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Química Geral · Chemical kinetics

Mecanismos de reação e a equação de Arrhenius

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O que você vai aprender

  • Assign the molecularity of an elementary step and write its rate law from its coefficients.
  • Derive the rate law a mechanism predicts from its rate-determining step, using a fast pre-equilibrium to remove intermediates.
  • Distinguish intermediates from catalysts and check that the steps of a mechanism add up to the overall equation.
  • Use collision theory and energy profiles to explain activation energy, orientation and the effect of temperature.
  • Calculate Ea, k or T with the Arrhenius equation, from two temperatures or from the slope of an Arrhenius plot.
  • Explain how homogeneous, heterogeneous and enzyme catalysts speed up reactions without changing ΔH or K.

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Elementary steps and molecularity

An elementary step is a single molecular event, written exactly as the particles collide or fall apart. Its molecularity is the number of reactant particles in that event: unimolecular (one particle, such as cyclopropane rearranging to propene, drawn), bimolecular (two) or termolecular (three). Three particles rarely meet at the same instant with the right energy and orientation, so termolecular steps are rare and higher ones are unknown. Only for an elementary step can you write the rate law from the equation: each reactant's coefficient becomes its order. NO + O₃ → NO₂ + O₂ is a bimolecular step with rate = k[NO][O₃]; a step 2 X + Y → products has rate = k[X]²[Y], so tripling [X] and halving [Y] changes its rate by 3² × ½ = 4.5. A unimolecular step has k in s⁻¹ and a bimolecular step k in M⁻¹ s⁻¹. The rate law of an overall reaction must be measured: a fractional order, such as the ½ order in Br₂ for the initial rate of H₂ + Br₂ → 2 HBr, proves at once that the reaction is not one elementary step.

Mechanisms and the rate-determining step

A mechanism is a sequence of elementary steps. Its steps must add up to the overall balanced equation, and it must predict the measured rate law. An intermediate is made in one step and used up in a later one, so it never appears in the overall equation; it can be a reactive species such as NO₃ or a stable molecule such as N₂O. When one step is much slower than the rest, it is the rate-determining step: product can appear no faster than that bottleneck allows. For NO₂ + CO → NO + CO₂ below 225 °C, the accepted mechanism is 2 NO₂ → NO₃ + NO (slow) followed by NO₃ + CO → NO₂ + CO₂ (fast). The slow step gives rate = k[NO₂]², which matches experiment, and CO does not appear because it reacts only after the slow step. Agreement with the rate law makes a mechanism plausible but never proves it. Order and molecularity are different ideas: 2 N₂O₅ → 4 NO₂ + O₂ is first order, yet no single step can turn two N₂O₅ into five product molecules.

A fast equilibrium before the slow step

If the slow step uses an intermediate, write the rate law of the slow step and then replace the intermediate's concentration using the equilibrium constant of the fast step before it. For 2 NO + Br₂ → 2 NOBr: NO + Br₂ ⇌ NOBr₂ (fast, K = k₁/k₋₁) and NOBr₂ + NO → 2 NOBr (slow, k₂). The slow step gives rate = k₂[NOBr₂][NO]; the equilibrium gives [NOBr₂] = K[NO][Br₂], so rate = k₂K[NO]²[Br₂], and the observed constant is k₂k₁/k₋₁. A product of the fast equilibrium can end up in the denominator: for ozone (drawn), O₃ ⇌ O₂ + O (fast) then O + O₃ → 2 O₂ (slow) gives [O] = K[O₃]/[O₂] and rate = k₂K[O₃]²/[O₂], so added O₂ slows the reaction. A final rate law contains only species you can add or measure, never an intermediate.

Collision theory and energy profiles

Molecules react only when they collide with at least the activation energy Ea and in a suitable orientation. The fraction of collisions with enough energy is about e^(−Ea/RT): for Ea = 50 kJ/mol at 298 K, Ea/RT = 50 000 ÷ (8.314 × 298) = 20.18 and the fraction is 1.7 × 10⁻⁹. Heating spreads the Maxwell–Boltzmann distribution to higher energies: the peak moves right and drops, the total area (the number of molecules) stays the same, and the tail beyond Ea grows sharply. Collision frequency rises only as √T, by 1.7 % from 300 K to 310 K, so the growing energetic fraction, not more collisions, is what speeds reactions up. The orientation factor p (k = p·Z·e^(−Ea/RT)) is the fraction of energetic collisions that are correctly aligned, often well below 1. On an energy profile each peak is a transition state and each valley between peaks is an intermediate. A step's barrier is measured from the valley it starts in, the highest transition state on the path sets the overall barrier, and for each step Ea(reverse) = Ea(forward) − ΔH, so an endothermic step always has Ea ≥ ΔH.

The Arrhenius equation

k = A e^(−Ea/RT), where A is the pre-exponential factor (same units as k), R = 8.314 J/(mol·K) and T is in kelvin. Taking logs gives ln k = ln A − (Ea/R)(1/T), so a plot of ln k against 1/T is a straight line of slope −Ea/R; with log₁₀ k the slope is −Ea/(2.303 R). For two temperatures, ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂). Worked example: k doubles between 300 K and 310 K, so Ea = R ln 2 ÷ (1/300 − 1/310) = 8.314 × 0.6931 ÷ (1.075 × 10⁻⁴ K⁻¹) = 53 600 J/mol ≈ 53.6 kJ/mol. Keep Ea in J/mol to match R, convert °C to kelvin, and use natural logs. A larger Ea makes k more sensitive to temperature, and comparing two reactions needs both A and Ea, because a large A can partly offset a high barrier.

Catalysis

A catalyst provides a different pathway with a lower activation energy and is regenerated, so a little of it does a lot of work. It lowers the forward and reverse barriers by the same amount, so ΔH and K are unchanged and equilibrium is only reached sooner. A catalyst is used in an early step and given back in a later one, whereas an intermediate is made first and used up later. Iodide catalyses the decomposition of hydrogen peroxide (drawn): H₂O₂ + I⁻ → H₂O + IO⁻ (slow), then H₂O₂ + IO⁻ → H₂O + O₂ + I⁻ (fast). I⁻ is the catalyst, IO⁻ the intermediate, and because I⁻ is in the slow step, rate = k[H₂O₂][I⁻]. A homogeneous catalyst is in the same phase as the reactants (I⁻ in water, Cl atoms destroying ozone in the stratosphere); a heterogeneous catalyst is a separate phase, usually a solid surface: in hydrogenation over nickel or platinum, H₂ and the alkene adsorb, H₂ splits into surface H atoms, the H atoms add one at a time and the alkane desorbs. Enzymes are protein catalysts that bind the substrate in an active site; at high substrate concentration every site is busy and the rate levels off.

What to carry forward

Write rate laws from coefficients only for elementary steps. For a mechanism, let the slow step set the rate law and use a fast pre-equilibrium to replace any intermediate. Reactions need energy above Ea and the right orientation; k = A e^(−Ea/RT) links rate to temperature, with Ea in J/mol and T in kelvin. Catalysts lower Ea for both directions and are regenerated; intermediates are made and then consumed.

Problemas de prática (40)

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Preveja o produto 5

Catalysis

  • Net result of the chlorine–ozone cyclePredict the net reaction of a catalytic cycle by adding its steps and cancelling the catalyst and intermediate.

Mechanisms and rate-determining steps

  • A fast equilibrium before the slow stepPredict a rate law by substituting a fast pre-equilibrium into the rate law of the slow step.
  • Rate law when the first step is slowPredict the rate law of a mechanism whose first step is rate-determining.
  • Ozone decomposition rate lawPredict a rate law with a negative order in a product that is formed in a fast pre-equilibrium.

Elementary steps and molecularity

  • Rate law of a termolecular stepWrite the rate law of a termolecular elementary step directly from its coefficients.

Classificar 5

Elementary steps and molecularity

  • Rate law and units for a unimolecular stepClassify the rate law and the units of k for a unimolecular elementary step.
  • Molecularity of a collision between two NO₂Classify an elementary step as unimolecular, bimolecular or termolecular by counting its reactant particles.

Catalysis

  • Spot the heterogeneous catalystClassify a catalysed process as homogeneous or heterogeneous by comparing the phases of catalyst and reactants.
  • Catalyst or intermediate?Classify the species of a catalysed mechanism as catalyst or intermediate from where each is consumed and formed.

Mechanisms and rate-determining steps

  • Find the intermediatesClassify the species in a three-step mechanism as reactants, products or intermediates by adding the steps.

Comparar 4

The Arrhenius equation

  • Rank rate constants from A and EaCompare rate constants at one temperature when both the pre-exponential factor and the activation energy differ.
  • Which rate is most sensitive to temperature?Compare how strongly the rate constants of reactions with different activation energies respond to the same temperature rise.

Catalysis

  • Hydrogenation on a metal surfaceOrder the events of heterogeneous catalytic hydrogenation of ethene on a metal surface.

Collisions and energy profiles

  • Rank the barriers of three stepsCompare the activation barriers of the individual steps of a three-step profile by measuring each from the valley where it starts.

Calcular 12

The Arrhenius equation

  • Activation energy from data at 25 °C and 65 °CCalculate an activation energy from rate constants measured at two Celsius temperatures, converting to kelvin first.
  • Activation energy of a rate that doublesCalculate the activation energy of a reaction whose rate constant doubles between two Celsius temperatures.
  • Rate constant at a higher temperatureCalculate a rate constant at a new temperature from one known rate constant and the activation energy.

Catalysis

  • Reverse barrier with a catalystCalculate the reverse activation energy of a catalysed reaction, using the fact that a catalyst leaves ΔH unchanged.
  • How hot to match the catalyst?Calculate the temperature at which an uncatalysed reaction would match the catalysed rate constant at 298 K, using the Arrhenius equation.
  • Rate constant of the iodide-catalysed stepCalculate the rate constant of a catalysed slow step from a measured consumption rate, accounting for the reactant used again in the fast step.

Collisions and energy profiles

  • How much more often do molecules collide?Calculate the percentage rise in collision frequency on heating, using its square-root dependence on absolute temperature.
  • Fraction of collisions with enough energyCalculate the fraction of collisions with at least the activation energy using the collision-theory factor e^(−Ea/RT).
  • Orientation factor from a measured rate constantCalculate the orientation (steric) factor p by comparing a measured rate constant with the collision-theory prediction.

Elementary steps and molecularity

  • Change concentrations in an elementary stepCalculate how the rate of a termolecular elementary step changes when two concentrations change at once.
  • Rate of the NO + O₃ stepCalculate the rate of a bimolecular elementary step from its rate constant and the concentrations of the two colliding species.

Mechanisms and rate-determining steps

  • Observed rate constant from a pre-equilibriumCalculate the observed rate constant of a pre-equilibrium mechanism from the elementary rate constants.

Encontre o erro 3

Elementary steps and molecularity

  • A rate law read from the balanced equationDiagnose the error of writing an overall reaction's rate law from its balanced equation.

Collisions and energy profiles

  • An impossible activation energyDiagnose an activation energy that is impossible because it is smaller than the step's endothermic ΔH.

Mechanisms and rate-determining steps

  • An intermediate left in the rate lawDiagnose a predicted rate law that still contains an intermediate and correct it with the pre-equilibrium constant.

Interprete as evidências 5

The Arrhenius equation

  • Activation energy from an Arrhenius plotInterpret the slope of a plot of ln k against 1/T as −Ea/R and calculate the activation energy.
  • Activation energy from a log₁₀ plotInterpret the slope of a plot of log₁₀ k against 1/T, including the factor ln 10 = 2.303, to obtain Ea.

Collisions and energy profiles

  • The energy distribution at a higher temperatureInterpret how the Maxwell–Boltzmann distribution changes when the temperature of a gas sample rises.
  • Overall barrier of a two-step profileRead the overall activation energy of a two-step reaction from the highest transition state on its energy profile.

Elementary steps and molecularity

  • Which reaction cannot be one step?Interpret measured rate laws to decide which overall reaction cannot be a single elementary step.

Trabalhe de trás para frente 1

  • Which mechanism fits rate = k[NO₂]²?Work back from a measured rate law to the proposed mechanism that is consistent with it.

Planeje uma síntese 1

  • How hot for a tenfold faster reaction?Plan the reaction temperature needed for a target rate increase by solving the Arrhenius equation for T.

Explique por quê 4

Elementary steps and molecularity

  • First order but not unimolecularExplain why a first-order overall reaction need not be a single unimolecular step, distinguishing order from molecularity.

Collisions and energy profiles

  • Why most collisions do not reactExplain the two requirements of collision theory that make most molecular collisions unproductive.

Catalysis

  • Why an enzyme-catalysed rate levels offExplain the saturation of an enzyme-catalysed rate at high substrate concentration in terms of occupied active sites.

Mechanisms and rate-determining steps

  • Why the slow step sets the rateExplain why the slowest step of a sequence controls the overall rate of reaction.

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