Temperature dependence — Arrhenius equation k = A·e^(−Ea - RT): Worked examples

This page is the "no surprises" workbook for the parent Arrhenius note ($k = A\,e^{-E_a/(RT)}$). The parent showed you the equation and two clean examples. Here we hunt down every kind of question the topic can ask — every unknown, every degenerate case, every sneaky exam twist — and solve each one from scratch.

Before we start, let us re-state the only two tools we ever need, so no symbol is used unfamiliar.

The scenario matrix

Every Arrhenius problem is really "which quantity is missing, and is any input at an edge?" Here is the full grid. Each example below is tagged with the cell it fills.

Cell	What is unknown	Twist / edge case	Example
A	$E_{a}$	two clean $(T, k)$ points	Ex 1
B	$k_{2}$	predict at a higher $T$	Ex 2
C	$k_{2}$	predict at a lower $T$ (cooling, $ln$ negative)	Ex 3
D	$E_{a}$	from a graph slope (many points, averages error)	Ex 4
E	$A$	back out the pre-exponential factor	Ex 5
F	$T$	find the temperature to hit a target $k$	Ex 6
G	degenerate: $E_{a} = 0$	what happens when there is no barrier	Ex 7
H	limiting: $T \to \infty$	ceiling of $k$ ; and $T \to 0$ floor	Ex 7
I	real-world word problem	Q10 / food spoilage doubling	Ex 8
J	exam twist	ratio of two reactions, or "how much faster"	Ex 9
K	edge: $E_{a} < 0$	cooling speeds up the reaction	Ex 10

Two sign facts to keep in your pocket, because they catch people out in cells B/C. The figure below draws them: it plots $k$ against $T$ for our running reaction and shows a burnt-orange arrow for heating (moving right and up) and a teal arrow for cooling (moving left and down) away from the same reference point.

Figure — Temperature dependence — Arrhenius equation k = A·e^(−Ea - RT)

Cell A — find $E_{a}$ from two points

Worked example Example 1 (Cell A)

A reaction has $k_{1} = 2.0 \times 1 0^{- 3} s^{- 1}$ at $T_{1} = 310 K$ and $k_{2} = 8.0 \times 1 0^{- 3} s^{- 1}$ at $T_{2} = 330 K$ . Find $E_{a}$ .

Forecast: $k$ rose 4× over a 20 K rise. Guess: a moderate barrier, tens of kJ/mol.

Step 1 — pick the two-temperature form. Why this step? $A$ is unknown and we don't need it; the two-point form deletes $A$ . $ln \frac{k _{2}}{k _{1}} = \frac{E _{a}}{R} (\frac{1}{T _{1}} - \frac{1}{T _{2}})$

Step 2 — compute the left side. Why this step? The ratio $k_{2} / k_{1} = 4$ , and $ln 4 = 1.3863$ .

Step 3 — compute the bracket in kelvin. Why this step? Both temperatures must be absolute or the Boltzmann factor is meaningless. $\frac{1}{310} - \frac{1}{330} = 0.0032258 - 0.0030303 = 1.9550 \times 1 0^{- 4} K^{- 1}$

Step 4 — solve for $E_{a}$ . $E_{a} = \frac{R l n ( k _{2} / k _{1} )}{( \frac{1}{T _{1}} - \frac{1}{T _{2}} )} = \frac{8.314 \times 1.3863}{1.9550 \times 1 0 ^{- 4}} \approx 5.90 \times 1 0^{4} J/mol = 59.0 kJ/mol$

Answer: $E_{a} \approx 59.0 kJ/mol$ — a moderate barrier, matching the forecast.

Verify: Units: $\frac{J/(mol\cdotK) \times (dimensionless)}{1/ K} = J/mol$ ✓. Plug back: $\frac{59000}{8.314} \times 1.955 \times 1 0^{- 4} = 1.387 \approx ln 4$ ✓.

Cell B — predict $k$ at a higher temperature

Worked example Example 2 (Cell B)

Using $E_{a} = 59.0 kJ/mol$ and $k_{1} = 2.0 \times 1 0^{- 3} s^{- 1}$ at $310 K$ , predict $k$ at $360 K$ .

Forecast: Heating by 50 K with a 59 kJ/mol barrier — I expect a jump of roughly an order of magnitude.

Step 1 — write the two-temperature form solved for $k_{2}$ . Why this step? $k_{2}$ is our unknown; everything else is known. $ln \frac{k _{2}}{k _{1}} = \frac{E _{a}}{R} (\frac{1}{T _{1}} - \frac{1}{T _{2}})$

Step 2 — bracket. $\frac{1}{310} - \frac{1}{360} = 0.0032258 - 0.0027778 = 4.4803 \times 1 0^{- 4} K^{- 1}$ . Why this step? Because $T_{2} > T_{1}$ the bracket is positive → $k$ grows, as the sign rule (and the orange arrow in the figure above) promised.

Step 3 — multiply by $E_{a} / R$ . $\frac{59000}{8.314} \times 4.4803 \times 1 0^{- 4} = 7096.5 \times 4.4803 \times 1 0^{- 4} = 3.1794$

Step 4 — exponentiate and multiply. Why this step? We had $ln (k_{2} / k_{1})$ ; undo the log with $e^{(\cdot)}$ . $k_{2} = k_{1} e^{3.1794} = 2.0 \times 1 0^{- 3} \times 24.03 = 4.81 \times 1 0^{- 2} s^{- 1}$

Answer: $k \approx 4.8 \times 1 0^{- 2} s^{- 1}$ — about 24× faster for a 50 K rise.

Verify: Sanity: $k_{2} > k_{1}$ ✓ (heating). Order of magnitude ~ $1 0^{- 2}$ matches the forecast.

Cell C — predict $k$ at a lower temperature (negative log)

Worked example Example 3 (Cell C)

Same reaction ( $E_{a} = 59.0 kJ/mol$ , $k_{1} = 2.0 \times 1 0^{- 3} s^{- 1}$ at $310 K$ ). What is $k$ in a fridge at $277 K$ (4 °C)?

Forecast: Cooling slows things down (for this normal $E_{a} > 0$ ), so $k$ should drop below $2.0 \times 1 0^{- 3}$ . The log will come out negative — that's the tell for cooling, matching the teal arrow in the figure above.

Step 1 — two-temperature form, $T_{2} = 277 K$ now below $T_{1}$ . Why this step? Keep $T_{1}$ as the known point; the sign will handle direction automatically.

Step 2 — bracket. Why this step? Now $1/ T_{2} = 1/277 = 0.0036101$ is larger than $1/ T_{1}$ , so the bracket is negative. $\frac{1}{310} - \frac{1}{277} = 0.0032258 - 0.0036101 = - 3.8430 \times 1 0^{- 4} K^{- 1}$

Step 3 — multiply. $ln \frac{k _{2}}{k _{1}} = 7096.5 \times (- 3.8430 \times 1 0^{- 4}) = - 2.7272$

Step 4 — exponentiate. $k_{2} = 2.0 \times 1 0^{- 3} \times e^{- 2.7272} = 2.0 \times 1 0^{- 3} \times 0.06544 = 1.31 \times 1 0^{- 4} s^{- 1}$

Answer: $k \approx 1.3 \times 1 0^{- 4} s^{- 1}$ — roughly 15× slower than at 310 K.

Verify: Negative log ✓ (cooling). $k_{2} < k_{1}$ ✓. This is exactly why refrigeration preserves food — see Example 8 for the Q10 Temperature Coefficient version.

Cell D — $E_{a}$ from a graph slope

The most reliable lab method: measure $k$ at many temperatures, plot $ln k$ against $1/ T$ , read the slope. The linear form $ln k = ln A - \frac{E _{a}}{R} \cdot \frac{1}{T}$ is a straight line $y = m x + c$ with slope $m = - E_{a} / R$ .

The figure below is exactly this Arrhenius plot: the plum line is the best fit, the orange dots are the five noisy measurements it averages, and the teal triangle marks the slope $m = - 6500 K$ that we read off. Notice the intercept label at the far left, where the line would meet the axis at $1/ T \to 0$ (i.e. $T \to \infty$ ) — that height is $ln A$ .

Worked example Example 4 (Cell D)

Reading the plot above, the best-fit Arrhenius line has slope $m = - 6500 K$ and intercept $c = 15.2$ . Find $E_{a}$ and $A$ .

Forecast: Slope magnitude 6500 K × 8.314 ≈ 54 kJ/mol. Intercept 15.2 is $ln A$ , so $A \approx e^{15} \sim$ a few million.

Step 1 — relate slope to $E_{a}$ . Why this step? From the linear form, the coefficient of $1/ T$ is exactly $- E_{a} / R$ ; the teal triangle in the figure is this slope. $m = - \frac{E _{a}}{R} \Rightarrow E_{a} = - m R = 6500 \times 8.314 = 5.40 \times 1 0^{4} J/mol$

Step 2 — relate intercept to $A$ . Why this step? At $1/ T = 0$ (infinite temperature) the line hits $ln k = ln A$ — the intercept marked on the figure. $ln A = 15.2 \Rightarrow A = e^{15.2} = 3.99 \times 1 0^{6} s^{- 1}$

Answer: $E_{a} = 54.0 kJ/mol$ , $A = 4.0 \times 1 0^{6} s^{- 1}$ .

Verify: $- m R$ has units $K \times J/(mol\cdotK) = J/mol$ ✓. A multi-point fit is better than two points because random scatter (the wobble of the orange dots off the plum line) averages out — connect to Maxwell-Boltzmann Distribution for why individual $k$ measurements fluctuate.

Cell E — back out the pre-exponential factor $A$

Worked example Example 5 (Cell E)

A reaction has $E_{a} = 75.0 kJ/mol$ and $k = 0.35 s^{- 1}$ at $T = 500 K$ . Find $A$ .

Forecast: The exponent $E_{a} / R T$ will be around 18, so $e^{- 18}$ is tiny (~ $1 0^{- 8}$ ); to still get $k = 0.35$ , $A$ must be huge — order $1 0^{7}$ – $1 0^{8}$ .

Step 1 — use the one-temperature form; solve for $A$ . Why this step? Only one $(T, k)$ point is given and we want $A$ , so the two-point form is useless here. $k = A e^{- E_{a} / (R T)} \Rightarrow A = k e^{+ E_{a} / (R T)}$

Step 2 — compute the exponent. Why this step? $E_{a} / (R T)$ is a dimensionless ratio of energies; keep everything in J. $\frac{E _{a}}{R T} = \frac{75000}{8.314 \times 500} = \frac{75000}{4157} = 18.043$

Step 3 — exponentiate and multiply. $A = 0.35 \times e^{18.043} = 0.35 \times 6.86 \times 1 0^{7} = 2.40 \times 1 0^{7} s^{- 1}$

Answer: $A \approx 2.4 \times 1 0^{7} s^{- 1}$ .

Verify: $A$ carries the same units as $k$ ( $s^{- 1}$ here, because this is first-order) since $e^{- E_{a} / R T}$ is dimensionless ✓. Plug back: $2.4 \times 1 0^{7} \times e^{- 18.043} = 2.4 \times 1 0^{7} \times 1.457 \times 1 0^{- 8} = 0.35$ ✓.

Cell F — solve for the temperature

Cell G & H — degenerate and limiting cases

These are the questions that expose whether you understand the equation rather than just plugging into it.

Worked example Example 7 (Cells G & H)

(G) Zero barrier. What is $k$ if $E_{a} = 0$ ? (H) Extreme temperatures. What is $lim_{T \to \infty} k$ and $lim_{T \to 0^{+}} k$ ?

Forecast: No barrier means every collision counts, so $k$ should just equal $A$ . Infinite $T$ means everyone can clear the wall, so $k \to A$ too. Freezing to $T \to 0$ means nobody can, so $k \to 0$ .

Step 1 — set $E_{a} = 0$ in the one-temperature form. Why this step? $E_{a} = 0$ makes the exponent zero, and $e^{0} = 1$ . $k = A e^{- 0/ (R T)} = A \cdot e^{0} = A$ So a barrier-free reaction runs at its maximum possible rate, the collision-limited rate $A$ . (A true $E_{a} = 0$ is rare; some radical recombinations approach it — see Catalysis for how catalysts lower $E_{a}$ toward this ideal.)

Step 2 — limit $T \to \infty$ . Why this step? As $T$ grows, $E_{a} / (R T) \to 0$ , so the exponent $\to 0$ and $e^{- E_{a} / R T} \to 1$ . $lim_{T \to \infty} k = A \cdot 1 = A$ $A$ is the ceiling on $k$ : you can never react faster than molecules collide. This is the intercept height in the Cell D figure.

Step 3 — limit $T \to 0^{+}$ . Why this step? As $T \to 0^{+}$ , $E_{a} / (R T) \to + \infty$ , so $e^{- \infty} \to 0$ . $lim_{T \to 0^{+}} k = A \cdot 0 = 0$ At absolute zero, no molecule has energy to cross the barrier — the reaction freezes.

Answer: $E_{a} = 0 \Rightarrow k = A$ ; $k \to A$ as $T \to \infty$ (the ceiling); $k \to 0$ as $T \to 0^{+}$ (the floor). The three edge results bracket every possible $k$ between $0$ and $A$ .

Verify: Check numerically with $A = 1 0^{13}, E_{a} = 50000$ : at $T = 1 0^{6} K$ , $e^{- 50000/ (8.314 \times 1 0^{6})} = e^{- 0.006} = 0.994$ , so $k = 9.94 \times 1 0^{12} \approx A$ ✓. At $T = 1 K$ , $e^{- 50000/8.314} = e^{- 6014} \approx 0$ ✓.

Cell I — real-world word problem (Q10 doubling)

A famous rule of thumb: many reaction rates roughly double for every 10 K rise. This is the Q10 Temperature Coefficient $= k_{T + 10} / k_{T}$ . Let's derive what $E_{a}$ makes $Q_{10} = 2$ .

Worked example Example 8 (Cell I)

Milk spoilage roughly doubles in rate between $4 °C$ (277 K) and $14 °C$ (287 K). Estimate the effective $E_{a}$ .

Forecast: "Doubling per 10 K" near room temperature is the classic case; textbooks quote $E_{a}$ around 40–60 kJ/mol for it.

Step 1 — two-temperature form with $k_{2} / k_{1} = 2$ . Why this step? "Doubling" is precisely $ln (k_{2} / k_{1}) = ln 2$ .

Step 2 — bracket. Why this step? Convert both temperatures to kelvin (they already are) and subtract the reciprocals. $\frac{1}{277} - \frac{1}{287} = 0.0036101 - 0.0034843 = 1.2578 \times 1 0^{- 4} K^{- 1}$

Step 3 — solve for $E_{a}$ . $E_{a} = \frac{R l n 2}{1.2578 \times 1 0 ^{- 4}} = \frac{8.314 \times 0.6931}{1.2578 \times 1 0 ^{- 4}} = \frac{5.763}{1.2578 \times 1 0 ^{- 4}} = 4.58 \times 1 0^{4} J/mol$

Answer: $E_{a} \approx 45.8 kJ/mol$ — squarely in the expected band.

Verify: Predict the doubling back: $\frac{45800}{8.314} \times 1.2578 \times 1 0^{- 4} = 5509.5 \times 1.2578 \times 1 0^{- 4} = 0.693 = ln 2$ ✓. So your fridge (Example 3) buying ~15× slowdown from 310→277 K is the same physics pushed harder.

Cell J — exam twist: "how many times faster?"

Worked example Example 9 (Cell J)

Two reactions run at the same 300 K. Reaction P has $E_{a} = 40 kJ/mol$ , reaction Q has $E_{a} = 80 kJ/mol$ . Assume equal $A$ . How many times faster is P than Q?

Forecast: Q's barrier is double P's, so its Boltzmann factor is far smaller — P should be many orders of magnitude faster.

Step 1 — write the ratio, same $T$ and same $A$ . Why this step? Equal $A$ cancels; only the exponents differ. $\frac{k _{P}}{k _{Q}} = \frac{A e ^{- E_{a, P} / (R T)}}{A e ^{- E_{a, Q} / (R T)}} = e^{(E_{a, Q} - E_{a, P}) / (R T)}$

Step 2 — plug in the energy difference. Why this step? $E_{a, Q} - E_{a, P} = 40000 J/mol$ is the "extra wall" Q must climb. $\frac{E _{a, Q} - E _{a, P}}{R T} = \frac{40000}{8.314 \times 300} = \frac{40000}{2494.2} = 16.037$

Step 3 — exponentiate. Why this step? We have $ln (k_{P} / k_{Q})$ ; undo the log to get the raw ratio. $\frac{k _{P}}{k _{Q}} = e^{16.037} = 9.2 \times 1 0^{6}$

Answer: P is about 9 million times faster than Q at 300 K.

Verify: Difference of exponents is dimensionless ✓. Cross-check with logs: $ln (9.2 \times 1 0^{6}) = 16.03 \approx 16.037$ ✓. This is why lowering $E_{a}$ via a catalyst gives such dramatic speed-ups — and why a thermodynamically favourable reaction can still be immeasurably slow if $E_{a}$ is high.

Cell K — the genuine exception: negative activation energy

Everywhere above, $E_{a} > 0$ and heating sped things up. But some real reactions (certain radical recombinations, and many multi-step reactions with a fast pre-equilibrium) show a measured $E_{a} < 0$ : they get slower when you heat them. The Arrhenius algebra handles this automatically — you just must not "correct" the sign by hand.

Worked example Example 10 (Cell K)

A reaction has an effective $E_{a} = - 12.0 kJ/mol$ and $k_{1} = 5.0 \times 1 0^{2} s^{- 1}$ at $T_{1} = 300 K$ . Predict $k$ at $T_{2} = 350 K$ .

Forecast: Negative $E_{a}$ means heating should slow it, so I expect $k_{2} < 500$ . And $ln (k_{2} / k_{1})$ should be negative even though we heated — the opposite of the normal sign rule.

Step 1 — two-temperature form, no sign patching. Why this step? Trust the formula; the negative $E_{a}$ carries its own sign. $ln \frac{k _{2}}{k _{1}} = \frac{E _{a}}{R} (\frac{1}{T _{1}} - \frac{1}{T _{2}})$

Step 2 — bracket (heating, so still positive). Why this step? $T_{2} > T_{1}$ , so the bracket is positive as always; the twist lives in $E_{a}$ , not here. $\frac{1}{300} - \frac{1}{350} = 0.0033333 - 0.0028571 = 4.7619 \times 1 0^{- 4} K^{- 1}$

Step 3 — multiply by $E_{a} / R$ (now negative). $ln \frac{k _{2}}{k _{1}} = \frac{- 12000}{8.314} \times 4.7619 \times 1 0^{- 4} = - 1443.4 \times 4.7619 \times 1 0^{- 4} = - 0.6873$

Step 4 — exponentiate. $k_{2} = 5.0 \times 1 0^{2} \times e^{- 0.6873} = 5.0 \times 1 0^{2} \times 0.5029 = 2.51 \times 1 0^{2} s^{- 1}$

Answer: $k \approx 2.5 \times 1 0^{2} s^{- 1}$ — heating halved the rate.

Verify: Heating gave a negative log ✓ (only possible because $E_{a} < 0$ ). $k_{2} < k_{1}$ ✓. On a Cell-D Arrhenius plot this reaction would have a line sloping the "wrong" way (positive slope $- E_{a} / R > 0$ ) — a dead giveaway of negative activation energy.

Recall Quick self-test

Which unknown does the two-temperature form solve without knowing $A$ ? ::: Either $E_{a}$ or a second $k$ , or a second $T$ — anything except $A$ itself, since $A$ cancels. If $ln (k_{2} / k_{1})$ comes out negative after heating, what does that imply? ::: Either you swapped a subscript, or the reaction genuinely has $E_{a} < 0$ (Cell K). What is $k$ when $E_{a} = 0$ ? ::: $k = A$ , the collision-limited maximum. As $T \to \infty$ , what does $k$ approach? ::: The ceiling $A$ . Why must $T$ be in kelvin? ::: $E_{a} / (R T)$ must be a clean dimensionless energy ratio; Celsius can be zero or negative and breaks it. Do the units of $k$ and $A$ change with reaction order? ::: Yes — $A$ always shares $k$ 's units, which depend on order (s⁻¹ for first-order, L·mol⁻¹s⁻¹ for second-order, etc.).