Chemistry | Kinetics

🚀 Welcome to Chemical Kinetics: Understanding How Fast Reactions Go! 🚀

Hello future Chemists! This chapter, Kinetics, is one of the most practical and fascinating areas of Chemistry. While thermodynamics tells us *if* a reaction is possible (spontaneous), kinetics tells us *how fast* that reaction actually happens.

Think about making toast: you know the reaction (browning) is possible, but if it took 10 hours, it wouldn't be very useful! Kinetics helps us understand and control the speed of chemical processes, from industrial synthesis to biological reactions in your body.

Don't worry if some of the math looks intimidating at first; we will break down every concept step-by-step. Let’s get started!

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1. Defining and Measuring Reaction Rate

The rate of reaction is simply how quickly the concentration of a reactant decreases or how quickly the concentration of a product increases over time.

1.1. The Basics of Rate

Key Definition: Rate of reaction is the change in concentration of a species (reactant or product) per unit time.

Mathematically, we can express the rate as:

\[ \text{Rate} = \frac{\Delta[\text{Concentration}]}{\Delta[\text{Time}]} \]

The standard units for rate of reaction are \(mol \, dm^{-3} \, s^{-1}\) (moles per cubic decimetre per second).

1.2. Average Rate vs. Instantaneous Rate

• Average Rate: This is the rate calculated over a large time interval. It’s easy to calculate, but doesn't tell us what's happening at a specific moment.
• Instantaneous Rate: This is the rate at a specific point in time. Reactions usually slow down as reactants are used up, so the instantaneous rate constantly changes. We find this by drawing a tangent to the concentration-time graph at the time of interest and calculating its gradient.

Quick Review: The rate measures speed, and the faster the reaction, the steeper the concentration-time graph will be.

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2. Collision Theory and Activation Energy

Why do reactions happen at all? The answer lies in the movement and collision of particles.

2.1. The Rules of Successful Collision

According to Collision Theory, reacting particles must collide with each other to form products. However, not all collisions lead to a reaction. A collision must satisfy two essential conditions:

1. Sufficient Energy: The collision must occur with energy equal to or greater than the Activation Energy (\(E_a\)).
2. Correct Orientation: The particles must align correctly so that the bonds that need to break and form are in the right position.

🧠 Memory Aid: The C.O.C. Trick
For a reaction to happen, you need a Collision with the correct Orientation and enough Concentrated energy (\(E_a\)).

2.2. Activation Energy (\(E_a\))

The Activation Energy (\(E_a\)) is the minimum amount of kinetic energy that particles must possess when they collide in order to start a reaction. It's the "energy hurdle" the reactants must jump over to become products.

Analogy: If you want to push a heavy box, you need to exert a certain minimum force to get it moving. That minimum force is your \(E_a\).

2.3. Factors Affecting Reaction Rate

Any factor that increases the frequency of successful collisions will increase the rate. The main factors are:

a) Concentration (or Pressure for Gases)

If you increase the concentration of reactants, there are more particles in the same volume. This leads to more frequent collisions, thus increasing the frequency of successful collisions and the rate.

• For gases, increasing pressure is the same as increasing concentration (forcing the same number of molecules into a smaller volume).

b) Temperature

Increasing the temperature has a massive effect on rate (often doubling the rate for every 10 °C rise!). This is because:

1. Particles move faster, leading to slightly more frequent collisions.
2. Crucially: A much larger *fraction* of particles will now possess energy greater than or equal to the \(E_a\). (We look at this next in the Boltzmann distribution!)

c) Surface Area

Only relevant for reactions involving solids. By breaking a solid into smaller pieces (e.g., powdering), you increase the surface area available for collisions, leading to a higher frequency of collisions and a faster rate.

d) Catalysts

Catalysts speed up the rate without being used up. They achieve this by providing an alternative reaction pathway with a lower \(E_a\).

Key Takeaway: Rate depends on collision frequency and, more importantly, the fraction of particles exceeding the Activation Energy.

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3. The Boltzmann Distribution

The Boltzmann Distribution is a graph showing the range of kinetic energies of molecules in a gas or liquid at a specific temperature. It is crucial for explaining the effects of temperature and catalysts.

3.1. Understanding the Curve

The curve shows:

• The area under the curve represents the total number of molecules.
• The curve starts at the origin (0 energy) because no molecules have zero energy.
• Most molecules have an average energy, but a small fraction possess very high energy.

The Effect of Temperature (T)

When you increase the temperature (from \(T_1\) to \(T_2\)):

1. The peak of the curve shifts to the right (molecules move faster, so the average energy increases).
2. The curve flattens (since the total area/number of molecules must remain the same).
3. The Most Important Part: The fraction of molecules with energy greater than or equal to \(E_a\) (the area under the curve to the right of \(E_a\)) increases significantly.

Imagine the Activation Energy (\(E_a\)) as a 'pass mark' on a test. By increasing the temperature, you haven't changed the pass mark, but you’ve improved everyone's scores, so many more students (molecules) pass the hurdle!

3.2. How Catalysts Affect the Distribution

A catalyst does not change the kinetic energy distribution of the molecules. Instead, a catalyst lowers the Activation Energy (\(E_a\)) required.

• On the Boltzmann diagram, this means the \(E_a\) line moves to the left.
• Even at the same temperature, lowering the \(E_a\) results in a much larger fraction of molecules now exceeding the new, lower energy hurdle, dramatically increasing the rate.

Did you know? Biological catalysts are called enzymes. They lower \(E_a\) so effectively that reactions vital for life can happen rapidly at body temperature (around 37 °C).

Key Takeaway: Temperature increases the proportion of high-energy molecules; a catalyst decreases the required energy barrier (\(E_a\)). Both increase the number of successful collisions.

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4. Rate Equations, Order of Reaction, and the Rate Constant

To quantify how concentration affects rate, we use a Rate Equation (sometimes called a Rate Law).

4.1. The General Rate Equation

For a general reaction: \[ aA + bB \rightarrow \text{Products} \]

The rate equation is defined as:

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

• \([A]\) and \([B]\): Molar concentrations of reactants A and B.
• \(k\): The Rate Constant. This value is constant for a specific reaction at a fixed temperature.
• \(m\) and \(n\): The Orders of Reaction with respect to A and B.

CRITICAL POINT: The orders \(m\) and \(n\) have NO necessary relationship to the stoichiometric coefficients \(a\) and \(b\) in the balanced equation. They must always be determined experimentally.

4.2. Order of Reaction

The order defines how the concentration of a reactant affects the rate. It is typically 0, 1, or 2.

a) Zero Order (e.g., \(m=0\))

\[ \text{Rate} \propto [A]^0 = 1 \]

Changing the concentration of A has no effect on the rate. The reaction rate is constant regardless of [A].

b) First Order (e.g., \(m=1\))

\[ \text{Rate} \propto [A]^1 \]

If you double [A], the rate doubles. If you triple [A], the rate triples. There is a direct proportionality.

c) Second Order (e.g., \(m=2\))

\[ \text{Rate} \propto [A]^2 \]

If you double [A], the rate quadruples (\(2^2 = 4\)). If you triple [A], the rate increases by nine times (\(3^2 = 9\)).

The Overall Order of the reaction is the sum of the individual orders: Overall Order \(= m + n\).

4.3. The Rate Constant (\(k\))

The rate constant (\(k\)) is a proportionality constant that links the rate of reaction to the concentrations of the reactants raised to the power of their orders.

What influences \(k\)? The value of \(k\) is only changed by two things:

1. Temperature: Increasing T significantly increases \(k\).
2. Catalysts: Adding a catalyst significantly increases \(k\).

Changing the concentration of reactants does not change \(k\).

Determining the Units of \(k\)

The units of \(k\) depend entirely on the overall order of the reaction. We must rearrange the rate equation:

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

Step-by-step Example (Overall Order = 2) Assume Rate = \(k [A]^1 [B]^1\). Overall order is 2. \[ k = \frac{\text{Rate}}{[A][B]} = \frac{(mol \, dm^{-3} \, s^{-1})}{(mol \, dm^{-3})(mol \, dm^{-3})} \]

Cancel out the units: \[ k = \frac{s^{-1}}{(mol \, dm^{-3})} = mol^{-1} \, dm^3 \, s^{-1} \]

Common Mistake Alert: Students often forget to flip the sign on the power of the concentration units when moving them to the numerator!

Key Takeaway: The rate equation is determined experimentally. The orders \(m\) and \(n\) tell us how sensitive the reaction rate is to changes in concentration.

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5. Determining the Rate Equation: The Initial Rate Method

Since we cannot predict the orders of reaction from stoichiometry, we must perform experiments. The Initial Rate Method is the standard procedure for finding \(m\) and \(n\).

5.1. The Principle

We measure the rate of reaction immediately after mixing the reactants (the initial rate) in several different experiments. We design these experiments so that the concentration of only one reactant is changed at a time.

5.2. Step-by-Step Procedure

Consider the reaction: \(A + B \rightarrow \text{Products}\). We want to find \(m\) and \(n\) in Rate = \(k [A]^m [B]^n\).

Step 1: Finding the Order (m) with respect to [A]

Compare two experiments (e.g., Experiment 1 and 2) where the concentration of B is kept constant, but the concentration of A is changed.

Example Data: | Exp. | Initial [A] | Initial [B] | Initial Rate | | :---: | :---: | :---: | :---: | | 1 | 0.10 | 0.10 | \(2.0 \times 10^{-4}\) | | 2 | 0.20 | 0.10 | \(8.0 \times 10^{-4}\) |

Analysis: When [A] doubles (0.10 to 0.20), the Rate quadruples (\(8.0/2.0 = 4\)).
Since \(2^m = 4\), the order \(m\) must be 2.

Step 2: Finding the Order (n) with respect to [B]

Compare two experiments (e.g., Experiment 1 and 3) where the concentration of A is kept constant, but the concentration of B is changed.

Example Data (using new Exp 3): | Exp. | Initial [A] | Initial [B] | Initial Rate | | :---: | :---: | :---: | :---: | | 1 | 0.10 | 0.10 | \(2.0 \times 10^{-4}\) | | 3 | 0.10 | 0.30 | \(6.0 \times 10^{-4}\) |

Analysis: When [B] triples (0.10 to 0.30), the Rate triples (\(6.0/2.0 = 3\)).
Since \(3^n = 3\), the order \(n\) must be 1.

Step 3: Writing the Complete Rate Equation

Rate = \(k [A]^2 [B]^1\) (or simply \(k [A]^2 [B]\)).

Step 4: Calculating the Rate Constant (\(k\))

Use the data from any experiment (e.g., Exp 1) and substitute the values into the completed rate equation.

\[ k = \frac{\text{Rate}}{[A]^2 [B]} \]

\[ k = \frac{2.0 \times 10^{-4} \, mol \, dm^{-3} \, s^{-1}}{(0.10 \, mol \, dm^{-3})^2 (0.10 \, mol \, dm^{-3})} \]

\[ k = \frac{2.0 \times 10^{-4}}{0.001} = 0.20 \]

Don't forget the units for a third-order reaction (2 + 1 = 3): \(mol^{-2} \, dm^6 \, s^{-1}\).
\(k = 0.20 \, mol^{-2} \, dm^6 \, s^{-1}\).

Encouragement: Calculations involving rate equations are highly systematic. Practice these steps, and you will find them become routine!

Key Takeaway: The Initial Rate Method allows us to isolate the effect of each reactant concentration on the overall reaction rate, giving us the experimental orders \(m\) and \(n\).