Orders of Reaction and Rate Laws from Experiment – Rate Laws and Measurements (Senior 6 Chemistry) | Ganzaa

Rate Laws and Measurements

Lessons

Senior 6 · Chemistry

Rate Laws and Measurements

5 lessons

Lesson 3 of 5

Orders of Reaction and Rate Laws from Experiment

25 min

Iodine is used up, yet it does not matter how much there is

In Lesson 2 the iodine in the propanone reaction fell by the same amount in every 5-minute interval. Doubling the iodine at the start would not change that rate either; doubling the propanone, or the acid, would double it. The balanced equation $\mathrm{CH_3COCH_3 + I_2 \rightarrow CH_3COCH_2I + HI}$ gives no hint of this. It tells you what reacts, not how the rate depends on each concentration. That dependence must be measured, and it is summed up in a rate equation.

Definition

Rate equation (rate law): an experimentally found equation linking the rate to the concentrations of the species that affect it, $\text{rate} = k[\mathrm{A}]^m[\mathrm{B}]^n$ . Example: for the iodination of propanone, $\text{rate} = k[\mathrm{CH_3COCH_3}][\mathrm{H^{+}}]$ .

Order with respect to a reactant: the power to which its concentration is raised in the rate equation ( $m$ for A, $n$ for B). It is usually 0, 1 or 2 and can be fractional.

Overall order: the sum of the individual orders, $m + n$ . Example: $\text{rate} = k[\mathrm{NO}]^2[\mathrm{H_2}]$ is second order in NO, first order in $\mathrm{H_2}$ , and third order overall.

What each order means

Order in A	Doubling [A] multiplies the rate by	Rate–concentration graph
0	$2^{0} = 1$ (no change)	horizontal line
1	$2^{1} = 2$	straight line through the origin
2	$2^{2} = 4$	upward curve through the origin
3	$2^{3} = 8$	steeper upward curve

Figure 3.1: The shape of a rate–concentration graph shows the order.

Units of the rate constant

The rate always has units $\mathrm{mol\,dm^{-3}\,s^{-1}}$ , so the unit of $k$ must make the two sides match. For overall order $n$ : $\text{unit of } k = (\mathrm{mol\,dm^{-3}})^{1-n}\,\mathrm{s^{-1}}$ .

Overall order	Rate equation (example)	Unit of $k$
0	$\text{rate} = k$	$\mathrm{mol\,dm^{-3}\,s^{-1}}$
1	$\text{rate} = k[\mathrm{A}]$	$\mathrm{s^{-1}}$
2	$\text{rate} = k[\mathrm{A}][\mathrm{B}]$	$\mathrm{dm^3\,mol^{-1}\,s^{-1}}$
3	$\text{rate} = k[\mathrm{A}]^2[\mathrm{B}]$	$\mathrm{dm^6\,mol^{-2}\,s^{-1}}$

The initial-rates method

Run the reaction several times. Between two runs, change the concentration of one reactant only, keeping the others and the temperature the same.
Measure the initial rate of each run (tangent at $t = 0$ , or a clock method), when the concentrations are known exactly.
Compare runs: if $[\mathrm{A}]$ is multiplied by $x$ and the rate by $y$ , then $y = x^{m}$ , so $m = \frac{\log y}{\log x}$ .
Write the rate equation, then substitute any run to find $k$ with its unit.

Worked example 1: the iodination of propanone

Problem. Use the initial rates below (illustrative data at 25 °C) to find the order with respect to each reactant, the rate equation and the rate constant.

Run	$[\mathrm{CH_3COCH_3}]$ / $\mathrm{mol\,dm^{-3}}$	$[\mathrm{I_2}]$ / $\mathrm{mol\,dm^{-3}}$	$[\mathrm{H^{+}}]$ / $\mathrm{mol\,dm^{-3}}$	Initial rate / $\mathrm{mol\,dm^{-3}\,s^{-1}}$
1	0.40	0.0050	0.20	$4.0 \times 10^{-6}$
2	0.80	0.0050	0.20	$8.0 \times 10^{-6}$
3	0.40	0.0100	0.20	$4.0 \times 10^{-6}$
4	0.40	0.0050	0.40	$8.0 \times 10^{-6}$

Solution.

Runs 1 and 2: propanone doubled, rate doubled: first order in propanone.

Runs 1 and 3: iodine doubled, rate unchanged: zero order in iodine.

Runs 1 and 4: $\mathrm{H^{+}}$ doubled, rate doubled: first order in $\mathrm{H^{+}}$ .

\text{rate} = k[\mathrm{CH_3COCH_3}][\mathrm{H^{+}}] \quad (\text{second order overall})

k = \frac{\text{rate}}{[\mathrm{CH_3COCH_3}][\mathrm{H^{+}}]} = \frac{4.0 \times 10^{-6}}{0.40 \times 0.20} = 5.0 \times 10^{-5}\,\mathrm{dm^3\,mol^{-1}\,s^{-1}}

$\mathrm{H^{+}}$ appears in the rate equation although it is not used up: it is a catalyst. Iodine is used up but does not appear: it must react only after the slow step (Lesson 5).

Worked example 2: when two concentrations change at once

Problem. For $\mathrm{2NO(g) + 2H_2(g) \rightarrow N_2(g) + 2H_2O(g)}$ (illustrative data):

Run	$[\mathrm{NO}]$ / $\mathrm{mol\,dm^{-3}}$	$[\mathrm{H_2}]$ / $\mathrm{mol\,dm^{-3}}$	Initial rate / $\mathrm{mol\,dm^{-3}\,s^{-1}}$
1	0.0050	0.0020	$1.25 \times 10^{-5}$
2	0.0100	0.0020	$5.0 \times 10^{-5}$
3	0.0100	0.0040	$1.0 \times 10^{-4}$
4	0.0150	0.0060	?

Find the rate equation, $k$ with units, and the missing rate.

Solution. Runs 1 and 2: [NO] × 2, rate × 4 = $2^{2}$ : second order in NO. Runs 2 and 3: $[\mathrm{H_2}]$ × 2, rate × 2: first order in $\mathrm{H_2}$ .

\text{rate} = k[\mathrm{NO}]^2[\mathrm{H_2}], \qquad k = \frac{1.25 \times 10^{-5}}{(0.0050)^2 \times 0.0020} = 250\,\mathrm{dm^6\,mol^{-2}\,s^{-1}}

Run 4: both concentrations change, so use the rate equation: $250 \times (0.0150)^2 \times 0.0060 = 3.4 \times 10^{-4}\,\mathrm{mol\,dm^{-3}\,s^{-1}}$ . Check by ratios from run 1: $3^2 \times 3 = 27$ , and $27 \times 1.25 \times 10^{-5} = 3.4 \times 10^{-4}$ .

Worked example 3: a fractional order

Problem. When the concentration of ethanal is doubled, the initial rate of its thermal decomposition, $\mathrm{CH_3CHO(g) \rightarrow CH_4(g) + CO(g)}$ , increases by a factor of 2.83. Find the order.

Solution. $2^{m} = 2.83$ , so $m = \frac{\log 2.83}{\log 2} = \frac{0.452}{0.301} = 1.50$ . The reaction is of order $\frac{3}{2}$ in ethanal, a sign that it goes by a chain mechanism of several steps. Orders are experimental numbers; they need not be whole numbers.

Warning

Common mistakes

Taking orders from the coefficients in the balanced equation. In $\mathrm{2NO + 2H_2}$ , the order in $\mathrm{H_2}$ is 1, not 2; in the propanone reaction the order in iodine is 0.
Comparing two runs in which two concentrations both change, and giving the whole rate change to one of them.
Giving $k$ without a unit, or with the unit of a different order.
Using rates measured late in a run. Only initial rates correspond to the concentrations you mixed.

Tip

Lay out an initial-rates answer as one line per reactant: "Runs __ and __: [X] × __, rate × __, so order in X = __." Examiners award a mark for each line, and the layout stops you comparing the wrong pair of runs.

Tony St John: Solving a Rate Law Using the Initial Rates Method. Watch how each order is found by choosing two runs in which only one concentration changes, and how k and its units follow.

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Practice

Define order of reaction with respect to a reactant and overall order.
Give the unit of $k$ for (a) $\text{rate} = k[\mathrm{A}]$ , (b) $\text{rate} = k[\mathrm{A}][\mathrm{B}]$ , (c) $\text{rate} = k[\mathrm{A}]^2[\mathrm{B}]$ .
For A + B → products: run 1, [A] = 0.10, [B] = 0.10, rate $2.0 \times 10^{-4}$ ; run 2, [A] = 0.20, [B] = 0.10, rate $8.0 \times 10^{-4}$ ; run 3, [A] = 0.10, [B] = 0.30, rate $2.0 \times 10^{-4}$ (concentrations in $\mathrm{mol\,dm^{-3}}$ , rates in $\mathrm{mol\,dm^{-3}\,s^{-1}}$ ). Find the rate equation and $k$ .
For $\text{rate} = k[\mathrm{X}][\mathrm{Y}]^2$ , by what factor does the rate change if [X] is doubled and [Y] is halved at the same time?
Tripling the concentration of C increases the initial rate by a factor of 5.2. Find the order with respect to C.
Sketch rate–concentration graphs for zero-, first- and second-order dependence, and say which one is a straight line through the origin.
Explain why the order with respect to iodine in the iodination of propanone is zero, even though iodine is a reactant.

Show answersHide answers

The power to which that reactant's concentration is raised in the experimentally determined rate equation; the overall order is the sum of these powers.
(a) $\mathrm{s^{-1}}$ ; (b) $\mathrm{dm^3\,mol^{-1}\,s^{-1}}$ ; (c) $\mathrm{dm^6\,mol^{-2}\,s^{-1}}$ .
Runs 1 and 2: [A] × 2, rate × 4, second order in A. Runs 1 and 3: [B] × 3, rate unchanged, zero order in B. $\text{rate} = k[\mathrm{A}]^2$ ; $k = \frac{2.0 \times 10^{-4}}{(0.10)^2} = 2.0 \times 10^{-2}\,\mathrm{dm^3\,mol^{-1}\,s^{-1}}$ .
$2 \times \left(\frac{1}{2}\right)^2 = \frac{1}{2}$ : the rate halves.
$m = \frac{\log 5.2}{\log 3} = 1.5$ .
Zero order: horizontal; first order: straight line through the origin; second order: curve through the origin that steepens.
Iodine reacts only in a fast step after the rate-determining step, so the rate of the slow step, and therefore of the whole reaction, does not depend on how much iodine is present.

Sources and credits

Video: Solving a Rate Law Using the Initial Rates Method, Tony St John, embedded from YouTube.
The text, tables, graphs, illustrative data, worked examples and Practice items were written and drawn by Ganzaa.

1Iodine is used up, yet it does not matter how much there is

2What each order means

3Units of the rate constant

4The initial-rates method

Iodine is used up, yet it does not matter how much there is

What each order means

Units of the rate constant

The initial-rates method