Chemistry · Ch 7 — Chemical Kinetics

Arrhenius Equation -- The Effect of Temperature on Reaction Rate

7.8

Arrhenius Equation -- The Effect of Temperature on Reaction Rate

The empirical starting point. Reaction rates generally rise as temperature rises (with only rare exceptions), though by how much varies reaction to reaction; a widely quoted rough rule near room temperature is that rate roughly DOUBLES for every $10^\circ$ C increase. This is demonstrated directly: magnesium granules in cold water (with phenolphthalein) show no colour change, while the identical setup in HOT water quickly turns pink as $Mg+2H_2O\rightarrow Mg(OH)_2+H_2\uparrow$ proceeds fast enough to be visible -- the SAME reaction, made observably faster purely by raising temperature. An even more extreme case: hydrogen and oxygen combine to form water only when an electric spark supplies the needed activation energy, despite the reaction being thermodynamically very favourable at room temperature all along.

The Arrhenius equation. Svante Arrhenius proposed that a rate constant's temperature-dependence follows

k=Ae^{-E_a/RT}

where A is the frequency factor (linked to how often reactant molecules collide; treated as essentially constant with temperature over ordinary ranges), R is the gas constant, $E_a$ the activation energy, and T the absolute temperature. This single equation packages together everything collision theory (Section 7.7) derived from first principles.

The graphical (straight-line) form. Taking natural logs of both sides:

\ln k=\ln A-\frac{E_a}{RT}

This matches $y=mx+c$ with $y=\ln k$ , $x=1/T$ , slope $=-E_a/R$ (NEGATIVE, since $E_a$ is normally positive), and intercept $=\ln A$ . So a genuine Arrhenius-obeying reaction gives a perfectly STRAIGHT line when $\ln k$ (or, equivalently, $\log k$ ) is plotted against $1/T$ -- and only ever with a negative slope; a curved plot, or a straight line with a POSITIVE slope, would both be inconsistent with simple Arrhenius behaviour holding over the full temperature range.

The two-temperature shortcut. If k is known at two different temperatures, $k_1$ at $T_1$ and $k_2$ at $T_2$ , writing the log form at each temperature and subtracting eliminates the unknown $\ln A$ entirely: …

Misc activity-mg-hot-cold-waterActivity -- reaction of magnesium with hot vs cold water

Worked out. Two test tubes A and B, each with 5 ml water plus a drop of phenolphthalein; magnesium granules added to cold water in A and hot water in B. Observation: tube B turns pink (indicating a basic solution has formed), tube A shows no colour change. Reaction: $Mg+2H_2O\rightarrow Mg(OH)_2+H_2\uparrow$ , occurring readily in hot water but not in cold -- demonstrating directly that raising temperature can push a reaction that is negligibly slow at room temperature to occur at an observable rate. A further textbook example: $H_2$ and $O_2$ combine to form $H_2O$ only when an electric spark is passed, even though the reaction is thermodynamically highly f …

Misc example-7Example 7 -- activation energy from rate constants at two temperatures

Worked out. The rate constants of a reaction at 400 K and 200 K are 0.04 and 0.02 s $^{-1}$ respectively. Calculate the activation energy. Book's solution: with $T_2=400$ K, $k_2=0.04$ ; $T_1=200$ K, $k_1=0.02$ : $\log\dfrac{0.04}{0.02}=\dfrac{E_a}{2.303\times8.314}\left(\dfrac{400-200}{400\times200}\right)$ . $\log2=0.301$ ; $\dfrac{400-200}{400\times200}=\dfrac{200}{80000}=\dfrac{1}{400}$ . So $0.301=\dfrac{E_a}{19.147}\times\dfrac{1}{400}\Rightarrow E_a=0.301\times19.147\times400\approx2305\ \text{J mol}^{-1}=2.305\ \text{kJ mo …

Misc example-8Example 8 -- activation energy from the slope of a log k vs 1/T plot

Worked out. $\log k=\log A-\dfrac{E_a}{2.303R}\left(\dfrac{1}{T}\right)$ . A graph of $\log k$ vs $1/T$ gives a straight line with slope $-4000$ K. Calculate the activation energy. Book's solution: matching to $y=mx+c$ , $m=-\dfrac{E_a}{2.303R}\Rightarrow E_a=-m\times2.303R=-(-4000)\times2.303\times8.314=76{,}589\ \text{J mol}^{-1}=76.589\ \text{kJ mol}^{-1}$ . …

Misc evaluate-yourself-arrheniusEvaluate Yourself -- frequency factor from a rate constant and activation energy

Worked out. Book's practice box (no printed solution). For a first order reaction, the rate constant at 500 K is $8\times10^{-4}\ \text{s}^{-1}$ . Calculate the frequency factor, given the activation energy is $190\ \text{kJ mol}^{-1}$ . Working it through with $k=Ae^{-E_a/RT}$ : $\dfrac{E_a}{RT}=\dfrac{190{,}000}{8.314\times500}=45.71$ ; $\ln A=\ln k+\dfrac{E_a}{RT}=\ln(8\times10^{-4})+45.71=(-7.131)+45.71=38.58\Rightarrow A=e^{38.58}\approx6.3\times10^{16}\ \text{s}^{-1}$ (own solution, not printed in the textbook). …

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