Rate Equations (Rate = k[A]ᵐ[B]ⁿ), Rate-Determining Step & The Arrhenius Equation
Chemical kinetics investigates how the concentrations of individual reactants and temperature govern the speed of a reaction through the experimentally determined rate equation, Rate = k[A]ᵐ[B]ⁿ. Critically, the orders of reaction (m and n) cannot be deduced from the overall balanced stoichiometric equation; instead, they reflect the exact molecularity of the species participating in or up to the slowest step of the mechanism, known as the rate-determining step (RDS). Meanwhile, the rate constant k increases exponentially with temperature according to the Arrhenius equation, k = A e^(-Eₐ / RT), because a higher temperature exponentially increases the fraction of colliding particles with energy ≥ Eₐ on a Maxwell-Boltzmann distribution. Examiners heavily test initial-rate table deductions, rate-concentration vs concentration-time graphs, iodine clock calculations, and linear Arrhenius plots of ln(k) against 1/T.
- Rate Equation, Orders of Reaction & Units of k
- For a reaction A + B → Products, Rate = k[A]ᵐ[B]ⁿ, where Rate has units of mol dm⁻³ s⁻¹, k is the temperature-dependent rate constant, m is the order with respect to A, n is the order with respect to B, and Overall Order = m + n. Zero order (m = 0): changing [A] has no effect on rate. First order (m = 1): doubling [A] doubles the rate (×2¹ = ×2). Second order (m = 2): doubling [A] quadruples the rate (×2² = ×4); tripling [A] multiplies the rate by 9 (×3² = ×9). Units of k are found by rearranging k = Rate / ([A]ᵐ[B]ⁿ) and cancelling mol dm⁻³.
- Concentration-Time vs Rate-Concentration Graphs & Half-Life (t₁/₂)
- 1. Zero Order: Concentration vs Time is a straight line with a negative gradient (gradient = -k); Rate vs Concentration is a horizontal flat line. 2. First Order: Concentration vs Time is a curve with a constant half-life (t₁/₂ = ln(2) / k, independent of concentration); Rate vs Concentration is a straight line through the origin (gradient = k). 3. Second Order: Concentration vs Time is a steeper curve where successive half-lives increase as concentration halves; Rate vs Concentration is an upward-curving parabola (Rate ∝ [A]²), so Rate vs [A]² is a straight line through the origin.
- Initial Rates Method & Iodine Clock Reactions
- In an iodine clock reaction (e.g. H₂O₂ + 2I⁻ + 2H⁺ → I₂ + 2H₂O), a small, fixed amount of sodium thiosulfate (S₂O₃²⁻) and starch indicator are added. As I₂ is formed, S₂O₃²⁻ immediately reduces it back to I⁻ until all the S₂O₃²⁻ is used up, at which point free I₂ reacts with starch to turn the solution suddenly blue-black at time t. Initial rate ∝ 1 / t. By varying the initial concentration of one reactant while keeping others constant, orders are found.
Rate Equations (Rate = k[A]ᵐ[B]ⁿ), Rate-Determining Step & The Arrhenius Equation: Reaction Profile (ΔH & E_a) + Le Chatelier Yield Simulator
Toggle a catalyst on the enthalpy profile diagram or adjust temperature and pressure to see how dynamic equilibrium shifts.
Modelled on N₂(g) + 3H₂(g) ⇌ 2NH₃(g) (ΔH = −92 kJ/mol) (4 moles gas → 2 moles gas):
Higher pressure shifts equilibrium to the side with fewer gas moles (right, 4 → 2). Lower temperature favours the exothermic forward reaction.
Key Concepts & Exam Definitions (6)
- Rate Equation, Orders of Reaction & Units of k
- For a reaction A + B → Products, Rate = k[A]ᵐ[B]ⁿ, where Rate has units of mol dm⁻³ s⁻¹, k is the temperature-dependent rate constant, m is the order with respect to A, n is the order with respect to B, and Overall Order = m + n. Zero order (m = 0): changing [A] has no effect on rate. First order (m = 1): doubling [A] doubles the rate (×2¹ = ×2). Second order (m = 2): doubling [A] quadruples the rate (×2² = ×4); tripling [A] multiplies the rate by 9 (×3² = ×9). Units of k are found by rearranging k = Rate / ([A]ᵐ[B]ⁿ) and cancelling mol dm⁻³.
- Concentration-Time vs Rate-Concentration Graphs & Half-Life (t₁/₂)
- 1. Zero Order: Concentration vs Time is a straight line with a negative gradient (gradient = -k); Rate vs Concentration is a horizontal flat line. 2. First Order: Concentration vs Time is a curve with a constant half-life (t₁/₂ = ln(2) / k, independent of concentration); Rate vs Concentration is a straight line through the origin (gradient = k). 3. Second Order: Concentration vs Time is a steeper curve where successive half-lives increase as concentration halves; Rate vs Concentration is an upward-curving parabola (Rate ∝ [A]²), so Rate vs [A]² is a straight line through the origin.
- Initial Rates Method & Iodine Clock Reactions
- In an iodine clock reaction (e.g. H₂O₂ + 2I⁻ + 2H⁺ → I₂ + 2H₂O), a small, fixed amount of sodium thiosulfate (S₂O₃²⁻) and starch indicator are added. As I₂ is formed, S₂O₃²⁻ immediately reduces it back to I⁻ until all the S₂O₃²⁻ is used up, at which point free I₂ reacts with starch to turn the solution suddenly blue-black at time t. Initial rate ∝ 1 / t. By varying the initial concentration of one reactant while keeping others constant, orders are found.
- Reaction Mechanisms & The Rate-Determining Step (RDS)
- Most reactions occur via a multi-step mechanism where individual steps proceed at different speeds. The slowest step is the Rate-Determining Step (RDS), which acts as a kinetic bottleneck. Rule: The orders of reactants in the rate equation equal the stoichiometric number of moles of each reactant involved IN or before the rate-determining step. Any reactant that appears in the overall chemical equation with zero order (does not appear in the rate equation) must enter the mechanism in a fast step after the rate-determining step.
- The Arrhenius Equation: k = A e^(-Eₐ / RT)
- Relates the rate constant k to activation energy Eₐ (in J mol⁻¹!), gas constant R (8.31 J K⁻¹ mol⁻¹), and absolute temperature T (in K): k = A e^(-Eₐ / RT), where A is the pre-exponential (Arrhenius) factor linked to collision frequency and steric orientation. Taking natural logs gives the linear form: ln(k) = (-Eₐ / R) × (1 / T) + ln(A).
- Arrhenius Graphs: Plotting ln(k) against 1/T
- Plotting ln(k) (or ln(1/t)) on the y-axis against 1/T (in K⁻¹) on the x-axis gives a straight line of the form y = mx + c with a negative gradient: 1. Gradient m = -Eₐ / R, so Activation Energy Eₐ (in J mol⁻¹) = -Gradient × 8.31 (divide by 1000 to convert to kJ mol⁻¹!). 2. The y-intercept (at 1/T = 0) equals ln(A), so A = e^(y-intercept).
Quick-Fire Flashcards (6)
Click any card to check the answer, then mark it as Known to track your progress.
If Rate = k[A][B]², how does the rate change when both [A] and [B] are tripled, and what are the units of k?
How can you prove from a Concentration vs Time graph that a reaction is first order with respect to that reactant?
Why does propanone react with iodine in acid according to Rate = k[CH₃COCH₃][H⁺], with zero order with respect to [I₂]?
Why does a small increase in temperature (e.g. 10°C) cause a large increase in the rate constant k?
On an Arrhenius graph of ln(k) against 1/T (in K⁻¹), how do you calculate the activation energy Eₐ in kJ mol⁻¹ from the gradient?
How do you measure the continuous rate of a reaction that produces a gas, or one that consumes coloured iodine (I₂)?
Exam-Style Questions & Mark Schemes (4)
Write or think through your answer first, then click Check Mark Scheme to see the exact points examiners award marks for.
- Q1
The reaction between nitrogen monoxide and hydrogen at 1000 K is: 2NO(g) + 2H₂(g) → N₂(g) + 2H₂O(g). Initial rates were measured in three experiments: Exp 1 ([NO] = 0.010 M, [H₂] = 0.010 M, Rate = 6.0 × 10⁻⁵ mol dm⁻³ s⁻¹); Exp 2 ([NO] = 0.020 M, [H₂] = 0.010 M, Rate = 2.4 × 10⁻⁴ mol dm⁻³ s⁻¹); Exp 3 ([NO] = 0.020 M, [H₂] = 0.030 M, Rate = 7.2 × 10⁻⁴ mol dm⁻³ s⁻¹). Deduce the rate equation, calculate the value and units of the rate constant k, and propose a two-step mechanism consistent with the rate equation. (6 marks)
Click “Check Mark Scheme” to compare your answer, 1. Compare Exp 1 and Exp 2 ([H₂] constant at 0.010 M): [NO] doubles (×2) and Rate increases from 6.0 × 10⁻⁵ to 2.4 × 10⁻⁴ (×4 = ×2²), so the reaction is second order with respect to NO [1 mark]. 2. Compare Exp 2 and Exp 3 ([NO] constant at 0.020 M): [H₂] triples (×3) and Rate increases from 2.4 × 10⁻⁴ to 7.2 × 10⁻⁴ (×3 = ×3¹), so the reaction is first order with respect to H₂ [1 mark]. 3. Rate equation: Rate = k[NO]²[H₂] [1 mark]. 4. Substitute Exp 1 into the rate equation: k = Rate ÷ ([NO]²[H₂]) = (6.0 × 10⁻⁵) ÷ ((0.010)² × 0.010) = (6.0 × 10⁻⁵) ÷ (1.0 × 10⁻⁶) = 60 [1 mark]. 5. Units of k: (mol dm⁻³ s⁻¹) ÷ (mol dm⁻³)³ = mol⁻² dm⁶ s⁻¹ [1 mark]. 6. Two-step mechanism: Step 1 (Slow / RDS): 2NO + H₂ → N₂ + H₂O₂ (or N₂O + H₂O); Step 2 (Fast): H₂O₂ + H₂ → 2H₂O (or N₂O + H₂ → N₂ + H₂O) [1 mark].
- Q2
Two mechanisms occur for the nucleophilic substitution of bromoalkanes with hydroxide ions (OH⁻): Primary bromoethane (CH₃CH₂Br) obeys the rate equation Rate = k[CH₃CH₂Br][OH⁻], whereas tertiary 2-bromo-2-methylpropane ((CH₃)₃CBr) obeys the rate equation Rate = k[(CH₃)₃CBr]. Explain what these two rate equations reveal about the rate-determining steps of their mechanisms. (4 marks)
Click “Check Mark Scheme” to compare your answer, 1. For primary bromoethane, both [CH₃CH₂Br] and [OH⁻] appear in the rate equation as first order (overall second order, SN2) [1 mark]. 2. This proves that both a bromoethane molecule and a hydroxide ion collide together in a single, concerted rate-determining step (simultaneous C-Br bond breaking and C-OH bond forming via a 5-coordinate transition state) [1 mark]. 3. For tertiary 2-bromo-2-methylpropane, [OH⁻] does not appear in the rate equation (zero order with respect to OH⁻, overall first order, SN1) [1 mark]. 4. This proves that the slow rate-determining step involves only the heterolytic fission of the C-Br bond in (CH₃)₃CBr to form a planar tertiary carbocation intermediate, which then reacts with OH⁻ in a subsequent fast step [1 mark].
- Q3
A reaction has an activation energy Eₐ = 54.0 kJ mol⁻¹ and a rate constant k = 2.50 × 10⁻³ s⁻¹ at 298 K. Using the Arrhenius equation k = A e^(-Eₐ / RT) (where R = 8.31 J K⁻¹ mol⁻¹), calculate the value of the Arrhenius constant A and the value of the rate constant k at 318 K. (4 marks)
Click “Check Mark Scheme” to compare your answer, 1. Convert Eₐ to J mol⁻¹: Eₐ = 54,000 J mol⁻¹. Calculate Eₐ / (RT) at 298 K: 54,000 ÷ (8.31 × 298) = 54,000 ÷ 2476.38 = 21.806 [1 mark]. 2. Calculate e^(-21.806) = 3.386 × 10⁻¹⁰, then rearrange for A: A = k ÷ e^(-Eₐ/RT) = (2.50 × 10⁻³) ÷ (3.386 × 10⁻¹⁰) = 7.38 × 10⁶ s⁻¹ [1 mark]. 3. At T = 318 K, calculate Eₐ / (RT) = 54,000 ÷ (8.31 × 318) = 54,000 ÷ 2642.58 = 20.435 [1 mark]. 4. Calculate k at 318 K: k = (7.38 × 10⁶) × e^(-20.435) = (7.38 × 10⁶) × (1.334 × 10⁻⁹) = 9.85 × 10⁻³ s⁻¹ (nearly 4 times faster for a 20 K rise) [1 mark].
- Q4
A student plotted a graph of ln(k) on the y-axis against 1/T (in K⁻¹) on the x-axis for the decomposition of dinitrogen pentoxide (N₂O₅). The line of best fit had a gradient of -1.24 × 10⁴ K. Calculate the activation energy (Eₐ) of the reaction in kJ mol⁻¹ (R = 8.31 J K⁻¹ mol⁻¹) and sketch how adding a catalyst would change the line on this graph. (4 marks)
Click “Check Mark Scheme” to compare your answer, 1. From ln(k) = (-Eₐ / R)(1 / T) + ln(A), the gradient of the line equals -Eₐ / R [1 mark]. 2. Calculate Eₐ in J mol⁻¹: Eₐ = -Gradient × R = -(-1.24 × 10⁴ K) × 8.31 J K⁻¹ mol⁻¹ = 103,044 J mol⁻¹ [1 mark]. 3. Convert to kJ mol⁻¹ (÷ 1000): Eₐ = +103 kJ mol⁻¹ (to 3 s.f) [1 mark]. 4. Adding a catalyst provides an alternative pathway with a lower activation energy (Eₐ), so the line on the ln(k) vs 1/T graph would be less STEEP (shallower negative gradient) and shifted higher up (higher ln(k) at any given 1/T) [1 mark].
Common Exam Questions (4)
What is the most common mistake students make with the Arrhenius equation?
Forgetting to convert Activation Energy (Eₐ) from kJ mol⁻¹ into J mol⁻¹ (or vice versa). Because the gas constant R is 8.31 J K⁻¹ mol⁻¹ (in Joules), Eₐ must be in J mol⁻¹ whenever you substitute into k = A e^(-Eₐ / RT) or calculate Eₐ = -Gradient × R.
Why must only a very small, fixed amount of sodium thiosulfate be added in an iodine clock experiment?
The initial rate approximation (Rate ≈ Δ[I₂] / t) is only valid if the reaction has proceeded less than ~5-10% toward completion so that reactant concentrations have barely changed from their initial values. If too much thiosulfate is added, the blue-black colour appears too late after reactant concentrations have significantly fallen.
Can a catalyst appear in the rate equation?
Yes. If a homogeneous catalyst (such as H⁺(aq) in the iodination of propanone or ester hydrolysis) participates in the slow rate-determining step (or a fast equilibrium step prior to the RDS), its concentration [H⁺] appears directly in the rate equation.
How do you find the rate of reaction at a specific time t (e.g. t = 60 s) from a Concentration vs Time curve?
Use a ruler to draw a straight tangent touching the curve at exactly t = 60 s, then calculate the magnitude of the gradient of that tangent using Gradient = Δ(Concentration) / Δ(Time) in mol dm⁻³ s⁻¹.
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