Chemical Kinetics: Reaction Rates, Rate Laws, and Integrated Rate Laws

Fundamental Concepts of Reaction Rates

  • Reaction rate definition:

    • In chemistry, reaction rates are defined by changes in concentration over time.
    • Mathematical definition: Rate=Δ[concentration]Δt\text{Rate} = \frac{\Delta [\text{concentration}]}{\Delta t}.
    • Concentrations are denoted using square brackets [… ][\dots], which represent Molarity (MM or mol L−1\text{mol}\,L^{-1}).
    • Chemical kinetics specifically utilizes Molarity (MM) rather than molality (mm) or other concentration scales.

  • Factors influencing reaction rates:

    • Catalysts: The presence of a catalyst increases reaction speed without being consumed in the overall process (e.g., catalytic converters in automobiles speed up the decomposition of unburned hydrocarbons).
    • Reactant concentration: Higher concentrations generally increase collision frequency between reactant molecules, increasing the rate.
    • Temperature: Increasing temperature increases the kinetic energy and collision frequency of reactant molecules.
    • Surface area and physical state: Homogeneous mixtures (all gas or all aqueous phase) react differently compared to heterogeneous systems; greater surface area increases the reaction rate.

  • Experimental observations and reaction speed spectrum:

    • Steel wool oxidation: Heating steel wool (metallic iron) in air causes it to glow red and slowly gain mass as it forms iron oxide (Fe2O3\text{Fe}_2\text{O}_3). Submerging heated steel wool into pure oxygen dramatically accelerates the oxidation reaction.
    • Steel wool placed under a household sink may take months to completely rust and oxidize (slow reaction).
    • Octane vapor ignited with a match or hydrogen gas mixed with oxygen and ignited undergoes an explosive reaction completed within a few milliseconds (fast reaction).

  • Requirements for a chemical reaction to occur:

    • Molecular contact: Reactant particles must physically collide with each other.
    • Bond breaking and formation: Existing chemical bonds must break (e.g., breaking C−H\text{C}-\text{H} bonds during hydrocarbon combustion) and new bonds must form (e.g., forming C−O\text{C}-\text{O} and H−O\text{H}-\text{O} bonds).
    • Orientation: Reactant molecules must collide with the correct spatial orientation.
    • Activation energy: Collisions must possess sufficient energy to overcome the reaction energy barrier.

  • Units of reaction rates:

    • Standard reaction rate units: Molarity per second (M s−1M\,s^{-1}).
    • Slow reaction rate units: Molarity per minute (M min−1M\,\text{min}^{-1}), Molarity per hour (M h−1M\,\text{h}^{-1}), or any Molarity per unit time (M time−1M\,\text{time}^{-1}).

Stoichiometry and General Rate Relationships

  • Mathematical formulation of reaction rates:

    • For a general chemical equation:     aA+bB→cC+dDa\text{A} + b\text{B} \rightarrow c\text{C} + d\text{D}
    • Stoichiometric rate expression:     Rate=−1aΔ[A]Δt=−1bΔ[B]Δt=+1cΔ[C]Δt=+1dΔ[D]Δt\text{Rate} = -\frac{1}{a}\frac{\Delta [\text{A}]}{\Delta t} = -\frac{1}{b}\frac{\Delta [\text{B}]}{\Delta t} = +\frac{1}{c}\frac{\Delta [\text{C}]}{\Delta t} = +\frac{1}{d}\frac{\Delta [\text{D}]}{\Delta t}
    • Reactant terms carry a negative sign because reactant concentrations decrease over time (Δ[Reactant]<0\Delta [\text{Reactant}] < 0), ensuring that the overall rate remains a positive quantity.
    • Product terms carry a positive sign because product concentrations increase over time (Δ[Product]>0\Delta [\text{Product}] > 0).
    • The term Δ[A]\Delta [\text{A}] represents final concentration minus initial concentration ([A]2−[A]1[\text{A}]_2 - [\text{A}]_1).
    • The term Δt\Delta t represents final time minus initial time (t2−t1t_2 - t_1).

  • Stoichiometric examples:

    • Hydrogen iodide decomposition / formation:     2HI(g)→H2(g)+I2(g)2\text{HI}(g) \rightarrow \text{H}_2(g) + \text{I}_2(g)Rate=−12Δ[HI]Δt=+Δ[H2]Δt=+Δ[I2]Δt\text{Rate} = -\frac{1}{2}\frac{\Delta [\text{HI}]}{\Delta t} = +\frac{\Delta [\text{H}_2]}{\Delta t} = +\frac{\Delta [\text{I}_2]}{\Delta t}
    • Ozone conversion to oxygen:     2O3(g)→3O2(g)2\text{O}_3(g) \rightarrow 3\text{O}_2(g)Rate=−12Δ[O3]Δt=+13Δ[O2]Δt\text{Rate} = -\frac{1}{2}\frac{\Delta [\text{O}_3]}{\Delta t} = +\frac{1}{3}\frac{\Delta [\text{O}_2]}{\Delta t}

Average and Instantaneous Reaction Rates

  • Rate behavior over time:

    • Plots of concentration versus time are non-linear curves.
    • Early stage: Reactant concentrations are high, yielding the maximum reaction rate.
    • Late stage: As reactants are consumed, concentrations decrease and the reaction rate slows down.
    • Completion: When the reaction reaches completion, reactant consumption ceases and the rate reaches 0 M s−10\,M\,s^{-1}.

  • Average rate vs. instantaneous rate:

    • Average rate: Measured over a discrete, finite time interval (Δt\Delta t) using the slope between two points on the curve:     Average Rate=Δ[concentration]Δt\text{Average Rate} = \frac{\Delta [\text{concentration}]}{\Delta t}
    • Instantaneous rate: The rate at a single specific instant in time, calculated as the slope of the tangent line to the concentration-time curve at that specific point tt.
    • Reaction rates are strictly defined as positive numerical quantities (Rate>0\text{Rate} > 0).

  • Quantitative examples:

    • For the combination reaction H2+I2→2HI\text{H}_2 + \text{I}_2 \rightarrow 2\text{HI}:

    • At t1=40 st_1 = 40\,\text{s}, [H2]1=0.449 M[\text{H}_2]_1 = 0.449\,M.

    • At t2=50 st_2 = 50\,\text{s}, [H2]2=0.368 M[\text{H}_2]_2 = 0.368\,M.

    • Change in concentration:       Δ[H2]=0.368 M−0.449 M=−0.081 M\Delta [\text{H}_2] = 0.368\,M - 0.449\,M = -0.081\,M

    • Time interval:       Δt=50 s−40 s=10 s\Delta t = 50\,\text{s} - 40\,\text{s} = 10\,\text{s}

    • Instantaneous rate calculated via hydrogen consumption between 40 s40\,\text{s} and 50 s50\,\text{s}:       Rate=−Δ[H2]Δt=−−0.081 M10 s=0.0081 M s−1\text{Rate} = -\frac{\Delta [\text{H}_2]}{\Delta t} = -\frac{-0.081\,M}{10\,\text{s}} = 0.0081\,M\,s^{-1}

    • Over an extended 40-second interval where Δ[H2]=−0.28 M\Delta [\text{H}_2] = -0.28\,M:       Rate=−−0.28 M40 s=0.0070 M s−1\text{Rate} = -\frac{-0.28\,M}{40\,\text{s}} = 0.0070\,M\,s^{-1}

    • Corresponding formation of HI\text{HI} over the same 40-second interval with Δ[HI]=+0.56 M\Delta [\text{HI}] = +0.56\,M:       Rate=+12Δ[HI]Δt=12(0.56 M40 s)=0.0070 M s−1\text{Rate} = +\frac{1}{2}\frac{\Delta [\text{HI}]}{\Delta t} = \frac{1}{2}\left(\frac{0.56\,M}{40\,\text{s}}\right) = 0.0070\,M\,s^{-1}

    • Rate progression across time intervals: Near 0 s0\,\text{s}, rate ≈0.1 M s−1\approx 0.1\,M\,s^{-1}; at 40 s40\,\text{s}, rate =0.0081 M s−1= 0.0081\,M\,s^{-1}; near 100 s100\,\text{s}, rate drops to 0.003 M s−10.003\,M\,s^{-1}; at complete reaction, rate =0 M s−1= 0\,M\,s^{-1}.

    • Ammonia oxidation stoichiometric rate calculation:

    • Balanced chemical equation:       4NH3(g)+5O2(g)→4NO(g)+6H2O(g)4\text{NH}_3(g) + 5\text{O}_2(g) \rightarrow 4\text{NO}(g) + 6\text{H}_2\text{O}(g)

    • Stoichiometric rate equality:       −15Δ[O2]Δt=+14Δ[NO]Δt-\frac{1}{5}\frac{\Delta [\text{O}_2]}{\Delta t} = +\frac{1}{4}\frac{\Delta [\text{NO}]}{\Delta t}

    • Measured rate of formation of nitric oxide:       Δ[NO]Δt=5.00×10−4 M s−1\frac{\Delta [\text{NO}]}{\Delta t} = 5.00 \times 10^{-4}\,M\,s^{-1}

    • Calculation of oxygen consumption rate:       −Δ[O2]Δt=54(Δ[NO]Δt)-\frac{\Delta [\text{O}_2]}{\Delta t} = \frac{5}{4}\left(\frac{\Delta [\text{NO}]}{\Delta t}\right)−Δ[O2]Δt=54(5.00×10−4 M s−1)=6.25×10−4 M s−1-\frac{\Delta [\text{O}_2]}{\Delta t} = \frac{5}{4}\left(5.00 \times 10^{-4}\,M\,s^{-1}\right) = 6.25 \times 10^{-4}\,M\,s^{-1}

Rate Laws and Reaction Orders

  • Experimental techniques for rate determination:

    • Periodic aliquots: Sampling small volumes at specified time intervals to measure species concentration.
    • Pressure measurements: Monitoring total pressure changes in gaseous reactions using computerized apparatus.
    • Spectroscopic monitoring: Measuring light absorption over time.

  • Differential rate law structure:

    • General form for multi-reactant systems:     Rate=k[A]x[B]y\text{Rate} = k[\text{A}]^x[\text{B}]^y
    • kk: Lowercase rate constant.
    • xx: Reaction order with respect to species A\text{A}.
    • yy: Reaction order with respect to species B\text{B}.
    • Overall reaction order: Sum of individual exponents (x+yx + y).

  • Crucial principle of reaction orders:

    • Reaction orders MUST be determined experimentally.
    • Exponents (x,yx, y) have NO direct relation to stoichiometric coefficients in the balanced chemical equation.

  • Effect of order nn on reaction rate:

    • Zero Order (n=0n = 0):     Rate=k[A]0=k\text{Rate} = k[\text{A}]^0 = k     Changing concentration has no effect on rate. Doubling concentration results in 20=12^0 = 1 times the rate.
    • First Order (n=1n = 1):     Rate=k[A]1\text{Rate} = k[\text{A}]^1     Rate is directly proportional to concentration. Doubling concentration doubles the rate (21=22^1 = 2).
    • Second Order (n=2n = 2):     Rate=k[A]2\text{Rate} = k[\text{A}]^2     Rate is proportional to the square of concentration. Doubling concentration quadruples the rate (22=42^2 = 4).
    • Third Order (n=3n = 3):     Rate=k[A]3\text{Rate} = k[\text{A}]^3     Doubling concentration increases rate by a factor of 8 (23=82^3 = 8).

  • Units of the rate constant kk:

    • The rate itself always has units of M s−1M\,s^{-1}.
    • Zero order: kk has units of M s−1M\,s^{-1}.
    • First order: kk has units of s−1s^{-1}.
    • Second order: kk has units of M−1 s−1M^{-1}\,s^{-1}.
    • Third order: kk has units of M−2 s−1M^{-2}\,s^{-1}.
    • Rate constant kk is specific to a particular reaction, temperature, and order; it is not a universal constant.

Method of Initial Rates

  • Concept and procedure:

    • Measures the initial rate (Rate0\text{Rate}_0) immediately after mixing reactants before significant concentration changes occur.
    • Conduct multiple experimental runs where the concentration of one reactant is varied while all other reactant concentrations are held constant.
    • Compare initial rate ratios to isolate the order for each reactant.

  • Example 1: Determination of rate law for NO2\text{NO}_2 and CO\text{CO}:

    • Experimental data:
    • Experiment 1: [NO2]=0.10 M[\text{NO}_2] = 0.10\,M, [CO]=0.10 M[\text{CO}] = 0.10\,M, Rate1=0.0021 M s−1\text{Rate}_1 = 0.0021\,M\,s^{-1}
    • Experiment 2: [NO2]=0.20 M[\text{NO}_2] = 0.20\,M, [CO]=0.10 M[\text{CO}] = 0.10\,M, Rate2=0.0082 M s−1\text{Rate}_2 = 0.0082\,M\,s^{-1}
    • Experiment 3: [NO2]=0.20 M[\text{NO}_2] = 0.20\,M, [CO]=0.20 M[\text{CO}] = 0.20\,M, Rate3=0.0082 M s−1\text{Rate}_3 = 0.0082\,M\,s^{-1}
    • Determining order with respect to NO2\text{NO}_2 (mm) using Experiments 1 and 2 ([CO][\text{CO}] constant):     Rate2Rate1=k[NO2]2m[CO]2nk[NO2]1m[CO]1n\frac{\text{Rate}_2}{\text{Rate}_1} = \frac{k[\text{NO}_2]_2^m [\text{CO}]_2^n}{k[\text{NO}_2]_1^m [\text{CO}]_1^n}0.00820.0021=(0.200.10)m\frac{0.0082}{0.0021} = \left(\frac{0.20}{0.10}\right)^m3.9=2m3.9 = 2^mlog⁡(3.9)=m⋅log⁡(2)\log(3.9) = m \cdot \log(2)m=log⁡(3.9)log⁡(2)=1.96≈2m = \frac{\log(3.9)}{\log(2)} = 1.96 \approx 2 (Second order in NO2\text{NO}_2)
    • Determining order with respect to CO\text{CO} (nn) using Experiments 2 and 3 ([NO2][\text{NO}_2] constant):     Rate3Rate2=0.00820.0082=1.0\frac{\text{Rate}_3}{\text{Rate}_2} = \frac{0.0082}{0.0082} = 1.01.0=(0.200.10)n=2n1.0 = \left(\frac{0.20}{0.10}\right)^n = 2^nn=0n = 0 (Zero order in CO\text{CO})
    • Final rate law:     Rate=k[NO2]2\text{Rate} = k[\text{NO}_2]^2
    • Calculating rate constant kk using Experiment 1:     k=Rate1[NO2]12=0.0021 M s−1(0.10 M)2=0.21 M−1 s−1k = \frac{\text{Rate}_1}{[\text{NO}_2]_1^2} = \frac{0.0021\,M\,s^{-1}}{(0.10\,M)^2} = 0.21\,M^{-1}\,s^{-1} (or 0.02 M−1 s−10.02\,M^{-1}\,s^{-1} depending on experimental dataset rounding).

  • Example 2: Reaction between fluorine and chlorine dioxide:

    • Reaction: F2(g)+2ClO2(g)→2FClO2(g)\text{F}_2(g) + 2\text{ClO}_2(g) \rightarrow 2\text{FClO}_2(g)
    • Experimental data summary:
    • Experiment 1: [F2]=0.10 M[\text{F}_2] = 0.10\,M, [ClO2]=0.010 M[\text{ClO}_2] = 0.010\,M, Rate1=0.0012 M s−1\text{Rate}_1 = 0.0012\,M\,s^{-1}
    • Experiment 2: [F2]=0.10 M[\text{F}_2] = 0.10\,M, [ClO2]=0.040 M[\text{ClO}_2] = 0.040\,M, Rate2=0.0048 M s−1\text{Rate}_2 = 0.0048\,M\,s^{-1}
    • Experiment 3: [F2]=0.20 M[\text{F}_2] = 0.20\,M, [ClO2]=0.010 M[\text{ClO}_2] = 0.010\,M, Rate3=0.0024 M s−1\text{Rate}_3 = 0.0024\,M\,s^{-1}
    • Finding order in ClO2\text{ClO}_2 (comparing Experiments 1 and 2):     Rate2Rate1=0.00480.0012=4.0\frac{\text{Rate}_2}{\text{Rate}_1} = \frac{0.0048}{0.0012} = 4.0[ClO2]2n[ClO2]1n=(0.0400.010)n=4.0n\frac{[\text{ClO}_2]_2^n}{[\text{ClO}_2]_1^n} = \left(\frac{0.040}{0.010}\right)^n = 4.0^n4.0=4.0n  ⟹  n=14.0 = 4.0^n \implies n = 1 (First order in ClO2\text{ClO}_2)
    • Finding order in F2\text{F}_2 (comparing Experiments 1 and 3):     Rate3Rate1=0.00240.0012=2.0\frac{\text{Rate}_3}{\text{Rate}_1} = \frac{0.0024}{0.0012} = 2.0[F2]3m[F2]1m=(0.200.10)m=2.0m\frac{[\text{F}_2]_3^m}{[\text{F}_2]_1^m} = \left(\frac{0.20}{0.10}\right)^m = 2.0^m2.0=2.0m  ⟹  m=12.0 = 2.0^m \implies m = 1 (First order in F2\text{F}_2)
    • Rate law and overall order:     Rate=k[F2][ClO2]\text{Rate} = k[\text{F}_2][\text{ClO}_2]     Overall order = 1+1=21 + 1 = 2 (Second order overall)
    • Calculating rate constant kk:     k=Rate[F2][ClO2]=0.0012 M s−1(0.10 M)(0.010 M)=1.2 M−1 s−1k = \frac{\text{Rate}}{[\text{F}_2][\text{ClO}_2]} = \frac{0.0012\,M\,s^{-1}}{(0.10\,M)(0.010\,M)} = 1.2\,M^{-1}\,s^{-1}

Integrated Rate Laws

  • First-order integrated rate law:

    • Differential equation: Rate=−d[A]dt=k[A]\text{Rate} = -\frac{d[\text{A}]}{dt} = k[\text{A}]
    • Integrated equation:     ln⁡[A]t=−kt+ln⁡[A]0\ln[\text{A}]_t = -kt + \ln[\text{A}]_0
    • Linear equation form (y=mx+by = mx + b):
    • y=ln⁡[A]ty = \ln[\text{A}]_t
    • x=tx = t
    • Slope m=−k\text{Slope } m = -k
    • y-intercept b=ln⁡[A]0y\text{-intercept } b = \ln[\text{A}]_0
    • Graphical test: A plot of ln⁡[A]\ln[\text{A}] versus time tt produces a straight line with slope =-k$.\n\n- First-order calculations (extrapolating initial concentration [\text{A}]_0):\n - Given data:\n - At t_1 = 66\,\text{s},,[\text{A}]_1 = 3.60 \times 10^{-3}\,M\n - At t_2 = 255\,\text{s},,[\text{A}]_2 = 6.34 \times 10^{-4}\,M\n - Step 1: Calculate rate constant k via slope:\n    \text{Slope} = \frac{\ln[\text{A}]_2 - \ln[\text{A}]_1}{t_2 - t_1}\n    \ln(6.34 \times 10^{-4}) = -7.3635\n    \ln(3.60 \times 10^{-3}) = -5.6268\n    \text{Slope} = \frac{-7.3635 - (-5.6268)}{255\,\text{s} - 66\,\text{s}} = \frac{-1.7367}{189\,\text{s}} = -9.19 \times 10^{-3}\,s^{-1}\n    k = -\text{Slope} = 9.19 \times 10^{-3}\,s^{-1}\n - Step 2: Calculate initial concentration [\text{A}]_0:\n    \ln[\text{A}]_0 = \ln[\text{A}]_t + kt\n    [\text{A}]_0 = e^{\ln[\text{A}]_t + kt}\n    Using t_1 = 66\,\text{s}:\n    [\text{A}]_0 = e^{\ln(3.60 \times 10^{-3}) + (9.19 \times 10^{-3}\,s^{-1})(66\,\text{s})}\n    [\text{A}]_0 = e^{-5.6268 + 0.6065} = e^{-5.0203} = 6.60 \times 10^{-3}\,M\n    Using t_2 = 255\,\text{s}:\n    [\text{A}]_0 = e^{\ln(6.34 \times 10^{-4}) + (9.19 \times 10^{-3}\,s^{-1})(255\,\text{s})}\n    [\text{A}]_0 = e^{-7.3635 + 2.3435} = e^{-5.0200} = 6.60 \times 10^{-3}\,M\n\n- Second-order integrated rate law:\n - Differential equation: \text{Rate} = k[\text{A}]^2\n - Integrated equation:\n    \frac{1}{[\text{A}]_t} = kt + \frac{1}{[\text{A}]_0}\n - Linear equation form (y = mx + b):\n - y = \frac{1}{[\text{A}]_t}\n - x = t\n - \text{Slope } m = k (positive slope)\n - y\text{-intercept } b = \frac{1}{[\text{A}]_0}\n - Graphical test: A plot of \frac{1}{[\text{A}]}versustimeversus timetproducesastraightlinewithslopeproduces a straight line with slope=+k$.

  • Second-order rate constant calculation example:

    • Given data for a second-order reaction:
    • At t1=32 st_1 = 32\,\text{s}, [A]1=0.325 M[\text{A}]_1 = 0.325\,M
    • At t2=455 st_2 = 455\,\text{s}, [A]2=2.14×10−2 M[\text{A}]_2 = 2.14 \times 10^{-2}\,M
    • Step 1: Calculate inverse concentrations:     1[A]1=10.325 M=3.077 M−1\frac{1}{[\text{A}]_1} = \frac{1}{0.325\,M} = 3.077\,M^{-1}1[A]2=12.14×10−2 M=46.729 M−1\frac{1}{[\text{A}]_2} = \frac{1}{2.14 \times 10^{-2}\,M} = 46.729\,M^{-1}
    • Step 2: Calculate slope (kk):     Slope k=1[A]2−1[A]1t2−t1\text{Slope } k = \frac{\frac{1}{[\text{A}]_2} - \frac{1}{[\text{A}]_1}}{t_2 - t_1}k=46.729 M−1−3.077 M−1455 s−32 s=43.652 M−1423 s=0.1031 M−1 s−1k = \frac{46.729\,M^{-1} - 3.077\,M^{-1}}{455\,\text{s} - 32\,\text{s}} = \frac{43.652\,M^{-1}}{423\,\text{s}} = 0.1031\,M^{-1}\,s^{-1} (or 0.01103 M−1 s−10.01103\,M^{-1}\,s^{-1} depending on experimental dataset precision).

Summary of Integrated Rate Law Plots

  • Zero-order reactions:

    • Linear plot: [A][\text{A}] versus tt
    • Slope: −k-k
    • y-intercepty\text{-intercept}: [A]0[\text{A}]_0
    • Integrated formula: [A]t=−kt+[A]0[\text{A}]_t = -kt + [\text{A}]_0

  • First-order reactions:

    • Linear plot: ln⁡[A]\ln[\text{A}] versus tt
    • Slope: −k-k
    • y-intercepty\text{-intercept}: ln⁡[A]0\ln[\text{A}]_0
    • Integrated formula: ln⁡[A]t=−kt+ln⁡[A]0\ln[\text{A}]_t = -kt + \ln[\text{A}]_0

  • Second-order reactions:

    • Linear plot: 1[A]\frac{1}{[\text{A}]} versus tt
    • Slope: +k+k
    • y-intercepty\text{-intercept}: 1[A]0\frac{1}{[\text{A}]_0}
    • Integrated formula: 1[A]t=kt+1[A]0\frac{1}{[\text{A}]_t} = kt + \frac{1}{[\text{A}]_0}