Chemical Kinetics: Reaction Rates, Rate Laws, Collision Theory, Arrhenius Equation, and Mechanisms

Chemical Kinetics & Stoichiometric Rate Math

  • Reaction Rate Definition: The change in concentration of reactants or products over time.

  • General Rate Formula: For aA+bB→cC+dDaA + bB \rightarrow cC + dD:   raterxn=−1aΔ[A]Δt=−1bΔ[B]Δt=1cΔ[C]Δt=1dΔ[D]Δt\text{rate}_{rxn} = -\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}

  • Why the Math Works:

    • Negative Signs: Reactant concentrations decrease over time (Δ[A]=[A]final−[A]initial<0\Delta[A] = [A]_{final} - [A]_{initial} < 0). The negative sign ensures the overall reaction rate is always a positive number.

    • Stoichiometric Coefficients (1a,1b,etc.\frac{1}{a}, \frac{1}{b}, \text{etc.}): Dividing by coefficients normalizes consumption/formation rates so that raterxn\text{rate}_{rxn} remains identical regardless of which chemical species is measured.

  • Beer-Lambert Law for Spectrophotometry:   Abs=εlcAbs = \varepsilon l c

    • Why: Absorbance (AbsAbs) is directly proportional to concentration (cc). Rearranging to c=Absεlc = \frac{Abs}{\varepsilon l} allows continuous tracking of concentration changes in colored reactants or products over time.

Rate Laws & The Method of Initial Rates

  • Rate Law Equation:   raterxn=k[A]x[B]y\text{rate}_{rxn} = k [A]^x [B]^y

    • kk = rate constant (temperature-dependent).

    • x,yx, y = reaction orders determined experimentally.

    • Overall reaction order n=x+yn = x + y

  • Why: Reaction orders (x,yx, y) reflect how many reactant molecules must collide in the rate-determining step, not the overall balanced stoichiometric coefficients.

  • Method of Initial Rates Math:   To solve for order xx holding [B][B] constant across Trials 1 and 2:   rate1rate2=k[A]1x[B]1yk[A]2x[B]2y=([A]1[A]2)x\frac{\text{rate}_1}{\text{rate}_2} = \frac{k [A]_1^x [B]_1^y}{k [A]_2^x [B]_2^y} = \left(\frac{[A]_1}{[A]_2}\right)^x   ln⁡(rate1rate2)=xln⁡([A]1[A]2)  ⟹  x=ln⁡(rate1rate2)ln⁡([A]1[A]2)\ln\left(\frac{\text{rate}_1}{\text{rate}_2}\right) = x \ln\left(\frac{[A]_1}{[A]_2}\right) \implies x = \frac{\ln\left(\frac{\text{rate}_1}{\text{rate}_2}\right)}{\ln\left(\frac{[A]_1}{[A]_2}\right)}

  • Dimensional Analysis for Rate Constant kk Units:   units of k=rateconcentrationn=M s−1Mn=M1−n s−1\text{units of } k = \frac{\text{rate}}{\text{concentration}^n} = \frac{M\,s^{-1}}{M^n} = M^{1-n}\,s^{-1}

    • Zero-Order (n=0n=0): M s−1M\,s^{-1}

    • First-Order (n=1n=1): s−1s^{-1}

    • Second-Order (n=2n=2): M−1 s−1M^{-1}\,s^{-1}

Collision Theory & Arrhenius Equation Math

  • Arrhenius Equation Forms:

    • Exponential: k=Ae(−EaRT)k = A e^{\left(-\frac{E_a}{R T}\right)}

    • Linear (y=mx+by = mx + b): ln⁡(k)=−EaR(1T)+ln⁡(A)\ln(k) = -\frac{E_a}{R}\left(\frac{1}{T}\right) + \ln(A)

    • Two-Point Equation: ln⁡(k2k1)=EaR(1T1−1T2)\ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)

    • Constants: R=8.314 J K−1 mol−1R = 8.314\,J\,K^{-1}\,mol^{-1}, TT in Kelvin (KK).

    • Problem: The rate constant for a reaction is k1=2.7×10−4 M−1 s−1k_1 = 2.7 \times 10^{-4}\,M^{-1}\,s^{-1} at T1=600 KT_1 = 600\,K and k2=3.5×10−3 M−1 s−1k_2 = 3.5 \times 10^{-3}\,M^{-1}\,s^{-1} at T2=650 KT_2 = 650\,K. Calculate the activation energy (EaE_a) for this reaction. (Use R=8.314 J K−1 mol−1R = 8.314\,J\,K^{-1}\,mol^{-1}).

      Solution:

      1. Select the Two-Point Arrhenius Equation:ln⁡(k2k1)=EaR(1T1−1T2)\ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)

      2. Substitute the given values:ln⁡(3.5×10−3 M−1 s−12.7×10−4 M−1 s−1)=Ea8.314 J K−1 mol−1(1600 K−1650 K)\ln\left(\frac{3.5 \times 10^{-3}\,M^{-1}\,s^{-1}}{2.7 \times 10^{-4}\,M^{-1}\,s^{-1}}\right) = \frac{E_a}{8.314\,J\,K^{-1}\,mol^{-1}}\left(\frac{1}{600\,K} - \frac{1}{650\,K}\right)

      3. Calculate the logarithmic term and temperature term:ln⁡(12.963)≈2.562\ln(12.963) \approx 2.5621600 K−1650 K≈1.282×10−4 K−1\frac{1}{600\,K} - \frac{1}{650\,K} \approx 1.282 \times 10^{-4}\,K^{-1}

      4. Ea≈166150 J mol−1=166 kJ mol−1E_a \approx 166150\,J\,mol^{-1} = 166\,kJ\,mol^{-1}

  • The Exponential Term (e−EaRTe^{-\frac{E_a}{RT}}): Represents the fraction of molecular collisions possessing kinetic energy ≥Ea\ge E_a. As temperature (TT) increases, this fraction grows exponentially, dramatically increasing kk.

  • The Frequency Factor (AA): Represents collision frequency and molecular orientation probability.

  • Linear Graph: Plotting ln⁡(k)\ln(k) vs. 1T\frac{1}{T} gives a straight line with slope m=−EaRm = -\frac{E_a}{R}. Activation energy is derived by Ea=−m×RE_a = -m \times R

Integrated Rate Laws (IRL) & Half-Life Derivations

  • Zero-Order Reactions (n=0n = 0):

    • Differential Rate Law: rate=−Δ[A]Δt=k\text{rate} = -\frac{\Delta[A]}{\Delta t} = k

    • Integrated Rate Law: [A]t=−kt+[A]0[A]_t = -k t + [A]_0

    • Linear Plot: [A][A] vs. tt (Slope = −k-k, y-intercept = [A]0[A]_0)

    • Half-Life Derivation: Setting [A]t=[A]02[A]_t = \frac{[A]_0}{2} at t=t1/2t = t_{1/2}:     [A]02=−kt1/2+[A]0  ⟹  kt1/2=[A]02  ⟹  t1/2=[A]02k\frac{[A]_0}{2} = -k t_{1/2} + [A]_0 \implies k t_{1/2} = \frac{[A]_0}{2} \implies t_{1/2} = \frac{[A]_0}{2k}

    • Why: Reaction speed is constant regardless of concentration. Half-life decreases over time because less reactant remains to be consumed at that fixed rate.

  • First-Order Reactions (n=1n = 1):

    • Differential Rate Law: rate=−Δ[A]Δt=k[A]\text{rate} = -\frac{\Delta[A]}{\Delta t} = k [A]

    • Integrated Rate Law: ln⁡([A]t)=−kt+ln⁡([A]0)\ln([A]_t) = -k t + \ln([A]_0) or [A]t=[A]0e−kt[A]_t = [A]_0 e^{-kt}

    • Linear Plot: ln⁡([A])\ln([A]) vs. tt (Slope = −k-k, y-intercept = ln⁡([A]0)\ln([A]_0))

    • Half-Life Derivation: Setting [A]t=[A]02[A]_t = \frac{[A]_0}{2} at t=t1/2t = t_{1/2}:     ln⁡([A]02)−ln⁡([A]0)=−kt1/2  ⟹  ln⁡(12)=−kt1/2  ⟹  t1/2=ln⁡(2)k≈0.693k\ln\left(\frac{[A]_0}{2}\right) - \ln([A]_0) = -k t_{1/2} \implies \ln\left(\frac{1}{2}\right) = -k t_{1/2} \implies t_{1/2} = \frac{\ln(2)}{k} \approx \frac{0.693}{k}

    • Why: Reaction speed drops proportionally with concentration. The fractional rate of decay stays constant, making half-life independent of initial concentration [A]0[A]_0

  • Second-Order Reactions (n=2n = 2):

    • Differential Rate Law: rate=−Δ[A]Δt=k[A]2\text{rate} = -\frac{\Delta[A]}{\Delta t} = k [A]^2

    • Integrated Rate Law: 1[A]t=kt+1[A]0\frac{1}{[A]_t} = k t + \frac{1}{[A]_0}

    • Linear Plot: 1[A]\frac{1}{[A]} vs. tt (Slope = +k+k, y-intercept = 1[A]0\frac{1}{[A]_0})

    • Half-Life Derivation: Setting [A]t=[A]02[A]_t = \frac{[A]_0}{2} at t=t1/2t = t_{1/2}:     1[A]02−1[A]0=kt1/2  ⟹  2[A]0−1[A]0=kt1/2  ⟹  t1/2=1k[A]0\frac{1}{\frac{[A]_0}{2}} - \frac{1}{[A]_0} = k t_{1/2} \implies \frac{2}{[A]_0} - \frac{1}{[A]_0} = k t_{1/2} \implies t_{1/2} = \frac{1}{k [A]_0}

    • Why: Rate slows exponentially as reactants deplete. Half-life increases as concentration drops because molecular collisions become significantly less frequent.

Reaction Mechanisms & Pre-Equilibrium Math

  • Rate-Determining Step (RDS): The overall reaction rate is limited by the slowest elementary step (highest EaE_a).

  • Why Substitute Intermediates: Reaction intermediates cannot appear in final rate laws because they are short-lived and difficult to measure experimentally.

  • Pre-Equilibrium Approximation Math:

    • Step 1 (fast, reversible): 2NO⇌N2O22 NO \rightleftharpoons N_2O_2     ratefwd=k1[NO]2,raterev=k−1[N2O2]\text{rate}_{fwd} = k_1 [NO]^2, \quad \text{rate}_{rev} = k_{-1} [N_2O_2]     Setting rates equal at equilibrium:     k1[NO]2=k−1[N2O2]  ⟹  [N2O2]=k1k−1[NO]2k_1 [NO]^2 = k_{-1} [N_2O_2] \implies [N_2O_2] = \frac{k_1}{k_{-1}} [NO]^2

    • Step 2 (slow, RDS): N2O2+H2→N2O+H2ON_2O_2 + H_2 \rightarrow N_2O + H_2O     raterxn=k2[N2O2][H2]\text{rate}_{rxn} = k_2 [N_2O_2] [H_2]

    • Substitution:     raterxn=k2(k1k−1[NO]2)[H2]=kobs[NO]2[H2]\text{rate}_{rxn} = k_2 \left(\frac{k_1}{k_{-1}} [NO]^2\right) [H_2] = k_{obs} [NO]^2 [H_2]     where kobs=k1k2k−1k_{obs} = \frac{k_1 k_2}{k_{-1}}