Determination of rate equation (A-level only)

What you'll learn

How to obtain reaction rates from concentration–time graphs.
How initial-rate experiments and rate–concentration graphs reveal reaction order.
How to build a rate equation and calculate the rate constant, including its units.
How orders of reaction can give evidence about the rate-determining step in a mechanism.

The big idea: rate equations come from experiments

This is one of the A-Level only kinetics ideas: the balanced chemical equation tells you the overall reacting ratio, but it does not usually tell you how the rate depends on concentration.

Definition

Rate of reaction

The rate of reaction is the change in concentration of a reactant or product per unit time, usually in mol dm⁻³ s⁻¹. For a reactant being used up, the concentration gradient is negative, so the rate is normally quoted as a positive value.

For a reaction involving reactants A and B, a typical rate equation is:

rate=k[A]m[B]n\text{rate}=k[A]^m[B]^n

Square brackets mean concentration in mol dm⁻³.

Definition

Order and rate constant

The order with respect to A is the power $m$ in the rate equation. The order with respect to B is $n$ . The overall order is $m + n$ . The rate constant, $k$ , is the constant of proportionality at a fixed temperature.

Key Idea

Do not use the balanced equation for orders

Orders of reaction are experimentally determined. They are not copied from the stoichiometric coefficients in the overall equation unless the data proves they match.

Example

Predicting a rate change

For the rate equation $rate=k[A][B]2\text{rate}=k[A][B]^2$ , work out what happens to the rate if $[A]$ is doubled and $[B]$ is halved.

Convert the concentration changes into factors: $[A]$ changes by a factor of 2, while $[B]$ changes by a factor of $12\frac{1}{2}$ .
Apply the powers in the rate equation: the rate factor is $\times \left(\frac{1}{2}\right)^2$ .
Simplify the factor: $\times \frac{1}{4}=\frac{1}{2}$ , so the rate halves.

Getting rates from concentration–time graphs

A concentration–time graph shows how the concentration of one substance changes as the reaction proceeds. The rate at any instant is found from the gradient of a tangent to the curve at that time.

The tangent at time zero gives the initial rate. Initial rates are especially useful because the concentrations are still very close to the starting concentrations.

Concentration-time graph showing tangents used to find initial rate and rate at a later time

For a reactant:

graph\text{rate}=-\text{gradient of concentration-time graph}

The minus sign is used because the reactant concentration decreases, giving a negative gradient.

Example

Calculating rate from a tangent

A tangent to a reactant concentration–time curve passes through two convenient points: 0 s, 0.092 mol dm⁻³ and 40 s, 0.052 mol dm⁻³. Calculate the rate at that time.

Calculate the gradient of the tangent: $s−1\text{gradient}=\frac{0.052-0.092}{40-0}=-1.00 \times 10^{-3}\ \text{mol dm}^{-3}\ \text{s}^{-1}$ .
Since the graph is for a reactant being used up, take the negative of the gradient: $rate=−(−1.00×10−3)\text{rate}=-(-1.00 \times 10^{-3})$ .
The rate is $1.00 \times 10^{-3}\ \text{mol dm}^{-3}\ \text{s}^{-1}$ .

Common Mistake

Using a chord instead of a tangent

A chord between two points on the curve gives an average rate over that time interval. An instantaneous rate needs a tangent drawn at the exact time.

Required practical 7: measuring rates

Key Idea

Required practical 7

You need to be able to describe measuring rate by both an initial rate method and a continuous monitoring method, while controlling variables such as temperature and total volume.

Initial rate method

In an initial rate method, you run several experiments with different starting concentrations and measure a rate close to the start of each reaction.

A common approach is a clock reaction, such as an iodine clock. If each experiment is timed until the same visible endpoint, the same amount of product has formed each time. Therefore:

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

where $t$ is the time to reach the fixed endpoint.

Example

Using clock reaction times

In two clock experiments, only the initial concentration of A is changed. In experiment 1, $dm−3[A]=0.100\ \text{mol dm}^{-3}$ and the time is 80.0 s. In experiment 2, $dm−3[A]=0.200\ \text{mol dm}^{-3}$ and the time is 40.0 s. Deduce the order with respect to A.

Compare the initial rates using reciprocal time: $rate2rate1=1/40.01/80.0=2\frac{\text{rate}_2}{\text{rate}_1}=\frac{1/40.0}{1/80.0}=2$ .
Compare the concentrations: $[A]$ doubles from 0.100 mol dm⁻³ to 0.200 mol dm⁻³.
Doubling $[A]$ doubles the rate, so the reaction is first order with respect to A.

Continuous monitoring method

In a continuous monitoring method, one experiment is followed throughout. You might measure gas volume with a gas syringe, mass loss on a balance, absorbance with a colorimeter, pH, or conductivity.

This gives a concentration–time or amount–time graph. You can then draw tangents to find rates at different times.

Tip

Practical control

Keep temperature constant, use the same total volume where appropriate, and repeat measurements. Rate is very temperature-sensitive, so small temperature changes can spoil the pattern.

Deducing order from rate–concentration data

To find the order with respect to one reactant, compare experiments where only that reactant’s concentration changes.

The three common A-Level cases are zero order, first order and second order.

Rate-concentration graphs for zero, first and second order reactions

Zero order: changing the concentration has no effect on rate.
First order: doubling the concentration doubles the rate.
Second order: doubling the concentration quadruples the rate.

You can also recognise them from rate–concentration graphs:

Zero order gives a horizontal line.
First order gives a straight line through the origin.
Second order gives an upward curve; plotting rate against concentration squared would give a straight line.

Example

Deducing orders and calculating k

Initial-rate data are collected for a reaction involving A and B:

Experiment 1: $dm−3[A]=0.100\ \text{mol dm}^{-3}$ , $dm−3[B]=0.100\ \text{mol dm}^{-3}$ , rate $\times 10^{-5}\ \text{mol dm}^{-3}\ \text{s}^{-1}$ .

Experiment 2: $dm−3[A]=0.200\ \text{mol dm}^{-3}$ , $dm−3[B]=0.100\ \text{mol dm}^{-3}$ , rate $\times 10^{-4}\ \text{mol dm}^{-3}\ \text{s}^{-1}$ .

Experiment 3: $dm−3[A]=0.200\ \text{mol dm}^{-3}$ , $dm−3[B]=0.300\ \text{mol dm}^{-3}$ , rate $\times 10^{-4}\ \text{mol dm}^{-3}\ \text{s}^{-1}$ .

Compare experiments 1 and 2: $[A]$ doubles while $[B]$ stays constant, and the rate increases by a factor of 4. Therefore the order with respect to A is 2.
Compare experiments 2 and 3: $[B]$ triples while $[A]$ stays constant, and the rate also triples. Therefore the order with respect to B is 1.
Write the rate equation using these orders: $rate=k[A]2[B]\text{rate}=k[A]^2[B]$ .
Substitute experiment 2 into the rate equation: $1.00 \times 10^{-4}=k(0.200)^2(0.100)$ , so $\times 10^{-2}$ .
Work out the units from $units3\frac{\text{rate units}}{\text{concentration units}^{3}}$ , giving $\times 10^{-2}\ \text{dm}^{6}\ \text{mol}^{-2}\ \text{s}^{-1}$ .

Common Mistake

Changing two concentrations at once

If two reactant concentrations change between experiments, you cannot directly identify the order for one reactant. Choose pairs where all other concentrations are constant.

Units of the rate constant

The units of $k$ depend on the overall order.

order\text{units of }k=\frac{\text{units of rate}}{(\text{units of concentration})^{\text{overall order}}}

Useful results:

Overall order zero: mol dm⁻³ s⁻¹
Overall order one: s⁻¹
Overall order two: dm³ mol⁻¹ s⁻¹
Overall order three: dm⁶ mol⁻² s⁻¹

Special case: zero-order concentration–time graphs

For a zero-order reaction with respect to A:

rate=k[A]0=k\text{rate}=k[A]^0=k

So the rate is constant while the reaction conditions remain suitable. A concentration–time graph for A is a straight line, and for a reactant:

k=−gradientk=-\text{gradient}

Example

Finding k from a zero-order graph

A reactant concentration falls linearly from 0.120 mol dm⁻³ to 0.060 mol dm⁻³ in 300 s. Find $k$ .

Calculate the gradient of the concentration–time graph: $s−1\text{gradient}=\frac{0.060-0.120}{300}=-2.00 \times 10^{-4}\ \text{mol dm}^{-3}\ \text{s}^{-1}$ .
For a reactant, the rate is the negative of the gradient: $s−1\text{rate}=2.00 \times 10^{-4}\ \text{mol dm}^{-3}\ \text{s}^{-1}$ .
Since the reaction is zero order, $rate=k\text{rate}=k$ , so $\times 10^{-4}\ \text{mol dm}^{-3}\ \text{s}^{-1}$ .

Using orders to suggest a mechanism

A mechanism is a sequence of smaller steps that explains how a reaction happens. The rate-determining step is the slowest step; it limits the rate of the overall reaction.

Definition

Rate-determining step

The rate-determining step is the slowest step in a reaction mechanism. Species involved in or before this step can appear in the rate equation.

If a reactant is zero order, changing its concentration does not affect the rate, so it is unlikely to be involved in the rate-determining step under those conditions.

Example

Identifying a possible rate-determining step

For the overall reaction of X, Y and Z, experiments give $rate=k[X][Y]\text{rate}=k[X][Y]$ . Decide which proposed slow step is more consistent.

Mechanism A: X + Y → intermediate is slow, then intermediate + Z → products is fast.

Mechanism B: Y + Z → intermediate is slow, then intermediate + X → products is fast.

Interpret the rate equation: X is first order, Y is first order, and Z is zero order.
Mechanism A has X and Y in the slow step, which matches the species that affect the rate.
Mechanism B has Z in the slow step, but Z does not appear in the rate equation, so mechanism B is less consistent.

Common Mistake

Mechanisms are supported, not proved

A rate equation can rule out unsuitable mechanisms and support a possible one, but it does not prove that one mechanism is uniquely correct.

Exam technique

In the exam

When using data, compare two experiments where only one reactant concentration changes.
Convert concentration changes and rate changes into factors, then match them to zero, first or second order.
Always write the full rate equation before calculating $k$ .
Give units for $k$ , and remember they depend on the overall order.
For mechanisms, compare the species in the rate equation with the proposed rate-determining step.

Self review

Check yourself

If doubling $[A]$ causes no change in rate, what is the order with respect to A?
How would you find an initial rate from a concentration–time graph?
Why can’t you usually deduce the rate equation from the balanced chemical equation?

What you'll learn

How to obtain reaction rates from concentration–time graphs.
How initial-rate experiments and rate–concentration graphs reveal reaction order.
How to build a rate equation and calculate the rate constant, including its units.
How orders of reaction can give evidence about the rate-determining step in a mechanism.

The big idea: rate equations come from experiments

This is one of the A-Level only kinetics ideas: the balanced chemical equation tells you the overall reacting ratio, but it does not usually tell you how the rate depends on concentration.

Definition

Rate of reaction

For a reaction involving reactants A and B, a typical rate equation is:

rate=k[A]m[B]n\text{rate}=k[A]^m[B]^n

Square brackets mean concentration in mol dm⁻³.

Definition

Order and rate constant

Key Idea

Do not use the balanced equation for orders

Orders of reaction are experimentally determined. They are not copied from the stoichiometric coefficients in the overall equation unless the data proves they match.

Example

Predicting a rate change

For the rate equation $rate=k[A][B]2\text{rate}=k[A][B]^2$ , work out what happens to the rate if $[A]$ is doubled and $[B]$ is halved.

Convert the concentration changes into factors: $[A]$ changes by a factor of 2, while $[B]$ changes by a factor of $12\frac{1}{2}$ .
Apply the powers in the rate equation: the rate factor is $\times \left(\frac{1}{2}\right)^2$ .
Simplify the factor: $\times \frac{1}{4}=\frac{1}{2}$ , so the rate halves.

Getting rates from concentration–time graphs

The tangent at time zero gives the initial rate. Initial rates are especially useful because the concentrations are still very close to the starting concentrations.

Concentration-time graph showing tangents used to find initial rate and rate at a later time

For a reactant:

graph\text{rate}=-\text{gradient of concentration-time graph}

The minus sign is used because the reactant concentration decreases, giving a negative gradient.

Example

Calculating rate from a tangent

A tangent to a reactant concentration–time curve passes through two convenient points: 0 s, 0.092 mol dm⁻³ and 40 s, 0.052 mol dm⁻³. Calculate the rate at that time.

Calculate the gradient of the tangent: $s−1\text{gradient}=\frac{0.052-0.092}{40-0}=-1.00 \times 10^{-3}\ \text{mol dm}^{-3}\ \text{s}^{-1}$ .
Since the graph is for a reactant being used up, take the negative of the gradient: $rate=−(−1.00×10−3)\text{rate}=-(-1.00 \times 10^{-3})$ .
The rate is $1.00 \times 10^{-3}\ \text{mol dm}^{-3}\ \text{s}^{-1}$ .

Common Mistake

Using a chord instead of a tangent

A chord between two points on the curve gives an average rate over that time interval. An instantaneous rate needs a tangent drawn at the exact time.

Required practical 7: measuring rates

Key Idea

Required practical 7

You need to be able to describe measuring rate by both an initial rate method and a continuous monitoring method, while controlling variables such as temperature and total volume.

Initial rate method

In an initial rate method, you run several experiments with different starting concentrations and measure a rate close to the start of each reaction.

A common approach is a clock reaction, such as an iodine clock. If each experiment is timed until the same visible endpoint, the same amount of product has formed each time. Therefore:

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

where $t$ is the time to reach the fixed endpoint.

Example

Using clock reaction times

Compare the initial rates using reciprocal time: $rate2rate1=1/40.01/80.0=2\frac{\text{rate}_2}{\text{rate}_1}=\frac{1/40.0}{1/80.0}=2$ .
Compare the concentrations: $[A]$ doubles from 0.100 mol dm⁻³ to 0.200 mol dm⁻³.
Doubling $[A]$ doubles the rate, so the reaction is first order with respect to A.

Continuous monitoring method

This gives a concentration–time or amount–time graph. You can then draw tangents to find rates at different times.

Tip

Practical control

Keep temperature constant, use the same total volume where appropriate, and repeat measurements. Rate is very temperature-sensitive, so small temperature changes can spoil the pattern.

Deducing order from rate–concentration data

To find the order with respect to one reactant, compare experiments where only that reactant’s concentration changes.

The three common A-Level cases are zero order, first order and second order.

Rate-concentration graphs for zero, first and second order reactions

Zero order: changing the concentration has no effect on rate.
First order: doubling the concentration doubles the rate.
Second order: doubling the concentration quadruples the rate.

You can also recognise them from rate–concentration graphs:

Zero order gives a horizontal line.
First order gives a straight line through the origin.
Second order gives an upward curve; plotting rate against concentration squared would give a straight line.

Example

Deducing orders and calculating k

Initial-rate data are collected for a reaction involving A and B:

Experiment 1: $dm−3[A]=0.100\ \text{mol dm}^{-3}$ , $dm−3[B]=0.100\ \text{mol dm}^{-3}$ , rate $\times 10^{-5}\ \text{mol dm}^{-3}\ \text{s}^{-1}$ .

Experiment 2: $dm−3[A]=0.200\ \text{mol dm}^{-3}$ , $dm−3[B]=0.100\ \text{mol dm}^{-3}$ , rate $\times 10^{-4}\ \text{mol dm}^{-3}\ \text{s}^{-1}$ .

Experiment 3: $dm−3[A]=0.200\ \text{mol dm}^{-3}$ , $dm−3[B]=0.300\ \text{mol dm}^{-3}$ , rate $\times 10^{-4}\ \text{mol dm}^{-3}\ \text{s}^{-1}$ .

Compare experiments 1 and 2: $[A]$ doubles while $[B]$ stays constant, and the rate increases by a factor of 4. Therefore the order with respect to A is 2.
Compare experiments 2 and 3: $[B]$ triples while $[A]$ stays constant, and the rate also triples. Therefore the order with respect to B is 1.
Write the rate equation using these orders: $rate=k[A]2[B]\text{rate}=k[A]^2[B]$ .
Substitute experiment 2 into the rate equation: $1.00 \times 10^{-4}=k(0.200)^2(0.100)$ , so $\times 10^{-2}$ .
Work out the units from $units3\frac{\text{rate units}}{\text{concentration units}^{3}}$ , giving $\times 10^{-2}\ \text{dm}^{6}\ \text{mol}^{-2}\ \text{s}^{-1}$ .

Common Mistake

Changing two concentrations at once

If two reactant concentrations change between experiments, you cannot directly identify the order for one reactant. Choose pairs where all other concentrations are constant.

Units of the rate constant

The units of $k$ depend on the overall order.

order\text{units of }k=\frac{\text{units of rate}}{(\text{units of concentration})^{\text{overall order}}}

Useful results:

Overall order zero: mol dm⁻³ s⁻¹
Overall order one: s⁻¹
Overall order two: dm³ mol⁻¹ s⁻¹
Overall order three: dm⁶ mol⁻² s⁻¹

Special case: zero-order concentration–time graphs

For a zero-order reaction with respect to A:

rate=k[A]0=k\text{rate}=k[A]^0=k

So the rate is constant while the reaction conditions remain suitable. A concentration–time graph for A is a straight line, and for a reactant:

k=−gradientk=-\text{gradient}

Example

Finding k from a zero-order graph

A reactant concentration falls linearly from 0.120 mol dm⁻³ to 0.060 mol dm⁻³ in 300 s. Find $k$ .

Calculate the gradient of the concentration–time graph: $s−1\text{gradient}=\frac{0.060-0.120}{300}=-2.00 \times 10^{-4}\ \text{mol dm}^{-3}\ \text{s}^{-1}$ .
For a reactant, the rate is the negative of the gradient: $s−1\text{rate}=2.00 \times 10^{-4}\ \text{mol dm}^{-3}\ \text{s}^{-1}$ .
Since the reaction is zero order, $rate=k\text{rate}=k$ , so $\times 10^{-4}\ \text{mol dm}^{-3}\ \text{s}^{-1}$ .

Using orders to suggest a mechanism

A mechanism is a sequence of smaller steps that explains how a reaction happens. The rate-determining step is the slowest step; it limits the rate of the overall reaction.

Definition

Rate-determining step

The rate-determining step is the slowest step in a reaction mechanism. Species involved in or before this step can appear in the rate equation.

If a reactant is zero order, changing its concentration does not affect the rate, so it is unlikely to be involved in the rate-determining step under those conditions.

Example

Identifying a possible rate-determining step

For the overall reaction of X, Y and Z, experiments give $rate=k[X][Y]\text{rate}=k[X][Y]$ . Decide which proposed slow step is more consistent.

Mechanism A: X + Y → intermediate is slow, then intermediate + Z → products is fast.

Mechanism B: Y + Z → intermediate is slow, then intermediate + X → products is fast.

Interpret the rate equation: X is first order, Y is first order, and Z is zero order.
Mechanism A has X and Y in the slow step, which matches the species that affect the rate.
Mechanism B has Z in the slow step, but Z does not appear in the rate equation, so mechanism B is less consistent.

Common Mistake

Mechanisms are supported, not proved

A rate equation can rule out unsuitable mechanisms and support a possible one, but it does not prove that one mechanism is uniquely correct.

Exam technique

In the exam

When using data, compare two experiments where only one reactant concentration changes.
Convert concentration changes and rate changes into factors, then match them to zero, first or second order.
Always write the full rate equation before calculating $k$ .
Give units for $k$ , and remember they depend on the overall order.
For mechanisms, compare the species in the rate equation with the proposed rate-determining step.

Self review

Check yourself

If doubling $[A]$ causes no change in rate, what is the order with respect to A?
How would you find an initial rate from a concentration–time graph?
Why can’t you usually deduce the rate equation from the balanced chemical equation?

What you'll learn

The big idea: rate equations come from experiments

Getting rates from concentration–time graphs

Required practical 7: measuring rates

Initial rate method

Continuous monitoring method

Deducing order from rate–concentration data

Units of the rate constant

Special case: zero-order concentration–time graphs

Using orders to suggest a mechanism

Review Determination of rate equation (A-level only) by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Determination of rate equation (A-level only)

What you'll learn

The big idea: rate equations come from experiments

Getting rates from concentration–time graphs

Required practical 7: measuring rates

Initial rate method

Continuous monitoring method

Deducing order from rate–concentration data

Units of the rate constant

Special case: zero-order concentration–time graphs

Using orders to suggest a mechanism

Review Determination of rate equation (A-level only) by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Determination of rate equation (A-level only)

What you'll learn

The big idea: rate equations come from experiments

Getting rates from concentration–time graphs

Required practical 7: measuring rates

Initial rate method

Continuous monitoring method

Deducing order from rate–concentration data

Units of the rate constant

Special case: zero-order concentration–time graphs

Using orders to suggest a mechanism

Review Determination of rate equation (A-level only) by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Determination of rate equation (A-level only) Revision Guide

Determination of rate equation (A-level only)

What you'll learn

The big idea: rate equations come from experiments

Getting rates from concentration–time graphs

Required practical 7: measuring rates

Initial rate method

Continuous monitoring method

Deducing order from rate–concentration data

Units of the rate constant

Special case: zero-order concentration–time graphs

Using orders to suggest a mechanism

Review Determination of rate equation (A-level only) by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Determination of rate equation (A-level only) Revision Guide