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Rates, equilibrium and pH

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Question 61

An Arrhenius plot of ln⁡k \ln k\,lnk against 1T\displaystyle \frac{1}{T}T1​ is constructed for the thermal decomposition of a nitrogen oxide. Two points on the linear line of best fit are:

  • Point 1: 1T=2.4×10−3 K−1\displaystyle \frac{1}{T} = 2.4 \times 10^{-3}\text{ K}^{-1}T1​=2.4×10−3 K−1, ln⁡k=−3.5\ln k = -3.5lnk=−3.5
  • Point 2: 1T=3.6×10−3 K−1\displaystyle \frac{1}{T} = 3.6 \times 10^{-3}\text{ K}^{-1}T1​=3.6×10−3 K−1, ln⁡k=−18.5\ln k = -18.5lnk=−18.5

What is the activation energy, EaE_aEa​, for this reaction?

(Use gas constant R=8.314 J K−1 mol−1R = 8.314\text{ J K}^{-1}\text{ mol}^{-1}R=8.314 J K−1 mol−1)

+0.104 kJ mol−1+0.104\text{ kJ mol}^{-1}+0.104 kJ mol−1

−104 kJ mol−1-104\text{ kJ mol}^{-1}−104 kJ mol−1

+104 kJ mol−1+104\text{ kJ mol}^{-1}+104 kJ mol−1

+1.04×105 kJ mol−1+1.04 \times 10^5\text{ kJ mol}^{-1}+1.04×105 kJ mol−1

Rates, equilibrium and pH Questions

  1. A Level
  2. /Chemistry
  3. /Rates, equilibrium and pH