Rates of Reaction and Collision Theory

AS · 12 min

Some reactions are over in a flash and others take years. Reaction kinetics is the study of how fast reactions go and why. This note defines rate of reaction, shows how rates are measured and calculated from experimental data (including from the gradient of a graph), and explains, using collision theory, why concentration, pressure and surface area affect the rate. The next note adds temperature and catalysts using the Boltzmann distribution. Explanations in terms of "frequency of effective collisions" are expected in every Paper 2, and rate experiments are a favourite of Paper 3.

Rate of reaction

Definition

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

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

With concentration in $\text{mol dm}^{-3}$ and time in seconds, rate has units of $\text{mol dm}^{-3}\ \text{s}^{-1}$ . In practice, any measurable quantity that changes in proportion to the amount reacted can be used, giving rates in $\text{cm}^3\ \text{s}^{-1}$ (gas volume), $\text{g s}^{-1}$ (mass) or $\text{mol s}^{-1}$ .

A rate is always given as a positive number: for a reactant, use the decrease in concentration; for a product, the increase.

How rate changes during a reaction

Most reactions are fastest at the start and slow down as they proceed, because the reactants are used up and their concentrations fall. The graph shows the volume of hydrogen collected against time when magnesium reacts with excess hydrochloric acid.

y = 60(1 - exp(-0.04x)) (0, 0) -- (25, 60) (5, 23.86) -- (55, 60)

(Horizontal axis: time / s; vertical axis: volume of $\ce{H2}$ / $\text{cm}^3$ . The curve is steep at first, then levels off at $60\ \text{cm}^3$ when the magnesium has all reacted. The two straight lines are tangents: the one through the origin gives the initial rate, and the other touches the curve at $t = 30\ \text{s}$ .)

The gradient of the curve at any time is the rate at that time.
The initial rate is the gradient of the tangent at $t = 0$ .
When the curve becomes horizontal, the reaction has stopped (a reactant is used up).

Method

Finding a rate from a graph

Draw a tangent to the curve at the time required: a straight line that touches the curve at that point only, with the same slope.
Choose two points on the tangent far apart (to reduce the uncertainty).
Gradient $= \dfrac{\Delta y}{\Delta x}$ , with units ( $\text{cm}^3\ \text{s}^{-1}$ or $\text{mol dm}^{-3}\ \text{s}^{-1}$ ).
For an average rate over a period, use the change in the quantity divided by the time taken (a chord, not a tangent).

On the graph above, the initial tangent goes from $(0, 0)$ to $(25, 60)$ : initial rate $= \dfrac{60}{25} = 2.4\ \text{cm}^3\ \text{s}^{-1}$ . The tangent at $30\ \text{s}$ goes from $(5, 23.9)$ to $(55, 60.0)$ : rate $= \dfrac{36.1}{50} = 0.72\ \text{cm}^3\ \text{s}^{-1}$ . The rate has fallen to under a third of its initial value.

Measuring rates

what changes	method	example
volume of gas produced	gas syringe, or collection over water in an inverted measuring cylinder	$\ce{Mg + 2HCl -> MgCl2 + H2}$ ; $\ce{2H2O2 -> 2H2O + O2}$
mass (a gas escapes)	conical flask on a balance, with cotton wool in the neck	$\ce{CaCO3 + 2HCl -> CaCl2 + H2O + CO2}$
colour	colorimeter measures absorbance, related to concentration	reactions of iodine or bromine (coloured)
a precipitate forms	time for a cross under the flask to disappear	$\ce{S2O3^2- + 2H+ -> SO2 + S + H2O}$
concentration of acid or other species	remove samples, quench (stop the reaction by cooling or dilution), titrate	ester hydrolysis
electrical conductivity	conductivity meter, when the number of ions changes

For "time to reach a fixed point" methods (such as the disappearing cross), the same fixed amount of reaction happens each time, so:

\text{rate} \propto \frac{1}{t}

Collision theory

For two particles to react, they must collide. But only a tiny fraction of collisions lead to reaction. For a collision to be effective:

the particles must collide with energy equal to or greater than the activation energy, $E_A$ , the minimum energy needed to break the bonds that must be broken;
the particles must collide with the correct orientation, so that the reacting parts of the molecules meet.

Definition

Frequency of collisions: the number of collisions per unit time (per second) between reacting particles.
Effective collision: a collision that results in a reaction (energy $\ge E_A$ and correct orientation).
Non-effective collision: a collision in which the particles bounce apart unchanged, because they have too little energy or the wrong orientation.

Key result

The rate of reaction depends on the frequency of effective collisions. Anything that increases the number of effective collisions per second increases the rate.

Effect of concentration

Increasing the concentration of a reactant in solution means there are more particles per unit volume. The particles collide more frequently. Since the same proportion of collisions has enough energy, the frequency of effective collisions increases, so the rate increases.

This also explains why reactions slow down: as reactants are used up, their concentrations fall, the frequency of effective collisions falls, and the rate decreases.

Effect of pressure

For reactions between gases, increasing the pressure (by compressing the same amount of gas into a smaller volume) has exactly the same effect as increasing concentration: more molecules per unit volume, so more frequent collisions and more frequent effective collisions, and a faster rate.

Effect of surface area

For a reaction involving a solid, only particles at the surface can collide with the other reactant. Breaking a solid into smaller pieces (or using a powder) increases the surface area exposed, so there are more frequent collisions and more frequent effective collisions. Powdered calcium carbonate reacts with acid much faster than marble chips of the same mass.

Tip

Temperature and catalysts also change the rate, but mainly by changing the proportion of collisions that are effective rather than their frequency. They are explained using the Boltzmann distribution in the next note.

Worked examples

Routine: average rate

In the reaction of magnesium with hydrochloric acid, $48.0\ \text{cm}^3$ of hydrogen was collected in the first $20.0\ \text{s}$ at room conditions. Calculate the average rate of production of hydrogen in (a) $\text{cm}^3\ \text{s}^{-1}$ , (b) $\text{mol s}^{-1}$ .

Solution

(a) $\text{rate} = \dfrac{48.0}{20.0} = 2.40\ \text{cm}^3\ \text{s}^{-1}$ .

(b) $n(\ce{H2}) = \dfrac{48.0}{24\,000} = 2.00 \times 10^{-3}\ \text{mol}$ . Rate $= \dfrac{2.00 \times 10^{-3}}{20.0} = 1.00 \times 10^{-4}\ \text{mol s}^{-1}$ .

Rate from concentration data

In a reaction $\ce{A -> B}$ , the concentration of A fell from $0.80\ \text{mol dm}^{-3}$ to $0.50\ \text{mol dm}^{-3}$ in the first $120\ \text{s}$ . Calculate the average rate of reaction over this time, and explain why the rate over the next $120\ \text{s}$ would be lower.

Solution

\text{rate} = \frac{0.80 - 0.50}{120} = 2.5 \times 10^{-3}\ \text{mol dm}^{-3}\ \text{s}^{-1}

Over the next $120\ \text{s}$ the concentration of A is lower, so there are fewer A particles per unit volume. Collisions are less frequent, so effective collisions are less frequent and the rate is lower.

Explaining a pressure effect

Explain why increasing the pressure increases the rate of the reaction $\ce{2NO(g) + O2(g) -> 2NO2(g)}$ .

Solution

At higher pressure the same number of gas molecules occupies a smaller volume, so there are more molecules per unit volume (a higher concentration). The molecules collide more frequently, so there are more effective collisions per unit time (collisions with energy $\ge E_A$ and the correct orientation). The rate increases.

A clock experiment

In the reaction between sodium thiosulfate and hydrochloric acid, the time for a cross under the flask to disappear was measured at different thiosulfate concentrations:

$[\ce{S2O3^2-}]$ / mol dm⁻³	0.050	0.100	0.150	0.200
time / s	100	50	33	25

Calculate the relative rate ( $1/t$ ) for each, and describe the relationship between rate and concentration.

Solution

$1/t$ / $\text{s}^{-1}$ : $0.010$ , $0.020$ , $0.030$ , $0.040$ .

The relative rate is directly proportional to the concentration of thiosulfate: doubling the concentration doubles the rate. More thiosulfate ions per unit volume means more frequent effective collisions with $\ce{H+}$ ions.

Exam-hard: converting mass loss into a rate

$50.0\ \text{cm}^3$ of hydrochloric acid (an excess) was added to calcium carbonate chips on a balance. The mass fell by $0.44\ \text{g}$ in the first $60\ \text{s}$ .

(a) Calculate the average rate of reaction over this time in terms of the decrease in concentration of hydrochloric acid, in $\text{mol dm}^{-3}\ \text{s}^{-1}$ . (b) The experiment is repeated with the same mass of powdered calcium carbonate. Sketch how the graph of mass loss against time would differ and explain.

Solution

(a) The mass loss is the carbon dioxide that escapes: $n(\ce{CO2}) = \dfrac{0.44}{44.0} = 0.0100\ \text{mol}$ .

$\ce{CaCO3 + 2HCl -> CaCl2 + H2O + CO2}$ : $n(\ce{HCl}) = 2 \times 0.0100 = 0.0200\ \text{mol}$ used.

Change in concentration $= \dfrac{0.0200}{0.0500} = 0.400\ \text{mol dm}^{-3}$ .

\text{rate} = \frac{0.400}{60} = 6.7 \times 10^{-3}\ \text{mol dm}^{-3}\ \text{s}^{-1}

(b) The curve is steeper at the start and levels off sooner, but at the same final mass loss (the same mass of $\ce{CaCO3}$ reacts with excess acid). The powder has a much larger surface area, so collisions between acid particles and the solid are more frequent, giving more frequent effective collisions and a faster rate. The total amount of $\ce{CO2}$ is unchanged because the amount of the limiting reactant is the same.

Measuring the effect of concentration on rate

Method A: gas syringe (magnesium with hydrochloric acid).

Measure a fixed volume of acid of known concentration into a conical flask connected to a gas syringe.
Add a measured length (or mass) of clean magnesium ribbon, seal quickly and start the stopwatch.
Record the volume of gas every 10 s until no more gas is produced.
Plot volume against time; draw a tangent at $t = 0$ to find the initial rate. Repeat with different acid concentrations (made by dilution with distilled water, keeping the total volume constant).

Method B: disappearing cross (thiosulfate with acid).

Put a fixed volume of sodium thiosulfate solution (diluted with water to vary concentration, total volume constant) in a flask on a cross drawn on paper.
Add a fixed volume of hydrochloric acid and start timing.
Stop timing when the cross is no longer visible through the cloudy (sulfur) precipitate. Calculate $1/t$ .

Variables: independent: concentration; dependent: volume of gas over time, or time for the cross to disappear; controlled: temperature, total volume, mass and surface area of Mg, the same cross and observer.

Errors and improvements:

error	improvement
gas escapes before the bung is inserted	place Mg in a separate small tube inside the flask and tip it over after sealing
magnesium has an oxide coating	clean with emery paper
judging when the cross disappears is subjective	use a light sensor or colorimeter; the same person judges every run
temperature changes (the reactions are exothermic)	use a water bath; record the temperature
reaction time for starting and stopping the stopwatch	use longer times (more dilute solutions) so reaction-time error is a small percentage

How Paper 3 asks about it: drawing a line of best fit or a smooth curve; drawing a tangent and calculating its gradient; calculating $1/t$ ; identifying the independent, dependent and controlled variables; suggesting improvements to reduce a named error.

Watch out

"Higher concentration means more collisions." Say more frequent collisions (more collisions per unit time); the total number of collisions is not the point.
Forgetting "effective". The rate depends on the frequency of effective collisions. A full answer says that more frequent collisions lead to more frequent effective collisions.
"Higher concentration gives more product." It gives the same amount of product faster, if the extra reactant is not the limiting reagent.
Using a chord instead of a tangent for the rate at an instant.
Rate as a negative number. Rate is positive; express it as the decrease in reactant or increase in product.

Exam tip

Define: rate of reaction is the change in concentration of a reactant or product per unit time.
For concentration and pressure questions, the marking points are usually: (1) more particles per unit volume; (2) more frequent collisions; (3) more frequent effective collisions (or "more successful collisions per second").
For graphs, quote gradients with units, and show the two points used on the tangent.
In practical questions, link each improvement to the error it removes.

Summary

Rate of reaction: change in concentration of a reactant or product per unit time ( $\text{mol dm}^{-3}\ \text{s}^{-1}$ ).
Rate at any time = gradient of the tangent to a concentration (or volume) against time graph; initial rate from the tangent at $t = 0$ .
For clock reactions, rate $\propto 1/t$ .
Effective collisions need energy $\ge E_A$ and correct orientation.
Rate depends on the frequency of effective collisions.
Higher concentration, higher gas pressure and larger surface area all increase the frequency of collisions and hence of effective collisions.

Practice

Question

Define rate of reaction and give its usual units.
Explain the difference between an effective and a non-effective collision.
Explain why the reaction between zinc and sulfuric acid slows down as it proceeds.
Explain, in terms of collisions, why increasing the pressure increases the rate of a reaction between two gases.
$0.120\ \text{g}$ of magnesium ribbon reacted completely with excess acid in $80\ \text{s}$ . Calculate the average rate of the reaction in mol of magnesium per second, and the volume of hydrogen produced at room conditions.
In a disappearing-cross experiment, $0.20\ \text{mol dm}^{-3}$ thiosulfate took $25\ \text{s}$ and $0.10\ \text{mol dm}^{-3}$ took $50\ \text{s}$ . Calculate the relative rates and describe the relationship.
Explain why powdered calcium carbonate reacts faster with hydrochloric acid than the same mass of marble chips, and state whether the volume of gas produced would differ (acid in excess).
A tangent drawn to a graph of volume of oxygen against time at $t = 20\ \text{s}$ passes through $(0\ \text{s}, 10\ \text{cm}^3)$ and $(40\ \text{s}, 50\ \text{cm}^3)$ . Calculate the rate of reaction at $20\ \text{s}$ .
$50.0\ \text{cm}^3$ of hydrogen peroxide decomposed with a catalyst, producing $60\ \text{cm}^3$ of oxygen at room conditions in the first $30\ \text{s}$ : $\ce{2H2O2 -> 2H2O + O2}$ . Calculate the average rate of decrease of the concentration of hydrogen peroxide in $\text{mol dm}^{-3}\ \text{s}^{-1}$ .
$0.10\ \text{g}$ of magnesium is added to (A) $50.0\ \text{cm}^3$ of $1.0\ \text{mol dm}^{-3}$ HCl, (B) $50.0\ \text{cm}^3$ of $0.50\ \text{mol dm}^{-3}$ HCl and (C) $50.0\ \text{cm}^3$ of $0.10\ \text{mol dm}^{-3}$ HCl, at the same temperature. Calculate the final volume of hydrogen (room conditions) in each case, and sketch or describe the three volume–time curves, explaining their differences.

Answers

The change in concentration of a reactant or product per unit time; $\text{mol dm}^{-3}\ \text{s}^{-1}$ .
An effective collision results in reaction: the particles collide with energy equal to or greater than the activation energy and with the correct orientation. A non-effective collision has too little energy or the wrong orientation, and the particles bounce apart unchanged.
As the reaction proceeds, the concentration of sulfuric acid decreases (and the zinc surface shrinks), so there are fewer acid particles per unit volume; collisions with the zinc become less frequent, so effective collisions are less frequent, and the rate decreases.
At higher pressure there are more gas molecules per unit volume, so collisions are more frequent, leading to more frequent effective collisions (with energy $\ge E_A$ ) and a faster rate.
$n(\ce{Mg}) = \dfrac{0.120}{24.3} = 4.94 \times 10^{-3}\ \text{mol}$ . Rate $= \dfrac{4.94 \times 10^{-3}}{80} = 6.17 \times 10^{-5}\ \text{mol s}^{-1}$ . $n(\ce{H2}) = 4.94 \times 10^{-3}\ \text{mol}$ ; volume $= 4.94 \times 10^{-3} \times 24\,000 = 119\ \text{cm}^3$ .
$1/t$ : $\dfrac{1}{25} = 0.040\ \text{s}^{-1}$ and $\dfrac{1}{50} = 0.020\ \text{s}^{-1}$ . Doubling the concentration doubles the rate: rate is proportional to concentration.
The powder has a much larger surface area, so more calcium carbonate particles are exposed to the acid; collisions are more frequent, so effective collisions are more frequent and the rate is faster. The volume of gas is the same, because the same mass of calcium carbonate (the limiting reactant) reacts.
Gradient $= \dfrac{50 - 10}{40 - 0} = 1.0\ \text{cm}^3\ \text{s}^{-1}$ .
$n(\ce{O2}) = \dfrac{60}{24\,000} = 2.50 \times 10^{-3}\ \text{mol}$ ; $n(\ce{H2O2}) = 2 \times 2.50 \times 10^{-3} = 5.00 \times 10^{-3}\ \text{mol}$ . Concentration change $= \dfrac{5.00 \times 10^{-3}}{0.0500} = 0.100\ \text{mol dm}^{-3}$ . Rate $= \dfrac{0.100}{30} = 3.3 \times 10^{-3}\ \text{mol dm}^{-3}\ \text{s}^{-1}$ .
$n(\ce{Mg}) = \dfrac{0.10}{24.3} = 4.12 \times 10^{-3}\ \text{mol}$ , which needs $8.23 \times 10^{-3}\ \text{mol}$ HCl. (A) $0.050\ \text{mol}$ HCl and (B) $0.025\ \text{mol}$ : acid in excess, so Mg is limiting: $V = 4.12 \times 10^{-3} \times 24\,000 = 98.8\ \text{cm}^3$ for both. (C) $0.0050\ \text{mol}$ HCl: acid is limiting: $n(\ce{H2}) = 0.0025\ \text{mol}$ , $V = 60\ \text{cm}^3$ . Curves: A is steepest initially (highest concentration, most frequent effective collisions) and levels off first at $98.8\ \text{cm}^3$ ; B starts less steeply and levels off later at the same $98.8\ \text{cm}^3$ ; C is the least steep and levels off at only $60\ \text{cm}^3$ .

Rate of reaction

How rate changes during a reaction

Measuring rates

Collision theory

Effect of concentration

Effect of pressure

Effect of surface area

Worked examples

Practice

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