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

See the parent for where these come from: the topic note (Hindi-English explanation). Supporting ideas live in Activation Energy and Transition State Theory, Rate Laws and Rate Constants, Maxwell-Boltzmann Distribution, Catalysis, Thermodynamics vs Kinetics, and Q10 Temperature Coefficient.

State: In the equation $k = A e^{- E_{a} / (R T)}$ , name each of the four quantities $k$ , $A$ , $E_{a}$ , $T$ , give the SI units of $E_{a}$ , $R$ , $T$ , and explain in one sentence why the exponent $- E_{a} / (R T)$ must be a pure number (no units).

Recall Solution

$k$ = rate constant (units depend on reaction order).
$A$ = pre-exponential / frequency factor, same units as $k$ .
$E_{a}$ = activation energy — the energy barrier from reactants to the transition state. Units: $J mol^{- 1}$ .
$T$ = absolute temperature, units $K$ .
$R = 8.314 J mol^{- 1} K^{- 1}$ .

The exponent is $\frac{E _{a}}{R T}$ . Its units are: $\frac{J mol ^{- 1}}{( J mol ^{- 1} K ^{- 1} ) \cdot K} = \frac{J mol ^{- 1}}{J mol ^{- 1}} = dimensionless.$ An exponent must be dimensionless — you cannot raise $e$ to "5 joules". This is exactly why $R$ (per mole, per kelvin) is the right companion to $E_{a}$ (per mole) and $T$ (in kelvin).

Exercise 1.2

State: Two reactions X and Y have the same $A$ . Reaction X has $E_{a} = 30 kJ/mol$ ; reaction Y has $E_{a} = 90 kJ/mol$ . At the same temperature, which is faster, and why in words?

Recall Solution

The bigger $E_{a}$ makes the exponent $- E_{a} / (R T)$ more negative, so $e^{- E_{a} / (R T)}$ is smaller. With equal $A$ , a smaller Boltzmann factor means a smaller $k$ . Therefore X ( $E_{a} = 30$ ) is faster. Fewer molecules can clear a 90 kJ/mol wall than a 30 kJ/mol wall at the same temperature.

Level 2 — Application

Exercise 2.1

State: A reaction has $E_{a} = 50 kJ/mol$ and $A = 2.0 \times 1 0^{13} s^{- 1}$ . Find $k$ at $T = 300 K$ .

Recall Solution

Step 1 — what/why: Plug straight into $k = A e^{- E_{a} / (R T)}$ (single form, since we know $A$ ). Step 2 — the exponent (keep $E_{a}$ in joules to match $R$ ): $\frac{E _{a}}{R T} = \frac{50000}{8.314 \times 300} = \frac{50000}{2494.2} = 20.05.$ Step 3 — Boltzmann factor: $e^{- 20.05} = 1.96 \times 1 0^{- 9}$ . Step 4 — multiply: $k = 2.0 \times 1 0^{13} \times 1.96 \times 1 0^{- 9} = 3.9 \times 1 0^{4} s^{- 1} .$ Answer: $k \approx 3.9 \times 1 0^{4} s^{- 1}$ .

Exercise 2.2

State: Convert first, then compute. A reaction has $E_{a} = 75 kJ/mol$ , $A = 5.0 \times 1 0^{11} s^{- 1}$ , measured at $2 7^{\circ} C$ . Find $k$ .

Recall Solution

Step 1 — convert temperature: $T = 27 + 273.15 = 300.15 K \approx 300 K$ . Step 2 — exponent: $\frac{E _{a}}{R T} = \frac{75000}{8.314 \times 300} = 30.07.$ Step 3: $e^{- 30.07} = 8.7 \times 1 0^{- 14}$ . Step 4: $k = 5.0 \times 1 0^{11} \times 8.7 \times 1 0^{- 14} = 4.4 \times 1 0^{- 2} s^{- 1} .$ Answer: $k \approx 0.044 s^{- 1}$ .

Level 3 — Analysis

Exercise 3.1

State: A rate constant is $k_{1} = 0.001 s^{- 1}$ at $T_{1} = 300 K$ and $k_{2} = 0.010 s^{- 1}$ at $T_{2} = 350 K$ . Find $E_{a}$ .

Recall Solution

Step 1 — why the two-point form: we don't know $A$ , so use the form that cancels it. $ln \frac{k _{2}}{k _{1}} = \frac{E _{a}}{R} (\frac{1}{T _{1}} - \frac{1}{T _{2}}) .$ Step 2 — left side: $ln (0.010/0.001) = ln 10 = 2.3026$ . Step 3 — the $1/ T$ bracket: $\frac{1}{300} - \frac{1}{350} = 0.0033333 - 0.0028571 = 4.762 \times 1 0^{- 4} K^{- 1} .$ Step 4 — solve: $E_{a} = \frac{R l n ( k _{2} / k _{1} )}{\frac{1}{T _{1}} - \frac{1}{T _{2}}} = \frac{8.314 \times 2.3026}{4.762 \times 1 0 ^{- 4}} = 4.02 \times 1 0^{4} J/mol .$ Answer: $E_{a} \approx 40.2 kJ/mol$ — a moderate barrier, typical of many organic reactions.

Exercise 3.2

State: Five measurements are plotted as $ln k$ (vertical) against $1/ T$ (horizontal). The best-fit straight line has slope $- 5800 K$ and intercept $+ 12.6$ . Find $E_{a}$ and $A$ .

Recall Solution

Compare with $ln k = ln A - \frac{E _{a}}{R} \cdot \frac{1}{T}$ : slope $= - E_{a} / R$ , intercept $= ln A$ . See the figure — the line's steepness carries $E_{a}$ , its height at $1/ T = 0$ carries $A$ .

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

Figure s01 — Arrhenius plot: yellow dots are the five measured $(1/ T, ln k)$ points; the blue line is the best fit. The red arrow marks the "run" along $1/ T$ and the green arrow the "rise" in $ln k$ ; their ratio is the slope $= - E_{a} / R = - 5800 K$ . Extending the line back to $1/ T = 0$ gives the intercept $ln A = 12.6$ . (If the image fails to load: plot $ln k$ up, $1/ T$ across, draw the straight fit line — its downward slope encodes $E_{a}$ , its left-edge height encodes $A$ .)

Step 1 — activation energy: $E_{a} = - slope \times R = 5800 \times 8.314 = 4.82 \times 1 0^{4} J/mol \approx 48.2 kJ/mol .$ Step 2 — pre-exponential factor: $ln A = 12.6 \Rightarrow A = e^{12.6} = 2.96 \times 1 0^{5} s^{- 1} .$ Answer: $E_{a} \approx 48.2 kJ/mol$ , $A \approx 3.0 \times 1 0^{5} s^{- 1}$ .

Level 4 — Synthesis

Exercise 4.1

State: For the reaction with $E_{a} = 40.2 kJ/mol$ and $k = 0.001 s^{- 1}$ at $300 K$ , predict $k$ at $400 K$ .

Recall Solution

Step 1 — two-point form, unknown is $k_{2}$ , with $T_{1} = 300$ , $T_{2} = 400$ : $ln \frac{k _{2}}{0.001} = \frac{E _{a}}{R} (\frac{1}{300} - \frac{1}{400}) .$ Step 2 — the factor $E_{a} / R$ : $\frac{40200}{8.314} = 4835 K$ . Step 3 — the bracket: $\frac{1}{300} - \frac{1}{400} = 0.0033333 - 0.0025000 = 8.333 \times 1 0^{- 4}$ . Step 4 — combine: $ln (k_{2} /0.001) = 4835 \times 8.333 \times 1 0^{- 4} = 4.029$ . Step 5 — exponentiate: $k_{2} /0.001 = e^{4.029} = 56.2$ , so $k_{2} = 0.001 \times 56.2 = 0.056 s^{- 1} .$ Answer: $k \approx 0.056 s^{- 1}$ — a $\sim 56 \times$ jump for $+ 100 K$ . That extreme sensitivity is the whole point of Arrhenius behaviour.

Exercise 4.2

State: A catalyst lowers $E_{a}$ from $75 kJ/mol$ to $50 kJ/mol$ . Assuming $A$ is unchanged, by what factor does $k$ increase at $298 K$ ?

Recall Solution

Step 1 — why a ratio: $A$ cancels because it's the same for both, so $\frac{k _{cat}}{k _{uncat}} = \frac{A e ^{- E_{a, cat} / (R T)}}{A e ^{- E_{a, unc} / (R T)}} = e^{(E_{a, unc} - E_{a, cat}) / (R T)} .$ Step 2 — the energy gap: $E_{a, unc} - E_{a, cat} = 75000 - 50000 = 25000 J/mol$ . Step 3 — divide by $R T$ : $\frac{25000}{8.314 \times 298} = 10.09$ . Step 4 — exponentiate: $e^{10.09} = 2.4 \times 1 0^{4}$ . Answer: the catalyst speeds the reaction by a factor of $\approx 2.4 \times 1 0^{4}$ (about 24000×). A 25 kJ/mol dent in the barrier is enormous because it sits inside an exponential.

Level 5 — Mastery

Exercise 5.1

State: A reaction's rate constant doubles when temperature rises from $300 K$ to $310 K$ (this "doubling per 10 K" is the $Q_{10}$ rule, see Q10 Temperature Coefficient). Find $E_{a}$ .

Recall Solution

Step 1 — set up: $k_{2} / k_{1} = 2$ , $T_{1} = 300$ , $T_{2} = 310$ . $ln 2 = \frac{E _{a}}{R} (\frac{1}{300} - \frac{1}{310}) .$ Step 2 — the bracket: $\frac{1}{300} - \frac{1}{310} = 0.0033333 - 0.0032258 = 1.0753 \times 1 0^{- 4}$ . Step 3 — solve: $E_{a} = \frac{R l n 2}{1.0753 \times 1 0 ^{- 4}} = \frac{8.314 \times 0.6931}{1.0753 \times 1 0 ^{- 4}} = 5.36 \times 1 0^{4} J/mol .$ Answer: $E_{a} \approx 53.6 kJ/mol$ . So the famous "rate doubles per 10 °C" corresponds to a specific barrier near room temperature — it is not a universal law.

Exercise 5.2

State: Two reactions have equal $A$ . Reaction P: $E_{a} = 40 kJ/mol$ . Reaction Q: $E_{a} = 80 kJ/mol$ . At what temperature (if any) do they have the same $k$ ? If never, explain which is always faster and how the gap between them changes as $T$ rises. (Use the figure.)

Recall Solution

Step 1 — form the ratio (equal $A$ cancels): $\frac{k _{P}}{k _{Q}} = e^{(E_{a, Q} - E_{a, P}) / (R T)} = e^{40000/ (R T)} .$ Step 2 — when equal? We'd need $k_{P} / k_{Q} = 1$ , i.e. $e^{40000/ (R T)} = 1$ , i.e. $40000/ (R T) = 0$ . That requires $T \to \infty$ . For any finite $T$ the exponent is positive, so $k_{P} / k_{Q} > 1$ . Answer: they are never exactly equal at finite $T$ ; P (lower $E_{a}$ ) is always faster. Step 3 — how the gap shrinks: as $T$ grows, $40000/ (R T)$ shrinks toward 0, so $k_{P} / k_{Q} \to 1$ . Look at the figure: the two Arrhenius lines have different slopes (steeper for Q) and only meet at $1/ T = 0$ (infinite $T$ ). At high $T$ both barriers are "easy," so the low-barrier advantage fades.

Figure s02 — two Arrhenius lines with equal intercept ( $ln A = 20$ ): green is P ( $E_{a} = 40$ kJ/mol, gentler slope) and red is Q ( $E_{a} = 80$ kJ/mol, steeper slope). The white dashed segment shows the vertical gap at a finite temperature — P always sits above Q. The yellow dot marks the only meeting point, at $1/ T = 0$ (infinite $T$ ). (If the image fails to load: draw two straight lines starting together at the left edge $1/ T = 0$ ; the steeper one is the high- $E_{a}$ reaction, and it stays below the other for every real temperature.)

Numeric check of the shrinking gap: at $300 K$ , $k_{P} / k_{Q} = e^{40000/ (8.314 \cdot 300)} = e^{16.04} = 9.2 \times 1 0^{6}$ ; at $1000 K$ , $k_{P} / k_{Q} = e^{40000/ (8.314 \cdot 1000)} = e^{4.81} = 123$ . Still faster, but the margin collapsed from millions to hundreds.

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

Level 1 — Recognition

Exercise 1.1

Exercise 1.2

Level 2 — Application

Exercise 2.1

Exercise 2.2

Level 3 — Analysis

Exercise 3.1

Exercise 3.2

Level 4 — Synthesis

Exercise 4.1

Exercise 4.2

Level 5 — Mastery

Exercise 5.1

Exercise 5.2