Rate laws

Thermodynamics says a mixture of hydrogen and oxygen should be water, and the mixture will sit in a flask for a thousand years without becoming any, so the question of how fast is a separate science with its own measurements.

The previous five lessons answered where a reaction ends up. None of them can be adapted to say when, because free energy depends only on the initial and final states and a rate depends entirely on the route between them. This lesson starts that second theory from the plainest possible measurement: the composition of a mixture as a function of time.

A rate everyone can agree on

Take $2 \to 4 N +$ . Nitrogen dioxide appears four times as fast as oxygen does, so "the rate" is ambiguous until we say the rate of what. The fix is the extent of reaction from the first lesson. Every amount obeys $n_{i} = n_{i, 0} + ν_{i} ξ$ , so dividing each rate of change by the stoichiometric number gives one quantity that all of them share:

v = \frac{1}{V} \frac{d ξ}{d t} = \frac{1}{ν_{i}} \frac{d [i]}{d t}

For the reaction above, $v = - \frac{1}{2} d [] / d t = \frac{1}{4} d [N] / d t = d [] / d t$ . The sign convention is built in, since $ν$ is negative for reactants, so $v$ comes out positive for a reaction running forward. Its units are mol dm⁻³ s⁻¹.

The division by volume matters when the volume changes, and is a nuisance rather than a principle. Everything below is at constant volume, where concentrations behave.

Measuring one

A rate is a slope, so measuring one means following a concentration in time without disturbing the mixture. The methods divide by how fast the reaction is.

Anything with a coloured species is followed by absorbance, since the Beer-Lambert law makes absorbance proportional to concentration, and a spectrophotometer samples continuously without removing anything. A reaction that changes the number of moles of gas is followed by pressure at constant volume, which is how the decomposition of dinitrogen pentoxide was originally studied. Reactions that change the ion count are followed by conductivity, and those involving a chiral substance by optical rotation.

Slower reactions can be sampled instead: withdraw an aliquot, quench it by chilling or by destroying the catalyst, and analyse at leisure. Faster ones need the stopped-flow method, which mixes two streams in about a millisecond and watches the mixing chamber, or, below that, flash photolysis, which starts the reaction with a light pulse. Manfred Eigen, Ronald Norrish and George Porter shared the 1967 Nobel Prize for pushing the accessible range down to microseconds; laser techniques have since reached femtoseconds, fast enough to watch a bond break.

For the arithmetic, two experimental designs recur. In the initial rates method the reaction is run several times from different starting concentrations and only the first few per cent is used, so the concentrations are still known and no product has accumulated. In the integrated method one run is followed to substantial conversion and the whole curve is fitted. The next two sections take them in that order.

The rate law is an experimental fact

For many reactions the rate turns out to depend on concentrations as a product of powers:

v = k [A]^{a} [B]^{b}

Here $k$ is the rate constant, which depends on temperature but not on concentration, and the exponents are the orders: order $a$ with respect to A, and $a + b$ overall. The units of $k$ are whatever makes the equation dimensionally consistent, so they change with the overall order, which is a useful check on an answer.

The essential point, and the one most often got wrong, is that the orders are measured and not derived. They are not the stoichiometric coefficients. The decomposition of dinitrogen pentoxide has a coefficient of $2$ and is first order. The gas phase reaction of hydrogen with iodine has coefficients of $1$ and $1$ and happens to be first order in each, which for a century was taken as evidence that it was a single collision, wrongly. And $2 \to 3$ has the rate law $v = k []^{2} []^{- 1}$ , in which a product appears with a negative order, meaning oxygen inhibits its own formation. No reading of the balanced equation predicts that.

Some rate laws are not of this form at all. The formation of hydrogen bromide from its elements obeys

v = \frac{k [] [B]^{1 / 2}}{1 + k' [HBr] / [B]}

measured by Max Bodenstein in 1906, which has no single order and changes shape as the reaction proceeds. A rate law like that is not a nuisance: it is a fingerprint of the mechanism, and the eighth lesson shows how to read it.

Orders by initial rates

The method is to change one concentration at a time and see what the rate does. Doubling a concentration doubles a first order rate, quadruples a second order one, and leaves a zero order one alone.

Example. The oxidation of iodide by peroxodisulfate, $+ 2 \to 2 S +$ , gives these initial rates at $298$ K.

$[]$ / mol dm⁻³	$[]$ / mol dm⁻³	rate / mol dm⁻³ s⁻¹
$0.038$	$0.060$	$1.4 \times 10^{- 5}$
$0.076$	$0.060$	$2.8 \times 10^{- 5}$
$0.076$	$0.030$	$1.4 \times 10^{- 5}$

Find the rate law and $k$ .

Between the first two rows the iodide is fixed and the persulfate doubles, and the rate doubles: first order in persulfate. Between the second and third the persulfate is fixed and the iodide halves, and the rate halves: first order in iodide. So $v = k [] []$ , second order overall, and $k = 1.4 \times 10^{- 5} / (0.038 \times 0.060) = 6.1 \times 10^{- 3}$ dm³ mol⁻¹ s⁻¹. Note that the order in iodide is one although the equation needs two of them.

Now you. For $2 NO + \to 2 N$ at $298$ K, doubling $[NO]$ from $0.0100$ to $0.0200$ mol dm⁻³ at fixed $[] = 0.0100$ raises the rate from $7.1 \times 10^{- 3}$ to $2.84 \times 10^{- 2}$ mol dm⁻³ s⁻¹, and doubling $[]$ instead raises it to $1.42 \times 10^{- 2}$ . Give the rate law, the overall order and $k$ with its units.

Answer

Doubling nitric oxide multiplies the rate by four, so the order in it is two; doubling oxygen doubles the rate, so that order is one. Thus $v = k [NO]^{2} []$ , third order overall, and $k = 7.1 \times 10^{- 3} / ({0.0100}^{2} \times 0.0100) = 7.1 \times 10^{3}$ dm⁶ mol⁻² s⁻¹.

Integrating the first order law

The other approach follows one run to completion, which needs the rate law solved as a differential equation. First order is the case worth doing in full, because it recurs everywhere.

Write $- d [A] / d t = k [A]$ , separate the variables and integrate from the start to time $t$ :

\ln \frac{[A]}{[A]_{0}} = - k t so [A] = [A]_{0} e^{- k t}

Two consequences are worth naming. A plot of $\ln [A]$ against $t$ is a straight line of slope $- k$ , which is the test for first order behaviour. And the half-life, the time to fall to half of any starting value, is

t_{1 / 2} = \frac{\ln 2}{k}

independent of concentration. That independence is the signature: a first order reaction takes as long to go from $0.01$ to $0.005$ mol dm⁻³ as from $1.0$ to $0.5$ . It is why radioactive decay, which is first order by nature, has a half-life worth tabulating, and why a drug cleared by a first order process has a dose-independent elimination time.

Example. Dinitrogen pentoxide decomposes with $k = 4.87 \times 10^{- 3}$ s⁻¹ at $65$ °C. Starting from $0.1000$ mol dm⁻³, find the half-life and the concentration after $300$ s.

The half-life is $\ln 2 / k = 0.6931 / 4.87 \times 10^{- 3} = 142$ s. After $300$ s, $[A] = 0.1000 e^{- 4.87 \times 10^{- 3} \times 300} = 0.1000 e^{- 1.461} = 0.0232$ mol dm⁻³. As a check, $300$ s is a little over two half-lives, and two half-lives would leave $0.025$ .

Now you. For the same reaction and rate constant, how long does it take for $90$ per cent of the pentoxide to decompose, and does the answer depend on the starting concentration?

Answer

Ninety per cent gone leaves a tenth, so $k t = \ln 10 = 2.303$ and $t = 2.303 / 4.87 \times 10^{- 3} = 473$ s. It does not depend on the starting concentration, because only the ratio appears in the integrated law.

Zero and second order, and how to tell them apart

The same integration for other orders gives different straight lines, and comparing which plot is straight is how an order is assigned from a single run.

Order	Integrated law	Straight line	Half-life
$0$	$[A] = [A]_{0} - k t$	$[A]$ against $t$	$[A]_{0} / 2 k$
$1$	$\ln [A] = \ln [A]_{0} - k t$	$\ln [A]$ against $t$	$\ln 2 / k$
$2$	$1 / [A] = 1 / [A]_{0} + k t$	$1 / [A]$ against $t$	$1 / (k [A]_{0})$

The half-life column is the quickest diagnostic. A zero order reaction's successive half-lives get shorter, each half of the one before; a first order reaction's are all equal; a second order reaction's double each time. Watching a decay curve halve three times and noting whether the intervals grow, stay or shrink identifies the order before any plotting.

Zero order sounds strange but is common wherever something other than the reactant concentration is the bottleneck: a saturated enzyme, a saturated catalyst surface, a photochemical reaction limited by the light supply. Ethanol is cleared from human blood at close to zero order above about $0.02$ per cent by volume, because the alcohol dehydrogenase is saturated, which is why blood alcohol falls linearly rather than exponentially.

Example. These concentrations were recorded for a decomposition at constant temperature.

$t$ / s	$0$	$100$	$200$	$300$
$[A]$ / mol dm⁻³	$0.01000$	$0.00648$	$0.00479$	$0.00380$

Find the order and the rate constant.

Test first order: the ratio over the first interval is $0.648$ and over the second $0.739$ , which are not equal, so the concentration is not falling exponentially. Test second order by taking reciprocals: $100.0$ , $154.3$ , $208.6$ , $262.9$ dm³ mol⁻¹. Those rise by exactly $54.3$ every $100$ s, so $1 / [A]$ is linear in time and the reaction is second order with $k = 54.3 / 100 = 0.543$ dm³ mol⁻¹ s⁻¹.

Now you. Another run gives $0.1000$ , $0.0614$ , $0.0378$ , $0.0232$ and $0.0143$ mol dm⁻³ at $0$ , $100$ , $200$ , $300$ and $400$ s. Find the order and the rate constant.

Answer

Each interval multiplies the concentration by the same factor, $0.614$ , so the decay is exponential and the reaction is first order. Then $k = - \ln (0.614) / 100 = 4.88 \times 10^{- 3}$ s⁻¹, and the half-life is $\ln 2 / k = 142$ s, consistent with the concentration falling from $0.1000$ to about $0.05$ somewhere between $100$ and $200$ s.

What the orders are hiding

Two honest limits close the lesson.

The first is that every integrated law above ignores the reverse reaction, so it describes only the early part of a run, before enough product has built up for the mixture to notice equilibrium. Near equilibrium the net rate falls to zero while both directions continue at full speed, and the correct treatment subtracts the reverse rate. This is why kinetic runs are usually analysed over the first fraction of the reaction, and why a reaction with a small $K$ cannot be studied this way at all without special handling.

The second is more interesting. Nothing so far explains why simple integer orders should turn up at all, why the order in iodide should be one when the equation needs two, or where a $- 1$ or a $1 / 2$ comes from. A rate law is a summary of a mechanism, and the orders are evidence about which molecules meet in the slowest step. Extracting the mechanism from the rate law is the eighth lesson.

Before that, there is a variable the rate constant is far more sensitive to than any concentration. Raising the temperature of a mixture by ten degrees typically doubles its rate, which is a much steeper response than any concentration effect, and it is the subject of the next lesson.