Chemistry · General chemistry II · Concept

Finding rate laws from experimental data

A rate law, $rate = k {[A]}^{m} {[B]}^{n}$ , says how a reaction’s speed depends on each concentration. Find each order by changing one concentration at a time and comparing the initial rates, or by finding which plot of a concentration against time is straight, then solve for $k$ .

A concrete case

Change one thing and time it

Mix hydrogen peroxide, iodide ions, a little thiosulfate and some starch in a weakly acidic solution. The peroxide turns iodide into iodine, and the thiosulfate turns the iodine straight back. When the thiosulfate runs out, iodine meets the starch and the whole beaker turns dark blue at once. The faster the reaction, the sooner the blue appears.

Say a run with $0.010 M$ peroxide and $0.010 M$ iodide turns blue at $80 s$ . Double the peroxide to $0.020 M$ , change nothing else, and the blue comes at $40 s$ . The reaction ran twice as fast. Go back to $0.010 M$ peroxide and double the iodide instead, and the blue again comes at $40 s$ .

Double both, and the blue comes at $20 s$ , four times as fast as the first run. Yet the balanced equation, $H_{2} O_{2} + 2 I^{-} + 2 H^{+}$ $\to I_{2} + 2 H_{2} O$ , uses two iodide ions for each peroxide. Doubling the iodide still only doubled the speed. The equation says what reacts. Only an experiment says how the speed depends on each amount.

Three runs of the iodine clock (the times are illustrative)
Run	$[H_{2} O_{2}]$ $(M)$	$[I^{-}]$ $(M)$	Time to blue	Speed compared with run $1$
$1$	$0.010$	$0.010$	$80 s$	$\times 1$
$2$	$0.020$	$0.010$	$40 s$	$\times 2$
$3$	$0.010$	$0.020$	$40 s$	$\times 2$
$4$	$0.020$	$0.020$	$20 s$	$\times 4$

The rule

The rate law

The rate of a reaction is how fast a concentration changes, in molarity per second, $M / s$ . A rate law writes that rate as a rate constant, $k$ , times each reactant’s concentration raised to a power. The power on $[A]$ is the order in A, here $m$ , and the power on $[B]$ is the order in B, here $n$ . Their sum is the overall order.

The clock runs gave $rate = k [H_{2} O_{2}] [I^{-}]$ , first order in each reactant and second order overall. Orders are usually $0$ , $1$ or $2$ , and they are found only by experiment. $k$ depends on the temperature, not on the concentrations.

Compare two runs that differ only in $[A]$ . Dividing one rate law by the other cancels $k$ and every concentration that stayed the same. What is left is the ratio of the rates, equal to the ratio of the $[A]$ values raised to the order $m$ .

rate = k {[A]}^{m} {[B]}^{n}, \frac{{rate}_{2}}{{rate}_{1}} = {(\frac{{[A]}_{2}}{{[A]}_{1}})}^{m}

\begin{array}{c} rate = k {[A]}^{m} {[B]}^{n} \\ \frac{{rate}_{2}}{{rate}_{1}} = {(\frac{{[A]}_{2}}{{[A]}_{1}})}^{m} \end{array}

Units of $k$

$k$ takes whatever units make the rate come out in $M / s$ . With time in seconds, a first-order $k$ is in $s^{- 1}$ and a second-order $k$ is in $M^{- 1} s^{- 1}$ . Each further order adds one more $M^{- 1}$ , so the units of $k$ tell you the overall order at a glance.

Units of $k$ (time in seconds)
Overall order	Rate law form	Units of $k$
$0$	$rate = k$	$M s^{- 1}$
$1$	$rate$ $= k [A]$	$s^{- 1}$
$2$	$rate$ $= k {[A]}^{2}$ or $k [A] [B]$	$M^{- 1} s^{- 1}$
$3$	$rate$ $= k {[A]}^{2} [B]$	$M^{- 2} s^{- 1}$

Why it works

Why one change at a time works

Write the rate law for two runs and divide one by the other. If only $[A]$ changed, every other factor on the right is the same in both runs and cancels. $k$ cancels too, because the temperature was the same. One equation in one unknown is left, $({rate}_{2} / {rate}_{1})$ $= {({[A]}_{2} / {[A]}_{1})}^{m}$ .

Initial rates are measured because at the very start the concentrations are the ones you mixed. Later, some of each reactant is gone by an amount you would have to measure, and products may begin to react back.

The orders describe the slowest part of the reaction, which is why the coefficients cannot give them. The clock reaction happens in steps, and its slow step uses one peroxide and one iodide. A mechanism, a list of such steps, has to reproduce the orders an experiment finds.

A second experiment follows one reactant as it is used up. Each order gives a different curve. A first-order reaction loses the same fraction of A in every equal time, so $ln [A]$ falls by equal steps, a straight line. A second-order reaction slows more sharply as A runs down, and $1 / [A]$ rises by equal steps. A zero-order reaction loses the same amount every second, so $[A]$ itself falls in a straight line.

The method

From initial rates, step by step

1. Find two runs in which only one concentration changed.

2. Divide the later rate by the earlier rate, and the later concentration by the earlier concentration.

3. Find the power that turns the concentration ratio into the rate ratio. Doubling a concentration gives a rate ratio of $1$ , $2$ , $4$ or $8$ for orders $0$ , $1$ , $2$ or $3$ . When the ratio is not tidy, the order is $ln (rate ratio)$ $\div ln (concentration ratio)$ .

4. Repeat for each reactant, then write the rate law with every order in it.

5. Put one run’s rate and concentrations into the rate law, solve for $k$ and give its units. Do the same with a second run as a check.

m = \frac{ln ({rate}_{2} / {rate}_{1})}{ln ({[A]}_{2} / {[A]}_{1})}

Reading an order when one concentration doubles
Rate changes by	Order in that reactant	Why
$\times 1$	$0$	$2^{0} = 1$
$\times 2$	$1$	$2^{1} = 2$
$\times 4$	$2$	$2^{2} = 4$
$\times 8$	$3$	$2^{3} = 8$
$\times 2.83$	$1.5$ $(3 / 2)$	$2^{1.5}$ $\approx 2.83$

From one run followed over time

1. Make three columns from the data, $[A]$ , $ln [A]$ and $1 / [A]$ , each against time.

2. For equal time steps, look at how much each column changes from one reading to the next. The column whose changes stay the same is the straight one.

3. Read the order from the table below. A straight $[A]$ means zero order, a straight $ln [A]$ means first order and a straight $1 / [A]$ means second order.

4. Find $k$ from the straight column. It is minus the slope for $[A]$ and $ln [A]$ , and plus the slope for $1 / [A]$ .

5. Check with half-lives. Equal successive half-lives mean first order, and half-lives that double mean second order.

ln {[A]}_{t} = - k t + ln {[A]}_{0}

Integrated rate laws
Order	Straight-line plot	Slope	Half-life
$0$	$[A]$ vs $t$	$- k$	${[A]}_{0} / 2 k$
$1$	$ln [A]$ vs $t$	$- k$	$0.693 / k$
$2$	$1 / [A]$ vs $t$	$+ k$	$1 / (k {[A]}_{0})$

Where it breaks

The ratio method needs runs that differ in one concentration only. In the $NO$ example below, trials 2 and 3 differ in both $[NO]$ and $[O_{2}]$ . Their rates differ by a factor of $2$ , which is ${(2)}^{2} \times \frac{1}{2}$ , and reading it as order $1$ in $NO$ would be wrong.

Measured rates scatter. A ratio of $1.9$ or $2.1$ for a doubling still means first order. A ratio of $2.8$ that repeats in every pair is a real order of $3 / 2$ , not a messy $1$ or $2$ . Fractional orders occur, so compare several pairs before you round.

Coefficients are not orders, even when they happen to match. In $2 NO + O_{2} \to 2 {NO}_{2}$ the orders are $2$ and $1$ , like the coefficients. In $H_{2} O_{2} + 2 I^{-} + 2 H^{+}$ $\to I_{2} + 2 H_{2} O$ the order in iodide is $1$ , not $2$ . In ${NO}_{2} + CO \to NO + {CO}_{2}$ below about $225 ° C$ , the order in $CO$ is $0$ .

A rate law and its $k$ belong to one temperature. Runs at different temperatures cannot be divided to find an order, because $k$ changed between them.

The straight-plot test tracks one reactant. When a second reactant is in large excess, its concentration hardly changes, and the plot shows only the order in the first. Over less than one half-life, all three plots can look straight, so follow the reaction for two half-lives or more.

Check your answer

Find $k$ twice. The same $k$ from two different trials, within the precision of the data, shows the orders are right. In the $NO$ example, trials 1 and 2 both give $7.0 \times 10^{3} M^{- 2} s^{- 1}$ .

Predict a run you did not use. Put the third trial’s concentrations into your rate law with your $k$ , and the rate it gives should match the measured one.

Read the units of $k$ . They must make the rate come out in $M / s$ , and they tell you the overall order. $M^{- 2} s^{- 1}$ goes with third order overall.

In a Chemistry box, write each ratio on its own row with its units, such as $0.112 M / s \div 0.028 M / s$ $= 4.0$ . The Checker marks its arithmetic ✓. Then write the exponent equation, $4 = 2^{m}$ , on a row of its own and the order you found, $m = 2$ , under it. The order reads ✓, and $m = 3$ reads ✗.

Write the rate law on its own row. With no reaction above it, the Checker takes it as given. It says “A rate law, taken as given: reaction orders come from experiments.”

A $k$ worked out as an unnamed quotient under it, such as $0.028 M / s$ $\div ({(0.020 M)}^{2} (0.010 M))$ $= 7.0 \times 10^{3} M^{- 2} s^{- 1}$ , reads ✓ for its arithmetic and units only. Chalk does not check that the quotient comes from the rate law.

To have the law checked, write the trial’s rate and concentrations on rows of their own, then name $k$ . With the powers on the wrong concentrations, $(0.020 M) {(0.010 M)}^{2}$ , the row reads ✗. Written correctly, $k = 0.028 M / s$ $\div ({(0.020 M)}^{2} (0.010 M))$ $= 7.0 \times 10^{3} M^{- 2} s^{- 1}$ reads ✓ with “Matches the rate law in row 1 with the values above.”

With the numbers only inside the working, the row reads “?” and the note says “These numbers can’t be matched to the rate law in row 1: [NO] isn’t defined above.”

A predicted rate is checked against the law. The worked example’s box puts $k$ and the concentrations on rows of their own. Its last row, $rate$ $= (7.0 \times 10^{3} M^{- 2} s^{- 1}) {(0.020 M)}^{2} (0.020 M)$ $= 0.056 M / s$ , reads ✓, and the Checker says “Matches the rate law in row 5 with the values above.”

Whose rate, flooding, and fractional orders

In $2 NO + O_{2} \to 2 {NO}_{2}$ , $NO$ is used up twice as fast as $O_{2}$ . To give the reaction one rate, IUPAC divides each rate of change by the coefficient, so the rate is $- \frac{1}{2} d [NO] / d t$ $= - d [O_{2}] / d t$ $= \frac{1}{2} d [{NO}_{2}] / d t$ . A $k$ quoted for “the rate of loss of NO” is twice the $k$ for the reaction’s rate, so a careful report says which rate it means.

When a reaction has two reactants, a chemist often floods it with one. With B a hundred times more concentrated than A, $[B]$ barely changes, and $rate = k [A] [B]$ behaves like $rate = k^{'} [A]$ with $k^{'} = k [B]$ . That is a pseudo-first-order reaction. Repeating it at several $[B]$ values gives $k$ from $k^{'}$ .

An order is defined only when the rate law is a product of powers of concentrations. Some reactions have rate laws with a sum in the denominator, and their orders change as the concentrations change. Fractional orders come from fast steps that split a molecule in two, such as ${Cl}_{2} ⇌ 2 Cl$ , which gives an order of $\frac{1}{2}$ in ${Cl}_{2}$ .

Common mistakes

Reading the orders from the coefficients. For $H_{2} O_{2} + 2 I^{-} + 2 H^{+}$ $\to I_{2} + 2 H_{2} O$ the rate law is $rate = k [H_{2} O_{2}] [I^{-}]$ , first order in iodide, not second. Coefficients count what reacts overall, while orders come from how the rate responds.
Comparing two runs in which two concentrations changed. Trials 2 and 3 of the $NO$ example differ in both $[NO]$ and $[O_{2}]$ , so their rate ratio of $2$ mixes both orders. Use trials 1 and 2 for $NO$ and trials 1 and 3 for $O_{2}$ .
Giving $k$ without units, or with the units of another order. For $rate = k {[NO]}^{2} [O_{2}]$ , $k = 7.0 \times 10^{3} M^{- 2} s^{- 1}$ . Three concentrations on the right need $M^{- 2}$ on $k$ for the rate to come out in $M / s$ .
Taking $k$ as the slope of a falling $ln [A]$ plot. The slope is $- k$ , so a slope of $- 0.01151 \min^{- 1}$ means $k = 0.01151 \min^{- 1}$ . A rate constant is never negative.
Choosing the order from one $R^{2}$ value. Over a short time $[A]$ , $ln [A]$ and $1 / [A]$ can each give $R^{2}$ above $0.98$ . Compare the changes in all three columns, over two half-lives or more.
Rounding a rate ratio of $2.83$ for a doubling to an order of $1$ or $2$ . $2^{1.5} = 2.83$ , so the order is $3 / 2$ . Orders can be fractions, and $ln 2.83 \div ln 2 = 1.50$ finds this one.

Key terms

Rate law: An equation linking reaction rate to reactant concentrations, such as $rate = k {[A]}^{2} [B]$ . The units of the rate constant $k$ depend on the overall order.
Reaction order: The exponent on a concentration in the rate law (the order in that reactant), or the sum of all the exponents (the overall order). Orders come from experiments, not from the balanced equation.
Overall order: The sum of the exponents on all the concentrations in a rate law. For $rate = k {[A]}^{m} {[B]}^{n}$ , the overall order is $m + n$ .
Initial rate: The reaction rate measured right at the start, before concentrations change much. Comparing initial rates from runs with different starting concentrations reveals the reaction orders.
Reaction rate: The rate of a reaction as written: the rate of change of a species’ concentration divided by its coefficient, with a minus sign for reactants, so every species gives the same value.
Integrated rate law: An equation for concentration versus time, derived from the rate law: $ln [A] = - k t + ln {[A]}_{0}$ for first order, $[A] = {[A]}_{0} - k t$ for zero order and $1 / [A] = k t + 1 / {[A]}_{0}$ for second order. Here $k$ is the rate constant, $t$ is the time and ${[A]}_{0}$ is the starting concentration. Each order gives its own straight-line plot.
Half-life: The time it takes for an amount to fall to half its starting value. For an amount that decays as $A_{0} e^{- k t}$ with $k > 0$ , it is $ln 2 \div k$ , whatever the starting amount. For a first-order reaction, $[A] = {[A]}_{0} e^{- k t}$ , so $t_{1 / 2} = ln 2 \div k$ , with $k$ the rate constant.

Work through an example

For $2 NO (g) + O_{2} (g) \to 2 {NO}_{2} (g)$ at a fixed temperature, three trials give these initial rates. Find the order in $NO$ , the order in $O_{2}$ , the rate law and $k$ with units.

Find a rate law from initial rates →

Order and $k$ from concentration–time data →

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