Reaction Kinetics: How Fast is Fast? (9701 A Level Chemistry Notes)
Hello future Chemists! Welcome to Reaction Kinetics, the chapter where we stop asking "will this reaction happen?" (Thermodynamics/Energetics) and start asking "how fast will this reaction happen?". This is a crucial topic that links fundamental concepts like energy and molecular structure to real-world industrial processes, like making ammonia or reducing car pollution.
Don't worry if the math looks complex initially. We will break down the rate laws and calculations step-by-step. Let’s dive into what makes reactions race or crawl!
Section 1: Defining Rate and Collision Theory (AS Foundation)
1.1 What is Rate of Reaction?
The rate of reaction is simply the change in concentration of a reactant or product per unit time.
Mathematically:
Rate \( = \frac{\text{Change in concentration}}{\text{Time taken}} \)
Units for rate are usually \(\text{mol dm}^{-3}\text{ s}^{-1}\) (or sometimes \(\text{mol dm}^{-3}\text{ min}^{-1}\)).
Quick Tip: Since reactants are used up, their concentration change is negative, so rate is often expressed as the positive magnitude of this change.
1.2 The Collision Theory – The Dating Game Analogy
For any chemical reaction to occur, particles must collide. However, just colliding isn't enough! Collision theory states that particles must satisfy two conditions to lead to a product:
- Sufficient Energy: The collision must have energy greater than or equal to the activation energy (\(E_a\)).
- Correct Orientation: The particles must collide in the correct geometric orientation.
Collisions that satisfy both criteria are called effective collisions (or successful collisions).
Analogy: Imagine two people (Reactant A and Reactant B) trying to start a relationship (Product).
- They must collide (meet).
- They need sufficient energy (be enthusiastic about dating).
- They need correct orientation (be compatible).
Key Terms Review
- Frequency of collisions: How often particles hit each other.
- Effective collisions: Collisions that lead to bond breaking/forming (i.e., meet $E_a$ and orientation requirements).
- Rate of reaction is directly proportional to the frequency of effective collisions.
1.3 Factors Affecting Reaction Rate (Qualitative)
We can explain the effect of common variables on reaction rate purely by considering how they change the frequency of effective collisions.
| Factor | Effect on Particles | Effect on Rate |
|---|---|---|
| Concentration (for solutions) | More particles per volume means shorter average distance between them. | Increases $E$ rate (more collisions overall). |
| Pressure (for gases) | Reducing the volume pushes gas particles closer together (increasing concentration). | Increases $E$ rate (more collisions overall). |
| Surface Area (for solids) | Breaking a solid into smaller pieces exposes more reacting sites. | Increases $E$ rate (more contact points for collisions). |
Key Takeaway (Section 1): Reaction rate depends on how often molecules collide effectively. Increasing concentration or surface area simply increases the *frequency* of all collisions, thus increasing the frequency of successful ones.
Section 2: Activation Energy and Temperature
2.1 Defining Activation Energy (\(E_a\))
The activation energy, \(\mathbf{E_a}\), is defined as the minimum energy required for a collision to be effective (i.e., to cause a reaction).
Analogy: Think of \(E_a\) as the height of a mountain pass. Reactant molecules need enough kinetic energy to climb over this "energy barrier" to turn into products.
We illustrate \(E_a\) using a Reaction Pathway Diagram (also known as an Enthalpy Profile Diagram):
(Sketching requirement: The diagram shows potential energy (y-axis) vs. reaction pathway (x-axis). Reactants start lower/higher than products depending on $\Delta H$. $E_a$ is the peak height measured from the Reactants level.)
2.2 The Boltzmann Distribution
To understand how temperature affects rate, we use the Boltzmann distribution. This curve shows the distribution of kinetic energies among molecules at a specific temperature.
(Sketching requirement: The curve shows Number of Molecules (y-axis) vs. Kinetic Energy (x-axis). It starts at zero, rises to a peak, and tails off, never hitting zero.)
The key features of the Boltzmann distribution:
- No molecules have zero energy.
- Most molecules have energy near the peak (the mean energy).
- Only a small fraction of molecules have energy greater than \(E_a\).
2.3 The Effect of Temperature on Rate
When you increase the temperature (e.g., from \(T_1\) to \(T_2\), where \(T_2 > T_1\)), two things happen:
- The peak of the Boltzmann curve shifts slightly to the right (molecules move faster, increasing collision frequency).
- Crucially: The area under the curve beyond the $E_a$ line increases significantly.
Explanation:
Raising the temperature increases the mean kinetic energy of the particles. While this slightly increases the total collision frequency, the major effect is the exponential increase in the number of molecules possessing energy equal to or greater than the activation energy (\(E_a\)).
More molecules meeting the energy requirement means a much higher frequency of effective collisions, leading to a much faster rate.
Key Takeaway (Section 2): Temperature is powerful because it drastically increases the *fraction* of molecules that are energetic enough to react (i.e., those exceeding the $E_a$).
Section 3: Catalysis (AS & A Level)
3.1 What is a Catalyst?
A catalyst is a substance that increases the rate of a chemical reaction without being chemically changed itself at the end of the reaction. The process is called catalysis.
How do they work?
A catalyst provides an alternative reaction mechanism (or pathway) that has a lower activation energy (\(E_a\)).
Analogy: If the reaction needed to climb a large mountain ($E_a$), the catalyst builds a tunnel through the mountain, which requires much less energy to get through.
Effect on Boltzmann Distribution:
By lowering \(E_a\), the proportion of molecules that now have sufficient energy to react ($\geq E_{a, \text{catalysed}}$) increases massively (represented by a much larger area under the Boltzmann curve beyond the new, lower \(E_a\)).
Did you know? Catalysts are vital in industry. For example, the Haber process uses iron to make ammonia, significantly speeding up production and saving huge amounts of energy.
Common Mistake Alert: Catalysts do NOT increase the energy of the reactant molecules, nor do they change the enthalpy change ($\Delta H$) of the reaction. They only change the *pathway* (and thus the $E_a$).
Key Takeaway (Section 3): Catalysts speed up reactions by offering a new, easier route (a different mechanism) which requires less minimum energy (lower \(E_a\)).
Section 4: Rate Equations and Orders of Reaction (A Level Focus)
4.1 Rate Equations
For a reaction \(a\text{A} + b\text{B} \rightarrow c\text{C} + d\text{D}\), the relationship between concentration and rate is expressed by the rate equation:
$$ \text{Rate} = k[\text{A}]^m[\text{B}]^n $$
Where:
- \([\text{A}]\) and \([\text{B}]\) are the concentrations of reactants.
- \(k\) is the rate constant.
- \(m\) and \(n\) are the orders of reaction with respect to A and B, respectively.
Important Rule: The orders \(m\) and \(n\) cannot be deduced from the stoichiometric coefficients (\(a\) and \(b\)) unless the reaction is a single-step (elementary) reaction. Orders must be determined experimentally!
4.2 Orders of Reaction
The order of reaction with respect to a specific reactant describes how its concentration affects the reaction rate. Orders are usually 0, 1, or 2.
- Zero Order (\(m = 0\)): Changing \([\text{A}]\) has no effect on the rate. (Rate \(\propto [\text{A}]^0\). Since \([\text{A}]^0 = 1\), the term disappears from the rate equation).
- First Order (\(m = 1\)): Doubling \([\text{A}]\) doubles the rate. (Rate \(\propto [\text{A}]^1\)).
- Second Order (\(m = 2\)): Doubling \([\text{A}]\) increases the rate by a factor of four (\(2^2\)). (Rate \(\propto [\text{A}]^2\)).
The overall order of reaction is the sum of the individual orders: \(\text{Overall Order} = m + n\).
4.3 The Rate Constant (\(k\))
The rate constant, \(\mathbf{k}\), is the proportionality constant in the rate equation.
- It is constant only if temperature remains constant.
- It is unique for a specific reaction at a specific temperature.
- The units of \(k\) depend entirely on the overall order of the reaction.
- Calculating Units of \(k\): Units of \(k = \text{(units of Rate)} / \text{(units of concentration)}^{\text{Overall Order}}\)
Example: If overall order is 2, Units of \(k = (\text{mol dm}^{-3}\text{ s}^{-1}) / (\text{mol dm}^{-3})^2 = \text{dm}^{3}\text{ mol}^{-1}\text{ s}^{-1}\).
4.4 Determining Order Experimentally (Initial Rates Method)
The most common method uses initial rates at different initial concentrations.
- Perform a series of experiments, measuring the initial rate.
- To find the order with respect to A, keep \([\text{B}]\) constant while varying \([\text{A}]\).
- Compare two experiments:
If \([\text{A}]\) doubles and Rate doubles: Order \(m=1\).
If \([\text{A}]\) doubles and Rate quadruples: Order \(m=2\).
If \([\text{A}]\) doubles and Rate stays the same: Order \(m=0\).
4.5 Determining Order Graphically (Concentration-Time Graphs)
The shape of a concentration-time graph (where concentration of reactant is plotted on the y-axis) tells us the order:
- Zero Order: Straight line sloping downwards. Rate is constant (gradient is constant).
- First Order: Exponential curve. Half-life is constant (see Section 5).
- Second Order: Exponential curve, but the concentration drops off much faster initially than first order.
We can also plot Rate-Concentration Graphs:
- Zero Order: Horizontal straight line (Rate independent of concentration).
- First Order: Straight line through the origin (Rate \(\propto [\text{A}]\)).
- Second Order: Curve (Rate \(\propto [\text{A}]^2\)).
Key Takeaway (Section 4): The rate equation ($\text{Rate} = k[\text{A}]^m[\text{B}]^n$) defines the kinetics of a reaction. Orders ($m, n$) must be found through experiments and are not necessarily equal to the coefficients in the overall balanced equation.
Section 5: Half-Life (\(t_{1/2}\)) and First Order Kinetics
5.1 Defining Half-Life
The half-life, \(\mathbf{t_{1/2}}\), is the time taken for the concentration of a reactant to decrease to half its initial value.
5.2 Half-Life in First Order Reactions
A unique and important feature of a first-order reaction is that its half-life, \(t_{1/2}\), is independent of the concentration of the reactant.
This means that if it takes 10 minutes for the concentration to drop from 1.0 \(\text{mol dm}^{-3}\) to 0.5 \(\text{mol dm}^{-3}\), it will take *another* 10 minutes to drop from 0.5 \(\text{mol dm}^{-3}\) to 0.25 \(\text{mol dm}^{-3}\).
Calculation using Half-Life:
For first order reactions, there is a simple relationship between \(k\) and \(t_{1/2}\):
$$ k = \frac{0.693}{t_{1/2}} $$
This equation is vital for calculations involving radioactive decay (which is always first order) and chemical kinetics problems where half-life data is provided.
Key Takeaway (Section 5): Only first-order reactions have a constant half-life, a key indicator used to determine their reaction order graphically or experimentally.
Section 6: Reaction Mechanisms and the Rate-Determining Step (RDS)
Most reactions occur in a sequence of small, simple steps, collectively known as the reaction mechanism.
6.1 Intermediates vs. Catalysts
When analyzing a multi-step mechanism, we often encounter species that are not the final products:
- Intermediate: Produced in an early step and then consumed in a later step. It does not appear in the overall balanced equation OR the rate equation.
- Catalyst: Consumed in an early step and then regenerated in a later step. It is present at the start and end of the reaction, and it may appear in the rate equation.
6.2 The Rate-Determining Step (RDS)
In a multi-step reaction, the individual steps often proceed at different speeds. The rate of the entire overall reaction is controlled by the slowest step, known as the rate-determining step (RDS).
Analogy: Think of a production line. If one machine (the RDS) can only process 10 items per hour, the entire factory can only produce 10 items per hour, even if every other machine can process 100 items per hour.
6.3 Linking RDS to the Rate Equation
The RDS is the key to deducing the rate equation:
The species involved in the RDS are the ONLY species that appear in the rate equation.
If the RDS involves 1 molecule of A and 2 molecules of B, the rate equation must be: $$ \text{Rate} = k[\text{A}]^1[\text{B}]^2 $$
If a proposed reaction mechanism results in a rate equation that matches the experimentally determined rate equation, the mechanism is considered consistent (though not definitively proven).
6.4 The Effect of Temperature on the Rate Constant (\(k\))
We established in Section 2 that increasing temperature significantly increases the reaction rate. In terms of the rate equation, this means that increasing temperature increases the value of the rate constant, \(k\).
The relationship is exponential: a small change in $T$ leads to a large change in $k$ (and rate) because it dramatically increases the fraction of successful collisions.
Key Takeaway (Section 6): The slowest step (RDS) controls the overall speed and determines which reactants appear in the rate equation. We suggest mechanisms by ensuring they produce a rate equation consistent with experimental data.
Section 7: Advanced Catalyst Mechanisms (A Level Depth)
A Level kinetics requires deeper understanding of how homogeneous and heterogeneous catalysts work, using specific examples.
7.1 Homogeneous vs. Heterogeneous
A catalyst is classified by its physical state relative to the reactants:
| Type | Definition | Mode of Action |
|---|---|---|
| Homogeneous | Catalyst and reactants are in the same phase (usually liquid/aqueous). | The catalyst reacts to form an intermediate, which then reacts to form the product, regenerating the catalyst later on. |
| Heterogeneous | Catalyst and reactants are in different phases (e.g., solid catalyst, gaseous reactants). | Provides a solid surface for the reaction to occur. |
7.2 Mechanism of Heterogeneous Catalysis
Heterogeneous catalysis involves four main steps occurring on the surface of the solid catalyst:
- Adsorption: Reactant molecules stick to the surface of the catalyst, often at specific "active sites," due to temporary intermolecular forces or weak chemical bonds.
- Bond Weakening: The interaction with the catalyst surface weakens the bonds within the reactant molecules, lowering the activation energy required for them to break.
- Reaction: The weakened reactants interact and form the product molecules on the surface.
- Desorption: The product molecules leave (desorb) the surface, freeing up the active site for the next reactant molecules.
Examples of Heterogeneous Catalysis:
- Iron (Fe) in the Haber Process: $\text{N}_2(\text{g})$ and $\text{H}_2(\text{g})$ adsorb onto the solid iron surface, where bonds are broken, allowing ammonia ($\text{NH}_3$) to form.
- Palladium, Platinum, and Rhodium in Catalytic Converters: These metals (coated onto ceramic supports) catalyze the removal of harmful gases like $\text{NO}$ and $\text{CO}$ from car exhaust, turning them into harmless $\text{N}_2$ and $\text{CO}_2$.
7.3 Mechanism of Homogeneous Catalysis
Homogeneous catalysis works via sequential intermediate steps, often involving transition metal ions which can easily change oxidation state.
Example: Catalysis of Persulfate ($\text{S}_2\text{O}_8^{2-}$) and Iodide ($\text{I}^-$)
The uncatalyzed reaction: $$ \text{S}_2\text{O}_8^{2-} (\text{aq}) + 2\text{I}^- (\text{aq}) \rightarrow 2\text{SO}_4^{2-} (\text{aq}) + \text{I}_2 (\text{aq}) $$ This reaction is slow because the two negative ions repel each other (high $E_a$).
We use $\text{Fe}^{2+}$ ions as a homogeneous catalyst. The mechanism involves two easier steps:
Step 1 (Oxidation of $\text{Fe}^{2+}$): The catalyst reacts with one reactant.
$$
\text{S}_2\text{O}_8^{2-} (\text{aq}) + 2\text{Fe}^{2+} (\text{aq}) \rightarrow 2\text{SO}_4^{2-} (\text{aq}) + 2\text{Fe}^{3+} (\text{aq})
$$
Step 2 (Regeneration of $\text{Fe}^{2+}$): The intermediate ($\text{Fe}^{3+}$) reacts with the second reactant, reforming the catalyst.
$$
2\text{Fe}^{3+} (\text{aq}) + 2\text{I}^- (\text{aq}) \rightarrow 2\text{Fe}^{2+} (\text{aq}) + \text{I}_2 (\text{aq})
$$
Both steps involve the reaction between oppositely charged ions ($\text{S}_2\text{O}_8^{2-}$ with $\text{Fe}^{2+}$ is the only one involving same-sign repulsion, but $\text{Fe}^{2+}$ is a smaller ion with lower charge density than $\text{I}^-$ so the repulsion is less severe than $\text{S}_2\text{O}_8^{2-}$ with $\text{I}^-$). The key is that the catalyst is used in Step 1 and reformed in Step 2, providing a pathway of lower $E_a$.
Key Takeaway (Section 7): Homogeneous catalysts work by changing oxidation state in intermediate steps, while heterogeneous catalysts work by providing a surface to weaken bonds (adsorption).