Chemical Kinetics

A reaction $\text{A}+\text{B}\to \text{P}$ is studied experimentally, and the rate law turns out to be $\text{rate}=k{\left[\text{A}\right]}^2$, with no dependence on $\text{B}$ at all. Yet $\text{B}$ is definitely consumed. How?

Because $\text{B}$ enters after the rate-determining step.

Suppose the mechanism is

$\text{A}+\text{A}\stackrel{\text{slow}}{\to }\text{I},\text{I}+\text{B}\stackrel{\text{fast}}{\to }\text{P}$

The overall rate is set entirely by the slow step, which involves only $\text{A}$. Once the intermediate forms, $\text{B}$ mops it up instantly, so supplying more $\text{B}$ changes nothing. The stoichiometry says one $\text{B}$ per product; the kinetics says $\text{B}$ is invisible.

A rate law is an experimental fact about the mechanism, not a restatement of the balanced equation. Every difficult question in this chapter lives in that gap, and the first move is always to stop reading coefficients as exponents.

1. Order, molecularity, and what each means

For an elementary step alone are the two the same. For anything multi-step they need not agree, and a fractional order is positive proof that the reaction is not elementary.

The units of $k$ give the order away immediately:

$k\text{has units of}{\text{mol}}^{1-n}{\text{L}}^{n-1}{\text{s}}^{-1}$

so a zero-order constant is in mol L${}^{-1}$ s${}^{-1}$, a first-order one in s${}^{-1}$, and a second-order one in L mol${}^{-1}$ s${}^{-1}$.

Illustration 1

The rate constant of a reaction has units of L mol${}^{-1}$ s${}^{-1}$. Deduce the order, and state whether the reaction must be bimolecular.

Matching against ${\text{mol}}^{1-n}{\text{L}}^{n-1}{\text{s}}^{-1}$ gives $n-1=1$, so $n=2$: the reaction is second order.

It need not be bimolecular. Second order overall could arise from a single elementary two-body collision, or from a multi-step mechanism whose slow step happens to involve two molecules.

Order is measured; molecularity is proposed. Only for a step already known to be elementary can one be inferred from the other.

2. Integrated rate laws, and reading the order off a graph

The half-life behaviour is the fastest diagnostic of all. In general $t_{1/2}\propto {\left[\text{A}\right]}_0^{1-n}$, so a half-life independent of concentration means first order, one proportional to concentration means zero order, and one inversely proportional means second order.

Illustration 2

A reaction is $30\%$ complete in $30$ minutes and follows first-order kinetics. Find the rate constant and the time for $75\%$ completion.

$k=\frac{2.303}{30}\log \frac{100}{70}=\frac{2.303\times 0.1549}{30}=0.0119$ min${}^{-1}$

For $75\%$: $t=\frac{2.303}{0.0119}\log 4=193.5\times 0.602=116.5$ min

Seventy-five per cent is exactly two half-lives, so the answer could also have been read off as $2\times \frac{0.693}{0.0119}=116.5$ min — a useful check that costs nothing.

Illustration 3

The half-life of a reaction doubles when the initial concentration is halved. Find the order.

$t_{1/2}\propto {\left[\text{A}\right]}_0^{1-n}$

Halving the concentration doubles the half-life, so $t_{1/2}\propto {\left[\text{A}\right]}_0^{-1}$, giving $1-n=-1$.

$n=2$: the reaction is second order.

This method needs no rate constant and no graph. Two half-lives at two starting concentrations determine the order outright.

3. Pseudo-order reactions

If one reactant is present in large excess, its concentration barely changes and can be absorbed into the rate constant:

$\text{rate}=k\left[\text{A}\right]\left[\text{B}\right]\approx k^{\prime }\left[\text{A}\right]\text{with}{k}^{\prime }=k{\left[\text{B}\right]}_0$

The reaction is genuinely second order but behaves as first order, which makes it far easier to study. Hydrolysis of an ester in dilute aqueous acid and inversion of cane sugar are the standard cases: water is both solvent and reactant, at about $55$ M, so its concentration is effectively constant.

Illustration 4

The hydrolysis of methyl acetate in $0.1$ M hydrochloric acid has a pseudo-first-order constant of $2.5\times 10^{-3}$ min${}^{-1}$. Find the true second-order constant with respect to water, taking $\left[{\text{H}}_2\text{O}\right]=55$ M.

$k=\frac{k^{\prime }}{\left[{\text{H}}_2\text{O}\right]}=\frac{2.5\times 10^{-3}}{55}$

$k=4.5\times 10^{-5}$ L mol${}^{-1}$ min${}^{-1}$

Note that the acid is a catalyst, not a reactant. It appears in $k^{\prime }$ but is not consumed, which is why doubling the acid doubles the observed constant without changing the order.

4. Arrhenius: getting the activation energy

$k=A{e}^{-E_a/RT},\ln \frac{k_2}{k_1}=\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)$

A plot of $\ln k$ against $1/T$ is a straight line of slope $-E_a/R$ and intercept $\ln A$. The exponential is what makes rates so temperature-sensitive: the familiar rule that a $10$ K rise near room temperature roughly doubles a rate corresponds to a very ordinary activation energy.

Illustration 5

A reaction's rate doubles when the temperature rises from $300$ K to $310$ K. Find the activation energy.

$\ln 2=\frac{E_a}{8.314}\left(\frac{1}{300}-\frac{1}{310}\right)=\frac{E_a}{8.314}\times 1.075\times 10^{-4}$

$E_a=\frac{0.693\times 8.314}{1.075\times 10^{-4}}=5.36\times 10^4$ J mol${}^{-1}$

About $54$ kJ mol${}^{-1}$.

This is why the doubling rule is so widely quoted. A great many ordinary reactions have activation energies in the range $50$ to $60$ kJ mol${}^{-1}$, and all of them obey it approximately.

5. Collision theory and the fraction that reacts

Not every collision reacts. Two conditions must both be met: the colliding pair must carry at least $E_a$ of energy along the line of approach, and they must be correctly oriented.

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

where $Z$ is the collision frequency and $P$ the steric factor, typically far below one for anything but the simplest molecules.

The fraction of collisions energetic enough is $e^{-E_a/RT}$, and because that fraction is tiny, a modest temperature rise multiplies it substantially even though the average energy has barely moved. The rate rises steeply not because molecules move much faster, but because the thin tail beyond $E_a$ fattens sharply.

Illustration 6

Estimate the fraction of collisions with enough energy to react at $300$ K for $E_a=50$ kJ mol${}^{-1}$, and at $310$ K.

At $300$ K: $\frac{E_a}{RT}=\frac{50000}{8.314\times 300}=20.05$, so the fraction is $e^{-20.05}=1.96\times 10^{-9}$.

At $310$ K: $\frac{E_a}{RT}=19.40$, giving $e^{-19.40}=3.75\times 10^{-9}$.

The ratio is $1.9$, close to a doubling.

Two molecules in a thousand million, rising to nearly four. The absolute numbers are minute but the ratio is what sets the rate change.

6. Mechanisms: the slow step and the steady state

For a mechanism with a clear rate-determining step, the rate law is written from that step alone, and any intermediate appearing in it is then eliminated using the fast pre-equilibrium before it.

When no single step is clearly slowest, the steady-state approximation is used instead: the concentration of a reactive intermediate is assumed constant, so

$\frac{d\left[\text{I}\right]}{dt}=0$

which gives an equation for $\left[\text{I}\right]$ in terms of the reactants. Substituting it into the rate expression eliminates the intermediate. Intermediates must never appear in a final rate law, because they cannot be measured.

Illustration 7

A proposed mechanism is ${\text{A}}_2\rightleftharpoons 2\text{A}$ (fast) followed by $\text{A}+\text{B}\to \text{P}$ (slow). Derive the rate law.

Rate $=k_2\left[\text{A}\right]\left[\text{B}\right]$, but $\text{A}$ is an intermediate.

The fast equilibrium gives $K=\frac{{\left[\text{A}\right]}^2}{\left[{\text{A}}_2\right]}$, so $\left[\text{A}\right]=\sqrt{K\left[{\text{A}}_2\right]}$.

Rate $=k_2\sqrt{K}{\left[{\text{A}}_2\right]}^{1/2}\left[\text{B}\right]$

A half-order appears. Fractional orders are the signature of a dissociative pre-equilibrium, and they are also proof that the reaction cannot be elementary.

Illustration 8

Nitrogen dioxide and carbon monoxide react by the mechanism $2{\text{NO}}_2\to {\text{NO}}_3+\text{NO}$ (slow) followed by ${\text{NO}}_3+\text{CO}\to {\text{NO}}_2+{\text{CO}}_2$ (fast). Write the overall equation and the rate law.

Adding the steps and cancelling ${\text{NO}}_3$ and one ${\text{NO}}_2$:

${\text{NO}}_2+\text{CO}\to \text{NO}+{\text{CO}}_2$

The slow step involves only nitrogen dioxide, so

Rate $=k{\left[{\text{NO}}_2\right]}^2$

Carbon monoxide is absent from the rate law, even though it appears with a coefficient of one in the overall equation — precisely the situation posed in the hook.

7. Catalysis

A catalyst provides an alternative path with a lower activation energy. It does not change $\Delta H$, does not change $\Delta G$, and therefore does not change the equilibrium constant. What it changes is how fast equilibrium is reached — and it accelerates the forward and reverse reactions by exactly the same factor.

$\frac{k_{cat}}{k_{uncat}}=e^{\left(E_a-E_a^{\prime }\right)/RT}$

Homogeneous catalysts share the phase of the reactants, as an acid does in ester hydrolysis. Heterogeneous catalysts do not, and work by adsorbing reactants onto a surface, which both concentrates them and weakens their bonds. Enzymes are extraordinarily selective heterogeneous-like catalysts operating in solution.

Illustration 9

A catalyst lowers the activation energy of a reaction from $75$ to $50$ kJ mol${}^{-1}$. Find the factor by which the rate increases at $300$ K.

$\frac{k_{cat}}{k_{uncat}}=e^{\left(75000-50000\right)/\left(8.314\times 300\right)}=e^{10.02}$

$=2.25\times 10^4$

Over twenty thousand times faster, from a $25$ kJ mol${}^{-1}$ reduction. The exponential dependence is why quite small catalytic effects transform a reaction from unusable to practical.

Illustration 10

Explain why a catalyst cannot shift an equilibrium, in terms of both thermodynamics and kinetics.

Thermodynamically, $K$ depends only on $\Delta G^{\circ }$, which is fixed by the initial and final states. A catalyst changes neither, so $K$ is untouched.

Kinetically, the catalyst lowers the same barrier for both directions, since forward and reverse reactions pass over the identical transition state. Both rate constants rise by the same factor, and their ratio, which is $K$, is unchanged.

The two arguments must agree, and their agreement is a good check that a proposed catalytic mechanism is legitimate rather than a perpetual-motion machine in disguise.

Illustration 11

For a first-order reaction, $99\%$ completion takes how many times as long as $90\%$?

$t_{90}=\frac{2.303}{k}\log 10=\frac{2.303}{k}$

$t_{99}=\frac{2.303}{k}\log 100=\frac{4.606}{k}$

Exactly twice as long.

And $99.9\%$ takes three times as long. Each additional factor of ten in completeness costs the same fixed interval, which is the defining property of exponential decay.

Illustration 12

The rate constants of a reaction at $300$ K and $350$ K are $2\times 10^{-3}$ and $3.2\times 10^{-2}$ s${}^{-1}$. Find $E_a$ and the pre-exponential factor.

$\ln 16=\frac{E_a}{8.314}\left(\frac{1}{300}-\frac{1}{350}\right)=\frac{E_a}{8.314}\times 4.762\times 10^{-4}$

$E_a=\frac{2.773\times 8.314}{4.762\times 10^{-4}}=4.84\times 10^4$ J mol${}^{-1}$, about $48$ kJ mol${}^{-1}$

$\ln A=\ln k+\frac{E_a}{RT}=-6.21+\frac{48400}{8.314\times 300}=-6.21+19.41=13.20$, so $A=5.4\times 10^5$ s${}^{-1}$

8. Parallel and consecutive reactions

Two multi-path situations recur, and both have clean results.

Parallel reactions. If $\text{A}$ can become either $\text{B}$ or $\text{C}$ by competing first-order routes,

$-\frac{d\left[\text{A}\right]}{dt}=\left(k_1+k_2\right)\left[\text{A}\right],\frac{\left[\text{B}\right]}{\left[\text{C}\right]}=\frac{k_1}{k_2}$

The product ratio is fixed by the ratio of rate constants and is the same at every instant, so it never changes as the reaction proceeds. The overall half-life uses the sum, $t_{1/2}=\frac{0.693}{k_1+k_2}$, and the observed activation energy is a weighted average of the two.

Because the ratio depends on the two rate constants, and they respond differently to temperature, raising the temperature favours the route with the larger activation energy — which is how a synthesis is steered towards one product.

Consecutive reactions. For $\text{A}\to \text{B}\to \text{C}$, the intermediate rises, peaks and falls, with the maximum at

$t_{max}=\frac{\ln \left(k_2/k_1\right)}{k_2-k_1}$

If $k_2\gg k_1$ the intermediate never accumulates and the first step is rate determining; if $k_1\gg k_2$ it piles up and the second step controls.

Illustration 13

A compound decomposes by two parallel first-order paths with $k_1=3\times 10^{-3}$ and $k_2=1\times 10^{-3}$ s${}^{-1}$. Find the overall half-life and the percentage of each product.

$t_{1/2}=\frac{0.693}{4\times 10^{-3}}=173$ s

$\frac{\left[\text{B}\right]}{\left[\text{C}\right]}=\frac{3}{1}$, so the products are $75\%$ $\text{B}$ and $25\%$ $\text{C}$.

Those percentages hold from the first instant. Sampling early or late gives the same ratio, which is what distinguishes parallel routes from consecutive ones.

Illustration 14

For $\text{A}\to \text{B}\to \text{C}$ with $k_1=0.01$ and $k_2=0.05$ min${}^{-1}$, find when the intermediate reaches its maximum.

$t_{max}=\frac{\ln \left(0.05/0.01\right)}{0.05-0.01}=\frac{\ln 5}{0.04}=\frac{1.609}{0.04}$

$t_{max}=40.2$ min

Since the second step is five times faster, $\text{B}$ never builds up much. The first step is rate determining, and the concentration of the intermediate stays low throughout — precisely the condition the steady-state approximation assumes.

Summary

• A rate law is experimental, not read off the balanced equation. A reactant absent from it enters after the rate-determining step.
• Order is measured and can be zero or fractional; molecularity applies to a single elementary step and is a small positive integer.
• The units of $k$ give the order at once: mol L${}^{-1}$ s${}^{-1}$, s${}^{-1}$ and L mol${}^{-1}$ s${}^{-1}$ for zero, first and second.
• Zero order plots $\left[\text{A}\right]$, first plots $\ln \left[\text{A}\right]$, second plots $\frac{1}{\left[\text{A}\right]}$ — whichever is straight names the order.
• $t_{1/2}\propto {\left[\text{A}\right]}_0^{1-n}$: independent of concentration means first order, proportional means zero, inverse means second.
• Pseudo-order: a large excess of one reactant is absorbed into $k^{\prime }=k{\left[\text{B}\right]}_0$, making a second-order reaction behave as first.
• $k=A{e}^{-E_a/RT}$; a $10$ K rise doubling the rate near $300$ K corresponds to about $54$ kJ mol${}^{-1}$.
• $\ln k$ against $1/T$ is linear with slope $-E_a/R$ and intercept $\ln A$.
• Reaction needs both energy above $E_a$ and correct orientation, hence $k=PZ{e}^{-E_a/RT}$ with $P$ well below one.
• The reactive fraction $e^{-E_a/RT}$ is minute — about two in a thousand million — but doubles for a $10$ K rise.
• Write the rate law from the slow step, then eliminate intermediates using the preceding fast equilibrium or the steady state.
• Fractional orders signal a dissociative pre-equilibrium and prove the reaction is not elementary.
• A catalyst lowers $E_a$, leaves $\Delta H$, $\Delta G$ and $K$ untouched, and accelerates both directions by the same factor.
• Parallel routes: $t_{1/2}=\frac{0.693}{k_1+k_2}$ and the product ratio $k_1/k_2$ is constant at all times; heating favours the higher-$E_a$ route.
• Consecutive: the intermediate peaks at $t_{max}=\frac{\ln \left(k_2/k_1\right)}{k_2-k_1}$.
• For first order, $99\%$ takes exactly twice as long as $90\%$ — each further factor of ten costs a fixed interval.