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CBSE Class 12 Chemistry 2018 Solved Paper

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Question : 13 of 26
Marks: +1, -0
A first order reaction is 50%50\% completed in 4040 minutes at 300 K300\,\mathrm{K} and in 20 minutes at 320 K320\,\mathrm{K}. Calculate the activation energy of the reaction.
   (Given :   log⁡2=0.3010,log⁡4=0.6021,\;\text{ (Given : }\;\log 2=0.3010, \log 4=0.6021, R=8.314  JK−1mol−1   )   R=8.314\;\mathrm{JK}^{-1}\mathrm{mol}^{-1}\;\text{ ) }\;
Solution:  
  k2=0.69320   ,   \;k_2=\frac{0.693}{20}\;\text{ , }\;
  k1=0.69340\;k_1=\frac{0.693}{40}
  log⁡k2k1=Ea2.303R[1T1−1T2]\;\log\frac{k_2}{k_1}=\frac{E_a}{2.303R}\left[\frac{1}{T_1}-\frac{1}{T_2}\right]
  k2k1=2\;\frac{k_2}{k_1}=2
  log⁡2=Ea2.303×8.314[320−300320×300]\;\log2=\frac{E_a}{2.303\times8.314}\left[\frac{320-300}{320\times300}\right]
  Ea=27663.8J/mol   or   27.66kJ/mol\;E_a=27663.8\mathrm{J}/\mathrm{mol}\;\text{ or }\;27.66\mathrm{kJ}/\mathrm{mol}
  \;
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