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# Chapter 06: Reaction Kinetics

Chapter 06: Reaction Kinetics 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.

Hence, exercise caution when attempting to apply the equation, t₁/₂ =
ₗₙ₂
.
In this example, t₁/₂ =
ₗₙ₂
=
ₗₙ₂
.

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: Essentially refers to surface area An increase in the surface area of a reactant will increase rate of reaction due to an increase in the frequency of effective collisions between reactant molecules Effect of concentration (for gas - pressure) : When concentration/pressure increases, the particles become closer to eachother Number of collisions increases, causing frequency of effective collisions to increase consequently hence rate of reaction will increase Note: effect of pressure is only applicable to gases as solids and liquids are not compressible thus pressure has little effect on their rates of reaction Effect of temperature: When temperature increases, molecules gain kinetic energy and move faster This increases the number of molecules having energy greater than or equal to the activation energy Hence, the frequency of effective collisions increases, increasing rate of reaction as a result Boltzman distribution curve (must know how to draw): Effect of a catalyst: A catalyst increases the rate of reaction by providing an alternative reaction pathway that requires a lower activation energy This increases the number of reactant molecules having energy greater than or equal to the activation energy Hence frequency of effective collisions increases causing rate of reaction to increase Boltzman distribution curve: 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
• The product, manganese(II) ion Mn²⁺ is an autocatalyst
• At the start, the reaction starts slowly because no Mn²⁺ ion is produced yet
• As the reaction progresses, Mn²⁺ ions are formed and the rate increases
• As the reaction nears completion, rate decreases as the reactants are gradually used up
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)
• The product, manganese(II) ion Mn²⁺ is an autocatalyst
• At the start, the reaction starts slowly because no Mn²⁺ ion is produced yet
• As the reaction progresses, Mn²⁺ ions are formed and the rate increases
• As the reaction nears completion, rate decreases as the reactants are gradually used up

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 