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The Arrhenius equation, k = A·exp(−Ea/RT), is the quantitative link between temperature and the rate constant of a chemical reaction. Proposed by Svante Arrhenius in 1889 on empirical grounds, the equation captures something deep: only those molecules in the high-energy tail of the Maxwell–Boltzmann distribution can react, and the fraction of such molecules rises exponentially with T. Two parameters carry all the chemistry — Ea, the energy barrier that must be surmounted in the transition state, and A, the pre-exponential factor that aggregates collision frequency and steric (orientational) requirements. Taking natural logarithms gives the workhorse form ln k = ln A − Ea/RT, a straight-line relationship in ln k vs 1/T whose gradient yields Ea directly. This lesson develops graphical Ea determination, the two-temperature shortcut, predictions of k at new temperatures, the quantitative effect of catalysts, and the bridge between the Boltzmann fraction and the Arrhenius exponential.
Spec mapping (AQA 7405): This lesson maps to §3.1.9 (rate equations), specifically the requirement that students understand the Arrhenius equation k = A·exp(−Ea/RT) and its logarithmic form ln k = ln A − Ea/RT, and can determine Ea from a plot of ln k against 1/T. It draws directly on lesson 0 (collision theory and the Maxwell–Boltzmann distribution) where the exp(−Ea/RT) factor originates, lesson 1 (rate equations and the meaning of the rate constant k) where the quantity being predicted is defined, and lesson 3 (reaction mechanisms — different elementary steps have different activation energies, and the rate-determining step's Ea is the one measured macroscopically). It also draws on §3.1.4 (energetics, A2) for the underlying concept of activation energy and on §3.1.5 (the foundation kinetics topic) for the qualitative effect of temperature on rate. Refer to the official AQA specification document for the exact wording of each section.
Assessment objectives: Stating the Arrhenius equation and the physical meanings of A and Ea are AO1 recall items. Rearranging the equation into logarithmic form and computing Ea from k(T) data — either graphically (gradient of ln k vs 1/T) or via the two-temperature shortcut — is AO2 and appears regularly on Paper 1 and Paper 3. AO3 work includes assessing how the two-temperature shortcut formula is derived from the log form, evaluating the quantitative effect of a catalyst on k by separating the Ea contribution from the A contribution, and using the Arrhenius framework to predict k at a new temperature given Ea and a single (k, T) datum. Synoptic AO3 questions may ask candidates to connect the Boltzmann fraction exp(−Ea/RT) to the Arrhenius exponential, or to compare empirical Arrhenius behaviour with the rigorous transition-state-theory result.
The rate constant k of an elementary or rate-determining step depends on absolute temperature according to:
k = A·exp(−Ea/RT)
where:
The exponential term exp(−Ea/RT) is the Boltzmann fraction: the fraction of molecules in a thermalised population whose kinetic energy exceeds Ea. As T increases, −Ea/RT becomes less negative (closer to zero), so the exponential factor grows toward unity, k grows, and the reaction speeds up. The growth is steeply non-linear in T because the exponential is supercharged: a modest rise in T translates into a large change in k.
Key Point: The Arrhenius equation has only two adjustable parameters per reaction: A and Ea. Once these are known, k can be predicted at any temperature within the range where the reaction mechanism is unchanged. Because A and Ea are themselves weakly temperature-dependent (and at the rigorous level non-constant — see transition-state theory below), Arrhenius behaviour is strictly empirical. For exam purposes A and Ea are treated as temperature-independent constants.
Taking natural logarithms of both sides of k = A·exp(−Ea/RT) gives:
ln k = ln A − Ea/(RT)
Rearranging to match the y = mx + c form of a straight line:
ln k = (−Ea/R) × (1/T) + ln A
with:
A graph of ln k against 1/T is therefore a straight line with negative gradient. The gradient times −R recovers the activation energy; the intercept exponentiated recovers the pre-exponential factor:
Note that the y-intercept is the value of ln k when 1/T = 0, which corresponds to T → ∞. This is an extrapolation of the line — never an experimentally accessible data point — but the algebraic intercept of the best-fit straight line is what determines ln A.
Exam Tip: Mark schemes routinely require students to state explicitly that the gradient equals −Ea/R (with the minus sign). A response that writes "gradient = Ea/R" will lose the AO1 mark even if the numerical answer for Ea comes out correctly because the candidate quietly negated twice.
The decomposition of dinitrogen pentoxide, 2 N₂O₅(g) → 4 NO₂(g) + O₂(g), is a first-order reaction. The first-order rate constant has been measured at five temperatures:
| T / K | 1/T / 10⁻³ K⁻¹ | k / s⁻¹ | ln k |
|---|---|---|---|
| 273 | 3.663 | 7.87 × 10⁻⁷ | −14.05 |
| 298 | 3.356 | 3.46 × 10⁻⁵ | −10.27 |
| 308 | 3.247 | 1.35 × 10⁻⁴ | −8.91 |
| 318 | 3.145 | 4.98 × 10⁻⁴ | −7.60 |
| 328 | 3.049 | 1.50 × 10⁻³ | −6.50 |
(Data adapted from textbook tabulations of the N₂O₅ decomposition; the exact values used historically are those associated with the early-twentieth-century studies of Lewis and others.)
Step 1 — convert. Compute 1/T in K⁻¹ for each row and ln k for each row, as shown in the table.
Step 2 — plot. Plot ln k (y-axis) against 1/T (x-axis). The five points lie on a straight line.
Step 3 — gradient. Compute the gradient using two well-separated points (always choose extremes, not adjacent points, for the cleanest answer). Using (1/T, ln k) = (3.663 × 10⁻³, −14.05) and (3.049 × 10⁻³, −6.50):
gradient = (−6.50 − (−14.05)) / (3.049 × 10⁻³ − 3.663 × 10⁻³) = 7.55 / (−6.14 × 10⁻⁴) = −1.23 × 10⁴ K
Step 4 — extract Ea.
Ea = −gradient × R = −(−1.23 × 10⁴) × 8.31 = 1.02 × 10⁵ J mol⁻¹ = 102 kJ mol⁻¹
This is in good agreement with the literature value for the activation energy of N₂O₅ decomposition (~103 kJ mol⁻¹), confirming that the gas-phase decomposition follows a single rate-determining elementary step over this temperature range.
Step 5 — extract A. The y-intercept is the value of ln k when 1/T = 0. Extrapolating the straight line (or rearranging at one data point): from (1/T, ln k) = (3.356 × 10⁻³, −10.27),
ln A = ln k + Ea/(RT) = −10.27 + (1.02 × 10⁵)/(8.31 × 298) = −10.27 + 41.18 = 30.91
A = e^30.91 ≈ 2.7 × 10¹³ s⁻¹
This A-factor is typical for a unimolecular gas-phase decomposition: ~10¹³ s⁻¹ is the rule-of-thumb frequency factor for a "loose" transition state.
If only two data points are available, plotting a line is statistically meaningless (any two points define a line trivially). Instead, the two-temperature shortcut is derived directly from the log form:
At T₁: ln k₁ = ln A − Ea/(RT₁) At T₂: ln k₂ = ln A − Ea/(RT₂)
Subtracting the first from the second eliminates ln A:
ln(k₂/k₁) = (Ea/R) × (1/T₁ − 1/T₂)
Note the order: the numerator inside the logarithm is k₂ (the rate constant at T₂) and the bracket on the right is (1/T₁ − 1/T₂) — same subscripts, opposite order. The sign convention is so error-prone that examiners often phrase the question to allow either ordering, but you must keep your own working internally consistent.
Worked example. At 300 K, k₁ = 1.4 × 10⁻³ s⁻¹. At 320 K, k₂ = 5.6 × 10⁻³ s⁻¹. Calculate Ea.
ln(k₂/k₁) = ln(5.6 × 10⁻³ / 1.4 × 10⁻³) = ln(4.0) = 1.386
1/T₁ − 1/T₂ = 1/300 − 1/320 = 3.333 × 10⁻³ − 3.125 × 10⁻³ = 2.083 × 10⁻⁴ K⁻¹
Ea = R × ln(k₂/k₁) / (1/T₁ − 1/T₂) = (8.31 × 1.386) / (2.083 × 10⁻⁴) = 11.52 / 2.083 × 10⁻⁴
Ea = 55 300 J mol⁻¹ = 55.3 kJ mol⁻¹
A 20 K rise quadrupling the rate constant is consistent with Ea in the 50–60 kJ mol⁻¹ range — physically reasonable for many room-temperature gas- or solution-phase reactions.
Exam Tip: Convert Ea to J mol⁻¹ (not kJ mol⁻¹) before any substitution into the Arrhenius equation, because R is in J K⁻¹ mol⁻¹. Most students who lose marks on this topic in Paper 1 forget that 55.3 kJ mol⁻¹ = 55 300 J mol⁻¹, drop a factor of 1000, and get answers 1000× off.
The same two-temperature formula rearranges to give k at a new T given Ea and a single anchor (k₁, T₁).
A reaction has Ea = 50.0 kJ mol⁻¹ = 5.00 × 10⁴ J mol⁻¹, with k₁ = 2.0 × 10⁻² s⁻¹ at T₁ = 298 K. Predict k₂ at T₂ = 308 K.
ln(k₂/k₁) = (Ea/R) × (1/T₁ − 1/T₂) = (5.00 × 10⁴ / 8.31) × (1/298 − 1/308) = 6017 × (3.356 × 10⁻³ − 3.247 × 10⁻³) = 6017 × (1.09 × 10⁻⁴) = 0.656
k₂/k₁ = e^0.656 = 1.927
k₂ = 1.927 × 2.0 × 10⁻² = 3.85 × 10⁻² s⁻¹
The rate constant has nearly doubled for a 10 K rise from 298 K to 308 K — consistent with the "rate doubles for a 10 K rise" rule of thumb that is genuinely a consequence of Arrhenius behaviour when Ea is roughly 50–55 kJ mol⁻¹. The rule is reliable only over this approximate Ea range and only over modest temperature intervals; for reactions with much higher or lower Ea, the same 10 K rise produces a much larger or much smaller k₂/k₁ ratio.
A catalyst opens an alternative reaction pathway with a lower activation energy. The pre-exponential factor A may also change slightly (different transition-state geometry, different steric demands), but in almost every case the Ea contribution dominates by orders of magnitude. The dramatic acceleration on adding a catalyst is therefore almost entirely an Ea effect — easy to quantify.
Worked example. Consider a hypothetical reaction with Ea (uncatalysed) = 150 kJ mol⁻¹ and Ea (catalysed) = 75 kJ mol⁻¹, both at 298 K. Assume A is unchanged (a reasonable first approximation). The ratio of catalysed to uncatalysed rate constants is:
k_cat / k_uncat = exp(−Ea_cat/RT) / exp(−Ea_uncat/RT) = exp[(Ea_uncat − Ea_cat)/(RT)] = exp[(150 000 − 75 000) / (8.31 × 298)] = exp[75 000 / 2476] = exp(30.3) = 1.4 × 10¹³
A halving of Ea has accelerated the reaction by a factor of roughly 10¹³ at room temperature. This is why enzymes and industrial catalysts work: a barrier reduction of just 40–80 kJ mol⁻¹ converts a reaction that takes geological time into one that occurs in milliseconds. Conversely, the same calculation explains why reactions with Ea > 200 kJ mol⁻¹ effectively do not proceed at room temperature without a catalyst — the Boltzmann fraction of sufficiently energetic molecules is vanishingly small.
Common Misconception: A catalyst does not change ΔH or the equilibrium position — it changes only the kinetics. The Arrhenius equation predicts k; it says nothing about thermodynamic driving force. A catalyst lowers Ea for both the forward and the reverse reaction by the same amount, so K_eq is preserved.
The exp(−Ea/RT) term in the Arrhenius equation is not a coincidence. From the Maxwell–Boltzmann distribution of molecular speeds (lesson 0 of this course), the fraction of molecules in a thermalised gas-phase population whose kinetic energy exceeds Ea is, to a very good approximation,
fraction(E ≥ Ea) ≈ exp(−Ea/RT)
(more rigorously, the integral of the Maxwell–Boltzmann distribution from Ea to infinity gives a slightly more complex expression involving the complementary error function, but in the high-Ea limit relevant to chemistry the exponential dominates).
When the rate of reaction is proportional to the rate at which sufficiently energetic collisions occur, the rate constant inherits this exponential dependence on temperature. The pre-exponential factor A then captures the rate at which collisions of any energy occur, weighted by orientation. Thus Arrhenius behaviour is a macroscopic manifestation of the microscopic Boltzmann distribution. This is the conceptual bridge between collision theory (microscopic, molecular) and macroscopic rate constants (laboratory observable).
The same exp(−Ea/RT) factor appears in the partition functions of statistical thermodynamics, in the equilibrium constant via the van 't Hoff equation, and in transition-state theory via the standard Gibbs energy of activation. Recognising this thread of exponentials running through physical chemistry is one of the marks of a strong A-level candidate.
The classic Required Practical 3 context (AQA RP3, effect of temperature on rate) generates real Arrhenius data. Two common school-laboratory experiments are:
For both, the principal sources of uncertainty are temperature drift in the bath (control by stirring and a tight tolerance, ±0.5 °C is achievable) and timing precision near the endpoint. Quoting Ea to three significant figures from school data is generous; ±10% is realistic. Cross-checking with a literature value (the H₂O₂ decomposition has Ea ≈ 75 kJ mol⁻¹ uncatalysed, much lower with MnO₂ or catalase) anchors the analysis.
Question 1. [13 marks total]
(a) State the Arrhenius equation and explain the physical meaning of the pre-exponential factor A and the activation energy Ea. [2 marks]
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