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Concept 05 / 06Chemical Kinetics

Effect of Temperature: Arrhenius Equation, Activation Energy and Catalysis

Why reaction rates increase with temperature, the Arrhenius equation k = A exp(-Ea/RT) and its logarithmic and two-temperature forms, graphical determination of activation energy, the energy profile with the activated complex, temperature coefficient, and how catalysts work.

Figure 1
For most chemical reactions, the rate constant nearly doubles for every 10 K10\,\mathrm{K} rise in temperature (temperature coefficient kT+10kT≈2\frac{k_{T+10}}{k_{T}} \approx 2 to 33). Activation energy. Reactants must pass over an energy barrier to become products. At the top of the barrier lies the activated complex (transition state), an unstable arrangement in which old bonds are partly broken and new ones partly formed. The activation energy EaE_{a} is the minimum extra energy (above the average energy of reactants) needed to reach it (see figure). Ea(forward)−Ea(backward)=ΔHE_{a}(\text{forward}) - E_{a}(\text{backward}) = \Delta H. Arrhenius equation (1889). k=Ae−Ea/RTk = Ae^{-E_{a}/RT}, where AA is the pre-exponential (frequency) factor and e−Ea/RTe^{-E_{a}/RT} is the fraction of molecules with energy at least EaE_{a}. Logarithmic form: ln⁡k=ln⁡A−EaRT\ln k = \ln A - \frac{E_{a}}{RT}; a plot of ln⁡k\ln k against 1T\frac{1}{T} is a straight line with slope −EaR-\frac{E_{a}}{R} and intercept ln⁡A\ln A (or log⁡k\log k against 1T\frac{1}{T} with slope −Ea2.303R-\frac{E_{a}}{2.303R}). Two temperatures: log⁡k2k1=Ea2.303R(1T1−1T2)\log\frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303R}\left(\frac{1}{T_{1}} - \frac{1}{T_{2}}\right). Higher temperature increases the fraction of molecules with enough energy, and this effect is exponential, so a modest temperature rise gives a large rate increase. Reactions with larger EaE_{a} are more sensitive to temperature. As T→∞T \to \infty, k→Ak \to A. Catalysts. A catalyst increases the rate without being consumed, by providing an alternative pathway with lower activation energy (see figure). Consequences: it does not change ΔH\Delta H or ΔG\Delta G of the reaction; it speeds up forward and backward reactions equally, so equilibrium is reached faster but KK is unchanged; it cannot make a non-spontaneous reaction (ΔG>0\Delta G > 0) occur; a small amount is usually enough, and it is recovered unchanged at the end. The intermediate complex theory: the catalyst combines with reactants to form a temporary unstable intermediate, which decomposes to products and regenerates the catalyst. Inhibitors (negative catalysts) slow reactions. Key Takeaways: - k=Ae−Ea/RTk = Ae^{-E_{a}/RT}; slope of ln⁡k\ln k vs 1T\frac{1}{T} is −EaR-\frac{E_{a}}{R}. - log⁡k2k1=Ea2.303R(1T1−1T2)\log\frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303R}\left(\frac{1}{T_{1}} - \frac{1}{T_{2}}\right). - Ea(f)−Ea(b)=ΔHE_{a}(f) - E_{a}(b) = \Delta H. - Catalysts lower EaE_{a} for both directions; they do not change ΔH\Delta H, ΔG\Delta G or KK. - Pitfall: a catalyst does not change the position of equilibrium, only how fast it is reached. Brain-Teaser 1: The rate constant of a reaction doubles when the temperature rises from 298 K298\,\mathrm{K} to 308 K308\,\mathrm{K}. Find EaE_{a}. Solution: log⁡2=Ea2.303×8.314(1298−1308)\log 2 = \frac{E_{a}}{2.303 \times 8.314}\left(\frac{1}{298} - \frac{1}{308}\right), with 1298−1308=1091784\frac{1}{298} - \frac{1}{308} = \frac{10}{91784}. Ea=0.301×19.15×9178410=52.9 kJ mol−1E_{a} = \frac{0.301 \times 19.15 \times 91784}{10} = 52.9\,\mathrm{kJ\,mol^{-1}}. Brain-Teaser 2: For a reaction, Ea(forward)=80 kJ mol−1E_{a}(\text{forward}) = 80\,\mathrm{kJ\,mol^{-1}} and ΔH=−30 kJ mol−1\Delta H = -30\,\mathrm{kJ\,mol^{-1}}. What is EaE_{a} for the reverse reaction? If a catalyst lowers the forward barrier by 20 kJ mol−120\,\mathrm{kJ\,mol^{-1}}, what happens to the reverse barrier? Solution: Ea(b)=Ea(f)−ΔH=80+30=110 kJ mol−1E_{a}(b) = E_{a}(f) - \Delta H = 80 + 30 = 110\,\mathrm{kJ\,mol^{-1}}. The catalyst lowers the peak itself, so the reverse barrier also falls by 2020, to 90 kJ mol−190\,\mathrm{kJ\,mol^{-1}}; ΔH\Delta H stays −30-30. Brain-Teaser 3: The rate constant of a first-order reaction is k=4.5×1011e−28000/T s−1k = 4.5 \times 10^{11}e^{-28000/T}\,\mathrm{s^{-1}}. Find EaE_{a}, AA, and the temperature at which k=1.0×10−4 s−1k = 1.0 \times 10^{-4}\,\mathrm{s^{-1}}. Solution: Comparing with k=Ae−Ea/RTk = Ae^{-E_{a}/RT}: A=4.5×1011 s−1A = 4.5 \times 10^{11}\,\mathrm{s^{-1}} and EaR=28000 K\frac{E_{a}}{R} = 28000\,\mathrm{K}, so Ea=28000×8.314=232.8 kJ mol−1E_{a} = 28000 \times 8.314 = 232.8\,\mathrm{kJ\,mol^{-1}}. For k=10−4k = 10^{-4}: 28000T=ln⁡4.5×101110−4=ln⁡(4.5×1015)=36.04\frac{28000}{T} = \ln\frac{4.5 \times 10^{11}}{10^{-4}} = \ln(4.5 \times 10^{15}) = 36.04, so T=777 KT = 777\,\mathrm{K}.