Chapter 06: Reaction Kinetics
Chapter 06: Reaction Kinetics
A-LEVEL H2 CHEMISTRY
Chapter 6:
Reaction Kinetics
In this topic, we will learn about how reaction rates are investigated, the different factors that affect the rate of reaction as well as the theories behind the rate of reactions.
Here is the list of definitions you should memorise:
- Rate of reaction
- Rate equation
- Rate constant
- Order of reaction/overall order of reaction
- Half life of a reaction
- Catalyst
Reminders and clarification for students:
- The rate constant, k, is only dependent on temperature and activation energy but independent of the concentration of the reactant
- The rate equation, rate constant and order of reaction can only be obtained experimentally and CANNOT be deduced theoretically from the overall stoichiometric equation
- The rate equation may or may not include all the reactants in the stoichiometric equation
Order of reaction
Here is a summary table detailing the various order of reactions that students should familiarise themselves with:
| Order of Rection Zero Order |
|
|---|---|
| Description |
|
| Graph | 
|
| Order of Rection First order |
|
|---|---|
| Description |
|
| Graph | 
|
| Order of Rection Second order |
|
|---|---|
| Description |
|
| Graph |
|
| Order of reaction | Description | Graph |
|---|---|---|
| Zero order |
|
|
| First order |
|
|
| Second order |
|
|
Deducing orders of reaction
Most questions you encounter under this chapter would probably ask you to deduce the orders of reaction in some way or another, by giving you experimental data in different forms.
Half-life method
Questions which want to test you on this method would typically present you with a concentration-time graph or guide you to plotting such a graph from the data given.
Here is an example of such a question,
Worked example 1
Look at the following equation:
2H₂O₂(aq) → 2H₂O(l) + O₂(g)
A solution of dilute hydrogen peroxide, H₂O₂, with an original concentration of 3.0 mol dm⁻³ was put in a bottle contaminated with transition metal ions, which act as catalyst for the decomposition of H₂O₂. The rate of decomposition was measured by withdrawing 10 cm³ of portions at time intervals of 5 minutes and subsequently titrating them with acidified 0.1 mol dm⁻³ of KMnO₄(aq).
The ionic equation for the reaction between H₂O₂ (aq) and KMnO₄ (aq) as follows,
5H₂O₂ + 2MnO₄⁻ + 6 H⁺ → 2Mn²⁺ + 8H₂O + 5O₂
The following results were obtained.
| Time / min | Volume of 0.1 mol dm⁻³ KMnO₄/ cm³ |
|---|---|
| 0 | 30.0 |
| 5 | 23.4 |
| 15 | 14.2 |
| 20 | 11.1 |
| 25 | 8.7 |
| 30 | 6.8 |
Find the order of reaction with respect to H₂O₂. Hence, write an expression for the rate equation.
Step 1: Since the volume of KMnO₄ is proportional to the concentration of H₂O₂, plotting the graph of volume of KMnO₄ in the reaction mixture against the time elapsed will help us find the order of reaction with respect to H₂O₂.
Step 2: Analyse if the graph is linear or not. A linear graph would indicate 0 order with respect to H₂O₂. But since this graph isn’t linear, we can rule that out.
Step 3: Hence, find at least 2 half lives of H₂O₂. Through “working lines” on the graph and labelling the half lives, prove if they are constant or not. If they are constant, order of reaction with respect to H₂O₂ is 1.
From 30cm³ to 15cm³, 1ˢᵗ t₁/₂= 14 min
From 15cm³ to 7.5cm³, 2ⁿᵈ t₁/₂ = 13 min
Since both half-lives are about constant at 13.5 minutes, it is first-order with respect to H₂O₂. Hence,
Rate = k[H₂O₂]
Pseudo-order reactions
You might notice that in the question, one of the reactants is at a much smaller concentration than the other. Typically, when concentration of reactant A is more than 10 times smaller or bigger than the concentration of reactant B, the reaction carried out is a pseudo-order reaction.
For example, looking at this reaction:
CH₃Br + OH⁻ → CH₃OH + Br⁻ , where rate = k[CH₃Br][OH⁻]
Let’s say that the concentration of CH₃Br was 0.0100 mol dm⁻³ while that of OH⁻ was 0.100 mol dm⁻³ initially. Notice that the concentration of CH₃Br is 10 times smaller than that of OH⁻.
When the reaction comes to completion, there will be no more CH₃Br but 0.099 mol dm⁻³ of OH⁻ left. Since there is only a 10% change in the concentration of OH⁻, we can say that its concentration is virtually unchanged.
As a result, OH⁻ can be seen as a constant and the rate law is essentially,
Rate = k₁ [CH₃Br], where k₁ = k[OH⁻]
Thus, this reaction is known as a pseudo first-order reaction, where first-order kinetics are observed because one reactant is in large excess.
Do note that k₁ varies with the concentration of OH⁻, thus it is not the same as k and neither is it a true constant like k.
Make sure you are not substituting the wrong values for k.
Initial rates method
This method might be a little tedious, as it can involve the sketching of tangents and calculating their gradients. The initial rates method involves multiple runs of the same experiment but with varying concentrations. The initial rates of each reaction will then be obtained through either drawing the tangent to the concentration-time graph at time = 0, or calculating the average rate over a relatively short period of time.
Tips for drawing perfect tangents
❏ Ensure the tangent coincides as much as possible with the graph at t=0.
❏ Use a long ruler to draw the tangent. Short rulers are harder to manage when drawing long lines.
❏ Extrapolate both ends of the tangent to meet the axis of the graph.
❏ When calculating the gradient, try to use coordinates as far as possible from each other. A good way to do this is to use the coordinates where the tangent meets the axis, as they are usually very far from each other and easier to read.
Factors affecting the rate of reaction
There are several factors affecting the rate of reaction. It is crucial that students familiarise themselves with these factors especially the effect of temperature as well as the effect of a catalyst. Questions covering the factors that affect rate of reaction will be standard qualitative questions that can be easily answered through identifying the right factor and stating the explanation behind each factor. This table would summarise what one has to know for the following section:
Effect of particle size:
|
Effect of concentration (for gas - pressure) :
|
Effect of temperature:
|
Effect of a catalyst:
|
When dealing with factors affecting rate of reaction, catalyst is one of the most commonly tested factors. Students are expected to know the definition of a catalyst as well as the features of a catalyst. Below is an energy profile diagram comparing a catalysed and uncatalysed reaction. This diagram should be memorised by students as it is commonly tested in the examinations.
There are mainly 4 different types of catalyst namely heterogeneous catalyst, homogeneous catalyst, autocatalyst and biological catalyst (enzymes). Examinations frequently ask students to identify the type of catalyst and describe the mechanism/workings of each of the catalyst hence it is crucial that students familiarise themselves with the explanations for each type of catalyst.
1. Heterogeneous catalyst - reactants and catalyst in DIFFERENT physical state
The table below shows some examples of heterogeneous catalyst :
| Example Haber process |
|
|---|---|
| Reaction | N₂(g) + 3H₂(g) → 2NH₃(g) |
| Catalys | Fe or Fe₂O₃ |
| Example Use of catalytic converters to remove oxides of nitrogen in the exhaust gases from car engine |
|
|---|---|
| Reaction |
2NO(g) → N(₂g) + O₂(g)
2NO₂(g) → N₂(g) + 2O₂(g) |
| Catalys | Rh (s) & Pt (s) |
| Example Hydrogenation of ethene |
|
|---|---|
| Reaction | C₂H₄(g) +H₂(g) → C₂H₆(g) |
| Catalys | Pd, Pt or Ni |
| Example | Reaction | Catalys |
|---|---|---|
| Haber process | N₂(g) + 3H₂(g) → 2NH₃(g) | Fe or Fe₂O₃ |
| Use of catalytic converters to remove oxides of nitrogen in the exhaust gases from car engine |
2NO(g) → N₂(g) + O₂(g)
2NO₂(g) → N₂(g) + 2O₂(g) |
Rh (s) & Pt (s) |
| Hydrogenation of ethene | C₂H₄(g) +H₂(g) → C₂H₆(g) | Pd, Pt or Ni |
Let’s look at the Haber process to see how a heterogeneous catalyst works.
2. Homogeneous catalyst - reactants and catalyst in the SAME physical state
- Take part in reaction process by being converted into an intermediate and the product is subsequently formed with the regeneration of the catalyst.
Worked example 3
Describe the reaction process between peroxodisulfate ions S₂O₈²⁻ and iodide ions, I⁻ and explain why the use of a catalyst is required.
Solution:
In the presence of the catalyst Fe²⁺, the reaction proceeds via a two-step mechanism, which involves the approach of oppositely-charged species.
Step 1 : 2Fe²⁺(aq) + S₂O₈²⁻(aq) → 2Fe³⁺(aq) + 2SO₄²⁻(aq)
Step 2 : 2Fe³⁺(aq) + 2I⁻(aq) → 2Fe²⁺(aq) + I₂(aq)
Overall equation : 2I⁻(aq) + S₂O₈²⁻(aq) → I₂(aq) + 2SO₄²⁻(aq)
A catalyst is required as the uncatalysed reaction is very slow due to the involvement of 2 negatively charged ions. The repulsion between the 2 negatively charged ions would cause the reaction to have high activation energy hence with the catalyst, the approach of oppositely charged species would lower the activation energy and increase the rate of reaction.
3. Autocatalyst - product of a reaction acting as its own catalyst
- In such reactions, initial reaction is slow however rate begins to increase with the formation of the product/autocatalyst
- Rate will eventually reach a maximum as more of the product/autocatalyst is formed
There are mainly two graphs that students should take note of:
One typical example of an autocatalytic reaction is the reaction between MnO₄⁻ and C₂O₄²⁻ in acidic medium. Students should take note of this particular reaction, familiarising themselves with the features behind such reactions as examinations might test reactions similar to the such and students would be expected to apply concepts learnt into a novel situation.
| Example |
|---|
| Reaction between MnO₄⁻ and C₂O₄²⁻ in acidic medium
2MnO₄⁻ (aq) + 5C₂O₄²⁻ (aq) + 16H⁺ (aq) → 2 Mn²⁺ (aq) + 10CO₂ (g) + 8H₂O (l) |
| Explanation with respect to above graph |
|
| Example | Explanation with respect to above graph |
|---|---|
|
Reaction between MnO₄⁻ and C₂O₄²⁻ in acidic medium
2MnO₄⁻ (aq) + 5C₂O₄²⁻ (aq) + 16H⁺ (aq) → 2 Mn²⁺ (aq) + 10CO₂ (g) + 8H₂O (l) |
|
4. Biological catalyst (enzymes) - proteins which catalyse SPECIFIC chemical reactions
(they are neither homogeneous nor heterogeneous in nature, but are colloidal)
- Certain important properties of enzymes include selectivity in reactions and high sensitivity to temperature and pH changes
Students will be expected to know how to draw the enzyme catalysed reaction graph which is as such:
Explanation of graph :
There are 2 sections of the graph, mainly at high concentration of substrate and low concentration of substrate.
At low [substrate], the active sites are not fully filled and the rate is directly proportional to [substrate]. The reaction is hence first order with respect to [substrate]. However at high [substrate], the active sites are fully filled and any increase in substrate concentration will have no effect on the reaction rate thus the reaction is zero order with respect to [substrate].
Continue Reading on Chapter 8: Chemical Equilibria