The Arrhenius Equation

Spec Mapping — OCR H432 Module 5.1.1 — How fast? The Arrhenius equation $k = Ae^{-E_a/(RT)}$k=Ae−Ea​/(RT) relating the rate constant of a reaction to absolute temperature, the pre-exponential (frequency) factor and the activation energy, the linearised form $\ln k = \ln A - E_a/(RT)$lnk=lnA−Ea​/(RT), the graphical determination of $E_a$Ea​ from a plot of $\ln k$lnk against $1/T$1/T (gradient $-E_a/R$−Ea​/R, intercept $\ln A$lnA), and the physical interpretation in terms of the Maxwell–Boltzmann distribution (refer to the official OCR H432 specification document for exact wording).

This lesson is the quantitative apex of OCR Module 5.1.1 — it tells us how much the rate constant changes when we change the temperature, and gives a graphical recipe for measuring the activation energy of any reaction. Svante Arrhenius proposed in 1889 (work that contributed to his 1903 Nobel Prize in Chemistry) that the rate constant depends on temperature according to $k = Ae^{-E_a/(RT)}$k=Ae−Ea​/(RT). The exponential term is the Boltzmann factor — the fraction of molecular collisions with enough energy to react — while the pre-exponential factor $A$A captures collision frequency and geometric requirements. Taking logarithms linearises the relationship, so a single plot of $\ln k$lnk against $1/T$1/T yields the activation energy directly from the gradient. We work through two complete examples — finding $E_a$Ea​ from a five-point dataset, and using the two-point form to compute the multiplicative factor for $k$k across a temperature jump — and link the physics back to the Maxwell–Boltzmann distribution from AS.

Key Equation: the Arrhenius equation $k = A\,e^{-E_a/(RT)}$k=Ae−Ea​/(RT) with linearised form $\ln k = \ln A - \frac{E_a}{R}\cdot\frac{1}{T}$lnk=lnA−REa​​⋅T1​ A plot of $\ln k$lnk (y) against $1/T$1/T (x) is a straight line with gradient $-E_a/R$−Ea​/R and intercept $\ln A$lnA. $R = 8.314\,\text{J K}^{-1}\,\text{mol}^{-1}$R=8.314J K−1mol−1; $T$T in kelvin; $E_a$Ea​ usually quoted in $\text{kJ mol}^{-1}$kJ mol−1 (divide J mol$^{-1}$−1 by 1000).

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The Arrhenius equation

In 1889, Svante Arrhenius proposed that the rate constant depends on temperature according to:

$k = A\,e^{-E_a/(RT)}$k=Ae−Ea​/(RT)

where:

• $k$k = rate constant (units depend on overall order — see Lesson 7).
• $A$A = pre-exponential factor (also called frequency factor), same units as $k$k.
• $E_a$Ea​ = activation energy in J mol$^{-1}$−1 — the minimum energy a colliding pair must have to react.
• $R$R = gas constant $= 8.314\,\text{J K}^{-1}\,\text{mol}^{-1}$=8.314J K−1mol−1.
• $T$T = temperature in kelvin.

The equation expresses two physical ideas:

1. $e^{-E_a/(RT)}$e−Ea​/(RT) is the Boltzmann factor — the fraction of colliding pairs with energy $\geq E_a$≥Ea​ at temperature $T$T. This is the probability that a collision has enough energy to react.
2. $A$A is the pre-exponential factor, which captures the frequency of collisions and the geometric requirement that they have the correct orientation. It is only weakly temperature-dependent compared with the exponential.

Multiplying the two gives the rate constant: $k =$k= (collisions per unit time per unit concentration product) $\times$× (probability that each collision reacts).

Why $T$T affects $k$k so dramatically

Even though $T$T appears only inside the exponent, a small change in $T$T changes $-E_a/(RT)$−Ea​/(RT) a lot — because $E_a$Ea​ is typically tens or hundreds of kJ mol$^{-1}$−1 while $RT$RT is only a few kJ mol$^{-1}$−1. A 10 K rise near room temperature roughly doubles $k$k for a typical reaction with $E_a \approx 50\,\text{kJ mol}^{-1}$Ea​≈50kJ mol−1.

A worked sanity check:

• At 298 K, with $E_a = 50\,000\,\text{J mol}^{-1}$Ea​=50000J mol−1: $-E_a/(RT) = -50000/(8.314 \times 298) = -20.18$−Ea​/(RT)=−50000/(8.314×298)=−20.18. Boltzmann factor $\approx 1.72 \times 10^{-9}$≈1.72×10−9.
• At 308 K: $-E_a/(RT) = -50000/(8.314 \times 308) = -19.52$−Ea​/(RT)=−50000/(8.314×308)=−19.52. Boltzmann factor $\approx 3.31 \times 10^{-9}$≈3.31×10−9.

Ratio $3.31/1.72 \approx 1.92$3.31/1.72≈1.92 — $k$k roughly doubles, matching the rule of thumb.

The linearised form

Taking natural logarithms of both sides of $k = Ae^{-E_a/(RT)}$k=Ae−Ea​/(RT):

$\ln k = \ln A + \ln\!\left(e^{-E_a/(RT)}\right) = \ln A - \frac{E_a}{RT}$lnk=lnA+ln(e−Ea​/(RT))=lnA−RTEa​​

Rearranged in the form $y = mx + c$y=mx+c:

$\ln k = -\frac{E_a}{R}\cdot\frac{1}{T} + \ln A$lnk=−REa​​⋅T1​+lnA

with:

• $y = \ln k$y=lnk
• $x = 1/T$x=1/T
• gradient $m = -E_a/R$m=−Ea​/R
• y-intercept $c = \ln A$c=lnA

So a plot of $\ln k$lnk (y) against $1/T$1/T (x) is a straight line with gradient $-E_a/R$−Ea​/R and intercept $\ln A$lnA. This is the standard OCR method for determining $E_a$Ea​ experimentally.

Determining $E_a$Ea​ graphically — the standard recipe

1. Run the reaction at several different temperatures (typically 5–8 for a good plot).
2. At each temperature, determine $k$k using any appropriate method (initial rates, gradient of concentration-time graph, half-life of a first-order reaction).
3. Tabulate $\ln k$lnk and $1/T$1/T (with $T$T in kelvin).
4. Plot $\ln k$lnk (y) against $1/T$1/T (x).
5. Draw a line of best fit by eye or by linear regression.
6. Measure the gradient (units: K).
7. $E_a = -\text{gradient} \times R$Ea​=−gradient×R (in J mol$^{-1}$−1; divide by 1000 for kJ mol$^{-1}$−1).

Worked Example 1 — finding $E_a$Ea​ from a five-point dataset

A reaction is studied at five temperatures, with these $k$k values:

$T$T / K	$1/T$1/T / K$^{-1}$−1	$k$k / s$^{-1}$−1	$\ln k$lnk
300	$3.33 \times 10^{-3}$3.33×10−3	$1.0 \times 10^{-3}$1.0×10−3	$-6.91$−6.91
310	$3.23 \times 10^{-3}$3.23×10−3	$2.0 \times 10^{-3}$2.0×10−3	$-6.21$−6.21
320	$3.13 \times 10^{-3}$3.13×10−3	$3.9 \times 10^{-3}$3.9×10−3	$-5.55$−5.55
330	$3.03 \times 10^{-3}$3.03×10−3	$7.6 \times 10^{-3}$7.6×10−3	$-4.88$−4.88
340	$2.94 \times 10^{-3}$2.94×10−3	$1.4 \times 10^{-2}$1.4×10−2	$-4.27$−4.27

Plot $\ln k$lnk vs $1/T$1/T — the line is approximately straight. Take gradient from the extreme points:

$\text{gradient} = \frac{\Delta \ln k}{\Delta (1/T)} = \frac{-4.27 - (-6.91)}{2.94 \times 10^{-3} - 3.33 \times 10^{-3}} = \frac{2.64}{-3.9 \times 10^{-4}} \approx -6770\,\text{K}$gradient=Δ(1/T)Δlnk​=2.94×10−3−3.33×10−3−4.27−(−6.91)​=−3.9×10−42.64​≈−6770K

Therefore:

$$\begin{aligned} E_a &= -\text{gradient} \times R \\ &= -(-6770) \times 8.314 \\ &= 56\,300\,\text{J mol}^{-1} \\ &= 56.3\,\text{kJ mol}^{-1} \end{aligned}$$Ea​​=−gradient×R=−(−6770)×8.314=56300J mol−1=56.3kJ mol−1​

Typical A-Level activation energies lie between 30 and 200 kJ mol$^{-1}$−1, so 56 kJ mol$^{-1}$−1 is a sensible value.

Finding the pre-exponential factor $A$A

The y-intercept of the $\ln k$lnk vs $1/T$1/T plot is $\ln A$lnA. For the data above, extrapolating the line to $1/T = 0$1/T=0 gives an intercept of approximately $\ln A \approx 15.6$lnA≈15.6, so $A \approx e^{15.6} \approx 6 \times 10^{6}\,\text{s}^{-1}$A≈e15.6≈6×106s−1. The pre-exponential factor has the same units as $k$k (here $\text{s}^{-1}$s−1).

Interpreting $E_a$Ea​ and $A$A

• Large $E_a$Ea​ → reaction is very sensitive to temperature (a small change in $T$T gives a big change in $k$k). Reactions with high $E_a$Ea​ are slow at room temperature and speed up dramatically on heating.
• Small $E_a$Ea​ → reaction is relatively insensitive to $T$T — $k$k is large and only weakly dependent on $T$T.
• Large $A$A → frequent, correctly oriented collisions.
• Small $A$A → most collisions have the wrong geometry — typical of reactions with large or stereochemically demanding molecules.
• A catalyst reduces $E_a$Ea​, so $k$k becomes much larger at the same temperature. $A$A also changes slightly, but the dominant effect is the exponential's response to lower $E_a$Ea​.

Mermaid — effects of $T$T, $E_a$Ea​ and catalyst on $k$k

graph TD
  A[Arrhenius k = A exp -Ea/RT] --> B[Higher T -> smaller -Ea/RT -> larger Boltzmann factor -> larger k]
  A --> C[Lower Ea -> smaller -Ea/RT -> larger Boltzmann factor -> larger k]
  A --> D[Catalyst -> lower Ea -> larger k same T]
  B --> E[Reaction faster]
  C --> E
  D --> E
  A --> F[Linearise: plot ln k vs 1/T to extract Ea]

Link with the Maxwell–Boltzmann distribution

Recall from AS that the Maxwell–Boltzmann distribution shows the spread of molecular kinetic energies in a sample. At higher $T$T, the curve shifts to higher energies and flattens — more molecules have energy above $E_a$Ea​. The fraction with energy $\geq E_a$≥Ea​ is approximately $e^{-E_a/(RT)}$e−Ea​/(RT) for an idealised distribution, which is where the Arrhenius exponential term comes from physically.

A catalyst reduces $E_a$Ea​, shifting the "threshold" energy to the left. A much larger fraction of molecules now have sufficient energy — explaining quantitatively why $k$k rises so steeply when a catalyst is added.

Worked Example 2 — two-point Arrhenius (ratio form)

A reaction has $E_a = 80\,\text{kJ mol}^{-1}$Ea​=80kJ mol−1. By what factor does $k$k increase from 298 K to 308 K?

The two-point form (subtract $\ln k_1$lnk1​ from $\ln k_2$lnk2​):

$\ln\!\left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$ln(k1​k2​​)=−REa​​(T2​1​−T1​1​)

Substituting:

$$\begin{aligned} \frac{1}{T_2} &= \frac{1}{308} = 3.247 \times 10^{-3}\,\text{K}^{-1} \\ \frac{1}{T_1} &= \frac{1}{298} = 3.356 \times 10^{-3}\,\text{K}^{-1} \\ \Delta(1/T) &= -1.09 \times 10^{-4}\,\text{K}^{-1} \\ -\frac{E_a}{R} &= -\frac{80\,000}{8.314} = -9623\,\text{K} \\ \ln\!\left(\frac{k_2}{k_1}\right) &= -9623 \times (-1.09 \times 10^{-4}) = 1.049 \\ \frac{k_2}{k_1} &= e^{1.049} = 2.85 \end{aligned}$$T2​1​T1​1​Δ(1/T)−REa​​ln(k1​k2​​)k1​k2​​​=3081​=3.247×10−3K−1=2981​=3.356×10−3K−1=−1.09×10−4K−1=−8.31480000​=−9623K=−9623×(−1.09×10−4)=1.049=e1.049=2.85​

So $k$k increases by about $\times 2.85$×2.85 for a 10 K rise near room temperature at this $E_a$Ea​. Higher-$E_a$Ea​ reactions increase even more steeply; lower-$E_a$Ea​ reactions less so. The famous "rule of thumb" that $k$k doubles every 10 K is closest to true for $E_a \approx 50\,\text{kJ mol}^{-1}$Ea​≈50kJ mol−1.

Worked Example 3 — calculating $k$k at a new temperature

A first-order reaction has $A = 1.5 \times 10^{12}\,\text{s}^{-1}$A=1.5×1012s−1 and $E_a = 100\,\text{kJ mol}^{-1}$Ea​=100kJ mol−1. Calculate $k$k at 350 K.

$$\begin{aligned} -\frac{E_a}{RT} &= -\frac{100\,000}{8.314 \times 350} = -34.37 \\ e^{-34.37} &= 1.20 \times 10^{-15} \\ k &= A \times e^{-E_a/(RT)} \\ &= 1.5 \times 10^{12} \times 1.20 \times 10^{-15} \\ &= 1.8 \times 10^{-3}\,\text{s}^{-1} \end{aligned}$$−RTEa​​e−34.37k​=−8.314×350100000​=−34.37=1.20×10−15=A×e−Ea​/(RT)=1.5×1012×1.20×10−15=1.8×10−3s−1​

The half-life $t_{1/2} = \ln 2 / k = 0.693 / 0.0018 \approx 385\,\text{s}$t1/2​=ln2/k=0.693/0.0018≈385s, around 6 minutes — a reasonable rate for a reaction with $E_a = 100\,\text{kJ mol}^{-1}$Ea​=100kJ mol−1 at 350 K.

Two routes to $E_a$Ea​ — when to use which

OCR exam questions test two methods for extracting $E_a$Ea​: the two-temperature method (Worked Example 2) and the graphical (linearised) method (Worked Example 1). They give complementary information and you should know when to use each.

Two-temperature method. Use when you have only two $(k, T)$(k,T) pairs. The formula $\ln(k_2/k_1) = -(E_a/R)(1/T_2 - 1/T_1)$ln(k2​/k1​)=−(Ea​/R)(1/T2​−1/T1​) rearranges to $E_a = -R\ln(k_2/k_1)/(1/T_2 - 1/T_1)$Ea​=−Rln(k2​/k1​)/(1/T2​−1/T1​) — a single calculation. Advantages: fast, requires only a calculator. Disadvantages: no independent check that the Arrhenius equation actually fits — one anomalous data point can give a nonsensical $E_a$Ea​ with no internal warning. Use when data is scarce, or when you already trust the underlying physics.

Graphical (linearised) method. Use when you have at least 4–5 $(k, T)$(k,T) pairs. Plot $\ln k$lnk (y) against $1/T$1/T (x); the gradient is $-E_a/R$−Ea​/R and the intercept is $\ln A$lnA. Advantages: (i) uses all data via least-squares fitting; (ii) deviation from linearity immediately flags a mechanism change or non-Arrhenius behaviour; (iii) both $E_a$Ea​ and $A$A extracted simultaneously; (iv) the uncertainty in $E_a$Ea​ can be estimated from the scatter about the line. Disadvantages: more work, requires careful unit conversion (kelvin in $1/T$1/T, $\ln k$lnk in natural log). The graphical method is the gold standard for OCR PAG 9/10 write-ups; the two-temperature method is the exam shortcut for a 4-mark calculation.

Cross-check: the two methods must give consistent $E_a$Ea​ values. If your two-temperature calculation between (300, 320) K gives $E_a = 60\,\text{kJ mol}^{-1}$Ea​=60kJ mol−1 but the full graphical analysis over (300, 360) K gives $E_a = 75\,\text{kJ mol}^{-1}$Ea​=75kJ mol−1, the discrepancy probably signals curvature in the Arrhenius plot — perhaps a mechanism change at higher $T$T, or a violation of the constant-$E_a$Ea​ assumption. OCR examiners occasionally embed exactly this kind of inconsistency as an AO3 evaluation prompt.

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