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Concept 02 / 06Chemical Kinetics

Rate Law, Order, Molecularity and Units of the Rate Constant

Rate law and rate constant, order of reaction (overall and with respect to each reactant) as an experimental quantity, units of the rate constant for different orders, elementary and complex reactions, molecularity, the rate-determining step, and finding orders by the initial rate method.

Figure 1
Rate law. An experimentally determined expression relating rate to the concentrations of reactants. For aA+bB→a\mathrm{A} + b\mathrm{B} \rightarrow products, r=k[A]x[B]yr = k[\mathrm{A}]^{x}[\mathrm{B}]^{y}, where kk is the rate constant (rate when all concentrations are 1 M1\,\mathrm{M}), which depends on temperature but not on concentration. The exponents xx and yy need not equal the stoichiometric coefficients aa and bb: they are found by experiment. Example: 2NO+O2→2NO2\mathrm{2NO + O_{2} \rightarrow 2NO_{2}} has r=k[NO]2[O2]r = k[\mathrm{NO}]^{2}[\mathrm{O_{2}}]; but CHCl3+Cl2→CCl4+HCl\mathrm{CHCl_{3} + Cl_{2} \rightarrow CCl_{4} + HCl} has r=k[CHCl3][Cl2]1/2r = k[\mathrm{CHCl_{3}}][\mathrm{Cl_{2}}]^{1/2}. Order of reaction. The sum x+yx + y is the overall order; xx is the order with respect to A. Order can be 00, a positive integer, a fraction or even negative (see figure for 00, 11, 22). Units of kk for order nn: (mol L−1)1−n s−1(\mathrm{mol\,L^{-1}})^{1-n}\,\mathrm{s^{-1}}. Zero order: mol L−1 s−1\mathrm{mol\,L^{-1}\,s^{-1}}; first order: s−1\mathrm{s^{-1}}; second order: L mol−1 s−1\mathrm{L\,mol^{-1}\,s^{-1}}; third order: L2 mol−2 s−1\mathrm{L^{2}\,mol^{-2}\,s^{-1}}. Elementary and complex reactions. An elementary reaction occurs in a single step. A complex reaction proceeds through a sequence of elementary steps (the mechanism). The slowest step is the rate-determining step, which controls the overall rate. Example: decomposition of H2O2\mathrm{H_{2}O_{2}} catalysed by iodide in alkaline medium, 2H2O2→2H2O+O2\mathrm{2H_{2}O_{2} \rightarrow 2H_{2}O + O_{2}}, has r=k[H2O2][I−]r = k[\mathrm{H_{2}O_{2}}][\mathrm{I^{-}}] (first order in each), because the mechanism is H2O2+I−→H2O+IO−\mathrm{H_{2}O_{2} + I^{-} \rightarrow H_{2}O + IO^{-}} (slow), then H2O2+IO−→H2O+I−+O2\mathrm{H_{2}O_{2} + IO^{-} \rightarrow H_{2}O + I^{-} + O_{2}} (fast). Molecularity. The number of reacting species (atoms, ions or molecules) that must collide simultaneously in an elementary step: unimolecular (NH4NO2→N2+2H2O\mathrm{NH_{4}NO_{2} \rightarrow N_{2} + 2H_{2}O}), bimolecular (2HI→H2+I2\mathrm{2HI \rightarrow H_{2} + I_{2}}), termolecular (2NO+O2→2NO2\mathrm{2NO + O_{2} \rightarrow 2NO_{2}}). It is always a small whole number (never zero or fractional); molecularity above 33 is very rare because simultaneous collisions of many particles are improbable. Molecularity applies only to elementary reactions; for complex reactions it has no meaning, but the molecularity of the slowest step equals the order of the overall reaction (when that step involves only reactants). Initial rate method. Measure the initial rate for several starting concentrations, changing one reactant at a time. If doubling [A][\mathrm{A}] doubles the rate, the order in A is 11; if it quadruples, 22; if no change, 00. Key Takeaways: - r=k[A]x[B]yr = k[\mathrm{A}]^{x}[\mathrm{B}]^{y}; orders are experimental. - Units of kk: (mol L−1)1−n s−1(\mathrm{mol\,L^{-1}})^{1-n}\,\mathrm{s^{-1}}. - Molecularity: elementary steps only, a small whole number. - The slowest step determines the rate law. - Pitfall: the order of a reaction cannot be read from the balanced equation unless the reaction is elementary. Brain-Teaser 1: For A+B→\mathrm{A + B \rightarrow} products, initial rates were: [A]=0.1[\mathrm{A}] = 0.1, [B]=0.1[\mathrm{B}] = 0.1: 2.0×10−32.0 \times 10^{-3}; [A]=0.2[\mathrm{A}] = 0.2, [B]=0.1[\mathrm{B}] = 0.1: 4.0×10−34.0 \times 10^{-3}; [A]=0.1[\mathrm{A}] = 0.1, [B]=0.2[\mathrm{B}] = 0.2: 8.0×10−3 mol L−1 s−18.0 \times 10^{-3}\,\mathrm{mol\,L^{-1}\,s^{-1}}. Find the rate law and kk. Solution: Doubling [A] doubles the rate: first order in A. Doubling [B] quadruples it: second order in B. r=k[A][B]2r = k[\mathrm{A}][\mathrm{B}]^{2}, overall order 33. k=2.0×10−30.1×0.12=2.0 L2 mol−2 s−1k = \frac{2.0 \times 10^{-3}}{0.1 \times 0.1^{2}} = 2.0\,\mathrm{L^{2}\,mol^{-2}\,s^{-1}}. Brain-Teaser 2: Identify the order from the units of kk: (a) 2.3×10−5 L mol−1 s−12.3 \times 10^{-5}\,\mathrm{L\,mol^{-1}\,s^{-1}}; (b) 3×10−4 s−13 \times 10^{-4}\,\mathrm{s^{-1}}. Solution: (a) L mol−1 s−1\mathrm{L\,mol^{-1}\,s^{-1}} corresponds to n=2n = 2: second order. (b) s−1\mathrm{s^{-1}}: first order. Brain-Teaser 3: A reaction is second order in A. How does its rate change if [A] is (a) doubled, (b) reduced to half? Solution: r∝[A]2r \propto [\mathrm{A}]^{2}. (a) Doubling makes it 44 times faster. (b) Halving makes it 14\frac{1}{4} as fast.

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