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Question

A first order reaction is 50% completed in 30 minutes at 27oC and in 10 minutes at 47oC. Calculate the reaction rate constant at 27oC and the energy of activation of the reaction in kJ mol1.

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Solution

For first order reaction k=0.693t1/2
At 27oC, k27oC=0.69330=0.0231min1
At 47oC, k47oC=0.69310=0.0693min1
Now applying the following equation:
log10k1k2=Ea2.303×R(T2T1T2T1)
or log100.02310.0693=Ea2.303×8.314:(320300320×300)
or log100.3333=Ea19.1471×2096000
Ea=19.1471×9600020×log0.3333
=91906×(0.4772)
=43857 J mol1 =43.857 kJ mol1

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