Verified Solution • 2025–26 Session moderate

QJEEM13Chf35c: JEE Main 2013 Chemistry - Class 12 JEEM13Chf35c Solution: The rate of a reaction doubles when its temperature changes…

The rate of a reaction doubles when its temperature changes from 300K to 310K. Activation energy of such a reaction will be:(R = 8.314 JK–1 mol–1 and log 2 = 0.301)

Multiple Choice Options

  • A 48.6 kJ mol–1
  • B 58.5 kJ mol–1
  • C 60.5 kJ mol–1
  • D 53.6 kJ mol–1

Step-by-Step Verified Solution

Activation energy can be calculated from the equation

log⁡ k2log⁡ k1=Ea2.303R(1T1−1T2){{\log \,{k_2}} \over {\log \,{k_1}}} = {{{E_a}} \over {2.303R}}\left( {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right)

given k2k1=2  T2=310 K  T1=300 K{{{k_2}} \over {{k_1}}} = 2\,\,{T_2} = 310\,K\,\,{T_1} = 300\,K

=log⁡2=−Ea2.303×8.314(1310−1300)= \log 2 = {{ - {E_a}} \over {2.303 \times 8.314}}\left( {{1 \over {310}} - {1 \over {300}}} \right)

Ea=53598.6J/mol{E_a} = 53598.6J/mol

=53.6kJ/mol.= 53.6kJ/mol.

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