Rate of reaction

1. Overview

Some reactions are over in a fraction of a second, while others take hours or years. This section is about the rate of reaction: how quickly reactants are used up and products form, and why that changes. The explanation is collision theory: particles react only when they collide, and only some collisions lead to a reaction. You also need to turn experimental results, such as gas volumes, masses, concentrations or reaction times, into a rate with the correct units.

Key Definitions

• Rate of reaction: the change in concentration of a reactant or a product per unit time; with concentration in mol dm⁻³ and time in seconds, its unit is mol dm⁻³ s⁻¹.
• Frequency of collisions: the number of collisions between reactant particles per unit time (for example, per second) in a given volume.
• Effective collision: a collision between reactant particles that results in a reaction; the colliding particles have at least the activation energy, EA, and meet in a suitable orientation.
• Non-effective collision: a collision between reactant particles that does not result in a reaction, because the colliding particles have less than the activation energy, EA, or meet in an unsuitable orientation; they bounce apart unchanged.
• Collision theory: the model in which reactant particles must collide in order to react, so the rate of reaction depends on the frequency of effective collisions.
• Mean rate: the total change in concentration (or in another measured quantity) divided by the time taken for that change; on a graph it is the gradient of the chord joining the points at those two times, taken as positive for a reactant.
• Initial rate: the rate at the very start of a reaction, at time zero; it is the gradient of the tangent to the curve at t = 0, taken as positive for a reactant.

Content

What rate of reaction means

The rate of reaction is the change in concentration of a reactant or a product per unit time:

$$\text{rate} = \frac{\text{change in concentration}}{\text{time taken}}$$

Square brackets mean "concentration in mol dm⁻³": $[\text{HNO}_3]$ is the concentration of nitric acid. With time in seconds, a rate is in mol dm⁻³ s⁻¹.

• A rate is always positive. For a reactant, rate $= -\Delta[\text{reactant}]/\Delta t$; for a product, rate $= \Delta[\text{product}]/\Delta t$. Here $\Delta$ means "change in" (final minus initial) and $t$ is time.
• Say which substance. Rates follow the coefficients in the equation. In N₂(g) + 3H₂(g) → 2NH₃(g), hydrogen is used up three times as fast as nitrogen.
• Other units. Following a gas volume or a mass gives a rate in cm³ s⁻¹ or g s⁻¹. A rate per minute ÷ 60 = the rate per second.

Collisions: frequency, effective and non-effective

Reactant particles (molecules, atoms or ions) can react only when they collide. The frequency of collisions is the number of collisions per unit time in a given volume.

Most collisions do not lead to a reaction. Bonds must start to break, and that needs energy. The activation energy, $E_{\text{A}}$, is the minimum energy a collision needs to be effective (8.2 explains why only some collisions have it). In an effective collision the particles have at least $E_{\text{A}}$ and, for most reactions, meet in a suitable orientation, so products form. A collision with less energy, or the wrong orientation, is a non-effective collision: the particles bounce apart unchanged.

$$\text{frequency of effective collisions} = \text{frequency of collisions} \times \text{fraction of collisions that are effective}$$

So a reaction speeds up if collisions become more frequent, or if a larger fraction of them are effective.

Change	Frequency of collisions	Fraction of collisions that are effective	Where it is taught
Higher concentration (solutions)	increases	unchanged	this page
Higher pressure (gases)	increases	unchanged	this page
Larger surface area of a solid	increases	unchanged	this page, briefly
Higher temperature	increases slightly	increases greatly (the main effect)	8.2
Adding a catalyst	unchanged	increases (a route with a lower $E_{\text{A}}$)	8.3

How concentration affects the rate

Explain the effect as a four-step chain:

1. A higher concentration means more reactant particles in each unit volume.
2. The particles collide more often: the frequency of collisions increases.
3. The temperature is unchanged, so the particles' energies and $E_{\text{A}}$ are the same, and the same fraction of collisions is effective.
4. So the frequency of effective collisions increases, and the rate increases.

Concentration does not give the particles more energy and does not lower $E_{\text{A}}$. How much the rate changes when a concentration doubles depends on the reaction, and for a few reactants it does not change at all; that (orders of reaction) is A Level topic 26.1.

As a reaction goes, reactants are used up, so collisions become less frequent and the rate falls until the limiting reactant runs out. For a solid, only surface particles can be hit, so a powder reacts faster than a lump of the same mass.

To sketch a second curve, decide two things separately: the starting gradient shows the initial rate (steeper for the more concentrated acid), and the final level depends only on the amount of the limiting reactant. In the diagram the carbonate is limiting, so both curves level off at the same volume. If the acid were limiting instead, the same volume of more concentrated acid would also give more gas.

How pressure affects the rate of a gas reaction

For gases, changing the pressure changes the concentration. At a fixed temperature, $pV = nRT$ (4.1) gives $n/V = p/RT$, so the concentration of a gas is proportional to its pressure. Squeezing a gas mixture into half the volume at the same temperature doubles the pressure and doubles the concentration of every gas in it.

Then the same chain applies: more molecules per unit volume, so more frequent collisions; the same fraction is effective at the same temperature; so there are more effective collisions per second and the rate increases.

• Only gaseous reactants count. Liquids and solids hardly compress, so pressure has almost no effect on reactions between substances in solution. For a gas reacting at a solid surface, more gas molecules hit each unit area per second.
• Concentration, not total pressure, matters. If an unreactive gas is added while the volume stays fixed, the total pressure rises but each reactant's concentration is unchanged, so at the same temperature the rate is unchanged.
• Rate is not yield. Pressure may also shift an equilibrium (7.1), but that is about how much product, not how fast. Answer a rate question with collisions only.

Following a reaction in the laboratory

Follow one quantity that changes as the reaction goes, at a constant temperature.

Method	What you measure	Points to know
Gas collection	volume of gas at intervals, in a gas syringe or an upturned burette or measuring cylinder full of water	rate in cm³ s⁻¹; some CO₂ dissolves in the water, so early readings over water can be a little low
Loss in mass	mass of the flask and contents on a balance as a gas escapes	rate in g s⁻¹; a loose cotton-wool plug lets gas out but keeps acid spray in; fine for CO₂, but hydrogen's tiny molar mass makes the mass change too small to measure well
Colorimetry	absorbance of a coloured reactant or product	a calibration line converts absorbance to concentration
Sampling and titration	concentration in samples removed at intervals	quench each sample at once, for example in iced water, which cools and dilutes it
Time to a fixed point	time, $t$, for a fixed amount of product to form	$1/t$ is a measure of the rate (see below)

Calculating a rate from experimental data

Mean rate = change in the measured quantity ÷ time taken. On a graph, it is the gradient of the straight line, or chord, joining the two points.

Rate at one instant. Draw a tangent that just touches the curve at that time, choose two points on it that are far apart, and find its gradient, $\Delta y/\Delta x$. The initial rate is the gradient of the tangent at $t = 0$. For a reactant's concentration the gradient is negative, so the rate is minus the gradient.

Converting units to a rate of change of concentration:

• Gas volume to amount: $n = V/V_{\text{m}}$, with $V_{\text{m}}$ = 24.0 dm³ mol⁻¹ (24 000 cm³ mol⁻¹) at room conditions. If a question gives a temperature and pressure instead, use $pV = nRT$ (4.1).
• Mass to amount: $n = m/M$ ($m$ in g, $M$ in g mol⁻¹).
• Between substances: use the ratio of the coefficients in the equation.
• Per second to concentration per second: divide by the solution volume in dm³.
• Powers of ten: a heading "[X] / 10⁻³ mol dm⁻³" means each number shown is multiplied by 10⁻³.

Significant figures. The difference between two close readings has fewer s.f. than the readings: 2.02 g − 1.87 g = 0.15 g has only 2. Give a rate to the same number of s.f. as the least precise data used, or one more.

Worked example 1: mean rates from loss in mass

Question: A conical flask containing 80.0 cm³ of 1.25 mol dm⁻³ nitric acid stands on a balance. At $t = 0$, 20.0 g of marble chips (calcium carbonate, a large excess) are added and a loose cotton-wool plug is put in the neck. Assume the temperature stays constant.

CaCO₃(s) + 2HNO₃(aq) → Ca(NO₃)₂(aq) + CO₂(g) + H₂O(l)

time / s	0	30	60	90	120	180	240	300
mass of flask and contents / g	186.42	185.83	185.39	185.08	184.84	184.55	184.40	184.31

Calculate (a) the mean rate of production of CO₂ in the first 60 s, in g s⁻¹ and mol s⁻¹; (b) the mean rate at which $[\text{HNO}_3]$ falls in the first 60 s; (c) the mean rate of production of CO₂ from 180 s to 240 s, in g s⁻¹. Explain the difference. ($A_{\text{r}}$: C 12.0, O 16.0)

Step 1: mass of CO₂. Taking CO₂ as the only substance that leaves the flask, the loss in mass in the first 60 s is 186.42 − 185.39 = 1.03 g.

Step 2: rate in g s⁻¹, then mol s⁻¹. $M(\text{CO}_2) = 12.0 + 2(16.0) = 44.0$ g mol⁻¹. $$\frac{1.03}{60} = 0.01717\ \text{g s}^{-1} \qquad \frac{0.01717}{44.0} = 3.902 \times 10^{-4}\ \text{mol s}^{-1}$$

Step 3: use the equation. 2 mol of HNO₃ are used per 1 mol of CO₂, so HNO₃ is used at $7.803 \times 10^{-4}$ mol s⁻¹.

Step 4: divide by the volume in dm³ (80.0 cm³ = 0.0800 dm³). $$\frac{7.803 \times 10^{-4}}{0.0800} = 9.75 \times 10^{-3}\ \text{mol dm}^{-3}\ \text{s}^{-1}$$

Step 5: the later interval. Mass lost by 180 s is 1.87 g and by 240 s is 2.02 g. $$\frac{2.02 - 1.87}{60} = \frac{0.15}{60} = 2.5 \times 10^{-3}\ \text{g s}^{-1}$$

Answer: (a) 0.0172 g s⁻¹, which is $3.90 \times 10^{-4}$ mol s⁻¹ of CO₂. (b) $9.75 \times 10^{-3}$ mol dm⁻³ s⁻¹. (c) $2.5 \times 10^{-3}$ g s⁻¹ (2 s.f., as 0.15 g has only 2), about 7 times slower.

Why it slows: by 180 s, 0.0850 mol of the 0.100 mol of HNO₃ has been used, so $[\text{HNO}_3]$ has fallen from 1.25 to about 0.19 mol dm⁻³, and hydrogen ions hit the surface of the chips far less often. Only about 4.3 g of the 20.0 g of calcium carbonate has reacted, so lost surface is a much smaller effect.

Worked example 2: rates from concentration data

Question: Buta-1,3-diene, C₄H₆, dimerises slowly when heated in a sealed vessel at constant temperature. Two gas molecules become one, so the pressure falls, and $[\text{C}_4\text{H}_6]$ is worked out from the pressure.

2C₄H₆(g) → C₈H₁₂(g)

time / s	0	200	400	600	800	1000
[C₄H₆] / 10⁻³ mol dm⁻³	18.00	14.80	12.57	10.92	9.66	8.65

(a) Calculate the mean rate of use of C₄H₆, and the mean rate of formation of C₈H₁₂, from 0 to 200 s. (b) Calculate the mean rate of use of C₄H₆ from 800 s to 1000 s. (c) The tangent to the curve at 400 s passes through (0 s, 16.4 × 10⁻³ mol dm⁻³) and (1000 s, 6.9 × 10⁻³ mol dm⁻³). Find the rate of use of C₄H₆ at 400 s.

Step 1: 0 to 200 s. Include the 10⁻³ from the heading. $$-\frac{\Delta[\text{C}_4\text{H}_6]}{\Delta t} = -\frac{(14.80 - 18.00) \times 10^{-3}}{200} = 1.60 \times 10^{-5}\ \text{mol dm}^{-3}\ \text{s}^{-1}$$

Step 2: use the equation. 2 mol of C₄H₆ give 1 mol of C₈H₁₂, so C₈H₁₂ forms at half that rate, $8.00 \times 10^{-6}$ mol dm⁻³ s⁻¹.

Step 3: 800 to 1000 s. $$-\frac{(8.65 - 9.66) \times 10^{-3}}{200} = 5.05 \times 10^{-6}\ \text{mol dm}^{-3}\ \text{s}^{-1}$$

Step 4: gradient of the tangent. $$\frac{(6.9 - 16.4) \times 10^{-3}}{1000 - 0} = -9.5 \times 10^{-6}\ \text{mol dm}^{-3}\ \text{s}^{-1}$$

Answer: (a) C₄H₆ is used at $1.60 \times 10^{-5}$ mol dm⁻³ s⁻¹; C₈H₁₂ forms at $8.00 \times 10^{-6}$ mol dm⁻³ s⁻¹. (b) $5.05 \times 10^{-6}$ mol dm⁻³ s⁻¹, about a third of (a). (c) $9.5 \times 10^{-6}$ mol dm⁻³ s⁻¹ (minus the gradient; 2 s.f., as read from a tangent).

The rate falls because $[\text{C}_4\text{H}_6]$ falls, so C₄H₆ molecules collide less often. The mean rate over the first 400 s, $1.36 \times 10^{-5}$ mol dm⁻³ s⁻¹, is higher than the rate at 400 s because it includes the faster start; the mean from 200 s to 600 s, $9.70 \times 10^{-6}$ mol dm⁻³ s⁻¹, is close to the tangent value.

Rates from reaction times

Sometimes you measure one time per run: how long a fixed amount of product takes to form. Every run ends at the same amount of product, so

$$\text{mean rate} = \frac{\text{fixed amount of product}}{t} \qquad\text{so}\qquad \text{rate} \propto \frac{1}{t}$$

$1/t$, in s⁻¹, is a relative rate for comparing runs: a long time means a slow rate. It reflects the initial rate if only a small fraction of the reactants is used by the end point.

The standard example is sodium thiosulfate with an acid:

S₂O₃²⁻(aq) + 2H⁺(aq) → S(s) + SO₂(aq) + H₂O(l)

Stand the beaker on a printed mark, look down through the mixture, and stop the clock when the sulfur hides the mark. Avoid breathing in the sulfur dioxide.

• Same total volume every run (make it up with water), so only the chosen concentration changes.
• Same beaker every run. The mark vanishes when a set amount of sulfur lies between eye and mark; in a wider beaker the liquid is shallower, so the time is longer though the rate is unchanged.
• Concentration in the mixture, not the stock: $c_{\text{mixture}} = c_{\text{stock}} \times V_{\text{stock}}/V_{\text{total}}$.
• Random errors: judging when the mark disappears, and changes in room temperature between runs.

Testing for proportionality. Plot $1/t$ against concentration. The rate is directly proportional to the concentration only if the points lie on a straight line through the origin.

Worked example 3: relative rates from reaction times

Question: A student times how long it takes to collect 15.0 cm³ of CO₂ when the same mass of marble chips of similar size (calcium carbonate, an excess) is added to mixtures of 1.50 mol dm⁻³ nitric acid and water. Each mixture has a total volume of 60.0 cm³, at the same temperature.

volume of 1.50 mol dm⁻³ HNO₃ / cm³	12.0	24.0	36.0	48.0
volume of water / cm³	48.0	36.0	24.0	12.0
$t$ / s	81	39	27	20

(a) Calculate $[\text{HNO}_3]$ in each mixture and $1/t$ for each run. (b) Is the rate directly proportional to $[\text{HNO}_3]$?

Step 1: concentrations. For the first run, $c = 1.50 \times \frac{12.0}{60.0} = 0.300$ mol dm⁻³. The others are 0.600, 0.900 and 1.20 mol dm⁻³.

Step 2: relative rates, compared with the first run.

$[\text{HNO}_3]$ / mol dm⁻³	0.300	0.600	0.900	1.20
$1/t$ / s⁻¹	0.0123	0.0256	0.0370	0.0500
$1/t$ ÷ first $1/t$	1.00	2.08	3.00	4.05

Answer: (a) As in the table. (b) Doubling, tripling and quadrupling the concentration multiplies $1/t$ by 2.08, 3.00 and 4.05: 2, 3 and 4 within the timing uncertainty (about 1 s, which is 5% of 20 s). So the results support direct proportionality; a plot of $1/t$ against $[\text{HNO}_3]$ is close to a straight line through the origin.

This works because 15.0 cm³ of CO₂ ($6.25 \times 10^{-4}$ mol at room conditions) uses only about 7% of the acid in the most dilute mixture, so each time reflects the rate near the start.

Key Equations

Rate of reaction ($[\ ]$ = concentration in mol dm⁻³; $t$ in s; rate in mol dm⁻³ s⁻¹, always positive): $$\text{rate} = -\frac{\Delta[\text{reactant}]}{\Delta t} = \frac{\Delta[\text{product}]}{\Delta t}$$

Mean rate from any measured quantity (for example cm³ or g; the rate is in that unit per second): $$\text{mean rate} = \frac{\text{change in quantity}}{\text{time taken}}$$

Rates of different substances (for $a\text{A} \rightarrow b\text{B}$, with coefficients $a$ and $b$): $$\text{rate of formation of B} = \frac{b}{a} \times \text{rate of use of A}$$

Amount from a gas volume or a mass ($V_{\text{m}}$ = 24.0 dm³ mol⁻¹ at room conditions; $m$ in g; $M$ in g mol⁻¹): $$n = \frac{V}{V_{\text{m}}} \qquad n = \frac{m}{M}$$

Concentration, and dilution ($c$ in mol dm⁻³; $V$ in dm³; in the dilution formula both volumes just need the same unit): $$c = \frac{n}{V} \qquad c_{\text{mixture}} = c_{\text{stock}} \times \frac{V_{\text{stock}}}{V_{\text{total}}}$$

Relative rate ($t$ = time for a fixed amount of reaction, in s): $$\text{rate} \propto \frac{1}{t}$$

Common Mistakes to Avoid

• Mistake: writing "more collisions" or "more successful collisions" with no idea of time. Fix: write "more frequent collisions" or "more effective collisions per second".
• Mistake: saying a higher concentration or pressure gives particles more energy, lowers $E_{\text{A}}$ or makes a larger fraction of collisions effective. Fix: at a fixed temperature only the collision frequency changes. In Worked example 3, quadrupling $[\text{HNO}_3]$ leaves the energies and $E_{\text{A}}$ unchanged.
• Mistake: answering a rate question with equilibrium position or yield. Fix: for a gas reaction at higher pressure, explain the rate with concentration and collision frequency; keep Le Chatelier's principle for questions about yield.
• Mistake: giving a mean rate when the rate at a particular time is asked for. Fix: draw the tangent at exactly that time (at $t = 0$ for the initial rate). In Worked example 2, the rate at 400 s is $9.5 \times 10^{-6}$ mol dm⁻³ s⁻¹, but the mean rate over the first 400 s is $1.36 \times 10^{-5}$ mol dm⁻³ s⁻¹.
• Mistake: using the stock concentration instead of the concentration in the mixture. Fix: allow for dilution. In Worked example 3, 12.0 cm³ of 1.50 mol dm⁻³ acid made up to 60.0 cm³ is 0.300 mol dm⁻³, not 1.50 mol dm⁻³.
• Mistake: skipping a step when converting a rate. Fix: go measured unit, then mol, then equation ratio, then divide by dm³. In Worked example 1, leaving out the factor of 2 gives $4.88 \times 10^{-3}$ instead of $9.75 \times 10^{-3}$ mol dm⁻³ s⁻¹, and dividing by 80.0 instead of 0.0800 gives an answer 1000 times too small.
• Mistake: calling a time "slower", or claiming proportion just because a line slopes upwards. Fix: a time is longer or shorter; a rate is faster or slower. Direct proportion needs a straight line through the origin: in Worked example 3, $1/t$ rises by 2.08, 3.00 and 4.05 times when the concentration rises 2, 3 and 4 times.

Exam Tips

• Define rate of reaction as the change in concentration of a reactant or product per unit time, and give a unit with every rate you calculate.
• Explain a concentration or pressure effect as a chain: more particles per unit volume; more frequent collisions; the same fraction effective (same temperature, same $E_{\text{A}}$); more effective collisions per unit time.
• Calculate a rate with each conversion on its own line, say which substance it refers to, and give as many s.f. as the data justify.
• Use a graph: for "the rate at" a time, draw a tangent there and use two points on it far apart; for "the mean rate between" two times, use the chord. Read any power of ten in an axis label.
• Suggest improvements to a timing experiment in concrete terms (same total volume, same beaker, same temperature), and name the random error, such as judging the end point, rather than "human error".