Chemical Kinetics

1. Check this before you revise anything

This was Unit 4 in the previous edition. Six chapters were cut from the Class 12 Chemistry book and the survivors renumbered, so Chemical Kinetics moved from 4 to 3. Unlike Electrochemistry, this chapter's internal cross-references were correctly updated, so nothing here points at the old numbering.

Thermodynamics and kinetics answer different questions. Thermodynamics says whether a reaction can happen; kinetics says how fast. The book's own example is the best one: thermodynamic data say diamond should turn into graphite, and it does, at a rate so slow that nobody has ever seen it.

This chapter has two separate question sets:

• Intext Questions, 9 of them, in boxes on pages 66, 71, 78 and 84.
• Exercises, 30 of them, at the end, numbered 3.1 to 3.30.

Both sets are numbered 3.1, 3.2, 3.3 and so on, so an intext question and an exercise can share a label. Identify a question by content, not by number.

The book answers eight of the nine intext questions — 3.1 to 3.6, 3.8 and 3.9 — and none of the thirty exercises. Intext 3.7 is the only one left blank, and it is descriptive. Every answer on this site is worked independently, and all eight checkable ones agree with the book.

Only zero and first order are integrated. The chapter states plainly that it derives integrated rate equations for zero and first order reactions only. Second order integration is not on the syllabus, though second order rate laws appear constantly in the exercises.

2. Rate of a Reaction (Textbook 3.1)

Rate is a change of concentration per unit time. For a reactant the change is negative, so a minus sign is attached to keep the rate positive.

Average rate is measured over an interval, instantaneous rate at a point, and the second is the slope of the tangent to a concentration-time curve.

Coefficients divide. For the general reaction

$a\text{A}+b\text{B}\to c\text{C}+d\text{D}$

$\text{Rate}=-\frac{1}{a}\frac{d[\text{A}]}{dt}=-\frac{1}{b}\frac{d[\text{B}]}{dt}=\frac{1}{c}\frac{d[\text{C}]}{dt}=\frac{1}{d}\frac{d[\text{D}]}{dt}$

This division is what makes the rate a single number rather than four different ones. Intext 3.2 is built entirely on it: A disappears at 0.010 mol L-1 min-1 but the reaction rate is 0.005, because the equation reads 2A gives products.

Units are mol L-1 s-1 for solutions, and bar min-1 or atm s-1 when a gas reaction is followed by pressure.

3. Rate Law, Order and Molecularity (Textbook 3.2)

The rate law must be measured, never predicted. For $a\text{A}+b\text{B}\to$ products the rate law is

$\text{Rate}=k[\text{A}{]}^x[\text{B}{]}^y$

where $x$ and $y$ come from experiment and may bear no relation at all to $a$ and $b$. Exercise 3.1 makes the point four times over: 3NO gives N2O is second order, and the H+ in the peroxide-iodide reaction does not enter the rate law despite a coefficient of 2.

Order is $x+y$. It may be zero, fractional or even negative.

Units of $k$ follow from the order. Rearranging the rate law:

$k=\frac{\text{Rate}}{(\text{concentration}{)}^n}=({\text{mol L}}^{-1}{)}^{1-n}{\text{s}}^{-1}$

Reading the order off the units is a habit worth building. Exercise 3.24 never uses the words "first order"; the s-1 is the only clue given.

Molecularity is the number of particles colliding in a single elementary step. The two ideas are constantly confused, so hold them apart:

For a complex reaction the slowest step sets the rate. The book's analogy is a relay team whose chances depend on its slowest runner. In the iodide-catalysed decomposition of H2O2, the first of two bimolecular steps is slow, so the overall reaction is first order in both H2O2 and I- even though iodide is regenerated.

Pseudo first order reactions are higher order reactions that behave as first order because one reactant is in vast excess. Hydrolysis of ethyl acetate and inversion of cane sugar are both really second order, but water barely changes concentration, so the rate depends on the ester or the sugar alone.

4. Integrated Rate Equations (Textbook 3.3)

Zero order. Rate is independent of concentration, so

$[\text{R}]=[\text{R}{]}_0-kt,t_{1/2}=\frac{[\text{R}{]}_0}{2k}$

A plot of $[\text{R}]$ against $t$ is a straight line of slope $-k$. Decomposition of NH3 on hot platinum and of HI on gold are the examples: the surface is saturated, so adding more gas changes nothing.

First order.

$k=\frac{2.303}{t}\log \frac{[\text{R}{]}_0}{[\text{R}]},[\text{R}]=[\text{R}{]}_0{e}^{-kt}$

A plot of $\log [\text{R}]$ against $t$ is a straight line of slope $-k/2.303$, and that straight line is the standard test for first order kinetics. Exercise 3.15 is exactly this test applied to nine data points.

The first order half-life is independent of starting amount, which is why radioactive decay is quoted as a half-life at all. Exercises 3.14 and 3.17 are radiocarbon dating and strontium-90 in bone, both worked as ordinary first order problems.

Any quantity proportional to concentration may be substituted, since only a ratio appears in the logarithm. Masses work in intext 3.5, percentages in Exercise 3.19, and partial pressures in Exercises 3.20 and 3.21.

Gas-phase problems need one extra step. When the total pressure is measured and one molecule becomes two, the reactant pressure is not the reading. With initial pressure $p_0$ and total pressure $P$,

$p_{\text{reactant}}=2p_0-P$

Both Exercise 3.20 and Exercise 3.21 turn on this line, and using $P$ directly gives an answer that is not even the right order of magnitude.

5. Temperature, Activation Energy and Catalysts (Textbook 3.4)

A rise of 10 K roughly doubles the rate constant near room temperature. The chapter's illustration is N2O5, whose half-life falls from 10 days at 0 °C to 5 hours at 25 °C to 12 minutes at 50 °C.

The Arrhenius equation puts a number on it:

$k=A{e}^{-E_a/RT}$

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

So a plot of $\ln k$ against $1/T$ is a straight line of slope $-E_a/R$ and intercept $\ln A$. Exercise 3.22 is this graph drawn from five temperatures.

For two temperatures, $A$ cancels:

$\log \frac{k_2}{k_1}=\frac{E_a}{2.303R}\left[\frac{T_2-T_1}{T_1{T}_2}\right]$

This one relation solves six of the thirty exercises — 3.22, 3.23, 3.26, 3.27, 3.28, 3.29 and 3.30 — usually by matching a given expression term by term against the Arrhenius form.

Why temperature matters so much is the Maxwell-Boltzmann distribution. The factor $e^{-E_a/RT}$ is the fraction of molecules with energy at least $E_a$. Heating broadens the curve and shifts its peak right, so that fraction grows steeply while the total area stays fixed at one. Intext 3.9 puts the fraction at $1.47\times 10^{-19}$ for HI decomposition at 581 K, which is why so few collisions succeed.

A catalyst lowers $E_a$ by offering an alternative path. It forms a temporary intermediate complex with the reactants and is released unchanged. Three limits are worth stating precisely:

• It does not alter $\Delta G$, so it cannot make a non-spontaneous reaction go.
• It does not change the equilibrium constant.
• It speeds the forward and backward reactions equally, so equilibrium is reached sooner at the same position.

A substance that slows a reaction is an inhibitor, not a catalyst.

6. Collision Theory (Textbook 3.5)

Collision theory treats molecules as hard spheres and was developed by Trautz and Lewis in 1916-18 from the kinetic theory of gases.

Collision frequency $Z$ is the number of collisions per second per unit volume. For a bimolecular reaction A + B gives products,

$\text{Rate}=Z_{\text{AB}}{e}^{-E_a/RT}$

Comparing with the Arrhenius equation shows that $A$ is related to collision frequency.

But not every energetic collision produces a reaction, because orientation matters. The chapter's example is bromoethane reacting with hydroxide: approach from the right side gives product, approach from the wrong side simply bounces. Collisions that have both enough energy and the right geometry are effective collisions.

The steric factor $P$ is introduced to account for this:

$\text{Rate}=P{Z}_{\text{AB}}{e}^{-E_a/RT}$

Threshold energy is defined in the chapter's own footnote as the activation energy plus the energy the reacting species already possess.

The theory has a stated limit. Treating molecules as hard spheres ignores their structure entirely, which is why $P$ has to be bolted on rather than predicted.

7. Where the printed chapter needs care

Exercise 3.3 gives a zero order rate constant in second order units. It states $k=2.5\times 10^{-4}$ mol-1 L s-1 for a reaction it has just declared zero order. Table 3.3 of the same chapter gives mol L-1 s-1 for zero order and mol-1 L s-1 for second order, so the printed unit contradicts the book's own table. The intended unit is mol L-1 s-1 and the numerical answers are unaffected.

The answer key to intext 3.2 contradicts itself. It reads "Rate of reaction = rate of disappearance of A = 0.005 mol litre-1 min-1". The number 0.005 is the rate of the reaction; the rate of disappearance of A is 0.010. The two are equal only when the coefficient is 1, and here it is 2.

Example 3.10 prints the wrong units for $E_a$ and $R$. It shows "209000 J mol L-1" and "8.314 J mol L-1 K-1". Activation energy is in J mol-1 and the gas constant in J mol-1 K-1; the litre has no business being there. The arithmetic and the answer, $k_2=6.36\times 10^{-3}$ s-1, are correct.

Example 3.3 rounds an intermediate and then ignores the rounding. It writes $A=0.02/0.012=1.61$. The quotient of those two printed numbers is 1.67; the answer 1.61 comes from the unrounded 0.01245. Carry the full value.

The table header in Exercise 3.21 reads "Time/s-1". Time is in seconds, so it should read "Time/s".

Exercise 3.22 is a graph question, so expect a range. A least-squares line through all five points gives $E_a=100.6$ kJ mol-1 and $A=1.4\times 10^{13}$ s-1; the two end points alone give 102.3 kJ mol-1 and $2.9\times 10^{13}$ s-1. The slope is robust, but the intercept is a long extrapolation to $1/T=0$ and is much more sensitive to how the line is drawn.

Summary

Chemical kinetics measures how fast a reaction goes, a question thermodynamics cannot answer. Rate is a concentration change per unit time, divided by the stoichiometric coefficient so that one number describes the reaction however it is monitored.

The rate law is the centre of the subject, and its exponents are experimental facts that cannot be read off a balanced equation. Order is their sum and may be fractional or zero; molecularity counts colliding particles in one elementary step and may not. For a complex reaction the slowest step governs, and a reactant in vast excess drops out of the rate law entirely, giving pseudo first order behaviour.

Integrating the rate law gives a straight-line test for order: $[\text{R}]$ against $t$ for zero order, $\log [\text{R}]$ against $t$ for first. The first order half-life $0.693/k$ is independent of the starting amount, which is what makes radioactive dating possible.

Temperature enters through the Arrhenius equation, whose exponential factor is the fraction of molecules carrying at least the activation energy. A plot of $\ln k$ against $1/T$ yields both $E_a$ and $A$, and a catalyst works by lowering $E_a$ alone, leaving the thermodynamics of the reaction untouched. Collision theory supplies the physical picture behind $A$, together with the reminder that energy is necessary but orientation is required too.