Kinetics

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Question 14
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The thermal isomerisation of cyclopropane to propene is a first-order reaction. A chemist studies this reaction at two elevated temperatures in a gas-phase reactor and obtains the following kinetic data:

  • Experiment 1: Temperature T1=750 KT_1 = 750\text{ K}T1​=750 K, Rate constant k1=1.80×10−4 s−1k_1 = 1.80 \times 10^{-4}\text{ s}^{-1}k1​=1.80×10−4 s−1
  • Experiment 2: Temperature T2=800 KT_2 = 800\text{ K}T2​=800 K, Rate constant k2=2.75×10−3 s−1k_2 = 2.75 \times 10^{-3}\text{ s}^{-1}k2​=2.75×10−3 s−1

The Arrhenius equation can be rearranged to calculate the activation energy, EaE_{\text{a}}Ea​, as follows:

ln⁡(k1k2)=EaR(1T2−1T1)\ln\left(\frac{k_1}{k_2}\right) = \frac{E_{\text{a}}}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)ln(k2​k1​​)=REa​​(T2​1​−T1​1​)

Calculate the value, in kJ mol−1\text{kJ mol}^{-1}kJ mol−1, of the activation energy, EaE_{\text{a}}Ea​, for this isomerisation reaction. Give your answer to 3 significant figures.

(The gas constant, R=8.31 J K−1 mol−1R = 8.31\text{ J K}^{-1}\text{ mol}^{-1}R=8.31 J K−1 mol−1)

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Kinetics Questions

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