Sign up freeSign in

Explore · Flashcard deck

AP Chemistry: Kinetics, Thermochemistry and Electrochemistry (High School)

34 cards · 9 topics · by Quizcard team · 0 hearts · Updated 2026-11-04

Rate LawsReaction MechanismsActivation Energy & CatalystsEnthalpy & CalorimetryHess's LawEntropyGibbs Free EnergyElectrochemical CellsFaraday's Law
Quiz me (15 questions) Copy to my decks Make your own flashcards

Shared from Quizcard. Flip through the cards below, or copy the deck to edit and study it in your own account — free.

  Space flips · arrows move · Esc exits

Rate Laws

What is a rate law?

Answer: An equation relating reaction rate to the concentrations of reactants, each raised to an experimentally determined power (order): rate = k[A]^m[B]^n.

Explanation

k is the rate constant; m and n are NOT taken from stoichiometric coefficients — they must be found experimentally.

How is reaction order determined experimentally?

Answer: By the method of initial rates: compare how rate changes when one reactant's concentration is varied while others are held constant.

Explanation

If doubling [A] quadruples rate, order in A is 2; if rate doubles, order is 1; if rate is unchanged, order is 0.

Zero-order reaction: rate law and units of k?

Answer: Rate = k; units of k are mol·L⁻¹·s⁻¹ (rate is independent of concentration).

First-order reaction: rate law, integrated form, and half-life?

Answer: Rate = k[A]; ln[A] = -kt + ln[A]₀; t₁/₂ = ln2/k. Half-life is constant, independent of concentration.

Explanation

Units of k: s⁻¹.

What does [A] vs. time look like for a first-order reaction?

Answer: An exponential decay curve — concentration falls by half every constant time interval (t₁/₂).

Explanation

Example: starting at 1.0 M with t1/2=1 s gives 0.5 M at t=1s, 0.25 M at t=2s, 0.125 M at t=3s.

Second-order reaction: rate law and units of k?

Answer: Rate = k[A]²; units of k are L·mol⁻¹·s⁻¹.

Explanation

Integrated form: 1/[A] = kt + 1/[A]₀, giving a straight line when 1/[A] is plotted vs t.

Reaction Mechanisms

What is the rate-determining step (RDS)?

Answer: The slowest elementary step in a multistep reaction mechanism; it controls the overall rate of reaction.

How do you write the overall rate law from a mechanism?

Answer: Use the rate law of the rate-determining step; if it contains an intermediate, substitute using a prior fast equilibrium step.

Explanation

Example: if RDS is Step2: X + C → D and X is an intermediate from fast Step1 A+B⇌X, express [X] via the equilibrium constant of Step1.

Activation Energy & Catalysts

What is activation energy (Ea)?

Answer: The minimum energy required for reactants to reach the transition state and form products.

Explanation

Higher Ea means a slower reaction at a given temperature.

Identify this diagram: what does it show about catalysts?

Answer: A reaction-coordinate (energy) diagram comparing catalyzed vs. uncatalyzed pathways — the catalyst lowers the activation energy peak but ΔH (start/end energy) is unchanged.

Explanation

Both curves start and end at the same energy levels; only the height of the peak (Ea) differs.

How do catalysts speed up reactions?

Answer: They provide an alternate mechanism with lower activation energy, without being consumed; they do not change ΔH or the equilibrium position.

Explanation

Catalysts increase both forward and reverse rates equally, so K is unaffected.

Write the Arrhenius equation.

Answer: k = A·e^(−Ea/RT)

Explanation

k = rate constant, A = frequency factor, Ea = activation energy, R = gas constant, T = temperature (K). Higher T or lower Ea increases k.

Enthalpy & Calorimetry

Define enthalpy of reaction (ΔH) and its sign conventions.

Answer: Heat absorbed or released at constant pressure. ΔH < 0 = exothermic (releases heat); ΔH > 0 = endothermic (absorbs heat).

Give the calorimetry equation and define each term.

Answer: q = mcΔT, where q = heat (J), m = mass (g), c = specific heat capacity (J/g·°C), ΔT = temperature change (°C).

Explanation

Example: 50.0 g water (c=4.18 J/g°C) heated from 20°C to 30°C: q = 50.0×4.18×10 = 2090 J.

What is specific heat capacity?

Answer: The amount of heat required to raise the temperature of 1 gram of a substance by 1°C.

Explanation

Water's high specific heat (4.18 J/g·°C) makes it resist temperature change.

Hess's Law

State Hess's Law.

Answer: The total enthalpy change for a reaction is the same whether it occurs in one step or several steps; ΔH is a state function.

What rules apply when combining reactions by Hess's Law?

Answer: If a reaction is reversed, flip the sign of ΔH; if a reaction is multiplied by a factor, multiply ΔH by that same factor.

Explanation

Add the adjusted ΔH values of all steps to get the overall ΔH.

What is standard enthalpy of formation (ΔH°f)?

Answer: The enthalpy change when 1 mole of a compound forms from its elements in their standard states; ΔH°f of an element in its standard state = 0.

Formula for ΔH°rxn using enthalpies of formation?

Answer: ΔH°rxn = Σ n·ΔH°f(products) − Σ n·ΔH°f(reactants)

Explanation

n = stoichiometric coefficients. Example: for CH4+2O2→CO2+2H2O, sum products' ΔHf minus reactants' ΔHf (O2=0).

Entropy

Define entropy (S).

Answer: A measure of the dispersal of energy/matter in a system, often described as the degree of disorder or number of microstates.

List the general sign rules for ΔS in a reaction.

Answer: ΔS > 0 when: gas moles increase, a solid/liquid becomes a gas, a solid dissolves, or particles/complexity increase (e.g., one molecule splits into more).

Explanation

ΔS < 0 for the reverse trends (gas→liquid/solid, fewer moles of gas).

Formula for ΔS°rxn?

Answer: ΔS°rxn = Σ n·S°(products) − Σ n·S°(reactants)

Explanation

Unlike ΔH°f, elements have nonzero standard molar entropy (S° ≠ 0).

Gibbs Free Energy

Write the Gibbs free energy equation.

Answer: ΔG = ΔH − TΔS

Explanation

ΔG = Gibbs free energy change, ΔH = enthalpy change, T = temperature (K), ΔS = entropy change.

How does the sign of ΔG indicate spontaneity?

Answer: ΔG < 0: spontaneous (forward, favored); ΔG > 0: nonspontaneous (reverse favored); ΔG = 0: system at equilibrium.

When is a reaction spontaneous at all temperatures vs. temperature-dependent?

Answer: ΔH<0,ΔS>0: always spontaneous. ΔH>0,ΔS<0: never spontaneous. ΔH<0,ΔS<0: spontaneous at low T. ΔH>0,ΔS>0: spontaneous at high T.

Explanation

Determined by the sign of ΔG = ΔH − TΔS across temperatures.

How does ΔG° relate to the equilibrium constant K?

Answer: ΔG° = −RT ln K

Explanation

R = 8.314 J/mol·K, T in Kelvin. Large K (products favored) gives a very negative ΔG°; K=1 gives ΔG°=0.

How does ΔG° relate to standard cell potential E°?

Answer: ΔG° = −nFE°

Explanation

n = moles of electrons transferred, F = Faraday's constant (96,485 C/mol e⁻). Positive E° gives negative ΔG° (spontaneous).

How are K and E°cell linked (combining the two Gibbs relations)?

Answer: Setting −nFE° = −RT ln K gives E° = (RT/nF) ln K — a larger K corresponds to a larger (more positive) E°.

Explanation

This connects thermodynamics (K), free energy (ΔG°), and electrochemistry (E°) into one consistent framework.

Electrochemical Cells

What is a galvanic (voltaic) cell?

Answer: An electrochemical cell that uses a spontaneous redox reaction (ΔG<0, E°>0) to generate electrical energy.

What is an electrolytic cell?

Answer: An electrochemical cell that uses an external electrical energy source to drive a nonspontaneous redox reaction (ΔG>0, E°<0).

Explanation

Used in electroplating and electrolysis (e.g., decomposing water).

Compare electrode charge/sign in galvanic vs. electrolytic cells.

Answer: Galvanic: anode is negative, cathode is positive. Electrolytic: anode is positive, cathode is negative (power source reverses polarity).

Explanation

In both cell types, oxidation always occurs at the anode and reduction always at the cathode.

Mnemonic for anode/cathode and oxidation/reduction?

Answer: "An Ox, Red Cat": Anode = Oxidation; Cathode = Reduction.

Explanation

Electrons flow from anode to cathode through the external circuit.

Formula for standard cell potential E°cell?

Answer: E°cell = E°cathode − E°anode (using standard reduction potentials for both).

Explanation

A positive E°cell indicates a spontaneous galvanic reaction.

Faraday's Law

Faraday's law: How many grams of Cu are deposited by 2.00 A of current for 965 s in electroplating (Cu²⁺ + 2e⁻ → Cu)?

Answer: ≈0.636 g Cu.

Explanation

Charge q=It=2.00×965=1930 C. Moles e⁻=1930/96485=0.0200 mol. Moles Cu=0.0200/2=0.0100 mol. Mass=0.0100×63.5 g/mol=0.635 g.

More Chemistry decks

  • AP Chemistry: Equilibrium, Acids and Bases36 cards · 12 topics
  • Chemistry: Stoichiometry and Chemical Reactions30 cards · 9 topics
  • Chemistry: Atomic Structure and Periodic Trends30 cards · 6 topics

All Chemistry decks → · Explore everything

Make your own deck like this

Upload a PDF, Word doc, or slides and Quizcard writes the cards for you, or build them by hand. Free to start, no credit card required.

Get startedRead the guides