Activation energy from Arrhenius plot; effect of catalyst: Exercises

This page is a self-test ladder. Each rung is harder than the last: from recognising the Arrhenius pieces to building new arguments from scratch. Solve first, then open the collapsible solution.

Everything here builds on the parent topic. If a symbol feels unfamiliar, revisit the Arrhenius Equation and Rate Constant Temperature Dependence notes.

Constants used throughout: $R = 8.314 J/(mol\cdotK)$ . Keep energies in J/mol inside the exponential (so kJ/mol × 1000).

Level 1 — Recognition

L1.1 — Read the slope

On an Arrhenius plot the best-fit line has slope $m = - 7500 K$ . What is $E_{a}$ in kJ/mol?

Recall Solution

WHAT: slope of $ln k$ vs $1/ T$ equals $- E_{a} / R$ . WHY: because $ln k = - \frac{E _{a}}{R} \cdot \frac{1}{T} + ln A$ , so the number multiplying $\frac{1}{T}$ is the slope. $E_{a} = - m \times R = - (- 7500) \times 8.314 = 62, 355 J/mol \approx 62.4 kJ/mol$ The minus sign on the slope and the minus in $- m$ cancel — $E_{a}$ comes out positive, as it must (you can't have a negative energy hill).

L1.2 — Which line is which?

Two straight lines are drawn on one Arrhenius plot (see figure). One reaction has $E_{a} = 40 kJ/mol$ , the other $E_{a} = 90 kJ/mol$ . Which line (steeper or shallower) belongs to the higher $E_{a}$ ?

Figure — Activation energy from Arrhenius plot; effect of catalyst

Recall Solution

Slope $= - E_{a} / R$ . A bigger $E_{a}$ makes the slope more negative, i.e. steeper downhill as you move right (toward larger $1/ T$ ). So the steeper line is the $E_{a} = 90 kJ/mol$ reaction (the red line in the figure); the shallower line is $40 kJ/mol$ .

L1.3 — True or false

"A catalyst makes $Δ H$ of the reaction smaller." True or false, and why?

Recall Solution

False. A catalyst lowers the activation energy (the height of the barrier), but the reactant and product energy levels are untouched, so $Δ H$ (product minus reactant energy) is unchanged. See Reaction Energy Diagrams. A catalyst changes how fast, never how far downhill.

Level 2 — Application

L2.1 — Ea from two data points

A reaction has $k = 1.0 \times 1 0^{- 3} s^{- 1}$ at $300 K$ and $k = 8.0 \times 1 0^{- 3} s^{- 1}$ at $340 K$ . Find $E_{a}$ .

Recall Solution

WHAT: with two temperatures we use the two-point Arrhenius form. WHY this form: subtract $ln k$ at the two temperatures; the $ln A$ cancels, leaving only $E_{a}$ . $ln \frac{k _{2}}{k _{1}} = - \frac{E _{a}}{R} (\frac{1}{T _{2}} - \frac{1}{T _{1}})$ $ln \frac{8.0 \times 1 0 ^{- 3}}{1.0 \times 1 0 ^{- 3}} = ln 8 = 2.0794$ $\frac{1}{340} - \frac{1}{300} = 0.0029412 - 0.0033333 = - 3.9216 \times 1 0^{- 4} K^{- 1}$ $2.0794 = - \frac{E _{a}}{8.314} \times (- 3.9216 \times 1 0^{- 4})$ $E_{a} = \frac{2.0794 \times 8.314}{3.9216 \times 1 0 ^{- 4}} = 44, 085 J/mol \approx 44.1 kJ/mol$

L2.2 — Predict a rate constant

For the reaction in L2.1 ( $E_{a} = 44.1 kJ/mol$ , $k = 1.0 \times 1 0^{- 3} s^{- 1}$ at $300 K$ ), predict $k$ at $360 K$ .

Recall Solution

Same two-point law, now solving for $k_{2}$ : $ln \frac{k _{2}}{k _{1}} = - \frac{E _{a}}{R} (\frac{1}{360} - \frac{1}{300})$ $\frac{1}{360} - \frac{1}{300} = 0.0027778 - 0.0033333 = - 5.5556 \times 1 0^{- 4}$ $ln \frac{k _{2}}{k _{1}} = - \frac{44 , 085}{8.314} \times (- 5.5556 \times 1 0^{- 4}) = 2.9457$ $\frac{k _{2}}{k _{1}} = e^{2.9457} = 19.02$ $k_{2} = 1.0 \times 1 0^{- 3} \times 19.02 = 1.90 \times 1 0^{- 2} s^{- 1}$

L2.3 — Catalyst rate boost

At $310 K$ a reaction has $E_{a} = 80 kJ/mol$ . A catalyst lowers it to $55 kJ/mol$ . Assuming $A$ is unchanged, by what factor does the rate rise?

Recall Solution

WHAT: divide catalysed by uncatalysed rate constant. The $A$ 's cancel. $\frac{k _{c a t}}{k _{u n c a t}} = e^{- (E_{a, c a t} - E_{a, u n c a t}) / R T} = e^{- (55, 000 - 80, 000) / (8.314 \times 310)}$ $= e^{25, 000/2577.3} = e^{9.700} = 1.63 \times 1 0^{4}$ A 25 kJ/mol cut multiplies the rate by roughly 16,300× — the tiny change in the exponent's top becomes enormous after exponentiating.

Level 3 — Analysis

L3.1 — Full plot with a best-fit slope

Four measurements are taken:

$T$ (K)	$k$ (s⁻¹)
290	0.0021
310	0.0089
330	0.0312
350	0.0946

Use the first and last points to estimate $E_{a}$ , and comment on whether the middle points support a straight line.

Recall Solution

Convert to $1/ T$ and $ln k$ :

$T$	$1/ T$ (K⁻¹)	$ln k$
290	0.00344828	$- 6.1658$
310	0.00322581	$- 4.7217$
330	0.00303030	$- 3.4674$
350	0.00285714	$- 2.3581$

Slope from endpoints: $m = \frac{- 2.3581 - ( - 6.1658 )}{0.00285714 - 0.00344828} = \frac{3.8077}{- 5.9113 \times 1 0 ^{- 4}} = - 6441 K$ $E_{a} = - m \times R = 6441 \times 8.314 = 53, 554 J/mol \approx 53.6 kJ/mol$ Straightness check: the plotted points (figure) fall on a near-perfect line — the middle points sit right on the endpoint line, so the single-slope $E_{a}$ is trustworthy. If they curved, $E_{a}$ itself would be temperature-dependent (a sign of a changing mechanism).

L3.2 — Back out the pre-exponential factor

Using $E_{a} = 53.6 kJ/mol$ and the point $T = 330 K$ , $k = 0.0312 s^{- 1}$ , find $A$ .

Recall Solution

From $ln k = ln A - \frac{E _{a}}{R T}$ , solve for $ln A$ : $ln A = ln k + \frac{E _{a}}{R T} = - 3.4674 + \frac{53 , 600}{8.314 \times 330}$ $= - 3.4674 + \frac{53 , 600}{2743.6} = - 3.4674 + 19.5364 = 16.069$ $A = e^{16.069} = 9.5 \times 1 0^{6} s^{- 1}$ (Order of magnitude $1 0^{6}$ – $1 0^{7}$ ; matching units to $k$ since this is first order.)

L3.3 — Same-A comparison

Two reactions share the same $A$ . At $298 K$ , reaction P ( $E_{a} = 60 kJ/mol$ ) has $k_{P} = 5.0 \times 1 0^{- 4} s^{- 1}$ . Reaction Q has $E_{a} = 48 kJ/mol$ . Find $k_{Q}$ .

Recall Solution

$\frac{k _{Q}}{k _{P}} = e^{- (E_{a, Q} - E_{a, P}) / R T} = e^{- (48, 000 - 60, 000) / (8.314 \times 298)}$ $= e^{12, 000/2477.6} = e^{4.8434} = 126.8$ $k_{Q} = 5.0 \times 1 0^{- 4} \times 126.8 = 6.34 \times 1 0^{- 2} s^{- 1}$

Level 4 — Synthesis

L4.1 — Catalyst on the plot

A reaction (uncatalysed $E_{a} = 90 kJ/mol$ ) is run with a catalyst that drops $E_{a}$ to 50 kJ/mol. Both have the same $A = 1.0 \times 1 0^{12} s^{- 1}$ . Sketch (and reason out) how the two Arrhenius lines relate, then compute the vertical gap $Δ (ln k)$ between them at $T = 320 K$ .

Recall Solution

Reasoning: both lines share the same intercept $ln A$ (same $A$ ), so they meet at $1/ T = 0$ (infinite temperature — the far left). The catalysed line is shallower ( $E_{a}$ smaller ⇒ slope less negative). Because it descends more gently, at any real finite temperature ( $1/ T > 0$ ) it sits above the uncatalysed line. The lines fan apart as $1/ T$ grows (i.e. as $T$ drops) — the catalyst helps most at low temperature. Gap at 320 K: $Δ (ln k) = ln k_{c a t} - ln k_{u n c a t} = - \frac{E _{a, c a t}}{R T} + \frac{E _{a, u n c a t}}{R T} = \frac{E _{a, u n c a t} - E _{a, c a t}}{R T}$ $= \frac{90 , 000 - 50 , 000}{8.314 \times 320} = \frac{40 , 000}{2660.5} = 15.03$ In plain rate terms: $k_{c a t} / k_{u n c a t} = e^{15.03} \approx 3.4 \times 1 0^{6}$ — a millions-fold speed-up. See Catalysis Mechanisms.

L4.2 — When does the catalyst stop helping?

Using L4.1's numbers, at what temperature would the two lines' rate ratio drop to a mere factor of 10 (i.e. $ln$ -gap $= ln 10 = 2.3026$ )? Interpret.

Recall Solution

Set the gap equal to $ln 10$ : $\frac{40 , 000}{8.314 T} = 2.3026 \Rightarrow T = \frac{40 , 000}{8.314 \times 2.3026} = \frac{40 , 000}{19.144} = 2089 K$ Interpretation: only at an absurdly high $\approx 2089 K$ does the catalyst's advantage shrink to a factor of 10. At ordinary temperatures the advantage is astronomically larger. This is why catalysts matter most for reactions we want to run cheaply near room temperature — connect to Industrial Catalysis.

Level 5 — Mastery

L5.1 — Temperature to match a catalyst

A reaction has $E_{a} = 100 kJ/mol$ . Rather than add a catalyst, an engineer wants to reach the same rate constant just by heating. The uncatalysed reaction runs at $300 K$ . A proposed catalyst would give the same rate as heating the uncatalysed reaction to $360 K$ . By how many kJ/mol must the catalyst lower $E_{a}$ to match that (assume $A$ unchanged)?

Recall Solution

Step 1 — target rate ratio from heating. Heating from 300 K to 360 K (same $E_{a}$ , same $A$ ): $ln \frac{k _{360}}{k _{300}} = - \frac{E _{a}}{R} (\frac{1}{360} - \frac{1}{300}) = - \frac{100 , 000}{8.314} \times (- 5.5556 \times 1 0^{- 4})$ $= 12028.6 \times 5.5556 \times 1 0^{- 4} = 6.6826$ So heating multiplies $k$ by $e^{6.6826}$ . Step 2 — match with a catalyst at 300 K. The catalyst must produce the same factor at 300 K: $ln \frac{k _{c a t}}{k _{u n c a t}} = \frac{E _{a, u n c a t} - E _{a, c a t}}{R \times 300} = 6.6826$ $E_{a, u n c a t} - E_{a, c a t} = 6.6826 \times 8.314 \times 300 = 16, 666 J/mol$ $Δ E_{a} \approx 16.7 kJ/mol$ Meaning: a catalyst dropping $E_{a}$ by just 16.7 kJ/mol at 300 K does the same job as heating 60 K — but with no fuel bill. This is the quantitative heart of why industry prefers catalysts to brute-force heating.

L5.2 — Curved Arrhenius plot

An experimenter finds their $ln k$ vs $1/ T$ plot is not straight — it curves so the apparent slope becomes less steep at high temperature. Give a physically consistent explanation in terms of $E_{a}$ , and state what a single straight-line fit would wrongly report.

Recall Solution

WHAT the curvature means: the local slope is $- E_{a} / R$ at each point. A less steep slope at high $T$ (small $1/ T$ , left side) means the effective $E_{a}$ is smaller at high temperature. WHY this happens: a reaction may proceed by two competing pathways (or a changing rate-determining step). At low $T$ the high-barrier path dominates the rate constant's temperature response; as $T$ rises, a lower-barrier channel takes over — so the mixture's apparent $E_{a}$ falls. (Transition State Theory's $T$ -dependence of $A$ can also add mild curvature — see Transition State Theory.) The error of a single fit: forcing one straight line through curved data returns an average slope — an $E_{a}$ that is neither the low- $T$ nor high- $T$ value, and that predicts $k$ badly at both extremes. Always inspect for curvature before quoting a single $E_{a}$ (ties back to L3.1's straightness check).

L5.3 — Reverse reaction & equilibrium sanity check

A catalyst lowers the forward barrier from 120 kJ/mol to 90 kJ/mol. The reaction is exothermic with $Δ H = - 40 kJ/mol$ . (a) What is the catalysed reverse activation energy? (b) Show the equilibrium constant is unchanged. (See Equilibrium vs Kinetics.)

Recall Solution

(a) On a reaction energy diagram, the reverse barrier = forward barrier − $Δ H$ measured from the product side. With $Δ H = - 40$ (products 40 kJ/mol below reactants): $E_{a, rev} = E_{a, fwd} - Δ H = 90 - (- 40) = 130 kJ/mol$ (Uncatalysed check: $120 - (- 40) = 160$ ; catalyst cut both forward and reverse by exactly 30 kJ/mol — $120 \to 90$ and $160 \to 130$ .) (b) The equilibrium constant tracks the difference of barriers: $K \propto \frac{k _{fwd}}{k _{rev}} = e^{- (E_{a, fwd} - E_{a, rev}) / R T}$ Uncatalysed: $E_{a, fwd} - E_{a, rev} = 120 - 160 = - 40$ . Catalysed: $90 - 130 = - 40$ . Identical. The catalyst subtracted 30 from both barriers, so their difference — and therefore $K$ — is untouched. This is the rigorous proof that catalysts change kinetics, never equilibrium.

Recall Quick self-quiz

Slope $= - m R$ ? ::: $E_{a} = - m \times R$ (positive $E_{a}$ from a negative slope $m$ ). Units inside the exponential? ::: Energy in J/mol, since $R$ is in J/(mol·K). Same $A$ , different $E_{a}$ — parallel lines? ::: No — same intercept, different slope ⇒ lines fan out from $1/ T = 0$ . Does a catalyst change $Δ H$ or $K$ ? ::: Neither; it lowers the shared barrier, changing only rate.