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Concept 04 / 06Chemical Kinetics

Pseudo First Order Reactions, Gas-Phase Kinetics and Determining Order

Pseudo first order reactions such as acid hydrolysis of esters and inversion of cane sugar, first-order kinetics followed by total pressure in gas-phase decompositions, and methods of determining order: initial rates, integrated rate law (graphical) and half-life methods.

Figure 1
Pseudo first order reactions. A reaction that is really of higher order behaves as first order when one reactant is present in large excess, so its concentration hardly changes. Acid hydrolysis of ethyl acetate: CH3COOC2H5+H2O→H+CH3COOH+C2H5OH\mathrm{CH_{3}COOC_{2}H_{5} + H_{2}O \xrightarrow{H^{+}} CH_{3}COOH + C_{2}H_{5}OH}. Rate =k′[CH3COOC2H5][H2O]= k'[\mathrm{CH_{3}COOC_{2}H_{5}}][\mathrm{H_{2}O}], but water is the solvent (about 55 M55\,\mathrm{M}) and effectively constant, so rate =k[CH3COOC2H5]= k[\mathrm{CH_{3}COOC_{2}H_{5}}] with k=k′[H2O]k = k'[\mathrm{H_{2}O}]. It is followed by titrating the acetic acid formed against NaOH. Inversion of cane sugar: C12H22O11+H2O→H+C6H12O6\mathrm{C_{12}H_{22}O_{11} + H_{2}O \xrightarrow{H^{+}} C_{6}H_{12}O_{6}} (glucose) +C6H12O6+ \mathrm{C_{6}H_{12}O_{6}} (fructose); rate =k[sucrose]= k[\mathrm{sucrose}]. It is followed with a polarimeter: dextrorotatory sucrose gives a laevorotatory mixture, hence "inversion". First-order gas-phase reactions followed by pressure. For A(g)→B(g)+C(g)\mathrm{A(g) \rightarrow B(g) + C(g)} with initial pressure p0p_{0}: if xx decomposes, pA=p0−xp_{\mathrm{A}} = p_{0} - x and total pt=p0+xp_{t} = p_{0} + x, so pA=2p0−ptp_{\mathrm{A}} = 2p_{0} - p_{t} and k=2.303tlog⁡p02p0−ptk = \frac{2.303}{t}\log\frac{p_{0}}{2p_{0} - p_{t}}. For other stoichiometries, express each partial pressure through the extent of reaction: for 2N2O5→4NO2+O2\mathrm{2N_{2}O_{5} \rightarrow 4NO_{2} + O_{2}}, if the pressure of N2O5\mathrm{N_{2}O_{5}} falls by 2x2x, the total pressure is pt=p0+3xp_{t} = p_{0} + 3x. Determining the order. Initial rate method: vary one concentration at a time (see previous concept). Integrated rate law (graphical) method: plot [A][\mathrm{A}], ln⁡[A]\ln[\mathrm{A}] and 1[A]\frac{1}{[\mathrm{A}]} against tt; the straight line identifies zero, first or second order. Substitution method: calculate kk from each data point using each integrated equation; the one that gives a constant kk is correct. Half-life method: t1/2∝[A]01−nt_{1/2} \propto [\mathrm{A}]_{0}^{1-n}, so from two runs, n=1+log⁡(t1/t2)log⁡([A]2/[A]1)n = 1 + \frac{\log(t_{1} / t_{2})}{\log([\mathrm{A}]_{2} / [\mathrm{A}]_{1})}. Successive half-lives are equal only for first order (see figure). Ostwald's isolation method: use large excess of all reactants but one, so the order in that one can be measured (pseudo order). Key Takeaways: - Pseudo first order: one reactant in large excess (ester hydrolysis, sugar inversion). - For A(g)→B(g)+C(g)\mathrm{A(g) \rightarrow B(g) + C(g)}: k=2.303tlog⁡p02p0−ptk = \frac{2.303}{t}\log\frac{p_{0}}{2p_{0} - p_{t}}. - Linear plots: [A][\mathrm{A}] (zero), ln⁡[A]\ln[\mathrm{A}] (first), 1[A]\frac{1}{[\mathrm{A}]} (second). - Half-life method: t1/2∝[A]01−nt_{1/2} \propto [\mathrm{A}]_{0}^{1-n}. - Pitfall: in pressure problems, use the partial pressure of the reactant, not the total pressure, in the rate equation. Brain-Teaser 1: SO2Cl2(g)→SO2(g)+Cl2(g)\mathrm{SO_{2}Cl_{2}(g) \rightarrow SO_{2}(g) + Cl_{2}(g)} is first order. Starting at 0.5 atm0.5\,\mathrm{atm}, the total pressure is 0.6 atm0.6\,\mathrm{atm} after 100 s100\,\mathrm{s}. Find kk. Solution: p(SO2Cl2)=2×0.5−0.6=0.4 atmp(\mathrm{SO_{2}Cl_{2}}) = 2 \times 0.5 - 0.6 = 0.4\,\mathrm{atm}. k=2.303100log⁡0.50.4=0.02303×0.0969=2.23×10−3 s−1k = \frac{2.303}{100}\log\frac{0.5}{0.4} = 0.02303 \times 0.0969 = 2.23 \times 10^{-3}\,\mathrm{s^{-1}}. Brain-Teaser 2: For a reaction, the half-life is 100 s100\,\mathrm{s} when the initial concentration is 0.2 M0.2\,\mathrm{M} and 50 s50\,\mathrm{s} when it is 0.4 M0.4\,\mathrm{M}. Find the order. Solution: Doubling [A]0[\mathrm{A}]_{0} halves t1/2t_{1/2}: t1/2∝[A]0−1t_{1/2} \propto [\mathrm{A}]_{0}^{-1}, so 1−n=−11 - n = -1 and n=2n = 2 (second order). Brain-Teaser 3: Why is the acid-catalysed hydrolysis of an ester first order, though two molecules react? Solution: The rate law is really r=k′[ester][H2O]r = k'[\mathrm{ester}][\mathrm{H_{2}O}], but water is present in huge excess (as solvent), so its concentration stays essentially constant and is absorbed into kk. The observed kinetics are first order in ester only (pseudo first order), while the molecularity of the key step is still two.