Chemical kinetics

What you'll learn

How reaction rates are measured, especially by sampling and quenching.
How experimental data is used to find orders of reaction and build a rate equation.
How kinetics gives evidence for a reaction mechanism and the rate-determining step.
How the Arrhenius equation links temperature, catalysts, activation energy and the rate constant.

1. Reaction rate: the basic idea

Chemical kinetics is the study of how fast reactions happen and what this tells us about the pathway from reactants to products.

Definition

Reaction rate

The rate of reaction is the change in concentration of a reactant or product per unit time. It is usually measured in mol dm⁻³ s⁻¹.

For a product, concentration increases, so the gradient of a concentration-time graph is positive. For a reactant, concentration decreases, so the gradient is negative; the rate is usually quoted as a positive value.

For a reaction such as:

\to cC+dD

the rate can be written as:

rate=−1aΔ[A]Δt=−1bΔ[B]Δt=1cΔ[C]Δt=1dΔ[D]Δt\text{rate}=-\frac{1}{a}\frac{\Delta[A]}{\Delta t}=-\frac{1}{b}\frac{\Delta[B]}{\Delta t}=\frac{1}{c}\frac{\Delta[C]}{\Delta t}=\frac{1}{d}\frac{\Delta[D]}{\Delta t}

The gradient of a tangent to a concentration-time curve gives the instantaneous rate at that time.

Example

Calculating an average rate

In a reaction, the concentration of iodine increases from 0.0000 mol dm⁻³ to $1.20 \times 10^{-3}$ mol dm⁻³ in the first 60.0 s.

Use the product concentration because iodine is increasing, so the rate is positive.
Calculate the concentration change: $dm−3\Delta[\text{I}_2]=1.20 \times 10^{-3}\ \text{mol dm}^{-3}$ .
Divide by the time interval: $s−1\text{rate}=\frac{1.20 \times 10^{-3}}{60.0}=2.00 \times 10^{-5}\ \text{mol dm}^{-3}\text{ s}^{-1}$ .

2. Measuring rate by sampling and quenching

Sometimes a reaction mixture must be analysed at different times. You remove small samples at known times, then stop the reaction in each sample before analysis.

Definition

Quenching

Quenching means stopping or greatly slowing a reaction at a known time so that the composition of the sample no longer changes before it is analysed.

A typical sampling and quenching method is:

Start the reaction and keep the temperature constant.
Remove equal-volume samples at recorded times.
Quench each sample immediately.
Analyse the sample, often by titration or colorimetry.
Use the concentrations found to plot a concentration-time graph.

Common quenching methods include rapid cooling, large dilution, neutralising an acid catalyst, or adding a reagent that removes one reactant.

Common Mistake

Analysing an unquenched sample

If the reaction continues while you are titrating or measuring the sample, the result no longer represents the concentration at the sampling time. The quench must be fast compared with the reaction.

3. Orders of reaction and the rate equation

The order with respect to a reactant tells you how the rate depends on the concentration of that reactant.

Definition

Order of reaction

The order with respect to a reactant is the power to which its concentration is raised in the experimentally determined rate equation. The overall order is the sum of these powers.

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

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

where $k$ is the rate constant, $m$ is the order with respect to A, and $n$ is the order with respect to B.

Zero order in A: changing $[A]$ has no effect on rate.
First order in A: doubling $[A]$ doubles the rate.
Second order in A: doubling $[A]$ makes the rate four times larger.

The shapes of rate graphs are a very useful way to recognise order.

Graphs showing concentration-time and rate-concentration patterns for zero, first and second order reactions

Key Idea

Orders come from experiments

Reaction orders are not normally found from the balanced equation. They are determined from experimental rate data.

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

Overall order 0: mol dm⁻³ s⁻¹
Overall order 1: s⁻¹
Overall order 2: dm³ mol⁻¹ s⁻¹
Overall order 3: dm⁶ mol⁻² s⁻¹

Example

Finding a rate equation from initial rates

The oxidation of iodide ions by hydrogen peroxide in acid can be studied using initial-rate data.

Experiment	Hydrogen peroxide concentration / mol dm⁻³	Iodide ion concentration / mol dm⁻³	Hydrogen ion concentration / mol dm⁻³	Initial rate / mol dm⁻³ s⁻¹
1	0.0100	0.100	0.100	2.50e-5
2	0.0200	0.100	0.100	5.00e-5
3	0.0100	0.200	0.100	5.00e-5
4	0.0100	0.100	0.200	5.00e-5

Compare experiments 1 and 2: $[H2O2][\text{H}_2\text{O}_2]$ doubles while the other concentrations stay constant; the rate doubles, so the reaction is first order in hydrogen peroxide.
Compare experiments 1 and 3: $[I−][\text{I}^-]$ doubles and the rate doubles, so the reaction is first order in iodide ions.
Compare experiments 1 and 4: $[H+][\text{H}^+]$ doubles and the rate doubles, so the reaction is first order in hydrogen ions.
Write the rate equation: $rate=k[H2O2][I−][H+]\text{rate}=k[\text{H}_2\text{O}_2][\text{I}^-][\text{H}^+]$ .
Use experiment 1 to find $k$ : $s−1k=\frac{2.50 \times 10^{-5}}{0.0100 \times 0.100 \times 0.100}=0.250\ \text{dm}^6\text{ mol}^{-2}\text{ s}^{-1}$ .

4. Specified practical: iodide and hydrogen peroxide clock reaction

The overall reaction is:

H2O2(aq)+2I−(aq)+2H+(aq)→I2(aq)+2H2O(l)\text{H}_2\text{O}_2(aq)+2\text{I}^-(aq)+2\text{H}^+(aq)\to \text{I}_2(aq)+2\text{H}_2\text{O}(l)

A common practical method uses sodium thiosulfate and starch. Iodine is produced slowly, but thiosulfate removes it:

I2(aq)+2S2O32−(aq)→2I−(aq)+S4O62−(aq)\text{I}_2(aq)+2\text{S}_2\text{O}_3^{2-}(aq)\to 2\text{I}^-(aq)+\text{S}_4\text{O}_6^{2-}(aq)

When the thiosulfate has been used up, iodine remains and forms a blue-black complex with starch. If the same small amount of thiosulfate is used each time, the same amount of iodine is needed to reach the endpoint, so the initial rate is proportional to $1/ t$ .

To determine order:

Keep temperature constant using a water bath.
Keep total volume constant by replacing solution with water when changing a reactant concentration.
Change only one reactant concentration at a time.
Measure the time for the blue-black colour to appear.
Use $1/ t$ as a value proportional to the initial rate.

Tip

Why use initial rates?

At the start of a reaction, concentrations are known accurately. Later on, several concentrations have changed, making the rate harder to interpret.

Important evaluation points:

Very short times have high percentage timing uncertainty.
Very long times may mean conditions drift, especially temperature.
The colour change is subjective; a colorimeter can reduce this uncertainty.
Hydrogen peroxide is an oxidising agent, acids are irritants, and iodine stains, so eye protection is needed.

5. Rate-determining step and mechanism

A mechanism is a sequence of elementary steps showing how reactants become products. An intermediate is made in one step and used up in a later step.

Definition

Rate-determining step

The rate-determining step is the slowest step in a reaction mechanism. It controls the overall rate, like the narrowest point in a funnel controls the flow.

For an elementary step, the rate depends on the particles involved in that step. Therefore, the experimental rate equation gives evidence about which species are involved before or during the rate-determining step.

Example

Using kinetics to test a mechanism

For the reaction $NO2(g)+CO(g)→NO(g)+CO2(g)\text{NO}_2(g)+\text{CO}(g)\to \text{NO}(g)+\text{CO}_2(g)$ , a proposed mechanism is:

Slow: $NO2(g)+NO2(g)→NO3(g)+NO(g)\text{NO}_2(g)+\text{NO}_2(g)\to \text{NO}_3(g)+\text{NO}(g)$
Fast: $NO3(g)+CO(g)→NO2(g)+CO2(g)\text{NO}_3(g)+\text{CO}(g)\to \text{NO}_2(g)+\text{CO}_2(g)$

Add the two steps and cancel the intermediate $NO3\text{NO}_3$ and the regenerated $NO2\text{NO}_2$ ; the overall equation is $NO2+CO→NO+CO2\text{NO}_2+\text{CO}\to \text{NO}+\text{CO}_2$ .
Use the slow step to predict the rate equation: two $NO2\text{NO}_2$ particles are involved, so $rate=k[NO2]2\text{rate}=k[\text{NO}_2]^2$ .
Compare with experiment: if the measured rate equation is $rate=k[NO2]2\text{rate}=k[\text{NO}_2]^2$ , the mechanism is consistent with the kinetic evidence.

Common Mistake

Kinetics supports, but does not prove

A rate equation can rule out an impossible mechanism, but it usually cannot prove that one mechanism is the only possible pathway.

6. The Arrhenius equation

Particles must collide with enough energy to react. The minimum energy needed is the activation energy, $E_a$ .

Definition

Arrhenius equation

The Arrhenius equation links the rate constant to temperature:

k=Ae^{-E_a/(RT)}

where $A$ is the frequency factor, $E_a$ is activation energy in J mol⁻¹, $R$ is 8.31 J K⁻¹ mol⁻¹, and $T$ is temperature in K.

Taking natural logarithms gives the straight-line form:

ln⁡k=ln⁡A−EaR⋅1T\ln k=\ln A-\frac{E_a}{R}\cdot\frac{1}{T}

So a graph of $ln⁡k\ln k$ against $1/ T$ has gradient $−EaR-\frac{E_a}{R}$ and intercept $ln⁡A\ln A$ .

A catalyst provides an alternative pathway with lower activation energy. This increases $k$ , but it does not change the enthalpy change or the equilibrium constant.

Arrhenius plot showing gradient and intercept, alongside an energy profile comparing catalysed and uncatalysed pathways

Example

Finding activation energy and frequency factor

For a first-order reaction, $\times 10^{-3}\ \text{s}^{-1}$ at 298 K and $\times 10^{-3}\ \text{s}^{-1}$ at 318 K.

Use the two-temperature Arrhenius form: $ln⁡(k2k1)=EaR(1T1−1T2)\ln\left(\frac{k_2}{k_1}\right)=\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)$ .
Substitute the values: $ln⁡(8.90×10−31.20×10−3)=2.00\ln\left(\frac{8.90 \times 10^{-3}}{1.20 \times 10^{-3}}\right)=2.00$ and $K−1\frac{1}{298}-\frac{1}{318}=2.11 \times 10^{-4}\ \text{K}^{-1}$ .
Rearrange for activation energy: $mol−1E_a=\frac{2.00 \times 8.31}{2.11 \times 10^{-4}}=7.89 \times 10^4\ \text{J mol}^{-1}=78.9\ \text{kJ mol}^{-1}$ .
Find $A$ using $ln⁡A=ln⁡k1+EaRT1\ln A=\ln k_1+\frac{E_a}{RT_1}$ : $ln⁡A=ln⁡(1.20×10−3)+7.89×1048.31×298=25.1\ln A=\ln(1.20 \times 10^{-3})+\frac{7.89 \times 10^4}{8.31 \times 298}=25.1$ .
Convert from $ln⁡A\ln A$ to $A$ : $s−1A=e^{25.1}=8.0 \times 10^{10}\ \text{s}^{-1}$ .

Common Mistake

Mixing J and kJ in Arrhenius calculations

Use $E_a$ in J mol⁻¹ when using $mol−1R=8.31\ \text{J K}^{-1}\text{ mol}^{-1}$ . Convert to kJ mol⁻¹ only at the end if needed.

Exam technique

In the exam

For orders, compare experiments where only one concentration changes; use the rate ratio and concentration ratio, not the balanced equation.
For graphs, remember that rate from a concentration-time curve is the gradient of a tangent, while an Arrhenius plot has gradient $−EaR-\frac{E_a}{R}$ .
For mechanisms, check that the steps add to the overall equation and that the slow step matches the experimental rate equation.

Self review

Check yourself

Why must a sampled reaction mixture be quenched before analysis?
If doubling $[A]$ quadruples the rate, what is the order with respect to A?
How does a catalyst affect $E_a$ , $k$ , and the position of equilibrium?

What you'll learn

How reaction rates are measured, especially by sampling and quenching.
How experimental data is used to find orders of reaction and build a rate equation.
How kinetics gives evidence for a reaction mechanism and the rate-determining step.
How the Arrhenius equation links temperature, catalysts, activation energy and the rate constant.

1. Reaction rate: the basic idea

Chemical kinetics is the study of how fast reactions happen and what this tells us about the pathway from reactants to products.

Definition

Reaction rate

The rate of reaction is the change in concentration of a reactant or product per unit time. It is usually measured in mol dm⁻³ s⁻¹.

For a reaction such as:

\to cC+dD

the rate can be written as:

rate=−1aΔ[A]Δt=−1bΔ[B]Δt=1cΔ[C]Δt=1dΔ[D]Δt\text{rate}=-\frac{1}{a}\frac{\Delta[A]}{\Delta t}=-\frac{1}{b}\frac{\Delta[B]}{\Delta t}=\frac{1}{c}\frac{\Delta[C]}{\Delta t}=\frac{1}{d}\frac{\Delta[D]}{\Delta t}

The gradient of a tangent to a concentration-time curve gives the instantaneous rate at that time.

Example

Calculating an average rate

In a reaction, the concentration of iodine increases from 0.0000 mol dm⁻³ to $1.20 \times 10^{-3}$ mol dm⁻³ in the first 60.0 s.

Use the product concentration because iodine is increasing, so the rate is positive.
Calculate the concentration change: $dm−3\Delta[\text{I}_2]=1.20 \times 10^{-3}\ \text{mol dm}^{-3}$ .
Divide by the time interval: $s−1\text{rate}=\frac{1.20 \times 10^{-3}}{60.0}=2.00 \times 10^{-5}\ \text{mol dm}^{-3}\text{ s}^{-1}$ .

2. Measuring rate by sampling and quenching

Sometimes a reaction mixture must be analysed at different times. You remove small samples at known times, then stop the reaction in each sample before analysis.

Definition

Quenching

Quenching means stopping or greatly slowing a reaction at a known time so that the composition of the sample no longer changes before it is analysed.

A typical sampling and quenching method is:

Start the reaction and keep the temperature constant.
Remove equal-volume samples at recorded times.
Quench each sample immediately.
Analyse the sample, often by titration or colorimetry.
Use the concentrations found to plot a concentration-time graph.

Common quenching methods include rapid cooling, large dilution, neutralising an acid catalyst, or adding a reagent that removes one reactant.

Common Mistake

Analysing an unquenched sample

If the reaction continues while you are titrating or measuring the sample, the result no longer represents the concentration at the sampling time. The quench must be fast compared with the reaction.

3. Orders of reaction and the rate equation

The order with respect to a reactant tells you how the rate depends on the concentration of that reactant.

Definition

Order of reaction

The order with respect to a reactant is the power to which its concentration is raised in the experimentally determined rate equation. The overall order is the sum of these powers.

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

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

where $k$ is the rate constant, $m$ is the order with respect to A, and $n$ is the order with respect to B.

Zero order in A: changing $[A]$ has no effect on rate.
First order in A: doubling $[A]$ doubles the rate.
Second order in A: doubling $[A]$ makes the rate four times larger.

The shapes of rate graphs are a very useful way to recognise order.

Graphs showing concentration-time and rate-concentration patterns for zero, first and second order reactions

Key Idea

Orders come from experiments

Reaction orders are not normally found from the balanced equation. They are determined from experimental rate data.

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

Overall order 0: mol dm⁻³ s⁻¹
Overall order 1: s⁻¹
Overall order 2: dm³ mol⁻¹ s⁻¹
Overall order 3: dm⁶ mol⁻² s⁻¹

Example

Finding a rate equation from initial rates

The oxidation of iodide ions by hydrogen peroxide in acid can be studied using initial-rate data.

Experiment	Hydrogen peroxide concentration / mol dm⁻³	Iodide ion concentration / mol dm⁻³	Hydrogen ion concentration / mol dm⁻³	Initial rate / mol dm⁻³ s⁻¹
1	0.0100	0.100	0.100	2.50e-5
2	0.0200	0.100	0.100	5.00e-5
3	0.0100	0.200	0.100	5.00e-5
4	0.0100	0.100	0.200	5.00e-5

Compare experiments 1 and 2: $[H2O2][\text{H}_2\text{O}_2]$ doubles while the other concentrations stay constant; the rate doubles, so the reaction is first order in hydrogen peroxide.
Compare experiments 1 and 3: $[I−][\text{I}^-]$ doubles and the rate doubles, so the reaction is first order in iodide ions.
Compare experiments 1 and 4: $[H+][\text{H}^+]$ doubles and the rate doubles, so the reaction is first order in hydrogen ions.
Write the rate equation: $rate=k[H2O2][I−][H+]\text{rate}=k[\text{H}_2\text{O}_2][\text{I}^-][\text{H}^+]$ .
Use experiment 1 to find $k$ : $s−1k=\frac{2.50 \times 10^{-5}}{0.0100 \times 0.100 \times 0.100}=0.250\ \text{dm}^6\text{ mol}^{-2}\text{ s}^{-1}$ .

4. Specified practical: iodide and hydrogen peroxide clock reaction

The overall reaction is:

H2O2(aq)+2I−(aq)+2H+(aq)→I2(aq)+2H2O(l)\text{H}_2\text{O}_2(aq)+2\text{I}^-(aq)+2\text{H}^+(aq)\to \text{I}_2(aq)+2\text{H}_2\text{O}(l)

A common practical method uses sodium thiosulfate and starch. Iodine is produced slowly, but thiosulfate removes it:

I2(aq)+2S2O32−(aq)→2I−(aq)+S4O62−(aq)\text{I}_2(aq)+2\text{S}_2\text{O}_3^{2-}(aq)\to 2\text{I}^-(aq)+\text{S}_4\text{O}_6^{2-}(aq)

To determine order:

Keep temperature constant using a water bath.
Keep total volume constant by replacing solution with water when changing a reactant concentration.
Change only one reactant concentration at a time.
Measure the time for the blue-black colour to appear.
Use $1/ t$ as a value proportional to the initial rate.

Tip

Why use initial rates?

At the start of a reaction, concentrations are known accurately. Later on, several concentrations have changed, making the rate harder to interpret.

Important evaluation points:

Very short times have high percentage timing uncertainty.
Very long times may mean conditions drift, especially temperature.
The colour change is subjective; a colorimeter can reduce this uncertainty.
Hydrogen peroxide is an oxidising agent, acids are irritants, and iodine stains, so eye protection is needed.

5. Rate-determining step and mechanism

A mechanism is a sequence of elementary steps showing how reactants become products. An intermediate is made in one step and used up in a later step.

Definition

Rate-determining step

The rate-determining step is the slowest step in a reaction mechanism. It controls the overall rate, like the narrowest point in a funnel controls the flow.

Example

Using kinetics to test a mechanism

For the reaction $NO2(g)+CO(g)→NO(g)+CO2(g)\text{NO}_2(g)+\text{CO}(g)\to \text{NO}(g)+\text{CO}_2(g)$ , a proposed mechanism is:

Slow: $NO2(g)+NO2(g)→NO3(g)+NO(g)\text{NO}_2(g)+\text{NO}_2(g)\to \text{NO}_3(g)+\text{NO}(g)$
Fast: $NO3(g)+CO(g)→NO2(g)+CO2(g)\text{NO}_3(g)+\text{CO}(g)\to \text{NO}_2(g)+\text{CO}_2(g)$

Add the two steps and cancel the intermediate $NO3\text{NO}_3$ and the regenerated $NO2\text{NO}_2$ ; the overall equation is $NO2+CO→NO+CO2\text{NO}_2+\text{CO}\to \text{NO}+\text{CO}_2$ .
Use the slow step to predict the rate equation: two $NO2\text{NO}_2$ particles are involved, so $rate=k[NO2]2\text{rate}=k[\text{NO}_2]^2$ .
Compare with experiment: if the measured rate equation is $rate=k[NO2]2\text{rate}=k[\text{NO}_2]^2$ , the mechanism is consistent with the kinetic evidence.

Common Mistake

Kinetics supports, but does not prove

A rate equation can rule out an impossible mechanism, but it usually cannot prove that one mechanism is the only possible pathway.

6. The Arrhenius equation

Particles must collide with enough energy to react. The minimum energy needed is the activation energy, $E_a$ .

Definition

Arrhenius equation

The Arrhenius equation links the rate constant to temperature:

k=Ae^{-E_a/(RT)}

where $A$ is the frequency factor, $E_a$ is activation energy in J mol⁻¹, $R$ is 8.31 J K⁻¹ mol⁻¹, and $T$ is temperature in K.

Taking natural logarithms gives the straight-line form:

ln⁡k=ln⁡A−EaR⋅1T\ln k=\ln A-\frac{E_a}{R}\cdot\frac{1}{T}

So a graph of $ln⁡k\ln k$ against $1/ T$ has gradient $−EaR-\frac{E_a}{R}$ and intercept $ln⁡A\ln A$ .

A catalyst provides an alternative pathway with lower activation energy. This increases $k$ , but it does not change the enthalpy change or the equilibrium constant.

Arrhenius plot showing gradient and intercept, alongside an energy profile comparing catalysed and uncatalysed pathways

Example

Finding activation energy and frequency factor

For a first-order reaction, $\times 10^{-3}\ \text{s}^{-1}$ at 298 K and $\times 10^{-3}\ \text{s}^{-1}$ at 318 K.

Use the two-temperature Arrhenius form: $ln⁡(k2k1)=EaR(1T1−1T2)\ln\left(\frac{k_2}{k_1}\right)=\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)$ .
Substitute the values: $ln⁡(8.90×10−31.20×10−3)=2.00\ln\left(\frac{8.90 \times 10^{-3}}{1.20 \times 10^{-3}}\right)=2.00$ and $K−1\frac{1}{298}-\frac{1}{318}=2.11 \times 10^{-4}\ \text{K}^{-1}$ .
Rearrange for activation energy: $mol−1E_a=\frac{2.00 \times 8.31}{2.11 \times 10^{-4}}=7.89 \times 10^4\ \text{J mol}^{-1}=78.9\ \text{kJ mol}^{-1}$ .
Find $A$ using $ln⁡A=ln⁡k1+EaRT1\ln A=\ln k_1+\frac{E_a}{RT_1}$ : $ln⁡A=ln⁡(1.20×10−3)+7.89×1048.31×298=25.1\ln A=\ln(1.20 \times 10^{-3})+\frac{7.89 \times 10^4}{8.31 \times 298}=25.1$ .
Convert from $ln⁡A\ln A$ to $A$ : $s−1A=e^{25.1}=8.0 \times 10^{10}\ \text{s}^{-1}$ .

Common Mistake

Mixing J and kJ in Arrhenius calculations

Use $E_a$ in J mol⁻¹ when using $mol−1R=8.31\ \text{J K}^{-1}\text{ mol}^{-1}$ . Convert to kJ mol⁻¹ only at the end if needed.

Exam technique

In the exam

For orders, compare experiments where only one concentration changes; use the rate ratio and concentration ratio, not the balanced equation.
For graphs, remember that rate from a concentration-time curve is the gradient of a tangent, while an Arrhenius plot has gradient $−EaR-\frac{E_a}{R}$ .
For mechanisms, check that the steps add to the overall equation and that the slow step matches the experimental rate equation.

Self review

Check yourself

Why must a sampled reaction mixture be quenched before analysis?
If doubling $[A]$ quadruples the rate, what is the order with respect to A?
How does a catalyst affect $E_a$ , $k$ , and the position of equilibrium?

What you'll learn

1. Reaction rate: the basic idea

2. Measuring rate by sampling and quenching

3. Orders of reaction and the rate equation

4. Specified practical: iodide and hydrogen peroxide clock reaction

5. Rate-determining step and mechanism

6. The Arrhenius equation

Review Chemical kinetics by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Chemical kinetics

What you'll learn

1. Reaction rate: the basic idea

2. Measuring rate by sampling and quenching

3. Orders of reaction and the rate equation

4. Specified practical: iodide and hydrogen peroxide clock reaction

5. Rate-determining step and mechanism

6. The Arrhenius equation

Review Chemical kinetics by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Chemical kinetics

What you'll learn

1. Reaction rate: the basic idea

2. Measuring rate by sampling and quenching

3. Orders of reaction and the rate equation

4. Specified practical: iodide and hydrogen peroxide clock reaction

5. Rate-determining step and mechanism

6. The Arrhenius equation

Review Chemical kinetics by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Chemical kinetics Revision Guide

Chemical kinetics

What you'll learn

1. Reaction rate: the basic idea

2. Measuring rate by sampling and quenching

3. Orders of reaction and the rate equation

4. Specified practical: iodide and hydrogen peroxide clock reaction

5. Rate-determining step and mechanism

6. The Arrhenius equation

Review Chemical kinetics by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Chemical kinetics Revision Guide