1. Rate Equations, Orders & The Rate Constant
Chemical kinetics examines the rate at which chemical reactions occur and the molecular pathways (mechanisms) they follow. For any general reaction:
A + B -> Products
The relationship between reactant concentrations and reaction rate is expressed mathematically by the rate equation:
Rate = k [A]^m [B]^n
- Order of reaction with respect to a reactant: The power to which the concentration of that reactant is raised in the rate equation (m for A, n for B).
- Overall order of reaction: The sum of individual reaction orders: Overall Order = m + n.
- Rate constant (k): The proportionality constant linking rate to reactant concentrations at a specific temperature. The value of k is independent of concentration but increases exponentially with temperature.
Orders of Reaction (0, 1, and 2)
- Zero Order (m = 0): Rate is independent of [A]. Doubling [A] has zero effect on rate (2^0 = 1).
- First Order (m = 1): Rate is directly proportional to [A]. Doubling [A] doubles the rate (2^1 = 2).
- Second Order (m = 2): Rate is proportional to [A]^2. Doubling [A] quadruples the rate (2^2 = 4); tripling [A] increases rate ninefold (3^2 = 9).
Deducing the Units of the Rate Constant (k)
Units of k vary depending on overall reaction order. Rearrange the rate equation to isolate k and substitute standard units (Rate in mol dm-3 s-1; Concentration in mol dm-3):
| Overall Order | Representative Rate Law | Rearranged Expression for k | Derived Units of k |
|---|---|---|---|
| Zero | Rate = k | k = Rate | mol dm-3 s-1 |
| First | Rate = k [A] | k = Rate / [A] | s-1 |
| Second | Rate = k [A]^2 | k = Rate / [A]^2 | mol-1 dm3 s-1 |
| Third | Rate = k [A]^2 [B] | k = Rate / ([A]^2 [B]) | mol-2 dm6 s-1 |
2. Determining Reaction Orders Experimentally
Reaction orders cannot be deduced from the balanced stoichiometric equation; they must be determined experimentally.
The Initial Rates Method
A series of experiments is carried out at constant temperature where the initial concentration of one reactant is varied while all other reactant concentrations are kept constant. The initial rate is measured for each run (e.g. from the initial gradient of a concentration-time graph or an iodine clock timer).
Continuous Monitoring & Half-Life Analysis
Progress of a reaction can be continuously monitored using:
- Gas syringe: Measuring volume of gas evolved over time (e.g. Mg + 2HCl -> MgCl2 + H2).
- Balance: Tracking mass loss over time for reactions producing heavy gases like CO2.
- Colorimetry: Measuring light absorbance as a coloured species is formed or consumed.
A graph of concentration against time provides immediate evidence of reaction order:
- Zero order: Constant negative gradient (straight line). Rate is constant regardless of concentration.
- First order: Exponential decay curve with a constant half-life (t_1/2). The time taken for reactant concentration to halve is independent of starting concentration: t_1/2 = ln 2 / k = 0.693 / k.
- Second order: Rapid initial drop followed by an extended shallow curve. Successive half-lives double with each halving of concentration.
3. The Rate-Determining Step (RDS)
Most chemical reactions proceed through a multi-step sequence of elementary collisions known as the reaction mechanism. One step in this sequence is significantly slower than all others; this is the rate-determining step (RDS).
Rules Connecting Rate Equations to Reaction Mechanisms
- Only species that take part in the rate-determining step, or in fast equilibrium steps that precede it, appear in the experimental rate equation.
- The order of reaction with respect to each reactant equals the stoichiometric coefficient of that species in (or prior to) the rate-determining step.
- Catalysts appear in the rate equation because they participate in the RDS, but they are regenerated in subsequent steps and do not appear in the overall chemical equation.
4. The Arrhenius Equation
The rate constant k increases exponentially with temperature because higher temperatures give a much larger fraction of molecules kinetic energy equal to or greater than the activation energy (Ea). This relationship is quantified by the Arrhenius equation:
k = A * e^(-Ea / (R * T))
Where:
- k: Rate constant (units depend on overall reaction order).
- A: Pre-exponential factor (frequency factor / Arrhenius constant), representing the frequency of collisions with correct orientation (same units as k).
- Ea: Activation energy in J mol-1.
- R: Gas constant = 8.314 J mol-1 K-1.
- T: Absolute temperature in Kelvin (K): T(K) = theta(degrees C) + 273.15.
The Logarithmic Form
Taking the natural logarithm (ln) of both sides converts the exponential equation into a linear equation matching y = mx + c:
ln k = (-Ea / R) * (1 / T) + ln A
Comparing this with y = mx + c:
- y: ln k
- x: 1 / T
- m (gradient): -Ea / R
- c (y-intercept): ln A
5. Arrhenius Graph Analysis
By measuring the rate constant k at several different temperatures and plotting ln k against 1/T, the activation energy and pre-exponential factor can be determined graphically:
The line on an Arrhenius plot has a negative gradient. When calculating Ea:
Ea = -(Gradient) * R = -(negative value) * 8.314 J mol-1 K-1.
The two negative signs cancel, giving a positive activation energy. Remember to divide by 1000 if the exam question requests Ea in kJ mol-1!
6. Worked Calculations
Exp 1: [A]=0.10, [B]=0.10, [C]=0.10 mol dm-3; Rate = 2.0 x 10^-4 mol dm-3 s-1
Exp 2: [A]=0.20, [B]=0.10, [C]=0.10 mol dm-3; Rate = 8.0 x 10^-4 mol dm-3 s-1
Exp 3: [A]=0.10, [B]=0.20, [C]=0.10 mol dm-3; Rate = 4.0 x 10^-4 mol dm-3 s-1
Exp 4: [A]=0.10, [B]=0.10, [C]=0.20 mol dm-3; Rate = 2.0 x 10^-4 mol dm-3 s-1
Deduce the rate equation, calculate the value of the rate constant k, and state its units.
Step 1: Determine reaction orders:
- Compare Exp 1 and Exp 2: [B] and [C] constant. [A] doubles (x2), Rate quadruples (x4). Since 2^2 = 4, the reaction is second order with respect to A.
- Compare Exp 1 and Exp 3: [A] and [C] constant. [B] doubles (x2), Rate doubles (x2). Since 2^1 = 2, the reaction is first order with respect to B.
- Compare Exp 1 and Exp 4: [A] and [B] constant. [C] doubles (x2), Rate remains unchanged (x1). Since 2^0 = 1, the reaction is zero order with respect to C.
Step 2: State the rate equation:
Rate = k [A]^2 [B]
Step 3: Calculate k using Experiment 1:
k = Rate / ([A]^2 [B]) = (2.0 x 10^-4) / ((0.10)^2 * 0.10) = (2.0 x 10^-4) / (1.0 x 10^-3) = 0.20
Step 4: Deduce units of k:
Units = (mol dm-3 s-1) / ((mol dm-3)^2 * (mol dm-3)) = (mol dm-3 s-1) / (mol3 dm-9) = mol-2 dm6 s-1
Final Answer: k = 0.20 mol-2 dm6 s-1
Calculate the activation energy (Ea) in kJ mol-1.
Step 1: State the relationship between gradient and activation energy:
Gradient = -Ea / R
Step 2: Rearrange for Ea:
Ea = -(Gradient) * R
Ea = -(-1.24 x 10^4 K) * 8.314 J mol-1 K-1 = +1.031 x 10^5 J mol-1
Step 3: Convert to kJ mol-1:
Ea = 1.031 x 10^5 / 1000 = 103.1 kJ mol-1
Exam-Style Practice Questions
Test your understanding of these core syllabus concepts with targeted questions and detailed explanations.
Question 1: A reaction has the rate equation Rate = k [NO]^2 [O2]. If the concentration of NO is doubled while the concentration of O2 is halved, how does the initial rate change?
Show Answer & Explanation
Correct Answer: A
Doubling [NO] quadruples the rate (2^2 = 4). Halving [O2] halves the rate (0.5^1 = 0.5). Combined effect: 4 * 0.5 = 2 times faster.
Question 2: What are the units of the rate constant k for a reaction that is first order overall?
Show Answer & Explanation
Correct Answer: B
k = Rate / [A] = (mol dm-3 s-1) / (mol dm-3) = s-1.
Question 3: A reaction mechanism consists of two steps: Step 1: NO2 + NO2 -> NO3 + NO (slow); Step 2: NO3 + CO -> NO2 + CO2 (fast). What is the expected rate equation?
Show Answer & Explanation
Correct Answer: B
The slow step is the rate-determining step. It involves two molecules of NO2 colliding, so the rate law is Rate = k [NO2]^2. Species in subsequent fast steps (CO) do not enter the rate equation.
Question 4: If temperature is increased from 300 K to 310 K, the rate of many reactions approximately doubles. Why does this occur?
Show Answer & Explanation
Correct Answer: C
The Maxwell-Boltzmann distribution shows that a modest 10 K temperature rise greatly increases the population of molecules possessing kinetic energy greater than or equal to Ea.