Chemical Kinetics – Lessons 19 & 20

Collision Theory

  • Fundamental assertion: every chemical reaction requires a successful collision between reactant particles.

    • “Successful” means:

    • Sufficient kinetic energy Ea\ge Ea (activation energy).

    • Correct molecular orientation so that the appropriate bonds can form / break.

  • Sequence of events (illustrated with CO+NO<em>2    CO</em>2+NO\text{CO} + \text{NO}<em>2 \;\rightarrow\; \text{CO}</em>2 + \text{NO}):

    1. Gas-phase molecules move randomly with a spectrum of speeds/kinetic energies.

    2. Low-energy or mis-oriented collisions → elastic bounce; molecules remain intact.

    3. High-energy, properly oriented collision → formation of an activated (transition-state) complex.

    • Simultaneous bond-breaking and bond-forming.

    • Extremely unstable; exists for only an instant.

    1. Activated complex relaxes → stable product molecules.

  • Energy Profile Diagram:

    • Vertical axis = Potential Energy; horizontal = Reaction Progress / Coordinate.

    • Peak corresponds to activated complex.

    • EaEa (activation energy) = energy gap between reactants and peak.

    • Only those reactant molecules with kinetic energy >Ea can surmount the barrier and form products.

Maxwell–Boltzmann Distribution & Factors Affecting Rate

Maxwell–Boltzmann Curve Basics

  • At any fixed temperature, molecules possess a distribution of kinetic energies.

  • Curve shape: rises sharply from zero energy, peaks (most probable energy), then decays with a long tail.

  • The area under the curve = total number of molecules.

  • Plotting a vertical line at EaEa divides:

    • Left region: molecules with E<Ea (cannot react upon collision).

    • Right region: molecules with E>Ea (can react if orientation is correct).

Parameters that Alter Reaction Rate

  1. Surface Area (for heterogeneous solids or large particles):

    • Smaller particle size ⇒ larger surface area per unit mass ⇒ more exposed reactant sites ⇒ higher collision frequency ⇒ faster rate.

  2. Concentration (solutions) / Pressure (gases):

    • Higher concentration or pressure ⇒ more particles per unit volume ⇒ higher collision frequency.

    • Maxwell–Boltzmann picture: curve does not move, but the number of collisions with E>Ea per unit time increases.

  3. Temperature:

    • Raising TT shifts the entire Boltzmann curve rightwards & flattens it.

    • Consequences:

      • Average kinetic energy increases.

      • Significantly larger fraction of molecules possess E>Ea.

      • Both collision frequency and fraction of successful collisions rise ⇒ rate increases markedly.

  4. Catalyst:

    • Provides an alternative reaction pathway with lower activation energy Ea<em>12Ea<em>12.

    • Boltzmann curve unchanged, but vertical EaEa line moves left ⇒ many more molecules have E> Ea_1 ⇒ higher successful-collision frequency.

Rate of Reaction

  • Generic stoichiometric equation: aA+bB    cC+dDaA + bB \;\rightarrow\; cC + dD.

  • Rate is expressed as time derivative of concentration: rate=1aΔ[A]Δt=1bΔ[B]Δt=1cΔ[C]Δt=1dΔ[D]Δtrate = -\frac{1}{a}\frac{\Delta[A]}{\Delta t} = -\frac{1}{b}\frac{\Delta[B]}{\Delta t} = \frac{1}{c}\frac{\Delta[C]}{\Delta t} = \frac{1}{d}\frac{\Delta[D]}{\Delta t}

    • Negative signs on reactants ensure positive rate values (reactant concentrations decrease).

Empirical Rate Law (rate equation)

  • Form: rate=k[A]m[B]nrate = k[A]^m[B]^n

    • [A],[B][A], [B] = instantaneous molar concentrations.

    • kk = rate constant (temperature dependent only).

    • m,nm, n = orders with respect to each reactant – obtained experimentally, NOT from stoichiometry (except for elementary steps).

  • Typical integer orders: 0, 1, 2 (can occasionally be non-integer or fractional).

Determining Reaction Order – Example (tert-butyl bromide hydrolysis)

  • Reaction: (CH3)3CBr+OH(CH3)3COH+Br(CH3)3CBr + OH^- \rightarrow (CH3)3COH + Br^-

  • Experimental data:

    • Doubling [(CH3)3CBr][(CH3)3CBr] doubles rate → first order in [(CH3)3CBr][(CH3)3CBr].

    • Doubling [OH][OH^-] leaves rate unchanged → zero order in [OH][OH^-].

  • Rate law: rate=k[(CH3)3CBr]1[OH]0=k[(CH3)3CBr]rate = k[(CH3)3CBr]^1[OH^-]^0 = k[(CH3)3CBr].

  • Overall order: 1+0=11+0 = 1 (first order overall).

Practice Dataset (A + B₂ → AB₂)

  • Use initial-rate method to determine:

    1. Order in [A][A].

    2. Order in [B2][B_2].

    3. Overall order (sum).

    4. Rate law.

    5. Calculate kk with units; predict rate change when both reactant concentrations are doubled.

    • (Dataset supplied in transcript: three experiments with specified [A], [B₂], rates.)

Integrated Rate Laws

  • Provide concentration–time relationships that integrate the differential rate laws.

Order

Differential Law

Integrated Form

Linear Plot (y vs x)

Slope mm

Intercept cc

0

rate=krate = k

C=kt+C0C = -kt + C_0

CC vs tt

k-k

C0C_0

1

rate=kCrate = kC

lnC=kt+lnC0\ln C = -kt + \ln C_0

lnC\ln C vs tt

k-k

lnC0\ln C_0

2

rate=kC2rate = kC^2

1C=kt+1C0\frac{1}{C} = kt + \frac{1}{C_0}

1/C1/C vs tt

kk

1/C01/C_0

  • Linearisation makes it easy to identify reaction order experimentally – plot each of the three suggested graphs and see which is linear.

Half-Life (t1/2t_{1/2})

  • Definition: time required for [reactant] to drop to half of C0C_0.

Order

t1/2t_{1/2} expression

Dependence on C0C_0

0

t1/2=C02kt{1/2}=\dfrac{C0}{2k}

directly proportional (longer half-life with higher initial concentration)

1

t1/2=ln2kt_{1/2}=\dfrac{\ln 2}{k}

independent of C0C_0 (constant)

2

t1/2=1kC0t{1/2}=\dfrac{1}{kC0}

inversely proportional (shorter half-life with higher C0C_0)

  • Utility: examining how half-life varies with C0C_0 reveals reaction order swiftly.

Worked Practice Highlights (mentioned in slides)

  1. Cyclopropane → propene (first order, k=6.7×104  s1,  C0=0.05  Mk = 6.7\times10^{-4}\;s^{-1},\;C_0 = 0.05\;M):

    • Use lnC=kt+lnC0\ln C = -kt + \ln C_0 to compute [cyclopropane] after 30 min.

    • Rearrange for tt when C=0.01  MC = 0.01\;M.

    • t1/2=ln2/kt_{1/2}=\ln 2/k (convert seconds to minutes).

  2. 2NOBr2NO+Br22\,NOBr\rightarrow 2\,NO + Br2 (second order; k=0.810  M1s1k = 0.810\;M^{-1}s^{-1}): compute [NOBr] after 10 min and half-life at given C0C0 using 1/C=kt+1/C01/C = kt + 1/C_0.

Reaction Mechanisms

  • Elementary step: single molecular-level event with its own rate law whose order equals stoichiometric coefficients.

  • Reaction mechanism: sequence of elementary steps summing to overall reaction.

  • Intermediate: produced in one step, consumed in a later step; not present initially or in final equation.

  • Catalyst: present at start and regenerated at end; participates but not consumed.

  • Rate-determining step (RDS): slowest elementary step; its rate law becomes overall rate law (assuming any preceding steps are at equilibrium or fast).

Example: Iodide-catalysed H2O2\text{H}2\text{O}2 Decomposition

Overall: H2O2+IH2O+IO\text{H}2\text{O}2 + I^- \rightarrow \text{H}2\text{O} + IO^- (Step 1, slow) H2O2+IOH2O+O2+I\text{H}2\text{O}2 + IO^- \rightarrow \text{H}2\text{O} + O_2 + I^- (Step 2, fast)

  • Intermediate: IOIO^-.

  • Catalyst: II^- (consumed Step 1, regenerated Step 2).

  • Observed rate law rate=k[H2O2][I]rate = k[H2O2][I^-] matches slow Step 1, confirming it as RDS.

Criteria for a Valid Mechanism

  1. Sum of elementary steps reproduces overall balanced equation.

  2. Predicted overall rate law (from RDS and any pre-equilibria) matches experimentally determined rate law.

Practice Scenarios from Slides

  1. Hypochlorite self-oxidation (two-step proposition): write individual rate equations, infer overall order from half-life constancy (constant t1/2t_{1/2} ⇒ first order), identify RDS accordingly.

  2. 2H2+2NON2+2H2O2H2 + 2NO \rightarrow N2 + 2H_2O with three alternative mechanisms:

    • For each, write rate law from slow step; compare to empirical rate=k[H2][NO]2rate = k[H_2][NO]^2 to decide which mechanism agrees (students should deduce Mechanism III etc.).

Arrhenius Equation

  • Quantifies temperature dependence of rate constant kk: k=AeEa/RTk = A e^{-Ea/RT} where:

    • AA = frequency factor (collision frequency × orientation factor).

    • EaEa = activation energy (kJ mol1^{-1}).

    • R=8.314×103  kJ mol1K1R = 8.314 \times 10^{-3}\;\text{kJ mol}^{-1}\,\text{K}^{-1}.

    • TT = absolute temperature (K).

Implications

  • Higher TT → exponent less negative → larger kk → faster reaction.

  • Higher EaEa (for given TT) → smaller kk → slower reaction.

  • Presence of catalyst lowers EaEa (and may slightly affect AA) → increases kk.

Linear Form (for data analysis)

  • Taking natural log: lnk=lnAEaR1T\ln k = \ln A - \frac{Ea}{R}\,\frac{1}{T}.

    • Plot of lnk\ln k vs 1/T1/T is straight line: slope =Ea/R= -Ea/R, intercept =lnA=\ln A.

Practice Example (Reactions P & Q)

  • Energy profiles provided: EaP=25  kJ mol1,  EaQ=50  kJ mol1EaP = 25\;\text{kJ mol}^{-1},\;EaQ = 50\;\text{kJ mol}^{-1} at T=298KT = 298\,K.

  • Plug into Arrhenius form to compute kP,kQkP, kQ at 298 K (illustrated in slides), then repeat at 348 K (298 K + 50 K increase) to show more pronounced increase for reaction with higher EaEa.

  • In presence of catalyst: EaEa decreases, kk increases (value of AA commonly similar but may vary slightly).

  • Apply collision theory to rationalise how surface area, concentration/pressure, temperature, catalysts influence reaction rate.

  • Formulate the empirical rate law, determine individual and overall reaction orders experimentally.

  • Utilise integrated rate laws and half-life relationships to classify