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Química Geral · Exam review

Revisão para a prova: cinética química

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O que você vai aprender

  • Relate the rates of change of different species through the balanced equation and find orders and rate constants, with units, from initial-rate data.
  • Use the zero-, first- and second-order integrated rate laws and half-lives to find concentrations, times, rate constants and the order from data, including gas-pressure and pseudo-first-order data.
  • Derive the rate law of a mechanism from its rate-determining step, using a fast pre-equilibrium to remove intermediates, and identify intermediates and catalysts.
  • Apply the two-temperature Arrhenius equation and Arrhenius plots to find rate constants, activation energies and half-lives at a new temperature.
  • Explain what a catalyst does and does not change, read energy profiles for catalysed and uncatalysed paths, and spot the common calculation traps.

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Rates and rate laws from data

Divide each rate of change by its coefficient to get one rate of reaction. For 2 N₂O₅ → 4 NO₂ + O₂, rate = −½ Δ[N₂O₅]/Δt = ¼ Δ[NO₂]/Δt = Δ[O₂]/Δt, so NO₂ appears four times as fast as O₂ forms. A rate law, rate = k[A]^m[B]^n, is found by experiment, never read from the coefficients. Method of initial rates: pick two runs in which only one concentration changes; then rate₂/rate₁ = (c₂/c₁)^m. Tripling [A] with a ninefold rate gives m = 2; an unchanged rate gives m = 0; a halved rate when a concentration doubles gives m = −1, which happens when that species is a product of a fast pre-equilibrium that comes before the slow step. When two concentrations change together, divide out the known order first. Then calculate k from any run and check it on the others. Units: k has units M^(1−n) s⁻¹ for overall order n (s⁻¹ for first order, M⁻¹ s⁻¹ for second, M⁻² s⁻¹ for third). Ester hydrolysis by hydroxide (ethyl acetate and OH⁻ are drawn) is a typical second-order case, first order in each reactant. Isolation: with every other reactant in large excess, its concentration is effectively constant, and the run shows only the order in the scarce reactant (a pseudo-order); its pseudo rate constant contains the excess concentrations.

Integrated rate laws and half-lives

Zero order: [A] = [A]₀ − kt, a straight line of [A] against t, t½ = [A]₀/2k (each half-life is half the one before). First order: ln[A] = ln[A]₀ − kt, a straight line of ln[A] against t, t½ = ln 2/k = 0.693/k, the same at every concentration. Second order in one reactant: 1/[A] = 1/[A]₀ + kt, a straight line of 1/[A] against t, t½ = 1/(k[A]₀), so each half-life is twice the one before. Worked example: a first-order reaction with k = 0.0231 min⁻¹ has t½ = 0.693/0.0231 = 30.0 min, so after 90 min, three half-lives, 12.5 % remains (e^(−0.0231 × 90) = 0.125). A second-order reaction with k = 0.50 M⁻¹ s⁻¹ and [A]₀ = 0.20 M has t½ = 1/(0.50 × 0.20) = 10 s, then 20 s, then 40 s. Keep k and t in the same time unit. For a gas, pressure is proportional to concentration at constant T and V: for A(g) → 2 B(g) starting with pure A at P₀, P_total = P₀ + (P₀ − P_A), so P_A = 2P₀ − P_total; azomethane (drawn), CH₃N=NCH₃(g) → C₂H₆(g) + N₂(g), behaves this way. Equal fractions lost in equal times mean first order; equal amounts lost mean zero order.

Mechanisms and the rate-determining step

An elementary step happens as written, so its rate law follows its molecularity: A → products is unimolecular (rate = k[A]); A + B is bimolecular (k[A][B]); termolecular steps are rare. The steps must add to the overall equation. An intermediate is made in one step and used in a later one; a catalyst is used first and regenerated. A slow first step sets the rate law directly: if A + B → C + D is slow and C + A → E is fast, rate = k₁[A][B], even though two A are used overall. When a slow step follows a fast equilibrium, write the slow step's rate law, then remove the intermediate with the equilibrium constant. For 2 NO + Br₂ → 2 NOBr with NO + Br₂ ⇌ NOBr₂ (fast, K) and NOBr₂ + NO → 2 NOBr (slow), [NOBr₂] = K[NO][Br₂], so rate = k₂K[NO]²[Br₂]. A dissociation such as Cl₂ ⇌ 2 Cl gives [Cl] = (K[Cl₂])^½ and a half order; a species produced in the pre-equilibrium appears in the denominator. Products formed in fast steps after the slow one still form at a rate fixed by the slow step times their stoichiometric count. A mechanism that matches the rate law is consistent, not proven; detecting a predicted intermediate is stronger support. Chloroform, CHCl₃ (drawn), reacts with Cl₂ through a Cl₂ ⇌ 2 Cl pre-equilibrium of this kind.

Temperature and the Arrhenius equation

k = A e^(−Ea/RT): only collisions with at least Ea, and a suitable orientation, react. Two temperatures: ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂), with T in kelvin, Ea in J/mol and R = 8.314 J mol⁻¹ K⁻¹. Worked example: Ea = 50.0 kJ/mol from 300 K to 310 K gives (50 000/8.314)(1/300 − 1/310) = 0.647, so k rises by e^0.647 = 1.91. A plot of ln k against 1/T is a straight line with slope −Ea/R, so Ea = −slope × R; a steeper line means a larger Ea and a stronger temperature dependence. Any quantity proportional to k can replace it in the ratio: an initial rate at the same concentrations, 1/t½ for a first-order reaction, or 1/(time to reach the same conversion). If the concentrations differ between the two temperatures, calculate k at each temperature first. Use the natural logarithm; log₁₀ needs a factor of 2.303.

Catalysis and energy profiles

A catalyst provides a different mechanism with a lower activation energy and is regenerated. It lowers the forward and the reverse barriers by the same amount, so it speeds up both directions equally: ΔH, ΔG°, K and the equilibrium composition do not change; equilibrium is only reached sooner. On an energy profile, Ea(reverse) = Ea(forward) − ΔH; for an exothermic reaction with Ea = 80 kJ/mol and ΔH = −30 kJ/mol, the reverse barrier is 110 kJ/mol. SO₂ and SO₃ are drawn. A homogeneous catalyst is in the same phase as the reactants (NO in the gas-phase oxidation of SO₂, Mn²⁺ ions in solution); a heterogeneous catalyst is a different phase, usually a solid surface (Pt, Pd, Ni, Fe). A catalyst can appear in the rate law when it takes part in the slow step. Enzymes are protein catalysts: their rate rises with temperature only until the protein unfolds (denatures), above about 40 °C for most human enzymes, and then activity falls.

Exam traps

Convert °C to kelvin and kJ to J before using R. Divide a measured rate of change by the coefficient before using a rate law written for the rate of reaction. Use ln, not log₁₀. Do not apply a constant half-life to a second- or zero-order reaction. A rate constant changes only with temperature (and catalyst), never with concentration. Read pressures carefully: total pressure is not reactant pressure. Never leave an intermediate in a final rate law. A pseudo rate constant is not the true k until the excess concentration is divided out.

What to carry forward

Orders come from experiment, half-lives tell the order, the slow step (with any fast pre-equilibrium) sets the rate law, and ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂) links temperatures. A catalyst lowers both barriers equally, so it changes rates but never K or ΔH. Check units, kelvin, J and stoichiometric factors in every calculation. Methyl isocyanide (drawn) isomerizes to acetonitrile, CH₃NC → CH₃CN, a classic first-order gas-phase reaction for half-life and Arrhenius problems.

Problemas de prática (40)

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Preveja o produto 4

Mechanisms and rate-determining steps

  • A half order from a dissociation pre-equilibriumPredict a fractional-order rate law from a mechanism with a fast dissociation equilibrium before the slow step.
  • Compress a gas reacting by a pre-equilibriumDerive a rate law from a fast pre-equilibrium mechanism and predict how compressing the gas changes the rate.
  • Rate law with a slow first stepPredict the rate law of a mechanism whose first step is rate-determining.

Rate laws from data

  • Double one reactant, quarter the otherPredict the factor by which a rate changes when two concentrations change at the same time in a known rate law.

Classificar 5

Arrhenius and catalysis

  • Adding a catalyst to an equilibrium mixtureClassify the effects of a catalyst on rates, activation energies, ΔH and the equilibrium constant.

Mechanisms and rate-determining steps

  • Manganese catalysis of cerium(IV) and thallium(I)Identify the catalyst, the intermediates and the rate law of a three-step catalytic mechanism with a slow first step.
  • Intermediates and transition statesDistinguish an intermediate from a transition state on an energy profile.
  • Catalyst or intermediate in SO₂ oxidation?Classify the species in a two-step gas-phase mechanism as catalyst or intermediate and identify the type of catalysis.

Integrated rate laws and half-life

  • Order from how the half-life scalesClassify a reaction's order from the way its half-life changes with the starting pressure.

Comparar 3

Arrhenius and catalysis

  • Rank the energies on a catalysed profileUse Ea(reverse) = Ea(forward) − ΔH to compare forward and reverse barriers with and without a catalyst.

Integrated rate laws and half-life

  • Which reaction has the most left?Compare the fractions remaining after a fixed time for reactions of different orders using their integrated rate laws.

Rate laws from data

  • Rank mixtures by initial rateRank reaction mixtures by initial rate using a rate law that is second order in one reactant.

Calcular 15

Arrhenius and catalysis

  • Activation energy from two half-livesConvert first-order half-lives at two temperatures into a rate-constant ratio and calculate the activation energy.
  • Activation energy when the concentrations differCalculate rate constants from rates measured at different concentrations and temperatures, then find the activation energy.
  • From 20 °C to 40 °CUse the two-temperature Arrhenius equation to find how much a rate constant increases between two Celsius temperatures.
  • Half-life after a 20 K riseUse the Arrhenius equation to find a rate constant at a new temperature and convert it into a first-order half-life.

Integrated rate laws and half-life

  • Shelf life of an aspirin solutionFind the time for a first-order reaction to reach a given fractional conversion from its half-life.
  • Half-life from total pressureConvert total-pressure readings into reactant pressure and use the first-order integrated rate law to find a half-life.
  • How long until the alcohol level falls?Apply the zero-order integrated rate law to a saturated enzyme process.
  • Ethylene formed from cyclobutaneCombine the first-order integrated rate law with reaction stoichiometry to find a product concentration.
  • Methyl isocyanide left after an hourUse the first-order integrated rate law with a unit conversion to find the percentage of reactant remaining.
  • Nitrosyl bromide left after 40 sUse the second-order integrated rate law to find a concentration after a given time.

Rate laws from data

  • Crystal violet half-life in dilute baseCombine pseudo-first-order rate constants measured at two hydroxide concentrations to find the true rate law and a half-life under new conditions.
  • Rate constant for ester saponificationFind both orders from initial-rate data with non-doubling concentration changes and calculate a second-order rate constant.
  • A rate constant with an inverse orderFind a negative reaction order from initial-rate data and use it to calculate the rate constant and its units.
  • How fast is iodide used up?Convert the rate of formation of one product into the rate of consumption of a reactant using the coefficients of the balanced equation.

Mechanisms and rate-determining steps

  • How fast does HCl form?Use a mechanism's rate-determining step and the stoichiometry of the overall reaction to calculate the rate of formation of a product.

Encontre o erro 3

Arrhenius and catalysis

  • Find the error in an activation energyDiagnose a logarithm error in a two-temperature Arrhenius calculation and give the corrected activation energy.

Rate laws from data

  • Find the error in a rate constantDiagnose a rate-constant calculation that confuses a reactant's rate of disappearance with the rate of reaction.

Integrated rate laws and half-life

  • Find the error in a three-half-life shortcutDiagnose the misuse of a constant half-life for a second-order reaction and correct the time.

Interprete as evidências 4

Arrhenius and catalysis

  • Two lines on an Arrhenius plotInterpret the slopes of two Arrhenius plots in terms of activation energy and temperature sensitivity.

Rate laws from data

  • Rate law for iron(III) and iodideDeduce a rate law from initial-rate data that includes a run in which two concentrations change together.
  • Is hydroxide in the rate law?Read a zero order and a first order from initial-rate data in which one concentration change leaves the rate unchanged.

Integrated rate laws and half-life

  • Order and half-life from a sucrose tableInterpret concentration–time data at equal intervals to find the order and the half-life of a reaction.

Trabalhe de trás para frente 1

  • Which mechanism fits rate = k[NO₂][F₂]?Choose the mechanism whose rate-determining step and pre-equilibria reproduce a measured rate law.

Planeje uma síntese 1

  • Isolate the order in iodidePlan a single kinetic run that isolates the order in one reactant by flooding the other.

Explique por quê 4

Mechanisms and rate-determining steps

  • What does a matching rate law show?Explain why agreement between a mechanism and a measured rate law supports but does not prove the mechanism.
  • Why does hydroxide slow the reaction?Explain a negative order in a species as the result of that species being produced in a fast pre-equilibrium.

Arrhenius and catalysis

  • Why an enzyme slows down when heatedExplain why an enzyme-catalysed rate falls above an optimum temperature even though rate constants normally rise with temperature.
  • A reaction that slows down when heatedExplain a negative apparent activation energy using a mechanism with an exothermic pre-equilibrium.

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