Simple rate equations, orders of reaction and rate constants - Chemistry (9701) - Cambridge A-Level

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Simple Rate Equations, Orders of Reaction, and Rate Constants

Welcome to Reaction Kinetics! This chapter is where we move beyond just *how fast* a reaction is and start exploring *why* it is that fast, and crucially, *what* controls its speed. Understanding rate equations is essential for predicting reaction behaviour and optimizing industrial processes.

Don't worry if this feels mathematically intensive—we will break down the formulas step-by-step using simple logic and real-world analogies!

1. The Rate Equation and Key Terms

What is the Rate Equation?

The rate of a reaction often depends on the concentration of the reactants. The rate equation (or rate law) mathematically relates the rate of reaction to the concentrations of the reactants.

For a general reaction:
$A + B \rightarrow Products$

The rate equation is generally written as:

$Rate = k [A]^m [B]^n$

Rate: The change in concentration of a reactant or product per unit time (usually in $\text{mol}\ \text{dm}^{-3}\ \text{s}^{-1}$).
$[\text{A}]$ and $[\text{B}]$: The concentrations of the reactants (usually in $\text{mol}\ \text{dm}^{-3}$).
$k$: The Rate Constant. This is a proportionality constant that links the rate of reaction to the concentrations of the reactants raised to their respective orders. The value of $k$ is constant for a specific reaction at a specific temperature.
$m$ and $n$: The Orders of Reaction with respect to reactants A and B, respectively.

The Most Important Rule: Orders are Experimental!

WARNING: The orders $m$ and $n$ cannot be deduced from the stoichiometric coefficients in the balanced chemical equation. They must be determined experimentally.

Example: For the reaction $2NO(g) + O_2(g) \rightarrow 2NO_2(g)$, the rate equation is found to be $Rate = k [NO]^2 [O_2]^1$. Notice the order for NO is 2, matching the stoichiometry, but this is a coincidence! You must rely only on experimental data.

Key Takeaway

The rate equation $Rate = k [A]^m [B]^n$ is the fundamental relationship describing how concentration affects reaction speed. The orders ($m$ and $n$) are always determined experimentally, not from the balanced equation.

2. Orders of Reaction (m and n)

The Order of Reaction with respect to a specific reactant is the power to which its concentration is raised in the rate equation. This tells us exactly how the rate changes when the concentration of that reactant changes.

A. Zero Order (Order = 0)

If a reaction is zero order with respect to reactant A, then $m=0$.

Effect: Changing the concentration of A has no effect on the rate.

In the Rate Equation: $[A]^0 = 1$, so the term disappears:
$Rate = k [A]^0 [B]^n = k [B]^n$

Analogy: Imagine washing cars in a factory. Adding more cars to the queue won't speed up the process if the speed is limited by the number of wash bays and staff (the catalyst or fixed conditions).

B. First Order (Order = 1)

If a reaction is first order with respect to reactant A, then $m=1$.

Effect: Rate is directly proportional to $[A]$. If you double $[A]$, the rate doubles.

In the Rate Equation:
$Rate \propto [A]^1$

Simple Trick: $\times 2$ concentration $\rightarrow \times 2$ rate.

C. Second Order (Order = 2)

If a reaction is second order with respect to reactant A, then $m=2$.

Effect: Rate is proportional to the square of $[A]$. If you double $[A]$, the rate increases by a factor of $2^2 = 4$.

In the Rate Equation:
$Rate \propto [A]^2$

Simple Trick: $\times 2$ concentration $\rightarrow \times 4$ rate. ($\times 3$ concentration $\rightarrow \times 9$ rate.)

D. Overall Order of Reaction

The Overall Order of Reaction is the sum of the individual orders with respect to all reactants in the rate equation.
Overall Order $= m + n + ...$

Example: If $Rate = k [A]^2 [B]^1$, the overall order is $2 + 1 = 3$. (Note: Syllabus limits individual orders to 0, 1, or 2.)

3. Determining Order of Reaction from Experimental Data

You need to be comfortable deducing orders using two main methods: the Initial Rates Method (using tables of data) and Graphical Methods.

Method 1: Initial Rates Method (Using Tables)

This method involves running the reaction multiple times, varying the initial concentration of only one reactant at a time while keeping others constant. We look at how the initial rate changes.

Step-by-Step Deduction:

Find two experiments where the concentration of the reactant you are investigating (e.g., A) changes, but the concentrations of all other reactants (e.g., B) remain constant.
Calculate the change ratio for the concentration of A and the initial rate.
Determine the order ($m$): Use the relationship:
$\frac{\text{Rate}_2}{\text{Rate}_1} = (\frac{[A]_2}{[A]_1})^m$

Example: Determining order $m$ for reactant A.

Expt.	$[\text{A}]\ \text{/}\ \text{mol}\ \text{dm}^{-3}$	$[\text{B}]\ \text{/}\ \text{mol}\ \text{dm}^{-3}$	Initial Rate $\text{/}\ \text{mol}\ \text{dm}^{-3}\ \text{s}^{-1}$
1	0.10	0.10	2.0 $\times 10^{-4}$
2	0.20	0.10	4.0 $\times 10^{-4}$

Compare Expt 1 and 2 (B is constant):

$[A]$ doubles (0.10 to 0.20).
Rate doubles (2.0 $\times 10^{-4}$ to 4.0 $\times 10^{-4}$).

$\times 2$ concentration $\rightarrow \times 2$ rate. Therefore, the reaction is First Order ($m=1$) with respect to A.

Method 2: Graphical Methods

We can use two types of graphs to deduce the order:

A. Rate-Concentration Graphs (Plotting Rate vs. $[\text{A}]$ or $[\text{A}]^2$)

These graphs show how the instantaneous rate changes as the concentration changes during a single experiment.

Zero Order ($m=0$): Plotting Rate vs. $[\text{A}]$ gives a horizontal straight line. The rate is constant regardless of $[\text{A}]$.
First Order ($m=1$): Plotting Rate vs. $[\text{A}]$ gives a straight line through the origin. Rate $\propto [\text{A}]$.
Second Order ($m=2$): Plotting Rate vs. $[\text{A}]$ gives a curve. To get a straight line, you must plot Rate vs. $[\text{A}]^2$.

B. Concentration-Time Graphs (Plotting $[\text{A}]$ vs. Time)

These graphs track the concentration of a reactant as it decreases over time.

Zero Order ($m=0$): Gives a straight line with a negative gradient. Concentration decreases linearly.
First Order ($m=1$): Gives an exponential decay curve. The concentration decreases quickly at first, then slows down.
Second Order ($m=2$): Gives a steeper curve than first order initially, but the rate of decrease slows down more dramatically as concentration drops.

Quick Review: Graphical Interpretation (LO 2c)

Concentration-Time Graph: Gradient is the rate. If the gradient is constant (straight line), it is Zero Order.
Rate-Concentration Graph: If Rate vs. [A] is a straight line through the origin, it is First Order.

4. The Rate Constant ($k$) and its Units

Once you know the orders of reaction, you can determine the numerical value of the rate constant, $k$.

Calculating $k$ (LO 4a)

To calculate $k$, simply rearrange the rate equation and substitute the experimental data (using any single set of data from your initial rates table):

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

Example: If the rate equation is $Rate = k [A]^1 [B]^0$ and from experiment 1, $Rate = 2.0 \times 10^{-4}\ \text{mol}\ \text{dm}^{-3}\ \text{s}^{-1}$ and $[A] = 0.10\ \text{mol}\ \text{dm}^{-3}$:

$k = \frac{2.0 \times 10^{-4}}{0.10^1} = 2.0 \times 10^{-3}$

Units of the Rate Constant, $k$

The units of $k$ depend entirely on the overall order of the reaction. This is often tested! You must ensure the units make the overall rate unit $\text{mol}\ \text{dm}^{-3}\ \text{s}^{-1}$.

General Units for $k$: $\text{(Concentration)}^{1 - \text{Overall Order}}\ \text{s}^{-1}$

We use the base units:
Rate unit: $\text{mol}\ \text{dm}^{-3}\ \text{s}^{-1}$
Concentration unit: $\text{mol}\ \text{dm}^{-3}$

Overall Order	Units of Rate Equation	Units of $k$	$k$ Calculation Example
0	$Rate = k$	$\text{mol}\ \text{dm}^{-3}\ \text{s}^{-1}$	$k = \frac{\text{mol}\ \text{dm}^{-3}\ \text{s}^{-1}}{1}$
1	$Rate = k [\text{A}]$	$\text{s}^{-1}$	$k = \frac{\text{mol}\ \text{dm}^{-3}\ \text{s}^{-1}}{\text{mol}\ \text{dm}^{-3}} = \text{s}^{-1}$
2	$Rate = k [\text{A}]^2$	$\text{dm}^{3}\ \text{mol}^{-1}\ \text{s}^{-1}$	$k = \frac{\text{mol}\ \text{dm}^{-3}\ \text{s}^{-1}}{(\text{mol}\ \text{dm}^{-3})^2} = \text{dm}^{3}\ \text{mol}^{-1}\ \text{s}^{-1}$
3	$Rate = k [\text{A}]^3$	$\text{dm}^{6}\ \text{mol}^{-2}\ \text{s}^{-1}$	$k = \frac{\text{mol}\ \text{dm}^{-3}\ \text{s}^{-1}}{(\text{mol}\ \text{dm}^{-3})^3} = \text{dm}^{6}\ \text{mol}^{-2}\ \text{s}^{-1}$

Memory Aid: If you forget the pattern, just plug in the units into $k = \frac{Rate}{[A]^m [B]^n}$ and simplify!

5. Half-Life ($t_{1/2}$) (LO 3)

The Half-Life ($t_{1/2}$) is the time taken for the concentration of a reactant to fall to half of its initial value.

Crucial Point: First-Order Half-Life (LO 3a)

For a reaction that is First Order overall, the half-life is constant. It does not depend on the initial concentration of the reactant.

Did you know? Radioactive decay always follows first-order kinetics, which is why we talk about the half-life of elements like Carbon-14—it’s always the same, regardless of how much C-14 is present.

Calculations using Half-Life (LO 3b, 4b)

For a first-order reaction, the half-life ($t_{1/2}$) is related to the rate constant ($k$) by the equation:

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

This allows you to calculate $k$ if you know $t_{1/2}$, or vice versa. Remember, this specific relationship only applies to First-Order Reactions.

Key Takeaway

The half-life method is a graphical way to confirm if a reaction is first order. If the time required for concentration to halve is the same regardless of the starting concentration, it is first order, and the formula $k = 0.693 / t_{1/2}$ applies.

6. Multi-Step Reactions and the Rate-Determining Step (RDS) (LO 5)

Most reactions don't happen in a single step; they happen via a series of elementary steps, called the Reaction Mechanism.

An Intermediate is a species formed in one step and used up in a subsequent step. It does not appear in the overall balanced equation OR the rate equation.
A Catalyst is used up in an early step and regenerated in a later step. A catalyst may appear in the rate equation (if it is involved in the RDS), but it does not appear in the overall balanced equation.

The Rate-Determining Step (RDS)

In a multi-step reaction, one step is much slower than all the others. This slowest step is the Rate-Determining Step (RDS) (or Rate-Limiting Step).

Analogy: Imagine driving through a construction zone. Your overall journey speed is limited by the slowest part—the bottleneck where only one lane is open. The RDS is the bottleneck in the chemical reaction.

Relating RDS to the Rate Equation (LO 5b, 5c)

The rate equation for the overall reaction is determined only by the reactants involved in the RDS.

The reactants involved in the RDS are the species whose concentrations appear in the rate equation.
The order of reaction with respect to a species in the RDS is equal to the stoichiometric coefficient of that species in the RDS equation. (e.g., if $2A$ collide in the RDS, the order is 2 for A).

RULE: The rate equation only includes species that are present at the start of the RDS (i.e., reactants or catalysts). It never includes intermediates.

Example: Suppose the overall reaction is $A + 2B \rightarrow Products$, but the mechanism is:

Step 1 (Fast): $A + B \rightarrow I$ (Intermediate)
Step 2 (Slow, RDS): $I + B \rightarrow Products$

The Rate Equation is determined by the slow step (RDS):
$Rate = k [\text{I}]^1 [\text{B}]^1$

However, intermediates (like I) cannot be measured and cannot appear in the final rate equation. Since Step 1 is fast, we assume it reaches equilibrium. If A and B make I, we substitute I based on the relationship from the fast step.

In A-Level Chemistry (9701), you must be able to predict the order that results from the given RDS, and suggest a mechanism that is consistent with the derived rate equation (LO 5a, 5b).

Key Takeaway

The RDS is the slowest step in the mechanism and dictates the overall rate equation. If a species appears in the rate equation, it must be involved in or before the RDS.

7. The Effect of Temperature on the Rate Constant ($k$) (LO 6)

We know that increasing temperature increases the reaction rate. Since the rate equation coefficients (the orders) don't change with temperature, the only factor that must change is the Rate Constant ($k$).

Increasing Temperature $\rightarrow$ Increases $k$ $\rightarrow$ Increases Rate.

You need to explain this change qualitatively using the Boltzmann distribution (linking back to section 8.2 of the AS syllabus).

Qualitative Explanation:

At a higher temperature, the molecules have a greater average kinetic energy.
Looking at the Boltzmann distribution curve, a significantly larger fraction of molecules now possess kinetic energy equal to or greater than the Activation Energy ($E_a$).
This results in a much greater frequency of effective collisions.
Since the rate depends on the frequency of effective collisions, the reaction rate increases dramatically, which is reflected in a larger value for the rate constant, $k$.

Important Note: Changing concentration, pressure, or adding a catalyst changes the rate but does not change the value of $k$ (except for a very specific type of surface reaction). Only temperature changes $k$.

Checklist: Common Mistakes to Avoid

Confusing Order and Stoichiometry: NEVER use coefficients from the balanced equation to write the rate law. Always use experimental data.
Units of $k$: Always derive the units for $k$ based on the overall order.
Effect of Temperature: Temperature affects $k$. Concentration and pressure affect the rate directly but do not change the numerical value of $k$.
RDS Species: Intermediates formed during the reaction cannot be in the rate equation; only initial reactants or catalysts can be.

Overall Order	Units of Rate Equation	Units of \(k\)	\(k\) Calculation Example
0	\(Rate = k\)	\(\text{mol}\ \text{dm}^{-3}\ \text{s}^{-1}\)	\(k = \frac{\text{mol}\ \text{dm}^{-3}\ \text{s}^{-1}}{1}\)
1	\(Rate = k [\text{A}]\)	\(\text{s}^{-1}\)	\(k = \frac{\text{mol}\ \text{dm}^{-3}\ \text{s}^{-1}}{\text{mol}\ \text{dm}^{-3}} = \text{s}^{-1}\)
2	\(Rate = k [\text{A}]^2\)	\(\text{dm}^{3}\ \text{mol}^{-1}\ \text{s}^{-1}\)	\(k = \frac{\text{mol}\ \text{dm}^{-3}\ \text{s}^{-1}}{(\text{mol}\ \text{dm}^{-3})^2} = \text{dm}^{3}\ \text{mol}^{-1}\ \text{s}^{-1}\)
3	\(Rate = k [\text{A}]^3\)	\(\text{dm}^{6}\ \text{mol}^{-2}\ \text{s}^{-1}\)	\(k = \frac{\text{mol}\ \text{dm}^{-3}\ \text{s}^{-1}}{(\text{mol}\ \text{dm}^{-3})^3} = \text{dm}^{6}\ \text{mol}^{-2}\ \text{s}^{-1}\)