DAT Formulas

Formula recall — Overall aerobic respiration

C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O + energy | Variables and units: C₆H₁₂O₆ = glucose; O₂ = oxygen; CO₂ = carbon dioxide; H₂O = water. The equation tracks matter; energy is captured mainly as ATP. Key condition: Modern estimates are usually about 30-32 ATP per glucose in eukaryotes, depending on shuttle and leak.

Application scenario — A question asks which reactants are consumed and which products form when glucose is completely oxidized with oxygen. Which relationship should you use?

Overall aerobic respiration: C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O + energy | Recognition cue: Modern estimates are usually about 30-32 ATP per glucose in eukaryotes, depending on shuttle and leak.

Formula recall — Exponential population growth

dN/dt = rN | Variables and units: N = population size; t = time; dN/dt = population change per time; r = intrinsic per-capita growth rate (time⁻¹). Key condition: N = population size; r = intrinsic per-capita growth rate.

Application scenario — A small population grows with unlimited resources and a constant per-capita growth rate. Which relationship should you use?

Exponential population growth: dN/dt = rN | Recognition cue: N = population size; r = intrinsic per-capita growth rate.

Formula recall — Logistic population growth

dN/dt = rN(1 - N/K) | Variables and units: N = population size; r = intrinsic growth rate; K = carrying capacity; (1 − N/K) = environmental resistance. Key condition: K = carrying capacity; growth slows as N approaches K.

Application scenario — A population initially grows quickly but slows as it approaches the environment's carrying capacity. Which relationship should you use?

Logistic population growth: dN/dt = rN(1 - N/K) | Recognition cue: K = carrying capacity; growth slows as N approaches K.

Formula recall — Hardy-Weinberg allele frequencies

p + q = 1 | Variables and units: p and q = frequencies of the two alleles; each ranges from 0 to 1. Key condition: For two alleles, p and q are their frequencies.

Application scenario — You are given the frequency of one of two alleles and need the other allele's frequency. Which relationship should you use?

Hardy-Weinberg allele frequencies: p + q = 1 | Recognition cue: For two alleles, p and q are their frequencies.

Formula recall — Hardy-Weinberg genotype frequencies

p² + 2pq + q² = 1 | Variables and units: p² = first homozygote; 2pq = heterozygote; q² = second homozygote; p + q = 1. Key condition: p² = homozygous dominant; 2pq = heterozygous; q² = homozygous recessive.

Application scenario — A recessive phenotype frequency is given and you need carrier or genotype frequencies. Which relationship should you use?

Hardy-Weinberg genotype frequencies: p² + 2pq + q² = 1 | Recognition cue: p² = homozygous dominant; 2pq = heterozygous; q² = homozygous recessive.

Formula recall — Product rule

P(A and B) = P(A) × P(B) | Variables and units: P(A), P(B) = probabilities of independent events; P(A and B) = probability both occur. Key condition: Use for independent events that must both occur.

Application scenario — Two independent genetic events must both occur in the same offspring. Which relationship should you use?

Product rule: P(A and B) = P(A) × P(B) | Recognition cue: Use for independent events that must both occur.

Formula recall — Sum rule

P(A or B) = P(A) + P(B) - P(A and B) | Variables and units: P(A or B) = probability at least one occurs; subtract P(A and B) to avoid double-counting overlap. Key condition: For mutually exclusive outcomes, the overlap term is zero.

Application scenario — A genetic outcome can occur through either of two routes and you need the probability of at least one route. Which relationship should you use?

Sum rule: P(A or B) = P(A) + P(B) - P(A and B) | Recognition cue: For mutually exclusive outcomes, the overlap term is zero.

Formula recall — Overall photosynthesis

6CO₂ + 6H₂O + light → C₆H₁₂O₆ + 6O₂ | Variables and units: CO₂ = carbon source; H₂O = electron source; light supplies energy; glucose stores chemical energy; O₂ is released from water. Key condition: The simplified net equation reverses the overall matter flow of respiration.

Application scenario — A question asks which matter enters and leaves during the simplified net photosynthesis reaction. Which relationship should you use?

Overall photosynthesis: 6CO₂ + 6H₂O + light → C₆H₁₂O₆ + 6O₂ | Recognition cue: The simplified net equation reverses the overall matter flow of respiration.

Formula recall — Acid dissociation

K_a = [H₃O⁺][A⁻]/[HA] | Variables and units: Kₐ = acid-dissociation constant; brackets = equilibrium molarity; HA = weak acid; A⁻ = conjugate base; H₃O⁺ = hydronium. Key condition: For HA + H₂O ⇌ H₃O⁺ + A⁻; omit liquid water.

Application scenario — Equilibrium concentrations of a weak acid, hydronium, and conjugate base are given and you need Kₐ. Which relationship should you use?

Acid dissociation: K_a = [H₃O⁺][A⁻]/[HA] | Recognition cue: For HA + H₂O ⇌ H₃O⁺ + A⁻; omit liquid water.

Formula recall — Base dissociation

K_b = [BH⁺][OH⁻]/[B] | Variables and units: K_b = base-dissociation constant; B = weak base; BH⁺ = conjugate acid; OH⁻ = hydroxide; brackets = equilibrium molarity. Key condition: For B + H₂O ⇌ BH⁺ + OH⁻.

Application scenario — A weak base establishes equilibrium in water and you need K_b from equilibrium concentrations. Which relationship should you use?

Base dissociation: K_b = [BH⁺][OH⁻]/[B] | Recognition cue: For B + H₂O ⇌ BH⁺ + OH⁻.

Formula recall — Conjugate acid-base relation

K_aK_b = K_w; pK_a + pK_b = pK_w | Variables and units: Kₐ and K_b belong to a conjugate pair; K_w = water ion-product; pK = −log K; at 25 °C, K_w = 1.0×10⁻¹⁴ and pK_w = 14.00. Key condition: Applies to a conjugate acid-base pair.

Application scenario — Kₐ of an acid is given and you need K_b of its conjugate base at 25 °C. Which relationship should you use?

Conjugate acid-base relation: K_aK_b = K_w; pK_a + pK_b = pK_w | Recognition cue: Applies to a conjugate acid-base pair.

Formula recall — Henderson-Hasselbalch

pH = pK_a + log([A⁻]/[HA]) | Variables and units: pH = buffer pH; pKₐ = −logKₐ; [A⁻] = conjugate-base concentration; [HA] = weak-acid concentration. A⁻/HA may also use mole ratios after reaction stoichiometry. Key condition: Use for a buffer containing a weak acid and its conjugate base.

Application scenario — A buffer contains known amounts of a weak acid and its conjugate base and you need its pH. Which relationship should you use?

Henderson-Hasselbalch: pH = pK_a + log([A⁻]/[HA]) | Recognition cue: Use for a buffer containing a weak acid and its conjugate base.

Formula recall — pH

pH = -log[H₃O⁺] | Variables and units: [H₃O⁺] = hydronium molarity; pH is dimensionless. Inverse: [H₃O⁺] = 10⁻ᵖᴴ. Key condition: [H₃O⁺] = 10^(-pH).

Application scenario — The hydronium concentration is given and you need acidity on the logarithmic pH scale. Which relationship should you use?

pH: pH = -log[H₃O⁺] | Recognition cue: [H₃O⁺] = 10^(-pH).

Formula recall — pKa and pKb

pK_a = -logK_a; pK_b = -logK_b | Variables and units: Kₐ/K_b = dissociation constants; pKₐ/pK_b = negative base-10 logarithms. Smaller pK means the corresponding acid/base is stronger. Key condition: Lower pK_a means a stronger acid; lower pK_b means a stronger base.

Application scenario — A dissociation constant is given and you need its pK value or need to rank acid/base strength. Which relationship should you use?

pKa and pKb: pK_a = -logK_a; pK_b = -logK_b | Recognition cue: Lower pK_a means a stronger acid; lower pK_b means a stronger base.

Formula recall — pOH

pOH = -log[OH⁻] | Variables and units: [OH⁻] = hydroxide molarity; pOH is dimensionless. Inverse: [OH⁻] = 10⁻ᵖᴼᴴ. Key condition: [OH⁻] = 10^(-pOH).

Application scenario — The hydroxide concentration is given and you need pOH. Which relationship should you use?

pOH: pOH = -log[OH⁻] | Recognition cue: [OH⁻] = 10^(-pOH).

Formula recall — Water relation at 25 °C

K_w = [H₃O⁺][OH⁻] = 1.0 × 10⁻¹⁴; pH + pOH = 14 | Variables and units: K_w = water ion-product; [H₃O⁺], [OH⁻] in mol/L. At 25 °C, K_w = 1.0×10⁻¹⁴ and pK_w = 14.00. Key condition: The numeric value changes with temperature, but the DAT typically assumes 25 °C.

Application scenario — You know pH or one ion concentration and need pOH or the other ion concentration at 25 °C. Which relationship should you use?

Water relation at 25 °C: K_w = [H₃O⁺][OH⁻] = 1.0 × 10⁻¹⁴; pH + pOH = 14 | Recognition cue: The numeric value changes with temperature, but the DAT typically assumes 25 °C.

Formula recall — Electronic transition energy

ΔE = E_final - E_initial | Variables and units: E_final and E_initial = energy levels in J per atom; ΔE > 0 means absorption and ΔE < 0 means emission. Photon magnitude is |ΔE|. Key condition: ΔE > 0 means absorption; ΔE < 0 means emission.

Application scenario — An electron changes from one stated energy level to another and you must determine whether energy is absorbed or emitted. Which relationship should you use?

Electronic transition energy: ΔE = E_final - E_initial | Recognition cue: ΔE > 0 means absorption; ΔE < 0 means emission.

Formula recall — Hydrogen energy level

Eₙ = -R_H/n² | Variables and units: Eₙ = energy of hydrogen level n in J/atom; R_H = 2.178×10⁻¹⁸ J; n = positive principal quantum number. Key condition: R_H = 2.178 × 10⁻¹⁸ J per H atom; n = principal level.

Application scenario — You need the energy of an electron in a particular principal level of a hydrogen atom. Which relationship should you use?

Hydrogen energy level: Eₙ = -R_H/n² | Recognition cue: R_H = 2.178 × 10⁻¹⁸ J per H atom; n = principal level.

Formula recall — Photon energy

E = hν = hc/λ | Variables and units: E = photon energy (J); h = 6.626×10⁻³⁴ J·s; ν = frequency (s⁻¹ or Hz); c = 3.00×10⁸ m/s; λ = wavelength (m). Key condition: h = 6.626 × 10⁻³⁴ J·s. Higher frequency means higher energy; longer wavelength means lower energy.

Application scenario — A photon wavelength or frequency is provided and you need its energy. Which relationship should you use?

Photon energy: E = hν = hc/λ | Recognition cue: h = 6.626 × 10⁻³⁴ J·s. Higher frequency means higher energy; longer wavelength means lower energy.

Formula recall — Wave relationship

c = νλ | Variables and units: c = 3.00×10⁸ m/s in vacuum; ν = frequency in Hz or s⁻¹; λ = wavelength in metres. Key condition: c = 3.00 × 10⁸ m/s; ν = frequency in s⁻¹; λ = wavelength in m.

Application scenario — Electromagnetic wavelength is given and you need frequency, or vice versa. Which relationship should you use?

Wave relationship: c = νλ | Recognition cue: c = 3.00 × 10⁸ m/s; ν = frequency in s⁻¹; λ = wavelength in m.

Formula recall — Cell potential and equilibrium

ΔG° = -RT lnK = -nFE°; therefore lnK = nFE°/(RT) | Variables and units: ΔG° = standard free energy (J/mol); R = 8.314 J·mol⁻¹·K⁻¹; T = K; K = equilibrium constant; n = mol e⁻ transferred; F = 96485 C/mol e⁻; E° = V. Key condition: A positive E° corresponds to K > 1 for the written reaction.

Application scenario — A standard electrochemical potential is given and you need the equilibrium constant or standard free energy. Which relationship should you use?

Cell potential and equilibrium: ΔG° = -RT lnK = -nFE°; therefore lnK = nFE°/(RT) | Recognition cue: A positive E° corresponds to K > 1 for the written reaction.

Formula recall — Electric charge

q = It | Variables and units: q = charge in coulombs; I = current in amperes (C/s); t = seconds. Key condition: q in coulombs; I in amperes (C/s); t in seconds.

Application scenario — A current passes through an electrolytic cell for a measured time and you need total charge. Which relationship should you use?

Electric charge: q = It | Recognition cue: q in coulombs; I in amperes (C/s); t in seconds.

Formula recall — Free energy from cell potential

ΔG° = -nFE°_cell | Variables and units: ΔG° = J/mol reaction; n = moles of electrons in the balanced reaction; F = 96485 C/mol e⁻; E°cell = volts (J/C). Key condition: n = moles of electrons; F = 96485 C/mol e⁻.

Application scenario — A balanced redox reaction and E°cell are given and you need ΔG°. Which relationship should you use?

Free energy from cell potential: ΔG° = -nFE°_cell | Recognition cue: n = moles of electrons; F = 96485 C/mol e⁻.

Formula recall — Nernst equation at 25 °C

E = E° - (0.0592 V/n)logQ | Variables and units: E and E° = cell potential in V; n = electrons transferred; Q = reaction quotient; 0.0592 V applies at 25 °C with log base 10. Key condition: n = electrons transferred; Q = reaction quotient.

Application scenario — A galvanic cell operates under nonstandard concentrations at 25 °C and you need its voltage. Which relationship should you use?

Nernst equation at 25 °C: E = E° - (0.0592 V/n)logQ | Recognition cue: n = electrons transferred; Q = reaction quotient.

Formula recall — Standard cell potential

E°_cell = E°_cathode - E°_anode | Variables and units: E°cell, E°cathode, E°anode in V. Use tabulated reduction potentials; do not multiply E° by stoichiometric coefficients. Key condition: Use tabulated reduction potentials. Cathode is reduction; anode is oxidation.

Application scenario — Two standard reduction potentials are given and you need the cell's standard voltage. Which relationship should you use?

Standard cell potential: E°_cell = E°_cathode - E°_anode | Recognition cue: Use tabulated reduction potentials. Cathode is reduction; anode is oxidation.

Formula recall — Equilibrium constant

K_c = [C]^c[D]^d / ([A]^a[B]^b) | Variables and units: [A] etc. = equilibrium molarity; lowercase a,b,c,d = balanced coefficients and exponents. Omit pure solids and liquids. Key condition: For aA + bB ⇌ cC + dD. Omit pure solids and liquids.

Application scenario — An equilibrium reaction and equilibrium concentrations are given and you need K_c. Which relationship should you use?

Equilibrium constant: K_c = [C]^c[D]^d / ([A]^a[B]^b) | Recognition cue: For aA + bB ⇌ cC + dD. Omit pure solids and liquids.

Formula recall — Gas equilibrium conversion

K_p = K_c(RT)^Δn_gas | Variables and units: K_p = pressure-based constant; K_c = concentration-based constant; R = gas constant; T = K; Δn_gas = gaseous-product coefficients minus gaseous-reactant coefficients. Key condition: Δn_gas = gaseous product coefficients - gaseous reactant coefficients.

Application scenario — K_c is known for a gaseous reaction and the question asks for K_p at a specified temperature. Which relationship should you use?

Gas equilibrium conversion: K_p = K_c(RT)^Δn_gas | Recognition cue: Δn_gas = gaseous product coefficients - gaseous reactant coefficients.

Formula recall — Reaction quotient

Q_c has the same form as K_c but uses current concentrations | Variables and units: Q has the same expression as K but uses current—not necessarily equilibrium—values. Compare Q with K to predict direction. Key condition: Q < K: forward; Q > K: reverse; Q = K: equilibrium.

Application scenario — Current concentrations are given and you must predict whether the reaction shifts forward or backward. Which relationship should you use?

Reaction quotient: Q_c has the same form as K_c but uses current concentrations | Recognition cue: Q < K: forward; Q > K: reverse; Q = K: equilibrium.

Formula recall — Solubility product

K_sp = product of dissolved-ion concentrations, each raised to its coefficient | Variables and units: K_sp = equilibrium constant for dissolution; brackets = equilibrium ion molarity; each exponent is the ion's coefficient. Omit the solid. Key condition: Example: CaF₂ ⇌ Ca²⁺ + 2F⁻, so K_sp = [Ca²⁺][F⁻]².

Application scenario — A sparingly soluble ionic solid dissolves and you need the ion-product expression or molar solubility relationship. Which relationship should you use?

Solubility product: K_sp = product of dissolved-ion concentrations, each raised to its coefficient | Recognition cue: Example: CaF₂ ⇌ Ca²⁺ + 2F⁻, so K_sp = [Ca²⁺][F⁻]².

Formula recall — Standard free energy and equilibrium

ΔG° = -RT ln K | Variables and units: ΔG° = standard free energy (J/mol); R = 8.314 J·mol⁻¹·K⁻¹; T = K; K = dimensionless equilibrium constant; ln = natural log. Key condition: K > 1 gives ΔG° < 0; K < 1 gives ΔG° > 0.

Application scenario — An equilibrium constant at a stated temperature is given and you need ΔG° or spontaneity under standard conditions. Which relationship should you use?

Standard free energy and equilibrium: ΔG° = -RT ln K | Recognition cue: K > 1 gives ΔG° < 0; K < 1 gives ΔG° > 0.

Formula recall — Maximum electrons in shell n

maximum electrons = 2n² | Variables and units: n = principal energy level (1,2,3…); 2n² = maximum number of electrons in that shell. Key condition: n is the principal energy level.

Application scenario — A principal shell number is given and you need its maximum electron capacity. Which relationship should you use?

Maximum electrons in shell n: maximum electrons = 2n² | Recognition cue: n is the principal energy level.

Formula recall — Avogadro's law

V₁/n₁ = V₂/n₂ | Variables and units: V₁,V₂ = gas volumes; n₁,n₂ = moles. Pressure and absolute temperature must stay constant. Key condition: Constant pressure and temperature.

Application scenario — Gas moles change while pressure and temperature remain constant; predict the new volume. Which relationship should you use?

Avogadro's law: V₁/n₁ = V₂/n₂ | Recognition cue: Constant pressure and temperature.

Formula recall — Boyle's law

P₁V₁ = P₂V₂ | Variables and units: P₁,P₂ = initial/final pressure; V₁,V₂ = initial/final volume. Temperature and moles stay constant. Key condition: Constant temperature and moles; pressure and volume are inverse.

Application scenario — A gas is compressed at constant temperature; pressure and volume change while moles remain fixed. Which relationship should you use?

Boyle's law: P₁V₁ = P₂V₂ | Recognition cue: Constant temperature and moles; pressure and volume are inverse.

Formula recall — Charles's law

V₁/T₁ = V₂/T₂ | Variables and units: V₁,V₂ = volumes; T₁,T₂ = absolute temperatures in K. Pressure and moles stay constant. Key condition: Constant pressure and moles; use kelvin.

Application scenario — A flexible container is heated at constant pressure and you need its new volume. Which relationship should you use?

Charles's law: V₁/T₁ = V₂/T₂ | Recognition cue: Constant pressure and moles; use kelvin.

Formula recall — Combined gas law

P₁V₁/T₁ = P₂V₂/T₂ | Variables and units: P = pressure; V = volume; T = kelvin; subscripts 1 and 2 are initial/final states. Moles stay constant. Key condition: Constant moles; all temperatures in kelvin.

Application scenario — A sealed gas sample changes pressure, volume, and temperature while moles stay constant. Which relationship should you use?

Combined gas law: P₁V₁/T₁ = P₂V₂/T₂ | Recognition cue: Constant moles; all temperatures in kelvin.

Formula recall — Dalton's law

P_total = ΣPᵢ | Variables and units: P_total = mixture pressure; Pᵢ = each gas's partial pressure. All pressures must use the same unit. Key condition: For an ideal gas mixture, total pressure is the sum of partial pressures.

Application scenario — Several gases share a container and you need total pressure from their partial pressures. Which relationship should you use?

Dalton's law: P_total = ΣPᵢ | Recognition cue: For an ideal gas mixture, total pressure is the sum of partial pressures.

Formula recall — Gay-Lussac's law

P₁/T₁ = P₂/T₂ | Variables and units: P₁,P₂ = pressures; T₁,T₂ = kelvin. Volume and moles stay constant. Key condition: Constant volume and moles; use kelvin.

Application scenario — A rigid sealed container is heated and you need the new pressure. Which relationship should you use?

Gay-Lussac's law: P₁/T₁ = P₂/T₂ | Recognition cue: Constant volume and moles; use kelvin.

Formula recall — General combined gas relation

P₁V₁/(n₁T₁) = P₂V₂/(n₂T₂) | Variables and units: P = pressure; V = volume; n = moles; T = kelvin. Use consistent units between states. Key condition: Use when pressure, volume, moles, and temperature may all change.

Application scenario — Pressure, volume, temperature, and amount of gas may all differ between two states. Which relationship should you use?

General combined gas relation: P₁V₁/(n₁T₁) = P₂V₂/(n₂T₂) | Recognition cue: Use when pressure, volume, moles, and temperature may all change.

Formula recall — Graham's law

r₁/r₂ = √(M₂/M₁) | Variables and units: r₁,r₂ = diffusion/effusion rates; M₁,M₂ = molar masses. The lighter gas is faster. Key condition: Effusion/diffusion rate is inversely proportional to the square root of molar mass.

Application scenario — You must compare how quickly two gases effuse using their molar masses. Which relationship should you use?

Graham's law: r₁/r₂ = √(M₂/M₁) | Recognition cue: Effusion/diffusion rate is inversely proportional to the square root of molar mass.

Formula recall — Ideal gas law

PV = nRT | Variables and units: P = pressure; V = volume; n = mol; T = K; R must match units: 0.08206 L·atm·mol⁻¹·K⁻¹ or 8.314 L·kPa·mol⁻¹·K⁻¹. Key condition: Common R: 0.08206 L·atm·mol⁻¹·K⁻¹ or 8.314 L·kPa·mol⁻¹·K⁻¹.

Application scenario — A single gas state gives three of P, V, n, and T and asks for the fourth. Which relationship should you use?

Ideal gas law: PV = nRT | Recognition cue: Common R: 0.08206 L·atm·mol⁻¹·K⁻¹ or 8.314 L·kPa·mol⁻¹·K⁻¹.

Formula recall — Ideal-gas density

ρ = PM/(RT) | Variables and units: ρ = gas density (usually g/L); P = pressure; M = molar mass (g/mol); R = matched gas constant; T = K. Key condition: P = pressure; M = molar mass; R = gas constant; T = kelvin.

Application scenario — Pressure, temperature, and molar mass are known and you need gas density, or density is used to identify molar mass. Which relationship should you use?

Ideal-gas density: ρ = PM/(RT) | Recognition cue: P = pressure; M = molar mass; R = gas constant; T = kelvin.

Formula recall — Partial pressure

Pᵢ = χᵢP_total | Variables and units: Pᵢ = partial pressure; χᵢ = mole fraction nᵢ/n_total; P_total = total mixture pressure. Key condition: χᵢ = mole fraction of gas i.

Application scenario — A gas's mole fraction and total mixture pressure are given and you need that gas's pressure. Which relationship should you use?

Partial pressure: Pᵢ = χᵢP_total | Recognition cue: χᵢ = mole fraction of gas i.

Formula recall — Pressure

P = F/A | Variables and units: P = pressure; F = perpendicular force; A = area. SI unit Pa = N/m². Key condition: Pressure is force per unit area.

Application scenario — A force is applied over a known surface area and you need pressure. Which relationship should you use?

Pressure: P = F/A | Recognition cue: Pressure is force per unit area.

Formula recall — Arrhenius equation

k = Ae^(-E_a/RT) | Variables and units: k = rate constant; A = frequency/orientation factor; Eₐ = activation energy (J/mol if R = 8.314); R = 8.314 J·mol⁻¹·K⁻¹; T = K. Key condition: A = frequency factor; E_a = activation energy; T = kelvin.

Application scenario — Activation energy and temperature are known and you need the rate constant's temperature dependence. Which relationship should you use?

Arrhenius equation: k = Ae^(-E_a/RT) | Recognition cue: A = frequency factor; E_a = activation energy; T = kelvin.

Formula recall — First-order half-life

t₁/₂ = ln2/k = 0.693/k | Variables and units: t₁/₂ = half-life; k = first-order rate constant (time⁻¹); ln2 = 0.693. It does not depend on [A]₀. Key condition: Independent of initial concentration.

Application scenario — A first-order process gives k and asks how long half the reactant takes to disappear. Which relationship should you use?

First-order half-life: t₁/₂ = ln2/k = 0.693/k | Recognition cue: Independent of initial concentration.

Formula recall — First-order integrated law

ln[A]_t = ln[A]_0 - kt | Variables and units: [A]₀ = initial concentration; [A]ₜ = concentration at time t; k = time⁻¹; t = time; ln = natural log. Key condition: Equivalent: [A]_t = [A]_0e^(-kt).

Application scenario — A first-order reaction asks for concentration after a specified time or for k from concentration data. Which relationship should you use?

First-order integrated law: ln[A]_t = ln[A]_0 - kt | Recognition cue: Equivalent: [A]_t = [A]_0e^(-kt).

Formula recall — General rate law

rate = k[A]^m[B]^n | Variables and units: rate = concentration/time; k = rate constant; [A],[B] = reactant concentrations; m,n = experimentally determined orders; overall order = m+n. Key condition: Orders m and n are determined experimentally; overall order = m + n.

Application scenario — Initial-rate experiments change reactant concentrations and you must determine orders or predict rate. Which relationship should you use?

General rate law: rate = k[A]^m[B]^n | Recognition cue: Orders m and n are determined experimentally; overall order = m + n.

Formula recall — Second-order half-life

t₁/₂ = 1/(k[A]_0) | Variables and units: t₁/₂ = half-life; k = concentration⁻¹·time⁻¹; [A]₀ = initial concentration. Key condition: Inversely proportional to initial concentration.

Application scenario — A second-order reaction gives k and initial concentration and asks for its half-life. Which relationship should you use?

Second-order half-life: t₁/₂ = 1/(k[A]_0) | Recognition cue: Inversely proportional to initial concentration.

Formula recall — Second-order integrated law

1/[A]_t = 1/[A]_0 + kt | Variables and units: [A]₀ and [A]ₜ = initial/current concentration; k = concentration⁻¹·time⁻¹; t = time. Key condition: Linear plot: 1/[A] vs t; slope = k.

Application scenario — A second-order reaction asks for concentration after time or produces a linear 1/[A] versus t plot. Which relationship should you use?

Second-order integrated law: 1/[A]_t = 1/[A]_0 + kt | Recognition cue: Linear plot: 1/[A] vs t; slope = k.

Formula recall — Two-temperature Arrhenius form

ln(k₂/k₁) = -E_a/R(1/T₂ - 1/T₁) | Variables and units: k₁,k₂ = rate constants; Eₐ = J/mol; R = 8.314 J·mol⁻¹·K⁻¹; T₁,T₂ = K. Key condition: Use to compare rate constants at two temperatures.

Application scenario — Rate constants at two temperatures are compared, or one k and Eₐ are used to find the other k. Which relationship should you use?

Two-temperature Arrhenius form: ln(k₂/k₁) = -E_a/R(1/T₂ - 1/T₁) | Recognition cue: Use to compare rate constants at two temperatures.