Chemistry (9701) | Rate of reaction

Chemistry 9701 Study Notes: Rate of Reaction (Kinetics)

Welcome to Reaction Kinetics! This is the exciting chapter where we stop just asking *if* a reaction happens, and start asking *how fast* it happens. Understanding reaction rate is crucial, whether you are trying to speed up industrial chemical production or slow down food spoilage. Don't worry if this seems tricky at first; we will break down the mechanisms and maths step-by-step!

1. Defining Reaction Rate and Collision Theory (AS Level)

The rate of reaction is all about how quickly reactants are used up, or how quickly products are formed.

1.1 Rate of Reaction Definition

The rate of reaction is the change in concentration of a reactant or product over time.

Rate \(= \frac{\text{Change in concentration}}{\text{Time taken}}\)

The units for rate are typically \(\text{mol dm}^{-3}\text{ s}^{-1}\) (or sometimes $\text{mol dm}^{-3}\text{ min}^{-1}$).

1.2 The Collision Theory

For two particles (atoms, ions, or molecules) to react, three conditions must be met, as described by the Collision Theory:

1. Collision: The particles must collide. (The more frequently they collide, the higher the reaction rate.)
2. Correct Orientation: They must collide in the correct geometric orientation (angle) for the necessary bonds to break and form.
3. Sufficient Energy: They must possess energy equal to or greater than the activation energy (\(E_a\)).

A collision that meets conditions 2 and 3 is called an effective collision. A collision that does not is a non-effective collision.

Key Takeaway: The overall reaction rate depends directly on the frequency of effective collisions.

1.3 Factors Affecting Reaction Rate (Qualitative)

Any factor that increases the frequency of effective collisions will increase the reaction rate.

A. Concentration/Pressure

• Effect: Increasing concentration (for solutions) or pressure (for gases) increases the reaction rate.
• Explanation: More particles are packed into the same volume, meaning the frequency of collisions increases. This leads directly to a higher frequency of effective collisions per unit time.
• Analogy: Imagine blindfolded people in a room. If you double the number of people (concentration), they will bump into each other twice as often.

B. Particle Size (Surface Area)

• Effect: Decreasing the particle size (using powder instead of lumps) increases the reaction rate.
• Explanation: Reactions involving solids only occur at the exposed surface. Smaller particles mean a much larger total surface area, exposing more reactant sites for collisions to occur.

2. Activation Energy and Temperature (AS Level)

Not all collisions are equal. Only those with enough energy overcome the Activation Energy barrier.

2.1 Activation Energy (\(E_a\))

Definition: The activation energy (\(E_a\)) is the minimum energy required for a collision between reactant particles to be effective (i.e., to result in a chemical reaction).

Analogy: $E_a$ is like the height of a fence. To get from the reactant field to the product field, you must jump over the fence. Only molecules with enough energy can jump.

You must be able to construct and interpret a reaction pathway diagram showing \(E_a\) and the enthalpy change ($\Delta H$) of the reaction (this links back to Chemical Energetics).

2.2 The Boltzmann Distribution

Since particles move randomly, they have a wide range of kinetic energies. The Boltzmann distribution curve shows how the kinetic energy is distributed among the molecules in a sample.

The distribution curve plots the number of molecules (y-axis) against the kinetic energy (x-axis).

• The area under the curve represents the total number of molecules.
• The curve starts at the origin (since no molecules have zero energy).
• The peak represents the most probable energy ($E_{mp}$).
• The curve never touches the x-axis (theoretically, there is no maximum energy).

Significance of \(E_a\) on the Boltzmann Distribution:

We mark the activation energy (\(E_a\)) on the x-axis. The molecules to the right of this line are the ones that have enough energy to react (i.e., they participate in effective collisions).

2.3 Effect of Temperature on Rate

Effect: Increasing the temperature significantly increases the reaction rate.

Explanation using the Boltzmann Distribution:

When temperature is increased (e.g., from \(T_1\) to \(T_2\)):

1. The peak of the curve shifts to the right (higher average kinetic energy).
2. The curve flattens (since the area under the curve must remain constant, representing the total number of molecules).
3. Crucially, the proportion of molecules with energy equal to or greater than \(E_a\) (the shaded area to the right of \(E_a\)) increases exponentially.

This massive increase in the proportion of high-energy molecules drastically increases the frequency of effective collisions, thus increasing the rate.

Did You Know? A rough rule of thumb is that for many reactions, increasing the temperature by $10^\circ\text{C}$ approximately doubles the reaction rate!

3. Catalysts and Catalysis (AS & A Level)

A catalyst offers an alternative, lower energy path for the reaction.

3.1 Defining a Catalyst

Definition: A catalyst is a substance that increases the rate of a chemical reaction by providing an alternative reaction pathway of lower activation energy (\(E_a\)), without being chemically changed itself at the end of the reaction.

The process by which a catalyst works is called catalysis.

Catalytic Effect via Reaction Pathway Diagram:

When you draw a reaction pathway diagram:

• The enthalpy change ($\Delta H$) remains the same (since the start and end points are unchanged).
• The curve representing the catalysed reaction has a lower maximum peak than the uncatalysed reaction, corresponding to a lower \(E_a\).

Catalytic Effect via Boltzmann Distribution:

When a catalyst is added, the activation energy (\(E_a\)) moves to the left on the kinetic energy axis. Even at the same temperature, this increases the shaded area, meaning a larger proportion of molecules now have sufficient energy to react, thus increasing the rate.

3.2 Types of Catalysis (A Level)

Catalysts are classified based on the physical state (phase) of the catalyst compared to the reactants.

A. Heterogeneous Catalysis

• Definition: The catalyst is in a different physical state from the reactants (e.g., solid catalyst, gaseous reactants).
• Mode of Action (Surface Catalysis):
1. Adsorption: Reactant molecules stick to the surface of the solid catalyst.
2. Bond Weakening: The catalyst surface weakens the bonds within the reactant molecules, making them more reactive.
3. Reaction: The molecules react on the surface due to the lower \(E_a\).
4. Desorption: The product molecules leave the surface of the catalyst.
• Examples:
  • Iron (\(\text{Fe}\)) in the Haber process (\(\text{N}_2\) and \(\text{H}_2\) to \(\text{NH}_3\)).
  • Platinum (\(\text{Pt}\)), Palladium (\(\text{Pd}\)), and Rhodium (\(\text{Rh}\)) in catalytic converters (converting harmful \(\text{NO}_x\) and unburnt hydrocarbons into \(\text{N}_2\), \(\text{CO}_2\), and \(\text{H}_2\text{O}\)).

B. Homogeneous Catalysis

• Definition: The catalyst is in the same physical state as the reactants (usually aqueous or gas phase).
• Mode of Action: The catalyst reacts with one reactant to form an intermediate species, which then reacts further to form the product and regenerate the catalyst. The catalyst is used up in Step 1 but reformed in Step 2.
• Examples:
  • \(\text{Fe}^{2+}\) or \(\text{Fe}^{3+}\) ions catalysing the reaction between iodide ions (\(\text{I}^-\)) and peroxodisulfate ions (\(\text{S}_2\text{O}_8^{2-}\)).
  • Atmospheric oxides of nitrogen (\(\text{NO}_x\)) acting as catalysts in the oxidation of atmospheric sulfur dioxide (\(\text{SO}_2\)).

4. Rate Equations, Order and Rate Constant (A Level)

To quantify how concentration affects rate, we use the rate equation.

4.1 The Rate Equation

The rate equation mathematically links the rate of reaction to the concentrations of the reactants. For a reaction \(a\text{A} + b\text{B} \rightarrow \text{products}\), the general form is:

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

Where:

• \(\text{Rate}\): Reaction rate (in $\text{mol dm}^{-3}\text{ s}^{-1}$).
• \([\text{A}]\) and \([\text{B}]\): Concentrations of reactants A and B (in $\text{mol dm}^{-3}$).
• \(k\): The rate constant (A measure of the efficiency of collisions, dependent only on temperature).
• \(m\) and \(n\): The order of reaction with respect to A and B, respectively.

4.2 Order of Reaction

The order of reaction ($m$ or $n$) is the power to which a reactant's concentration is raised in the rate equation. It is determined experimentally, not from the stoichiometric coefficients ($a$ and $b$) in the balanced equation.

The possible orders tested are 0, 1, or 2.

Overall Order: The sum of the individual orders ($m + n + \dots$).

• Zero Order (\(m=0\)): Rate is independent of the concentration of A. Doubling \([\text{A}]\) has no effect on the rate. (Rate \(\propto [\text{A}]^0\)).
• First Order (\(m=1\)): Rate is directly proportional to the concentration of A. Doubling \([\text{A}]\) doubles the rate. (Rate \(\propto [\text{A}]^1\)).
• Second Order (\(m=2\)): Rate is proportional to the square of the concentration of A. Doubling \([\text{A}]\) quadruples the rate ($2^2 = 4$). (Rate \(\propto [\text{A}]^2\)).

4.3 Determining Order from Experimental Data (Initial Rates Method)

The initial rates method involves running the reaction multiple times while keeping all variables constant except the concentration of one reactant.

Step-by-step example:

If we look at experiments 1 and 2, and we double \([\text{A}]\) but keep \([\text{B}]\) constant:

1. If the initial rate doubles: Order with respect to A is 1 ($2^1 = 2$).
2. If the initial rate quadruples: Order with respect to A is 2 ($2^2 = 4$).
3. If the initial rate remains the same: Order with respect to A is 0.

4.4 The Rate Constant (\(k\))

The rate constant \(k\) is calculated once the orders ($m$ and $n$) are known, by substituting the rate and concentration data from any single experiment into the rate equation.

Units of \(k\): The units of \(k\) depend entirely on the overall order of the reaction. The final units must ensure the overall rate unit is always $\text{mol dm}^{-3}\text{ s}^{-1}$.

$$\text{Unit of } k = \frac{\text{Unit of Rate}}{(\text{Unit of Concentration})^{\text{Overall Order}}}$$

• Overall Order 1: $k$ has units of \(\text{s}^{-1}\).
• Overall Order 2: $k$ has units of \(\text{dm}^3 \text{ mol}^{-1} \text{ s}^{-1}\).
• Overall Order 3: $k$ has units of \(\text{dm}^6 \text{ mol}^{-2} \text{ s}^{-1}\).

Memory Aid: For the units of $k$, the power of $\text{dm}$ is (Overall Order - 1), and the power of $\text{mol}$ is -(Overall Order - 1). Time is always $\text{s}^{-1}$.

4.5 Effect of Temperature on \(k\)

The rate constant, \(k\), is constant *only* if the temperature is constant.

Qualitative Effect: Increasing the temperature increases the value of \(k\).

Explanation: Higher temperature dramatically increases the number of particles that can overcome \(E_a\). Since $k$ is fundamentally a measure of the reaction speed at a fixed concentration, an increase in effective collisions means $k$ must increase.

5. Half-Life (\(t_{1/2}\))

5.1 Defining Half-Life

Definition: The half-life (\(t_{1/2}\)) is the time required for the concentration of a reactant to fall to half its initial value.

5.2 Half-Life for a First-Order Reaction

For a reaction that is first-order overall, the half-life exhibits a unique and crucial property:

The half-life (\(t_{1/2}\)) is independent of the initial concentration of the reactant.

Example: If it takes 50 seconds for the concentration to drop from $1.0\text{ mol dm}^{-3}$ to $0.5\text{ mol dm}^{-3}$, it will also take 50 seconds to drop from $0.5\text{ mol dm}^{-3}$ to $0.25\text{ mol dm}^{-3}$.

Calculation: For first-order reactions, the half-life and rate constant are linked by the equation:

$$k = \frac{0.693}{t_{1/2}}$$

This relationship allows you to calculate $k$ if $t_{1/2}$ is known, or vice versa, purely from data on a concentration-time graph.

6. Reaction Mechanisms and Rate-Determining Step (A Level)

Most reactions don't happen in a single step; they happen via a sequence of elementary steps called the reaction mechanism.

6.1 Intermediates and the RDS

Intermediate: A species that is formed in one step of the mechanism and used up in a subsequent step. Intermediates do *not* appear in the overall balanced equation or the rate equation.

Rate-Determining Step (RDS): This is the slowest step in the entire reaction mechanism. Since it is the bottleneck, the overall rate of the reaction is determined by the speed of this single, slow step.

Analogy: If you are building a house (the reaction), and waiting for the delivery of bricks (one step) takes 90% of the total time, then that delivery step is the RDS.

Note on Catalysts: A catalyst appears in the mechanism as a reactant in an early step and a product in a later step (it is regenerated).

6.2 Linking Rate Equation to Mechanism

The concentrations of only those species (reactants or catalysts) that are involved in the rate-determining step (RDS) appear in the rate equation.

Rules for Mechanism Deduction:

1. The reactants shown in the RDS must match the species whose concentrations appear in the experimental rate equation.
2. The stoichiometric coefficients of the reactants in the RDS must match the orders in the rate equation.

Example: If the overall reaction is \(\text{A} + \text{B} \rightarrow \text{P}\), but the experimental rate equation is $\text{Rate} = k [\text{A}]^2$.

This suggests the mechanism is not a single step. The slowest step (RDS) must involve two molecules of A colliding, and molecule B is likely involved in a subsequent fast step.