Kinetics: every key term you need (+ practice quiz)
90 flashcard terms for AP Chemistry Unit 5, written to match the course framework. Read them here, drill them as flashcards, or take the 27-question quiz. Free, no account needed.
Electrical work (wₑ = -nFE) and mechanical work (w = -PΔV) are both forms of energy transfer from chemical reactions.
Gibbs Energy and Electrochemistry
ΔG° = -nFE°; combines thermodynamics and electrochemistry to predict spontaneity of redox reactions.
Period 5 Big Picture
Thermodynamics and kinetics together determine if and how fast reactions occur. Spontaneity (ΔG), energy (ΔH), disorder (ΔS), and activation barriers (Ea) are interconnected in predicting and controlling chemical change.
Method of Initial Rates
Compare experiments where one concentration changes: if doubling [A] doubles rate, order 1; quadruples, order 2; no change, order 0. Sum gives overall order.
Units of k
Zero order: M/s; first order: s⁻¹; second order: M⁻¹s⁻¹. The units reveal the overall order of a rate law.
Integrated Rate Law Plots
Linear plots identify order: [A] vs t (zero), ln[A] vs t (first, slope −k), 1/[A] vs t (second, slope +k).
First-Order Half-Life
t½ = 0.693/k, independent of concentration; radioactive decay follows this. Second-order half-life = 1/(k[A]₀), lengthening as reactant depletes.
Rate-Determining Step
The slowest elementary step controls the overall rate; the rate law is written from that step's molecularity, substituting for intermediates if needed.
Intermediates vs. Catalysts
An intermediate is produced then consumed (appears in steps, not overall); a catalyst is consumed then regenerated and lowers activation energy.
Fast Equilibrium Pre-Step
If step 1 is a fast equilibrium and step 2 is slow, set forward rate = reverse rate for step 1 to express the intermediate in terms of reactants.
Molecularity
Unimolecular, bimolecular, termolecular elementary steps; termolecular steps are rare because three-body collisions are improbable.
Collision Theory Requirements
Reaction needs a collision with energy ≥ E_a and correct orientation; the fraction with sufficient energy rises exponentially with temperature.
Arrhenius Equation
k = A·e^(−Ea/RT); ln k vs 1/T is linear with slope −Ea/R. Larger E_a means k is more sensitive to temperature.
Two-Point Arrhenius
ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂) lets you find E_a from two rate constants or predict k at a new temperature.
Reaction Energy Profile
Peaks are transition states, valleys between them are intermediates; the tallest barrier from a valley marks the rate-determining step; ΔH is products minus reactants.
Catalysis Types
Homogeneous (same phase, e.g., acid catalysis), heterogeneous (surface adsorption, e.g., Pt in catalytic converters), enzymatic (active site lowers E_a).
Rate Law Is Experimental
Reaction orders cannot be read from overall stoichiometric coefficients — only from data or from a valid mechanism's slow step.
Relative Rates
For aA → bB, rate = −(1/a)Δ[A]/Δt = (1/b)Δ[B]/Δt; a product forming twice as fast as a reactant disappears indicates a 1:2 coefficient ratio.
Raising T does not change E_a; it increases the fraction of molecules with energy above E_a, so k rises roughly exponentially.
Extracting Orders from Ratios
Divide two rate expressions where only one concentration changes: if tripling [A] multiplies the rate by nine, the order in A is two because 3^m = 9.
Fractional and Zero Orders
A zero-order reactant does not appear in the rate law, which typically means a surface or enzyme is saturated so added reactant cannot increase the rate.
Half-Life Order Diagnostics
Only first-order half-life is constant. Zero-order half-lives shrink as reaction proceeds; second-order half-lives grow, which distinguishes the orders from data.
Linearized Plot Slopes
Zero order gives [A] versus t with slope −k; first order gives ln[A] with slope −k; second order gives 1/[A] with slope +k.
Estimating k from a Half-Life
For a first-order process, k = 0.693 ÷ t½. A 20-minute half-life corresponds to k ≈ 0.0347 min⁻¹ regardless of starting concentration.
Mechanism Validity Tests
A proposed mechanism must sum to the overall equation and its rate-determining step must reproduce the experimental rate law after substituting for intermediates.
Substituting Out an Intermediate
When a fast pre-equilibrium precedes the slow step, set forward and reverse rates equal and solve for the intermediate's concentration in terms of reactants.
Steady-State Intuition
A reactive intermediate stays at low, nearly constant concentration because it is consumed as fast as it forms, which is why it never appears in the final rate law.
Catalyst Signature in a Mechanism
A catalyst is consumed in an early step and regenerated later, so it appears among the reactants of one elementary step and the products of another.
Homogeneous vs. Heterogeneous Catalysis
Homogeneous catalysts share the reactants' phase, as with aqueous acid catalysis, while heterogeneous ones provide a surface, as in a platinum catalytic converter.
Enzyme Specificity and Saturation
Enzymes bind a specific substrate in an active site; once every site is occupied the rate becomes zero order in substrate and depends only on enzyme amount.
Arrhenius Plot Interpretation
Plot ln k against 1/T. The slope equals −Ea/R, so a steeper negative slope means a larger activation energy and greater temperature sensitivity.
Reading a Multi-Step Energy Profile
Each hump is a transition state and each valley between humps is an intermediate. The tallest hump measured from its preceding valley is the rate-determining step.
Why Catalysts Do Not Shift Equilibrium
A catalyst lowers the activation energy of forward and reverse paths equally, speeding both directions and leaving ΔH and the equilibrium position unchanged.
Temperature Effect Magnitude
Rate constants often roughly double per 10 °C rise because the exponential term e^(−Ea/RT) is extremely sensitive to temperature near room conditions.
Orientation Factor
Even sufficiently energetic collisions fail unless the molecules are aligned properly, which is why complex molecules react more slowly than simple ones at equal energy.
Relating Rates of Different Species
For a A + b B → c C, −(1/a)Δ[A]/Δt = −(1/b)Δ[B]/Δt = (1/c)Δ[C]/Δt, so the species with the largest coefficient changes fastest.