Chapter 3 · Chemical Kinetics

3.3 Temperature and Rate: the Arrhenius Equation and Collision Theory

Why does a 10 K rise often double a rate? The Arrhenius equation $k=Ae^{-E_{a}/RT}$, activation energy from two temperatures, the Maxwell–Boltzmann distribution, catalysts, and collision theory.

Food keeps for days in a refrigerator but spoils in hours on a hot kitchen shelf. A pressure cooker, which lets water boil at about $120\ ^{\circ}\mathrm{C}$ instead of $100\ ^{\circ}\mathrm{C}$, cooks dal in a third of the time. Even living things show the effect: crickets chirp faster on warm evenings, and a fever speeds up a patient's metabolism. For a great many reactions near room temperature, a rise of just $10\ \mathrm{K}$ roughly doubles the rate.

That is surprising. A $10\ \mathrm{K}$ rise from $300\ \mathrm{K}$ increases the average kinetic energy of the molecules by only about $3\%$. How can it double the rate? The answer is that reactions are not driven by average molecules. Only the small fraction of collisions with energy above a threshold, the activation energy, lead to reaction, and that small fraction is extremely sensitive to temperature. In this section you will meet the Arrhenius equation, which puts this idea into numbers, use it to find activation energies from experiments, see why catalysts work, and learn the simple model behind it all: collision theory.

Temperature, Activation Energy and the Arrhenius Equation

📘 Definition 3.7 — Temperature coefficient
The temperature coefficient of a reaction is the ratio of its rate constants at two temperatures $10\ \mathrm{K}$ apart, usually $308\ \mathrm{K}$ and $298\ \mathrm{K}$: $$\text{temperature coefficient}=\frac{k_{T+10}}{k_{T}}.$$ For many reactions near room temperature it lies between about $2$ and $3$. It is a rule of thumb, not a law: it is different for each reaction and changes with temperature.

In 1889 Svante Arrhenius proposed that molecules must first be raised to a high-energy, unstable arrangement before they can turn into products. Think of $\mathrm{H_{2}(g)+I_{2}(g)\rightarrow2HI(g)}$: as the molecules approach, the H–H and I–I bonds stretch and new H–I bonds start to form. For a moment the four atoms form an activated complex, a fleeting arrangement at the top of an energy "hill", which can fall either back to reactants or on to products.

📘 Definition 3.8 — Activation energy and activated complex

The activated complex (transition state) is the highest-energy arrangement of atoms along the path from reactants to products. It exists only for about $10^{-13}\ \mathrm{s}$ and cannot be isolated.

The activation energy $E_{a}$ is the minimum extra energy (per mole) that reacting molecules must have, above their average energy, to reach the activated complex. On an energy profile it is the height of the hill measured from the reactants. Every reaction, exothermic or endothermic, has $E_{a}\gt0$, which is why many favourable reactions (paper burning in air) need a spark or a flame to start.

📐 Law 3.6 — The Arrhenius equation

The rate constant of a reaction depends on the absolute temperature $T$ as

$$k=A\,e^{-E_{a}/RT},$$

where $E_{a}$ is the activation energy (in $\mathrm{J\,mol^{-1}}$), $R=8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}$, and $A$ is the pre-exponential factor (frequency factor), a constant for the reaction with the same units as $k$. The factor $e^{-E_{a}/RT}$ is the fraction of molecules (more exactly, of collisions) with energy at least $E_{a}$.

Taking logarithms,

$$\ln k=\ln A-\frac{E_{a}}{RT},\qquad \log k=\log A-\frac{E_{a}}{2.303\,RT}.$$

So a graph of $\ln k$ against $1/T$ is a straight line with slope $-E_{a}/R$ and intercept $\ln A$ (for $\log k$: slope $-E_{a}/2.303R$). The equation assumes that $A$ and $E_{a}$ do not change with temperature, which is a good approximation over moderate ranges of $T$.

✏️ Example 1 — What does "doubling every 10 K" mean for $E_{a}$?
(a) A reaction has a temperature coefficient of $2$. By what factor does its rate increase from $300\ \mathrm{K}$ to $340\ \mathrm{K}$? (b) If the rate constant exactly doubles from $300\ \mathrm{K}$ to $310\ \mathrm{K}$, find $E_{a}$.
(a) $340-300=40\ \mathrm{K}$ is four 10-K steps, each doubling the rate: $2^{4}=16$ times.
(b) Two-temperature form Subtracting the log form of Law 3.6 at the two temperatures (Relations 3.7 below): $$\log\frac{k_{2}}{k_{1}}=\frac{E_{a}}{2.303R}\left(\frac{T_{2}-T_{1}}{T_{1}T_{2}}\right).$$
(b) Solve $$E_{a}=\frac{2.303R\,T_{1}T_{2}}{T_{2}-T_{1}}\log\frac{k_{2}}{k_{1}}=\frac{2.303\times8.314\times300\times310}{10}\times0.3010=5.36\times10^{4}\ \mathrm{J\,mol^{-1}}=53.6\ \mathrm{kJ\,mol^{-1}}.$$
Meaning The "doubling rule" corresponds to activation energies around $50\ \mathrm{kJ\,mol^{-1}}$, typical of many reactions near room temperature. Reactions with a larger $E_{a}$ are even more sensitive to temperature.

Finding $E_{a}$ and $A$ from Experiments

📐 Relations 3.7 — Arrhenius equation at two temperatures

If the rate constants are $k_{1}$ at $T_{1}$ and $k_{2}$ at $T_{2}$ (same $A$ and $E_{a}$), subtracting $\log k_{1}=\log A-\dfrac{E_{a}}{2.303RT_{1}}$ from the same equation at $T_{2}$ removes $A$:

$$\log\frac{k_{2}}{k_{1}}=\frac{E_{a}}{2.303R}\left(\frac{1}{T_{1}}-\frac{1}{T_{2}}\right)=\frac{E_{a}}{2.303R}\left(\frac{T_{2}-T_{1}}{T_{1}T_{2}}\right).$$

Use $R=8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}$ with $E_{a}$ in joules, and temperatures in kelvin. Only the ratio $k_{2}/k_{1}$ matters, so rates, or reciprocals of the times taken for the same change, can be used in place of $k$.

✏️ Example 2 — $E_{a}$ and $A$ from two rate constants
A first-order gas-phase decomposition has $k=1.5\times10^{-3}\ \mathrm{s^{-1}}$ at $600\ \mathrm{K}$ and $3.0\times10^{-2}\ \mathrm{s^{-1}}$ at $700\ \mathrm{K}$. Find $E_{a}$ and $A$.
Ratio $\dfrac{k_{2}}{k_{1}}=\dfrac{3.0\times10^{-2}}{1.5\times10^{-3}}=20$, $\ \log20=1.301$.
$E_{a}$ $$E_{a}=\frac{2.303\times8.314\times600\times700}{700-600}\times1.301=1.05\times10^{5}\ \mathrm{J\,mol^{-1}}=105\ \mathrm{kJ\,mol^{-1}}.$$ (Unrounded: $104\,630\ \mathrm{J\,mol^{-1}}$.)
$A$ From $\log A=\log k+\dfrac{E_{a}}{2.303RT}$ at $600\ \mathrm{K}$: $\log A=\log(1.5\times10^{-3})+\dfrac{104\,630}{2.303\times8.314\times600}=-2.824+9.108=6.284$, so $A=1.9\times10^{6}\ \mathrm{s^{-1}}$.
Check At $700\ \mathrm{K}$: $\log k=6.284-\dfrac{104\,630}{2.303\times8.314\times700}=6.284-7.807=-1.523$, so $k=3.0\times10^{-2}\ \mathrm{s^{-1}}$ ✓.
✏️ Example 3 — Reading an Arrhenius plot
For a first-order reaction, a graph of $\log k$ against $1/T$ is a straight line with slope $-5.0\times10^{3}\ \mathrm{K}$ and intercept $12.0$. Find (a) $E_{a}$, (b) $A$, and (c) $k$ at $400\ \mathrm{K}$.
(a) Slope $=-\dfrac{E_{a}}{2.303R}$, so $E_{a}=2.303\times8.314\times5.0\times10^{3}=9.57\times10^{4}\ \mathrm{J\,mol^{-1}}=95.7\ \mathrm{kJ\,mol^{-1}}$.
(b) Intercept $=\log A=12.0$, so $A=1.0\times10^{12}\ \mathrm{s^{-1}}$.
(c) $\log k=12.0-\dfrac{5.0\times10^{3}}{400}=12.0-12.5=-0.5$, so $k=10^{-0.5}=0.32\ \mathrm{s^{-1}}$.
Warning If the graph uses $\ln k$ instead, the slope is $-E_{a}/R$ with no $2.303$. Always check which logarithm is on the axis.
✏️ Example 4 — Predicting a rate constant at a new temperature
A reaction has $E_{a}=50.0\ \mathrm{kJ\,mol^{-1}}$ and $k=2.0\times10^{-3}\ \mathrm{s^{-1}}$ at $300\ \mathrm{K}$. Find $k$ at $350\ \mathrm{K}$.
Substitute $$\log\frac{k_{2}}{k_{1}}=\frac{50\,000}{2.303\times8.314}\left(\frac{350-300}{300\times350}\right)=2611\times4.762\times10^{-4}=1.244.$$
Antilog $\dfrac{k_{2}}{k_{1}}=10^{1.244}=17.5$, so $k_{2}=17.5\times2.0\times10^{-3}=3.5\times10^{-2}\ \mathrm{s^{-1}}$.
Check $50\ \mathrm{K}$ is five 10-K steps; doubling five times would give $32$. The real factor ($17.5$) is smaller, because each 10-K step has a smaller effect at higher temperature ($1/T$ changes less).
Interactive 1 — Arrhenius Lab

Set the activation energy and the temperature. The left graph is the Arrhenius plot, $\ln k$ against $1000/T$: a straight line whose slope is $-E_{a}/R$. The right graph shows how $\log k$ climbs with $T$. Both mark the current temperature and $T+10\ \mathrm{K}$, and the readouts give the temperature coefficient $k_{T+10}/k_{T}$. Make $E_{a}$ larger: the line gets steeper and the reaction becomes more sensitive to temperature. Press Play to sweep the temperature.

75300
$k$ at $T$ ($\mathrm{s^{-1}}$)—
$e^{-E_{a}/RT}$—
temperature coefficient $k_{T+10}/k_{T}$—
slope $-E_{a}/R$ (K)—

The Maxwell–Boltzmann Distribution and Catalysts

In a gas (or a solution) the molecules do not all have the same kinetic energy. Collisions constantly pass energy around, so at any instant a few molecules are slow, many have middling energies and a few are very fast. Ludwig Boltzmann and James Clerk Maxwell worked out the shape of this spread. Plot the fraction of molecules with each energy against the energy and you get the Maxwell–Boltzmann distribution: a curve that rises from zero, peaks at the most probable energy, and then falls away in a long tail towards high energies.

Mark the activation energy $E_{a}$ on the energy axis. Only the molecules in the tail beyond $E_{a}$ can react when they collide. The area under that part of the curve, as a fraction of the total area, is the fraction of "reactive" molecules. When $E_{a}$ is many times $RT$, this fraction rises and falls with temperature almost exactly as the Arrhenius factor $e^{-E_{a}/RT}$ does. Now raise the temperature:

That is the secret of the doubling rule: temperature acts on a small, exponentially sensitive tail, not on the average.

Interactive 2 — Maxwell–Boltzmann Energy Distribution, $E_{a}$ and a Catalyst

The curve shows how the molecules of a gas share out their kinetic energy at temperature $T$. The shaded tail beyond the activation-energy line holds the molecules that can react. The faint curve is the distribution at $300\ \mathrm{K}$ for comparison. Raise $T$ and watch the peak drop and move right while the shaded fraction grows. Tick catalyst to lower the barrier (the green line): at the same temperature, many more molecules clear it. Press Play to heat and cool the gas. (To keep the tail visible, $E_{a}$ here is smaller than in most real reactions; real tails are far thinner but behave the same way.)

30010
fraction with $E\ge E_{a}$—
$e^{-E_{a}/RT}$—
compared with 300 K, no catalyst—
most probable energy $\tfrac12RT$—

A catalyst works on the other end of the problem. Instead of giving the molecules more energy, it offers a different pathway, with different intermediates, over a lower hill. The same Maxwell–Boltzmann tail, cut at a lower $E_{a}$, contains many more molecules.

📘 Definition 3.9 — Catalyst and its characteristics

A catalyst is a substance that increases the rate of a reaction without itself undergoing any permanent chemical change. It works by providing an alternative pathway (mechanism) with a lower activation energy: it combines with a reactant to form an intermediate, and is set free again in a later step (as $\mathrm{I^{-}}$ is in the decomposition of $\mathrm{H_{2}O_{2}}$, Section 3.1). A substance that slows a reaction is an inhibitor.

  • A small amount of catalyst can change the rate greatly, and it is recovered unchanged in mass and composition at the end.
  • A catalyst does not change $\Delta_{r}H$ or $\Delta_{r}G$ of the reaction, so it cannot make a non-spontaneous reaction ($\Delta_{r}G\gt0$) happen.
  • It lowers the hill for the forward and the backward reaction by the same amount, so it speeds both up equally. It does not change the equilibrium constant; it only helps the system reach equilibrium sooner.
  • Catalysts are often specific: enzymes, the catalysts of living cells, may act on only one reaction.
✏️ Example 5 — How much faster with a catalyst?
At $298\ \mathrm{K}$ a catalyst lowers the activation energy of a reaction from $100\ \mathrm{kJ\,mol^{-1}}$ to $75\ \mathrm{kJ\,mol^{-1}}$. Assuming $A$ is unchanged, by what factor does the rate constant increase?
Ratio $A$ cancels: $$\frac{k_{cat}}{k}=\frac{Ae^{-E_{a,cat}/RT}}{Ae^{-E_{a}/RT}}=e^{(E_{a}-E_{a,cat})/RT}.$$
Exponent $\dfrac{E_{a}-E_{a,cat}}{RT}=\dfrac{25\,000}{8.314\times298}=10.09$.
Factor $e^{10.09}=2.4\times10^{4}$ (or $10^{10.09/2.303}=10^{4.38}$). The reaction goes about $24\,000$ times faster.
Meaning Lowering $E_{a}$ by a quarter changes the rate by more than four powers of ten. The same reaction without the catalyst would have to be heated by roughly $100\ \mathrm{K}$ to speed up this much, which is why industry and living cells rely on catalysts rather than heat.

Collision Theory

Why should the rate be proportional to concentrations, and what is the pre-exponential factor $A$? Collision theory, developed by Max Trautz and William Lewis around 1916–1918, gives a simple picture. It treats reacting molecules as hard spheres and says that a reaction can happen only when two of them collide. Most collisions, however, are fruitless.

📘 Definition 3.10 — Collision frequency, effective collisions and the steric factor
  • The collision frequency $Z_{\mathrm{AB}}$ is the number of collisions between molecules of A and B per second per unit volume of the reaction mixture. It is proportional to $[\mathrm{A}][\mathrm{B}]$, and it is enormous: in a gas at $1\ \mathrm{bar}$ each molecule collides about $10^{9}$ times a second.
  • An effective collision is one that actually produces products. It needs two things: (i) enough energy, at least the threshold energy, so that bonds can break; and (ii) the proper orientation, so that the atoms that must bond come into contact.
  • The steric factor (probability factor) $P$ is the fraction of energetic-enough collisions that have a suitable orientation; $0\lt P\le1$.
📐 Law 3.8 — Rate from collision theory

For a bimolecular reaction $\mathrm{A+B\rightarrow products}$,

$$\text{rate}=P\,Z_{\mathrm{AB}}\,e^{-E_{a}/RT}.$$

$Z_{\mathrm{AB}}$ supplies the concentration dependence ($\propto[\mathrm{A}][\mathrm{B}]$), $e^{-E_{a}/RT}$ the fraction of collisions with enough energy, and $P$ the fraction of those with a suitable orientation. Comparing with Arrhenius, $k[\mathrm{A}][\mathrm{B}]=Ae^{-E_{a}/RT}[\mathrm{A}][\mathrm{B}]$, so the factor $A$ is essentially $P$ times the collision frequency per unit concentrations. Collision theory works well for reactions of atoms and simple molecules. For complex molecules the hard-sphere picture is too crude, and a much smaller $P$ is needed to make it fit.

Orientation in action. In $\mathrm{CH_{3}Br+OH^{-}\rightarrow CH_{3}OH+Br^{-}}$, the hydroxide ion must strike the carbon atom from the side opposite the bromine atom. A hydroxide ion that hits the bromine end, however hard, simply bounces off. You will meet this "back-side attack" again in Chapter 6 (Haloalkanes and Haloarenes).

✏️ Example 6 — How many collisions succeed?
In a gas-phase reaction $\mathrm{A+B\rightarrow products}$ at $500\ \mathrm{K}$, the collision frequency is $Z_{\mathrm{AB}}=1.0\times10^{32}$ collisions $\mathrm{L^{-1}\,s^{-1}}$, $E_{a}=80\ \mathrm{kJ\,mol^{-1}}$ and $P=0.20$. Find (a) the fraction of collisions with enough energy, (b) the fraction of all collisions that are effective, and (c) the rate in $\mathrm{mol\,L^{-1}\,s^{-1}}$ ($N_{A}=6.022\times10^{23}\ \mathrm{mol^{-1}}$).
(a) $\dfrac{E_{a}}{RT}=\dfrac{80\,000}{8.314\times500}=19.24$; $\ e^{-19.24}=4.4\times10^{-9}$. Only about four collisions in a billion have enough energy.
(b) $P\,e^{-E_{a}/RT}=0.20\times4.39\times10^{-9}=8.8\times10^{-10}$.
(c) Effective collisions $=1.0\times10^{32}\times8.77\times10^{-10}=8.8\times10^{22}\ \mathrm{L^{-1}\,s^{-1}}$. Dividing by $N_{A}$: $$\text{rate}=\frac{8.77\times10^{22}}{6.022\times10^{23}}=0.15\ \mathrm{mol\,L^{-1}\,s^{-1}}.$$
Meaning If every collision reacted, the gas would be used up in a tiny fraction of a second. The energy and orientation requirements are what make most reactions slow enough to study.
⚠️ Common mistakes with the Arrhenius equation

1. kJ and J. With $R=8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}$, $E_{a}$ must be in joules: $50\ \mathrm{kJ\,mol^{-1}}=50\,000\ \mathrm{J\,mol^{-1}}$.

2. Celsius. $T$ is always in kelvin. Using $25$ and $35$ instead of $298$ and $308$ gives nonsense.

3. $\log$ versus $\ln$. $\log\dfrac{k_{2}}{k_{1}}$ needs the $2.303$; $\ln\dfrac{k_{2}}{k_{1}}$ does not.

4. "A catalyst shifts equilibrium." It does not: it speeds up the forward and backward reactions equally.

5. "Higher temperature lowers $E_{a}$." It does not: $E_{a}$ is (nearly) constant; heating increases the fraction of molecules that can climb the same hill. Only a catalyst lowers the hill.

🚀 Beyond the board syllabus (JEE Main · NEET · JEE Advanced) — the backward hill

In JEE Main, NEET and JEE Advanced 2026 (activation energy and its calculation); not required by the CBSE syllabus.

The reverse reaction climbs the same hill from the product side. Reading the energy profile, $$E_{a}(\text{forward})-E_{a}(\text{backward})=\Delta_{r}H.$$ For an exothermic reaction ($\Delta_{r}H\lt0$) the backward barrier is the higher one. For example, if $E_{a}(\text{f})=120\ \mathrm{kJ\,mol^{-1}}$ and $\Delta_{r}H=-40\ \mathrm{kJ\,mol^{-1}}$, then $E_{a}(\text{b})=160\ \mathrm{kJ\,mol^{-1}}$. A catalyst lowers both barriers by the same amount, so their difference, $\Delta_{r}H$, is unchanged. Because the backward reaction has the larger $E_{a}$, raising $T$ speeds it up more than the forward one, which is the kinetic reason why heating shifts an exothermic equilibrium to the left.

Exercises

I
Temperature coefficient and the Arrhenius equation. Use $R=8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}$.
  • 1 A reaction has a temperature coefficient of $2$. By what factor does its rate increase when the temperature rises from $25\ ^{\circ}\mathrm{C}$ to $65\ ^{\circ}\mathrm{C}$?
  • 2 A reaction with a temperature coefficient of $3$ takes $54\ \mathrm{min}$ to reach a certain stage at $300\ \mathrm{K}$. How long does it take at $330\ \mathrm{K}$?
  • 3 Write the Arrhenius equation and state the meaning and units of each symbol. What happens to $k$ as $T\rightarrow\infty$?
  • 4 A graph of $\ln k$ against $1/T$ has slope $-1.2\times10^{4}\ \mathrm{K}$. Find $E_{a}$.
  • 5 For a first-order reaction, a graph of $\log k$ against $1/T$ has slope $-6.0\times10^{3}\ \mathrm{K}$ and intercept $13.0$. Find $E_{a}$, $A$ and $k$ at $500\ \mathrm{K}$.
  • 6 For $E_{a}=60\ \mathrm{kJ\,mol^{-1}}$, find the fraction $e^{-E_{a}/RT}$ at $300\ \mathrm{K}$ and at $310\ \mathrm{K}$, and their ratio.
  • 7 What would the Arrhenius equation predict for a reaction with $E_{a}=0$? Explain in terms of molecules.
  • 8 A first-order reaction has $k=(2.0\times10^{13}\ \mathrm{s^{-1}})\,e^{-1.5\times10^{4}\ \mathrm{K}/T}$. Find $E_{a}$ and $k$ at $500\ \mathrm{K}$.
1 $40\ \mathrm{K}$ is four 10-K steps: $2^{4}=16$ times.
2 Three steps: rate $\times3^{3}=27$, so the time is $\dfrac{54}{27}=2.0\ \mathrm{min}$ (time is inversely proportional to rate).
3 $k=Ae^{-E_{a}/RT}$: $k$ rate constant; $A$ pre-exponential (frequency) factor, same units as $k$; $E_{a}$ activation energy ($\mathrm{J\,mol^{-1}}$); $R$ gas constant ($8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}$); $T$ absolute temperature (K). As $T\rightarrow\infty$, $E_{a}/RT\rightarrow0$ and $k\rightarrow A$.
4 Slope $=-E_{a}/R$: $E_{a}=8.314\times1.2\times10^{4}=9.98\times10^{4}\ \mathrm{J\,mol^{-1}}\approx99.8\ \mathrm{kJ\,mol^{-1}}$.
5 $E_{a}=2.303\times8.314\times6.0\times10^{3}=1.15\times10^{5}\ \mathrm{J\,mol^{-1}}=115\ \mathrm{kJ\,mol^{-1}}$; $A=10^{13.0}=1.0\times10^{13}\ \mathrm{s^{-1}}$; $\log k=13.0-\dfrac{6.0\times10^{3}}{500}=1.0$, so $k=10\ \mathrm{s^{-1}}$.
6 $300\ \mathrm{K}$: $e^{-60\,000/2494}=e^{-24.06}=3.6\times10^{-11}$; $\ 310\ \mathrm{K}$: $e^{-60\,000/2577}=e^{-23.28}=7.8\times10^{-11}$; ratio $=2.2$. The fraction is tiny but more than doubles.
7 $k=A$ at every temperature: the rate would not depend on $T$. Every collision with the right orientation would react, since no energy barrier has to be overcome. (Some reactions of free radicals and ions come close to this.)
8 Compare with $Ae^{-E_{a}/RT}$: $\dfrac{E_{a}}{R}=1.5\times10^{4}\ \mathrm{K}$, so $E_{a}=1.5\times10^{4}\times8.314=1.25\times10^{5}\ \mathrm{J\,mol^{-1}}=125\ \mathrm{kJ\,mol^{-1}}$. At $500\ \mathrm{K}$: $k=2.0\times10^{13}\times e^{-30}=2.0\times10^{13}\times9.36\times10^{-14}=1.9\ \mathrm{s^{-1}}$.
II
The two-temperature equation. Use $R=8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}$.
  • 1 The rate constant of a reaction doubles when the temperature rises from $290\ \mathrm{K}$ to $300\ \mathrm{K}$. Find $E_{a}$.
  • 2 A reaction has $k=2.0\times10^{-4}\ \mathrm{s^{-1}}$ at $400\ \mathrm{K}$ and $3.0\times10^{-3}\ \mathrm{s^{-1}}$ at $450\ \mathrm{K}$. Find $E_{a}$.
  • 3 A reaction has $E_{a}=75.0\ \mathrm{kJ\,mol^{-1}}$ and $k=1.0\times10^{-3}\ \mathrm{s^{-1}}$ at $300\ \mathrm{K}$. Find $k$ at $320\ \mathrm{K}$.
  • 4 At what temperature will the rate constant of a reaction with $E_{a}=60.0\ \mathrm{kJ\,mol^{-1}}$ be ten times its value at $300\ \mathrm{K}$?
  • 5 Food science. Milk turns sour in $48\ \mathrm{h}$ at $25\ ^{\circ}\mathrm{C}$ but takes $10$ days in a refrigerator at $5\ ^{\circ}\mathrm{C}$. Estimate the activation energy of the souring process.
  • 6 Cooking. Rice cooks in $20\ \mathrm{min}$ in boiling water at $100\ ^{\circ}\mathrm{C}$. In a pressure cooker water boils at $120\ ^{\circ}\mathrm{C}$. Taking $E_{a}=60\ \mathrm{kJ\,mol^{-1}}$ for the cooking reactions, estimate the cooking time in the pressure cooker.
  • 7 Reactions X and Y have $E_{a}=50$ and $100\ \mathrm{kJ\,mol^{-1}}$. Find the factor by which each rate constant increases from $300$ to $310\ \mathrm{K}$. Which is more sensitive to temperature?
Check: activation energy from two temperatures
$\log(k_{2}/k_{1})$—
$E_{a}$ ($\mathrm{kJ\,mol^{-1}}$)—
$A$ (units of $k$)—
1 $E_{a}=\dfrac{2.303\times8.314\times290\times300}{10}\times\log2=50.1\ \mathrm{kJ\,mol^{-1}}$.
2 $\dfrac{k_{2}}{k_{1}}=15$, $\log15=1.176$: $E_{a}=\dfrac{2.303\times8.314\times400\times450}{50}\times1.176=81.1\ \mathrm{kJ\,mol^{-1}}$ (the check-lab above starts with these numbers).
3 $\log\dfrac{k_{2}}{k_{1}}=\dfrac{75\,000}{2.303\times8.314}\times\dfrac{20}{300\times320}=0.816$; ratio $=6.55$; $k=6.5\times10^{-3}\ \mathrm{s^{-1}}$.
4 $\log10=1=\dfrac{60\,000}{2.303\times8.314}\left(\dfrac{1}{300}-\dfrac{1}{T_{2}}\right)$, so $\dfrac{1}{T_{2}}=\dfrac{1}{300}-3.191\times10^{-4}=3.014\times10^{-3}\ \mathrm{K^{-1}}$ and $T_{2}=332\ \mathrm{K}$.
5 Rate $\propto1/\text{time}$: $\dfrac{k_{298}}{k_{278}}=\dfrac{240\ \mathrm{h}}{48\ \mathrm{h}}=5$. $E_{a}=\dfrac{2.303\times8.314\times278\times298}{20}\times\log5=55.4\ \mathrm{kJ\,mol^{-1}}$.
6 $\log\dfrac{k_{393}}{k_{373}}=\dfrac{60\,000}{2.303\times8.314}\times\dfrac{20}{373\times393}=0.428$; ratio $=2.68$. Time $\approx\dfrac{20}{2.68}=7.5\ \mathrm{min}$.
7 X: $10^{\frac{50\,000}{19.147}\times\frac{10}{93\,000}}=10^{0.281}=1.91$; Y: $10^{0.562}=3.64$. Y, with the larger $E_{a}$, is more sensitive: its Arrhenius line is steeper.
III
The Maxwell–Boltzmann distribution and catalysts.
  • 1 Describe how the Maxwell–Boltzmann distribution of molecular energies changes when the temperature of a gas is raised. Which feature does not change, and why?
  • 2 A $10\ \mathrm{K}$ rise near $300\ \mathrm{K}$ raises the average kinetic energy of molecules by only about $3\%$, yet it can double a rate. Explain.
  • 3 At $310\ \mathrm{K}$ (body temperature) an enzyme lowers the activation energy of a reaction from $80$ to $60\ \mathrm{kJ\,mol^{-1}}$. Assuming $A$ is unchanged, by what factor does the rate increase?
  • 4 Does a catalyst change (a) $\Delta_{r}H$, (b) $\Delta_{r}G$, (c) the equilibrium constant, (d) the time taken to reach equilibrium? Give reasons.
  • 5 True or false? Give a reason. (a) A catalyst is consumed in the reaction it catalyses. (b) A catalyst increases the rates of the forward and backward reactions by the same factor. (c) A catalyst can make a reaction with $\Delta_{r}G\gt0$ happen.
  • 6 A catalyst makes a reaction $1000$ times faster at $300\ \mathrm{K}$. Assuming $A$ is unchanged, by how much has it lowered $E_{a}$?
  • 7 For $E_{a}=50.0\ \mathrm{kJ\,mol^{-1}}$, find the fraction of molecules with enough energy, $e^{-E_{a}/RT}$, at $300\ \mathrm{K}$ and at $600\ \mathrm{K}$.
1 The peak becomes lower and moves to higher energy, and the curve spreads out, so the high-energy tail is much larger. The total area under the curve does not change, because it represents all the molecules, whose number is fixed.
2 The rate depends on the fraction of molecules in the tail with $E\ge E_{a}$, which is about $e^{-E_{a}/RT}$. Because $E_{a}$ is many times $RT$, this exponential fraction is very sensitive to $T$ and can double even when the average energy barely changes (Interactive 2).
3 $\dfrac{k_{cat}}{k}=e^{20\,000/(8.314\times310)}=e^{7.76}=2.3\times10^{3}$.
4 (a), (b), (c) No: these depend only on the initial and final states, which the catalyst does not change; it lowers the forward and backward barriers by the same amount. (d) Yes: equilibrium is reached sooner.
5 (a) False: it is regenerated, unchanged in mass and composition. (b) True: both barriers fall by the same amount, so (with $A$ unchanged) both rates rise by the same factor and $K$ is unchanged. (c) False: a catalyst cannot change $\Delta_{r}G$; it only speeds up reactions that are already feasible.
6 $1000=e^{\Delta E_{a}/RT}$, so $\Delta E_{a}=RT\ln1000=8.314\times300\times6.908=1.72\times10^{4}\ \mathrm{J\,mol^{-1}}=17.2\ \mathrm{kJ\,mol^{-1}}$.
7 $300\ \mathrm{K}$: $e^{-50\,000/2494}=e^{-20.05}=2.0\times10^{-9}$; $\ 600\ \mathrm{K}$: $e^{-10.02}=4.4\times10^{-5}$, more than $20\,000$ times larger.
IV
Collision theory.
  • 1 Define collision frequency and effective collision. Why does only a tiny fraction of collisions lead to reaction?
  • 2 What is the steric (probability) factor $P$? Use the reaction $\mathrm{CH_{3}Br+OH^{-}\rightarrow CH_{3}OH+Br^{-}}$ to explain why it is less than $1$.
  • 3 In a gas-phase reaction at $600\ \mathrm{K}$, $Z_{\mathrm{AB}}=5.0\times10^{31}\ \mathrm{L^{-1}\,s^{-1}}$, $E_{a}=100\ \mathrm{kJ\,mol^{-1}}$ and $P=0.50$. Find the rate in $\mathrm{mol\,L^{-1}\,s^{-1}}$.
  • 4 Use collision theory to explain why the rate of a bimolecular reaction is proportional to $[\mathrm{A}][\mathrm{B}]$.
  • 5 State one main limitation of collision theory.
1 Collision frequency: the number of collisions per second per unit volume. Effective collision: one that forms products. Most collisions either lack the energy to reach the activated complex (only a fraction $\approx e^{-E_{a}/RT}$ have enough) or meet with the wrong orientation.
2 $P$ is the fraction of sufficiently energetic collisions that have a suitable orientation. In the example, $\mathrm{OH^{-}}$ must hit the carbon atom on the side away from Br; collisions at the bromine end, or side-on, fail however energetic they are. So $P\lt1$.
3 $\dfrac{E_{a}}{RT}=\dfrac{100\,000}{8.314\times600}=20.05$, $e^{-20.05}=1.97\times10^{-9}$. Rate $=\dfrac{0.50\times5.0\times10^{31}\times1.97\times10^{-9}}{6.022\times10^{23}}=8.2\times10^{-2}\ \mathrm{mol\,L^{-1}\,s^{-1}}$.
4 The number of A–B collisions per second in a fixed volume is proportional to the number of A molecules and to the number of B molecules: doubling either doubles the collisions. So $Z_{\mathrm{AB}}\propto[\mathrm{A}][\mathrm{B}]$, and the rate (a fixed fraction of $Z_{\mathrm{AB}}$ at a given $T$) is too.
5 It treats molecules as hard spheres and ignores their structure, so it cannot predict $P$; for complex molecules the calculated rates are far too high unless an (empirical) small $P$ is introduced.
🎯 Exam corner BoardJEE MainJEE AdvancedNEET

E1. Board · 3 marks The rate constant of a reaction becomes three times larger when the temperature is raised from $310\ \mathrm{K}$ to $330\ \mathrm{K}$. Calculate the activation energy. ($R=8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}$, $\log3=0.4771$.)

Formula $\log\dfrac{k_{2}}{k_{1}}=\dfrac{E_{a}}{2.303R}\left(\dfrac{T_{2}-T_{1}}{T_{1}T_{2}}\right)$ (1 mark)
Substitute $0.4771=\dfrac{E_{a}}{2.303\times8.314}\times\dfrac{20}{310\times330}$ (1 mark)
Answer $E_{a}=\dfrac{0.4771\times2.303\times8.314\times102\,300}{20}=4.67\times10^{4}\ \mathrm{J\,mol^{-1}}=46.7\ \mathrm{kJ\,mol^{-1}}$ (1 mark)

E2. JEE Main Numerical value. At $300\ \mathrm{K}$ a catalyst lowers the activation energy of a reaction by $20\ \mathrm{kJ\,mol^{-1}}$, the pre-exponential factor being unchanged. The rate increases by a factor of $x\times10^{3}$. The value of $x$ (nearest integer) is ______. ($R=8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}$.)

Ratio $\dfrac{k_{cat}}{k}=e^{\Delta E_{a}/RT}=e^{20\,000/(8.314\times300)}=e^{8.019}$.
Evaluate $e^{8.019}=10^{8.019/2.303}=10^{3.482}=3.04\times10^{3}$.
Answer $\mathbf{3}$.

E3. NEET A graph of $\ln k$ against $1/T$ for a reaction is a straight line. Its slope is (A) $E_{a}/R$  (B) $-E_{a}/R$  (C) $-E_{a}/2.303R$  (D) $\ln A$

Idea $\ln k=\ln A-\dfrac{E_{a}}{R}\cdot\dfrac1T$ is $y=c+mx$ with $x=1/T$.
Answer (B). Option (C) is the slope of the $\log k$ plot; (D) is the intercept.

E4. JEE Main NEET Adding a catalyst to a reversible reaction does not change (A) the activation energy of the forward reaction  (B) the rate constant of the backward reaction  (C) the enthalpy change of the reaction  (D) the time needed to reach equilibrium

Idea A catalyst lowers both activation energies by the same amount, so both rate constants rise and equilibrium is reached sooner. The energies of reactants and products are unchanged.
Answer (C).

E5. JEE Advanced Integer type. Two reactions P and Q have the same pre-exponential factor. Their activation energies are $40\ \mathrm{kJ\,mol^{-1}}$ (P) and $80\ \mathrm{kJ\,mol^{-1}}$ (Q). At what temperature, in kelvin, does Q have the same rate constant that P has at $300\ \mathrm{K}$?

Equal $k$ With the same $A$, $Ae^{-E_{P}/RT_{1}}=Ae^{-E_{Q}/RT_{2}}$ requires $\dfrac{E_{P}}{T_{1}}=\dfrac{E_{Q}}{T_{2}}$.
Solve $T_{2}=T_{1}\dfrac{E_{Q}}{E_{P}}=300\times\dfrac{80}{40}=600\ \mathrm{K}$.
Answer $\mathbf{600}$. (At $300\ \mathrm{K}$ itself, P is faster by $e^{40\,000/(8.314\times300)}\approx9\times10^{6}$ times.)

E6. Board NEET Assertion–reason. Assertion (A): The rate of most reactions increases when the temperature is raised. Reason (R): The activation energy of a reaction decreases as the temperature increases. (A) Both A and R are true and R is the correct explanation of A. (B) Both A and R are true but R is not the correct explanation of A. (C) A is true but R is false. (D) A is false but R is true.

Check A True.
Check R False: $E_{a}$ is practically independent of $T$. Heating increases the fraction of molecules with energy at least $E_{a}$ (and the collision frequency); only a catalyst lowers $E_{a}$.
Answer (C).
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. Can a living "clock" — the chirping of a cricket — be described by the Arrhenius equation, and what "activation energy" does it have?

Why it matters. Cold-blooded animals cannot control their body temperature, so the chemical reactions that drive their muscles and nerves speed up on warm nights. Biologists studying how insects, fish and crops will respond to climate warming use exactly this kind of Arrhenius analysis, and food scientists use it to predict shelf lives at different storage temperatures.

What to do.

  1. Dolbear's classic rule of thumb (1897) for the snowy tree cricket of North America — an approximate rule, not a set of measurements — gives about $76$, $112$, $148$ and $184$ chirps per minute at $15$, $20$, $25$ and $30\ ^{\circ}\mathrm{C}$. Convert the temperatures to kelvin.
  2. Treat the chirp rate as a rate constant. Plot $\ln(\text{rate})$ against $1/T$, draw the best straight line and find an "activation energy" from its slope.
  3. Calculate $E_{a}$ separately for each 5-degree interval with the two-temperature equation. Are the three values the same? What does this tell you about whether chirping is controlled by one simple reaction?
  4. Find the ratio of chirp rates between $15$ and $25\ ^{\circ}\mathrm{C}$ and compare it with the "doubling per $10\ \mathrm{K}$" rule.

Tools: Spreadsheet (use LINEST or a trendline); Interactive 1 and the check-lab under Exercises II.

Hand in: Your plot, the overall $E_{a}$, the three interval values, and a short paragraph on what a single "activation energy" means for a whole organism.

Going further. Look up data on how the development time of a crop pest (for example, a moth or an aphid) depends on temperature, and estimate how a $2\ \mathrm{K}$ warming would change the number of generations per year.

Instructor note: least-squares slope $\approx-5.1\times10^{3}\ \mathrm{K}$, giving $E_{a}\approx43\ \mathrm{kJ\,mol^{-1}}$. Interval values fall from about $54$ to $40$ to $33\ \mathrm{kJ\,mol^{-1}}$: the Arrhenius plot curves, because chirping depends on many coupled processes (and Dolbear's rule is a straight line in temperature rather than an exponential). Rate ratio $15\rightarrow25\ ^{\circ}\mathrm{C}$: $148/76=1.95$ over $10\ \mathrm{K}$, close to the doubling rule.

🧪 Self-check quiz

0/5 correct

Q1. According to the Arrhenius equation, as $T\rightarrow\infty$ the rate constant $k$ approaches

Q2. A catalyst does not change

Q3. A reaction's rate doubles for every $10\ \mathrm{K}$ rise. Raising the temperature from $300$ to $330\ \mathrm{K}$ multiplies the rate by

Q4. When a gas is heated, its Maxwell–Boltzmann energy distribution

Q5. According to collision theory, a collision is effective only if the molecules

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