Answer please

Dear Student,
For a first order reaction 
ln K2ln K1=EaR(T2-T1T1T2)
logK2K1=Ea2.303R(T2-T1T1T2)  -------eq 1
The reaction is 50% complete that means the half life time is given and we can calculate k1 and k2 from it.
so K1=0.69320 
K1= 0.03456 min-1
putting the value to second half life time t1/2= 5 min at 470C we get,
K2=0.1386 min-1
putting the value of k1 and K2 and both the temperatures in eq 1 we get
solving it for E we get 6.65 KJ/mol 
option d

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