General Chemistry

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.

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): $[\text{Ar}]\,3d^6\,4s^2$
• Cl (Z=17): $[\text{Ne}]\,3s^2\,3p^5$

High-Yield Exceptions (Memorize These Two)

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.

Quantum Numbers & Electron Configuration 🎯

Periodic Trends — Reason, Don't Memorize

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

$Z_{eff} = Z - S$

• $Z$ = 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.

Summary Table

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

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

1. 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.

2. 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

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 $[\text{Ar}]\,3d^5$. To get the neutral atom, add back 2 electrons. For transition metals, they go back to 4s:
$M = [\text{Ar}]\,3d^5\,4s^2$
Counting electrons: $\text{Ar}$ = 18, plus $3d^5$ = 5, plus $4s^2$ = 2 → total 25 electrons → 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.

Key Takeaways — Part 1

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

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

Formal Charge

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) → $\text{FC} = 4 - 0 - 4 = 0$ ✓

Resonance

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.

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

Polarity

A molecule is polar if:

1. It has polar bonds AND
2. 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

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).

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).

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

Four-Step Stoichiometry Workflow

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

Limiting Reagent

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.

Percent Yield

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

Percent Composition

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

Solutions & Concentration

Key Concentration Units

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 $Na^{+}$, $K^{+}$, $NH_{4}^{+}$, $NO_{3}^{-}$, $C_{2}H_{3}O_{2}^{-}$ salts

Usually soluble: halides ($Cl^{-}$, $Br^{-}$, $I^{-}$) except AgX, $PbX_{2}$, $Hg_{2}X_{2}$

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

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

Stoichiometry & Solutions 🎯

Colligative Properties

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

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

Freezing Point Depression & Boiling Point Elevation

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

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.

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

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 $e^{-}$ pair)

Strong vs. Weak

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 ($K_a = 1.8 \times 10^{-5}$): $[\text{H}^+] = \sqrt{1.8 \times 10^{-5} \times 0.10} = \sqrt{1.8 \times 10^{-6}} \approx 1.34 \times 10^{-3} \text{ M}$ $\text{pH} \approx 2.87$

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!)
• Buffers resist pH change best when $\text{pH} \approx \text{p}K_a \pm 1$

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:

Buffer Capacity

A buffer resists pH change by:

• Adding acid $(H^{+})$: consumed by conjugate base → $\text{A}^- + \text{H}^+ \to \text{HA}$
• Adding base $(OH^{-})$: consumed by weak acid → $\text{HA} + \text{OH}^- \to \text{A}^- + \text{H}_2\text{O}$

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 $(HCO_{3}^{-})$.

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 ($\Delta H$)

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

Biological example: ATP hydrolysis ($\text{ATP} + \text{H}_2\text{O} \to \text{ADP} + P_i$) has $\Delta G°' \approx -30.5$ kJ/mol under standard biochemical conditions — spontaneous, drives unfavorable reactions when coupled.

Connecting $\Delta G°$ to Equilibrium

$\Delta G° = -RT\ln K$

$\ln K = -\frac{\Delta G°}{RT}$

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 $Q = K$: $\Delta G = 0$ (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
• $K_p = K_c(RT)^{\Delta n_{gas}}$ where $\Delta n_{gas}$ = moles gas products − moles gas reactants

Le Chatelier's Principle

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

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.

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

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

• $k$ = 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:

Worked Example:

• 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: $t_{1/2} = \frac{\ln 2}{k} = \frac{0.693}{k}$ — concentration-independent (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 $[A]_0$

Graphical Identification

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

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.

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)

1. Free element = 0 (e.g., $\text{O}_2$, $\text{Fe}$)
2. Monatomic ion = ionic charge (e.g., $\text{Na}^+ = +1$, $\text{Cl}^- = -1$)
3. F is always $-1$ in compounds
4. O is usually $-2$ (exception: peroxides $-1$; $\text{OF}_2$ = $+2$)
5. H is usually $+1$ (exception: metal hydrides $= -1$)
6. 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)

OIL RIG

Oxidation Is Loss of electrons | Reduction Is Gain of electrons

Balancing Redox Half-Reactions (Acidic Solution)

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

Galvanic & Electrolytic Cells

Key Vocabulary

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:

$Cu^{2+}$ 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 $E°_{cell}$ → $K > 1$ → 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)
• At equilibrium: $Q = K$, $E = 0$

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) → $H_{2}O$.