Worked examples — Activation energy from Arrhenius plot; effect of catalyst

This page is a drill of cases. The parent note (parent topic) gave you the Arrhenius Equation and the plot. Here we push it into every corner: normal data, two-point tricks, catalyst factors, the "temperature-that-does-what" twist, and the degenerate cases where the equation almost breaks.

Everything below only uses tools you already have: multiplication, division, the natural logarithm $\ln$, and its inverse $e^x$. If any of those still feel like magic, read the two boxes first.

Recall Reminder: what

$\ln$ and $e^x$ actually do $e^x$ is a "growth machine": feed it a number $x$, it hands back how many times bigger something got. $e^0=1$ (no change), $e^1\approx 2.718$, $e^{10}\approx 22026$ (huge). $\ln$ is the question that undoes it: "$\ln y$" asks "e to what power gives $y$?" So $\ln (e^x)=x$. That is the ONLY reason we ever take $\ln$ of the Arrhenius equation — to pull $E_a$ down out of the exponent where we can read it off a straight line.

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The scenario matrix

Below is every distinct kind of problem this topic can ask. Each worked example is tagged with the cell it covers.

#	Case class	What makes it different	Example
C1	Full data set → slope	Many $(T,k)$ points, best-fit line	Ex 1
C2	Two-point shortcut	Only two temperatures given, no plot	Ex 2
C3	Find $A$ (intercept)	Asked for pre-exponential factor	Ex 3
C4	Same $A$, compare two $E_a$	Ratio of rate constants	Ex 4
C5	Catalyst rate-enhancement	$\Delta E_a<0$, factor $\gg 1$	Ex 5
C6	Degenerate: $E_a=0$	Barrier-free reaction, limiting case	Ex 6
C7	Limit: $T\to \infty$ and $T\to 0$	What happens at the extremes	Ex 7
C8	Real-world word problem	Food spoilage, "how long?"	Ex 8
C9	Exam twist: find $T$	Solve for the unknown temperature	Ex 9

We hit C1–C9 below. Two sign-traps (positive vs negative exponent) recur in almost every one, so watch for the "Why this step?" notes.

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Example 1 — C1: Full data set, extract $E_a$ from the best-fit slope

Forecast: Guess — will $E_a$ come out in the tens of kJ/mol or the hundreds? (Hint: $k$ roughly triples every 20 K. That is a "gentle" temperature response, so bet on tens.)

Step 1 — turn the raw table into plot-coordinates. We need $\frac{1}{T}$ and $\ln k$.

Why this step? The Arrhenius equation is only a straight line in the variables $x=\frac{1}{T}$ and $y=\ln k$. Plot $k$ vs $T$ directly and you get a curve you cannot read a clean slope off (left panel of the figure); switch to $x=1/T,y=\ln k$ and the points fall on a line (right panel).

Figure — same data, two views. Left: raw $k$ vs $T$, a bending curve with no readable slope. Right: the Arrhenius plot $\ln k$ vs $1/T$; the four lavender dots line up, and the mint line is their best fit. The line's slope is $-E_a/R$.

$T$	$1/T$ (${\text{K}}^{-1}$)	$\ln k$
300	0.003333	$-5.404$
320	0.003125	$-4.160$
340	0.002941	$-3.062$
360	0.002778	$-2.083$

Step 2 — fit a line through all four points, not just two. With more than two measurements the honest method is least squares: find the single straight line $y=mx+c$ whose slope $m$ makes the total squared vertical gap to the four dots as small as possible. The closed-form slope is

$m=\frac{n\sum xy-\sum x\sum y}{n\sum x^2-{\left(\sum x\right)}^2},n=4.$

Feeding the four $(x=1/T,y=\ln k)$ pairs into this gives

$m\approx -5972\text{K}.$

Why this step? Real data has scatter; a two-point slope throws away two of your four measurements and trusts the endpoints blindly. Least squares uses every point, so a single bad reading can't hijack the answer. (Here the data is almost perfectly linear, so the quick endpoint slope below lands within $0.2\%$ of the fit.)

Step 2b — the exam shortcut (endpoints only). If you have no calculator with a fit function, use the two extreme points:

$m=\frac{\ln k_4-\ln k_1}{\frac{1}{T_4}-\frac{1}{T_1}}=\frac{-2.083-(-5.404)}{0.002778-0.003333}=\frac{3.321}{-0.000555}=-5984\text{K}$

Why this step? Slope is rise-over-run. The run is negative (as $T$ rises, $1/T$ falls) and the rise is positive — a genuinely downward-sloping line. Notice $-5984$ vs the best-fit $-5972$: agreement to three figures, which tells you the data really is linear.

Step 3 — convert slope to $E_a$.

$E_a=-m\times R=-(-5984)(8.314\text{J}{\text{mol}}^{-1}{\text{K}}^{-1})=49748\text{J/mol}\approx 49.7\text{kJ/mol}$

Why this step? Slope $=-E_a/R$, so $E_a=-mR$. The two minus signs multiply to a positive $E_a$ — activation energy is never negative.

Verify: Units: $\text{K}\times \text{J}{\text{mol}}^{-1}{\text{K}}^{-1}=\text{J/mol}$ ✓. Magnitude sits in the "tens of kJ/mol" band we forecast ✓. Sanity: a mid-range small-molecule reaction is 40–120 kJ/mol, so 49.7 is chemically believable.

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Example 2 — C2: The two-point shortcut (no graph at all)

Forecast: $k$ jumped 8× for a 40 K rise. Bigger jump than Ex 1 → guess a larger $E_a$.

Step 1 — write Arrhenius twice and divide.

$\frac{k_2}{k_1}=\frac{A{e}^{-E_a/R{T}_2}}{A{e}^{-E_a/R{T}_1}}=e^{-\frac{E_a}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)}$

Why this step? We do not know $A$. Dividing the two equations makes the unknown $A$ cancel exactly, leaving one equation in the one unknown $E_a$.

Step 2 — take $\ln$ to free $E_a$.

$\ln \frac{k_2}{k_1}=-\frac{E_a}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)=\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)$

Why this step? $\ln$ is the tool that pulls $E_a$ out of the exponent (see the recall box). I flipped the bracket's sign so the front sign becomes $+$ — cleaner, since $\frac{1}{T_1}>\frac{1}{T_2}$ makes the bracket positive.

Step 3 — plug in numbers.

$\ln \frac{2.0\times 10^{-3}}{2.5\times 10^{-4}}=\ln 8=2.079$ $\frac{1}{310}-\frac{1}{350}=0.0032258-0.0028571=3.687\times 10^{-4}{\text{K}}^{-1}$ $E_a=\frac{R\ln (k_2/k_1)}{\frac{1}{T_1}-\frac{1}{T_2}}=\frac{8.314\times 2.079}{3.687\times 10^{-4}}=46880\text{J/mol}\approx 46.9\text{kJ/mol}$

Why this step? We just solved the Step 2 equation for $E_a$ (multiply both sides by $R$, divide by the bracket) and substituted the two numbers we computed. $R$ stays in J/mol·K, and $E_a$ lands in J/mol — divide by 1000 to quote 46.9 kJ/mol.

Verify: Positive $E_a$ ✓. Comparable band to Ex 1 (both ~50 kJ/mol) ✓. The bracket is positive and $\ln 8$ positive, so $E_a>0$ automatically — a built-in sign check.

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Example 3 — C3: Recover the pre-exponential factor $A$

Forecast: $A$ is the rate constant a reaction would have if the barrier vanished — so $A\gg k$. Guess a big number.

Step 1 — start from the linear form and solve for $\ln A$.

$\ln k=\ln A-\frac{E_a}{RT}\Rightarrow \ln A=\ln k+\frac{E_a}{RT}$

Why this step? $\ln A$ is the y-intercept of the Arrhenius line — the value of $\ln k$ when $\frac{1}{T}=0$. We reconstruct it by adding back the bit the barrier subtracted.

Step 2 — plug in. ($E_a$ is already in J/mol, so no ×1000 needed here.)

$\frac{E_a}{RT}=\frac{49748}{8.314\times 300}=19.94$ $\ln A=\ln (0.0045)+19.94=-5.404+19.94=14.53$

Step 3 — undo the log.

$A=e^{14.53}=2.04\times 10^6{\text{s}}^{-1}$

Why this step? $e^x$ undoes $\ln$; we want $A$ itself, not $\ln A$.

Verify: $A\approx 2\times 10^6\gg k=0.0045$ ✓ (as forecast). Units of $A$ match $k$ (${\text{s}}^{-1}$, a first-order reaction) ✓.

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Example 4 — C4: Two reactions, same $A$, different $E_a$

Forecast: B's barrier is 25 kJ/mol taller. Taller barrier → slower. Guess $k_B$ is thousands of times smaller.

Step 1 — ratio, so $A$ cancels.

$\frac{k_B}{k_A}=e^{-(E_{a,B}-E_{a,A})/RT}$

Why this step? Same $A$ ⇒ dividing kills it (same trick as Ex 2).

Step 2 — convert to J/mol, then evaluate the exponent.

$E_{a,B}-E_{a,A}=(75-50)\frac{\text{kJ}}{\text{mol}}=25\frac{\text{kJ}}{\text{mol}}=25000\frac{\text{J}}{\text{mol}}$ $\frac{25000}{8.314\times 298}=10.09,\frac{k_B}{k_A}=e^{-10.09}=4.15\times 10^{-5}$

Why the ×1000 and the minus? The ×1000 matches the J-based $R$ (see the units box). B's exponent $-E_{a,B}/RT$ is more negative than A's, so the ratio is less than 1 — B is slower. The sign encodes "which barrier is taller."

Step 3 — multiply out to get $k_B$.

$k_B=k_A\times \frac{k_B}{k_A}=2.0\times 10^{-3}\times 4.15\times 10^{-5}=8.3\times 10^{-8}{\text{s}}^{-1}$

Why this step? We know $k_A$ and just found the ratio $k_B/k_A$; multiplying the known $k_A$ by that ratio isolates the target $k_B$. Units: ${\text{s}}^{-1}\times (\text{dimensionless ratio})={\text{s}}^{-1}$ ✓.

Verify: $1/(4.15\times 10^{-5})\approx 24000$, so B is ~24,000× slower ✓ — "thousands," as forecast. Extra 25 kJ/mol costs a factor $\approx 2.4\times 10^4$: the exponential's brutal sensitivity to $E_a$.

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Example 5 — C5: Catalyst rate-enhancement factor

Figure — energy landscape, catalyzed vs not. The coral hump is the uncatalyzed path (one tall barrier $E_a$ big). The mint path is the catalyzed route: two smaller humps. Crucially the reactant and product levels are identical in both — the lavender $\Delta H$ arrow is unchanged. The catalyst lowers the peak, never the endpoints.

Forecast: Barrier drops 25 kJ/mol — same size drop as Ex 4's gap, so expect a similar ~20,000× — but now it works for us.

Step 1 — write the ratio.

$\frac{k_{cat}}{k_{uncat}}=e^{-\Delta E_a/RT},\Delta E_a=E_{a,cat}-E_{a,uncat}$

Why this step? Catalyst only changes the exponent (barrier height). $\Delta E_a$ will be negative because the catalyst path is lower — that is the whole point of a catalyst.

Step 2 — convert to J/mol and evaluate.

$\Delta E_a=(60-85)\frac{\text{kJ}}{\text{mol}}=-25\frac{\text{kJ}}{\text{mol}}=-25000\frac{\text{J}}{\text{mol}}$ $-\frac{\Delta E_a}{RT}=-\frac{-25000}{8.314\times 300}=+10.02,\frac{k_{cat}}{k_{uncat}}=e^{10.02}=22400$

Why this step? The ×1000 matches $R$'s joules. Two minus signs (the $-$ in front, the negative $\Delta E_a$) make a positive exponent → factor $\gg 1$ → speed-up. If you ever get a factor $<1$ from a catalyst, you flipped a sign.

Verify: A 1-hour ($3600$ s) uncatalyzed reaction now takes $3600/22400\approx 0.16$ s ✓. Note in the figure: reactant and product energies are identical in both paths — the catalyst does not move $\Delta H$, only the peak (see Reaction Energy Diagrams and Catalysis Mechanisms).

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Example 6 — C6: Degenerate case, $E_a=0$

Forecast: No barrier to climb → guess that every collision counts, and temperature barely matters.

Step 1 — set $E_a=0$ in $k=A{e}^{-E_a/RT}$.

$k=A{e}^{-0/RT}=A{e}^0=A\cdot 1=A$

Why this step? Anything over $RT$ times zero is zero, and $e^0=1$. The exponential "gate" is fully open.

Step 2 — read the temperature dependence.

Since $k=A$ (a constant), $k$ is independent of $T$ in this idealisation.

Why this step? On the Arrhenius plot the slope is $-E_a/R=0$: a perfectly flat horizontal line. No barrier ⇒ no steepness.

Verify: Consistent with Collision Theory: with no energy requirement, the fraction of "successful" collisions is 1, so rate is limited only by how often molecules meet — captured by $A$. (Real barrier-free reactions still show mild $T$-dependence through $A$ itself, but the Arrhenius exponent contributes nothing.)

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Example 7 — C7: The two limits, $T\to \infty$ and $T\to 0$

Forecast: Hot enough → every molecule clears the bar. Cold enough → nobody does. Guess $k\to A$ (hot) and $k\to 0$ (cold).

Step 1 — the hot limit $T\to \infty$.

$\frac{E_a}{RT}\to \frac{E_a}{R\cdot \infty }=0\Rightarrow k\to A{e}^0=A$

Why this step? A fixed $E_a$ divided by an ever-growing $RT$ shrinks to 0; the gate opens fully. On the plot, $\frac{1}{T}\to 0$ (far left), and $\ln k\to \ln A$ — that's why the intercept is $\ln A$.

Step 2 — the cold limit $T\to 0^{+}$.

$\frac{E_a}{RT}\to \frac{E_a}{R\cdot 0^{+}}=+\infty \Rightarrow k\to A{e}^{-\infty }=0$

Why this step? Tiny $RT$ makes the exponent enormously negative; $e^{-\infty }=0$. On the plot, $\frac{1}{T}\to \infty$ (far right), $\ln k\to -\infty$ — the line dives off the bottom.

Verify: Both match the forecast and the Maxwell-Boltzmann Distribution picture: at high $T$ the high-energy tail is fat (almost everyone clears $E_a$); at $T\to 0$ the distribution collapses toward zero energy and essentially nobody has $\ge E_a$. $k$ is bounded between $0$ and $A$ for all $T>0$ ✓.

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Example 8 — C8: Real-world word problem (food spoilage)

Forecast: Warmer → faster spoiling → much shorter than 7 days. Guess a couple of days at most.

Step 1 — link "shelf life" to rate. Spoiling to the same visible endpoint means the same amount of reaction, so time $\propto 1/k$:

$\frac{t_{warm}}{t_{cold}}=\frac{k_{cold}}{k_{warm}}=e^{-\frac{E_a}{R}\left(\frac{1}{T_{cold}}-\frac{1}{T_{warm}}\right)}$

Why this step? Longer-lasting means slower reaction; time and rate are reciprocals. Ratio cancels $A$, so we never need it.

Step 2 — convert $E_a$ to J/mol and evaluate the exponent.

$E_a=75\frac{\text{kJ}}{\text{mol}}=75000\frac{\text{J}}{\text{mol}}$ $\frac{1}{277}-\frac{1}{298}=0.0036101-0.0033557=2.544\times 10^{-4}{\text{K}}^{-1}$ $\frac{E_a}{R}\times 2.544\times 10^{-4}=\frac{75000}{8.314}\times 2.544\times 10^{-4}=9021\times 2.544\times 10^{-4}=2.295$

Why this step? The ×1000 keeps units consistent with $R$ in J/mol·K; then $E_a/R$ (units K) times the bracket (units ${\text{K}}^{-1}$) gives a pure number $2.295$, exactly what the exponent must be.

Step 3 — assemble the shelf life. The cold side is slower, so $k_{cold}/k_{warm}=e^{-2.295}=0.1008$:

$t_{warm}=t_{cold}\times 0.1008=7\text{days}\times 0.1008=0.706\text{days}\approx 16.9\text{hours}$

Why this step? We wanted $t_{warm}$; from Step 1, $t_{warm}=t_{cold}\times (k_{cold}/k_{warm})$, so multiply the known 7-day cold life by the (small) rate ratio. The minus sign in $e^{-2.295}$ makes the ratio less than 1, i.e. a shorter warm shelf life — the physically sensible direction.

Verify: Under a day on the counter vs a week in the fridge ✓ — matches lived experience and the "much shorter" forecast. A ~21 K rise gave a ~10× rate change: the everyday "rule of thumb" that reactions roughly double or more per 10 °C.

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Example 9 — C9: Exam twist — solve for the temperature

Forecast: With a large $A$ but a big barrier, you need enough thermal energy to open the gate — likely a few hundred K.

Step 1 — take $\ln$ of Arrhenius and isolate $T$.

$\ln k=\ln A-\frac{E_a}{RT}\Rightarrow \frac{E_a}{RT}=\ln A-\ln k=\ln \frac{A}{k}$

Why this step? The unknown $T$ is trapped inside the exponent. $\ln$ frees it, and now it sits in one clean fraction.

Step 2 — rearrange for $T$ and convert $E_a$ to J/mol.

$T=\frac{E_a}{R\ln (A/k)},E_a=100\frac{\text{kJ}}{\text{mol}}=100000\frac{\text{J}}{\text{mol}}$

Why this step? We multiply both sides of Step 1 by $\frac{RT}{\ln (A/k)}$ to get $T$ alone; the ×1000 again matches $R$ in joules so the units cancel to kelvin.

Step 3 — plug in.

$\ln \frac{A}{k}=\ln \frac{1.0\times 10^{13}}{1.0}=\ln (10^{13})=29.93$ $T=\frac{100000}{8.314\times 29.93}=\frac{100000}{248.9}=401.8\text{K}$

Why this step? Straight substitution; note the units $\frac{\text{J/mol}}{(\text{J}{\text{mol}}^{-1}{\text{K}}^{-1})\times (\text{pure number})}=\text{K}$ ✓.

Verify: $T\approx 402$ K $=129$ °C — a "few hundred K," as forecast ✓. Back-substitute: $\frac{E_a}{RT}=\frac{100000}{8.314\times 401.8}=29.94$, and $A{e}^{-29.94}=10^{13}\times e^{-29.94}\approx 1.0{\text{s}}^{-1}$ ✓ (see Rate Constant Temperature Dependence).

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Recall Self-test (cover the right side)

Slope of an Arrhenius plot equals what? ::: $-E_a/R$ With four data points, what fitting method uses all of them? ::: Least-squares (best-fit) line Before plugging a kJ/mol value into $E_a/RT$, what must you do? ::: Multiply by 1000 to get J/mol (to match $R$ in J/mol·K) To find $E_a$ from only two $(T,k)$ points, what cancels? ::: The pre-exponential factor $A$ A catalyst dropping $E_a$ by 25 kJ/mol at 300 K speeds the reaction by roughly? ::: $\sim$22,000× As $T\to \infty$, $k\to$ ? ::: $A$ (the intercept, gate fully open) When $E_a=0$, the Arrhenius plot line looks like? ::: A flat horizontal line ($k=A$, no $T$-dependence) Does a catalyst change $K_{eq}$? ::: No — only how fast equilibrium is reached