IB Chemistry HL 2.2.12 Hl Arrhenius Equation Questions

Use real HL exam questions to apply the Arrhenius equation, interpret ln k against reciprocal temperature and calculate activation energy from gradients or paired temperatures.

Exam points

Apply the Arrhenius equation to calculate activation energy from rate constants measured at two absolute temperatures, with correct sign and units.
Use the slope of an ln k against 1/T plot to obtain activation energy and identify the correct axes, gradient and intercept.
Interpret the linear Arrhenius plot and explain how temperature changes k while activation energy remains a separate quantity.

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Study 2.2.12 (HL)—Arrhenius equation conceptsOpen the corresponding Concept lesson for explanations, examples and exam connections.

[Maximum number: 2]

Phenylethene (styrene) is produced from ethylbenzene in a gas-phase equilibrium.

Figure for Question IB Chemistry HL 2.2.12 Hl Arrhenius Equation Questions question 1 — IB Chemistry HL

The forward reaction is first order with respect to ethylbenzene.

\mathrm{C}_{8} \mathrm{H}_{10}(\mathrm{~g}) \rightleftharpoons \mathrm{C}_{8} \mathrm{H}_{8}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g})

Determine the activation energy of the reaction, $E_{\mathrm{a}}$ , in $\mathrm{kJ} \mathrm{mol}^{-1}$ , from the Arrhenius plot. Use sections 1 and 2 of the data booklet.

gradient =-8900
$\text { « } E_{\mathrm{a}}=-(-8900 \times 8.31)=+74000 \mathrm{~J} \mathrm{~mol}^{-1}=» 74 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Award [2] for correct final answer.

Accept $E_{a}$ in the range 66 to 76 «kJ $\mathrm{mol}^{-1}{ }^{-}{ }^{-}$

Do not accept $E_{a}$ in $\mathrm{J} \mathrm{mol}^{-1}$ .

Award [1 max] for -74 « $\mathrm{kJ} \mathrm{mol}^{-1}$ ».

Accept gradient in the range: -8000 to -9100 for M1.

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IB Chemistry HL 2.2.12 Hl Arrhenius Equation Questions question 1