - T\Delta S$ : memorize this and the four cases in the spontaneity table.
Watch units: $\Delta H$ is usually in kJ; $\Delta S$ in J/K — convert before calculating $T$ crossover.
$\Delta G° = -RT\ln K$ : negative $\Delta G°$ → $K > 1$ → products favored.
Le Chatelier: adding stress shifts system to relieve stress. Catalyst does NOT shift position.
$Q$ vs $K$ : $Q < K$ → forward; $Q > K$ → reverse.
Hess's Law: $\Delta H$ is a state function; add/subtract reactions algebraically.
Biological link: ATP hydrolysis ( $\Delta G° < 0$ ) drives coupled biosynthetic reactions.

$k$ = rate constant (temperature-dependent)
$m$ , $n$ = reaction orders in A and B — determined experimentally, NOT from stoichiometric coefficients
Overall order = $m + n$

Method of Initial Rates

Change one reactant concentration at a time and measure the effect on rate:

$[A]$ doubled	Rate doubles	→ 1st order in A ( $m=1$ )
$[A]$ doubled	Rate quadruples	→ 2nd order in A ( $m=2$ )
$[A]$ doubled	Rate unchanged	→ 0th order in A ( $m=0$ )

Experiment	$[A]$ (M)	$[B]$ (M)	Rate (M/s)
1	0.10	0.10	$2.0 \times 10^{-4}$
2	0.20	0.10	$8.0 \times 10^{-4}$
3	0.10	0.20	$4.0 \times 10^{-4}$

Exp 1→2: [A] doubles, rate × 4 → 2nd order in A
Exp 1→3: [B] doubles, rate × 2 → 1st order in B
Rate = $k[A]^2[B]$ ; overall 3rd order

Calculate $k$ : $k = \text{Rate}/([A]^2[B]) = 2.0 \times 10^{-4}/(0.01 \times 0.10) = 0.20$ $M^{-2}s^{-1}$

Integrated Rate Laws & Half-Lives

First-Order Reactions (Most Important on MCAT)

$\ln[A]_t = \ln[A]_0 - kt$

$[A]_t = [A]_0 e^{-kt}$

Half-life: $t_{1/2} = \frac{\ln 2}{k} = \frac{0.693}{k}$ — (constant half-life)

MCAT application: Radioactive decay, many drug elimination processes, and first-order enzyme reactions at low substrate are 1st order.

Zero-Order Reactions

$[A]_t = [A]_0 - kt$

Half-life: $t_{1/2} = \frac{[A]_0}{2k}$ — depends on initial concentration

Second-Order Reactions

$\frac{1}{[A]_t} = \frac{1}{[A]_0} + kt$

Half-life: $t_{1/2} = \frac{1}{k[A]_0}$ — inversely proportional to

Graphical Identification

Plot	Linear For	Slope
$[A]$ vs time	Zero order	$-k$
$\ln[A]$ vs time	First order	$-k$

Radioactive Decay Example

A radioactive isotope has $t_{1/2} = 5730$ years (Carbon-14). After 11,460 years (2 half-lives): $[A] = [A]_0 \left(\frac{1}{2}\right)^2 = \frac{[A]_0}{4}$

Only 25% of original remains.

Rate Laws & Integrated Equations 🎯

Arrhenius Equation & Reaction Mechanisms

Arrhenius Equation

$k = Ae^{-E_a/RT}$

$\ln k = \ln A - \frac{E_a}{RT}$

$A$ = frequency factor (collision frequency × proper orientation)
$E_a$ = activation energy (energy barrier)
$R = 8.314$ J/(mol·K)
$T$ = temperature in Kelvin

Two-temperature form (MCAT-friendly):

$\ln\frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$

Effects on Rate

Factor	Effect on Rate	Effect on $k$	Effect on $E_a$	Effect on $\Delta G$
↑ Temperature	Increases	Increases	No change	No change

Catalyst: provides alternative mechanism with lower $E_a$ . Does NOT change $\Delta G$ , $\Delta H$ , $K$ , or equilibrium position.

Reaction Mechanisms & Rate-Determining Step

An elementary mechanism consists of steps; the slowest step determines the overall rate.

Example — Overall reaction: $2\text{NO} + \text{O}_2 \to 2\text{NO}_2$

Step 1 (fast): $2\text{NO} \rightleftharpoons \text{N}_2\text{O}_2$
Step 2 (slow, rate-determining): $\text{N}_2\text{O}_2 + \text{O}_2 \to 2\text{NO}_2$

Rate = $k[\text{N}_2\text{O}_2][\text{O}_2]$

Since $[\text{N}_2\text{O}_2] = K_{eq}[\text{NO}]^2$ from fast equilibrium Step 1:

$\text{Rate} = k'[\text{NO}]^2[\text{O}_2]$

Intermediates (appear and are consumed during mechanism) do NOT appear in the overall rate law.

Arrhenius & Mechanism 🎯

Key Takeaways — Part 6

Rate law exponents are experimental — do NOT read them from the balanced equation.
Method of initial rates: vary one reactant, compute rate ratio to find order.
First-order half-life ( $t_{1/2} = 0.693/k$ ) is concentration-independent — diagnostic feature.
Graph trick: which plot is linear identifies the order (0=linear [A]; 1=ln[A]; 2=1/[A]).
Catalyst: lowers $E_a$ for both directions, increases rate, does NOT change equilibrium or $\Delta G$ .
Rate-determining step = slowest step = highest $E_a$ step.
Intermediates appear in mechanism steps but not in the overall rate law.

Monatomic ion = ionic charge (e.g.,

\text{Na}^+ = +1

\text{Cl}^- = -1

)

F is always

-1

in compounds

O is usually

-2

(exception: peroxides

-1

;

\text{OF}_2

+2

)

H is usually

+1

(exception: metal hydrides

= -1

)

Sum of oxidation states = overall charge of species

Example — $\text{Cr}_2\text{O}_7^{2-}$ :
$2x + 7(-2) = -2$ → $2x = 12$ → $x = +6$ (Cr is +6, a strong oxidizing agent)

Oxidation Is Loss of electrons | Reduction Is Gain of electrons

Term	Definition
Oxidation	Loss of electrons; increase in oxidation state
Reduction	Gain of electrons; decrease in oxidation state
Oxidizing agent	Gets reduced $(accepts e^{-})$ ; is itself oxidized
Reducing agent	Gets oxidized $(donates e^{-})$ ; is itself reduced

Balancing Redox Half-Reactions (Acidic Solution)

Split into oxidation and reduction half-reactions.
Balance atoms other than O and H.
Balance O by adding $\text{H}_2\text{O}$ .
Balance H by adding $\text{H}^+$ .
Balance charge by adding $e^-$ .
Multiply half-reactions so electrons cancel; add together.

Galvanic & Electrolytic Cells

Key Vocabulary

Term	Galvanic Cell	Electrolytic Cell
Purpose	Converts chemical energy → electrical	Converts electrical → chemical
$\Delta G$	Negative (spontaneous)	Positive (non-spontaneous)
$E°_{cell}$	Positive	Negative (or forced)
Anode charge	Negative	Positive
Cathode charge	Positive	Negative

Unchanging rule: Oxidation always at the anode; reduction always at the cathode.
Memory: AN-OX, RED-CAT (ANode = OXidation; REDuction = CAThode)

Standard Cell Potential

$E°_{cell} = E°_{cathode} - E°_{anode}$

The half-reaction with the more positive standard reduction potential is the cathode (gets reduced).

Example — Galvanic cell with Zn and Cu:

Half-reaction	$E°$ (V)
$\text{Cu}^{2+} + 2e^- \to \text{Cu}$

$Cu^{2+}$ has higher $E°$ → cathode (reduced). Zn → anode (oxidized).

$E°_{cell} = 0.34 - (-0.76) = +1.10\text{ V}$

Free Energy Connection

$\Delta G° = -nFE°_{cell}$

where $n$ = moles of electrons transferred, $F = 96{,}485$ C/mol (Faraday's constant)

Also: $\Delta G° = -RT\ln K$ , so a positive $E°_{cell}$ → → products favored.

Redox & Cell Potentials 🎯

Nernst Equation & Biological Redox

Nernst Equation

Cell potential changes with concentration. At 25°C:

$E = E° - \frac{0.0592}{n}\log Q$

General form:

$E = E° - \frac{RT}{nF}\ln Q$

What Q does:

$Q < 1$ (more reactants): $E > E°$ (reaction more favorable)
$Q > 1$ (more products): $E < E°$ (reaction less favorable)

Faraday's Law of Electrolysis

$\text{mass deposited} = \frac{M \cdot I \cdot t}{n \cdot F}$

where $M$ = molar mass, $I$ = current (amps), $t$ = time (seconds), $n$ = electrons per ion, $F = 96{,}485$ C/mol

Example: How long to deposit 0.635 g of Cu (M = 63.5 g/mol) at 1.00 A? $t = \frac{m \cdot n \cdot F}{M \cdot I} = \frac{0.635 \times 2 \times 96{,}485}{63.5 \times 1.00} = \frac{122{,}500}{63.5} \approx 1930\text{ s}$

Biological Redox (High-Yield MCAT Connection)

Electron carriers in cellular respiration:

$\text{NAD}^+$ (oxidized) / NADH (reduced) — 2 electrons + 1 proton
FAD (oxidized) / $\text{FADH}_2$ (reduced)

In the electron transport chain (ETC):

NADH is oxidized (donates electrons to Complex I)
Electrons move through protein complexes with decreasing energy
$\text{O}_2$ is the final electron acceptor (reduced to $\text{H}_2\text{O}$ )
Proton gradient drives ATP synthase

This is electrochemistry at its most biological: ETC = a series of redox couples, each with successively more positive $E°$ , driving spontaneous electron flow.

Nernst Equation & Biological Redox 🎯

Key Takeaways — Part 7

AN-OX, RED-CAT: Anode = Oxidation; Cathode = Reduction — in both galvanic AND electrolytic cells.
Galvanic: spontaneous, $E° > 0$ , $\Delta G < 0$ ; Electrolytic: non-spontaneous, requires external energy.
$E°_{cell} = E°_{cathode} - E°_{anode}$ : the half-reaction with higher $E°$ is at the cathode.
$\Delta G° = -nFE°$ : positive $E°$ → negative $\Delta G°$ → $K > 1$ .
Nernst equation: more product ( $Q$ ↑) → lower cell potential; at equilibrium $E = 0$ .
Oxidizing agent gets reduced; reducing agent gets oxidized.
ETC connection: NADH (reducing agent) → Complex I → $O_{2}$ (final oxidizing agent) → $H_{2}O$ .

C_{2}H_{3}O_{2}^{-}

i(solution freezes ΔTf lower than pure solvent)

\Delta T_b = K_b \cdot m \cdot i \qquad (\text{solution boils } \Delta T_b \text{ higher than pure solvent})

[H^{+}] = 1.8 \times 1 0^{- 5} \times 0.10 = 1.8 \times 1 0^{- 6} \approx 1.34 \times 1 0^{- 3} M

\text{pH} \approx 2.87

Buffers resist pH change best when

\text{pH} \approx \text{p}K_a \pm 1

Adding base $(OH^{-})$ : consumed by weak acid →

\text{HA} + \text{OH}^- \to \text{A}^- + \text{H}_2\text{O}

\Delta G°' \approx -30.5

Δ G °^{'} \approx - 30.5

K_{p} = K_{c} (R T)^{Δ n_{g a s}}

concentration-independent

At equilibrium:

Q = K

E = 0

63.5×1.000.635×2×96,485=

General Chemistry

General Chemistry - Complete Interactive Lesson

Part 1: Atomic Structure & Periodic Trends

General Chemistry for the MCAT

Quantum Numbers

Three Key Rules

Electron Configuration & MCAT Exceptions

Standard Filling

Periodic Trends — Reason, Don't Memorize

Worked Example: Connecting Electron Config to Properties

Key Takeaways — Part 1

Part 2: Chemical Bonding

General Chemistry for the MCAT

Types of Chemical Bonds

Electronegativity Difference → Bond Type

Part 3: Stoichiometry & Solutions

General Chemistry for the MCAT

Stoichiometry: The Mole Map

Part 4: Acids, Bases & Buffers

General Chemistry for the MCAT

Acid-Base Theories

Part 5: Thermodynamics & Equilibrium

General Chemistry for the MCAT

Enthalpy, Entropy, and Free Energy

Enthalpy ()

Part 6: Chemical Kinetics

General Chemistry for the MCAT

Rate Laws & Reaction Orders

The Rate Law

Part 7: Electrochemistry & Redox

General Chemistry for the MCAT

Oxidation States & Half-Reactions

Assigning Oxidation States — Rules (in priority order)

High-Yield Exceptions (Memorize These Two)

Transition Metal Cations: Remove 4s First

Diamagnetic vs. Paramagnetic

Summary Table

IE1IE_{1}IE1​ Exceptions (MUST Know for MCAT)

Ionization Energy Jump Logic (MCAT Favorite)

Formal Charge

Resonance

VSEPR & Molecular Geometry

Polarity

Intermolecular Forces (IMFs)

London Dispersion Forces (LDF)

Dipole-Dipole Forces

Hydrogen Bonding

MCAT Application — Boiling Point Comparisons

Key Takeaways — Part 2

Four-Step Stoichiometry Workflow

Limiting Reagent

Percent Yield

Percent Composition

Solutions & Concentration

Key Concentration Units

Dilution

Solubility Rules (MCAT High-Yield)

Colligative Properties

Freezing Point Depression & Boiling Point Elevation

Osmotic Pressure

Vapor Pressure Lowering (Raoult's Law)

Key Takeaways — Part 3

Strong vs. Weak

pH Calculations

Fundamental Relationships

Strong Acid/Base pH (Direct)

Weak Acid pH (Approximate Formula)

Henderson-Hasselbalch Equation (Buffer pH)

Titration Curves & Buffer Behavior

Strong Acid + Strong Base

Weak Acid + Strong Base (Most Common MCAT Type)

Buffer Capacity

Physiological Buffer: Bicarbonate

Key Takeaways — Part 4

Entropy (ΔS\Delta SΔS)

Gibbs Free Energy (ΔG\Delta GΔG)

Spontaneity Analysis — The Four Cases

Connecting ΔG°\Delta G°ΔG° to Equilibrium

Reaction at Non-Standard Conditions

Chemical Equilibrium

$IE_{1}$ Exceptions (MUST Know for MCAT)

Entropy ( $\Delta S$ )

Gibbs Free Energy ( $\Delta G$ )

Connecting $\Delta G°$ to Equilibrium