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 K=0 693t12 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 â€‹â€‹log0 13860 03456=Ea2 303(320-30096000) solving it for Ea we get 6 65 KJ/mol option d Regards!

Answer please

Dear Student,
For a first order reaction

\frac{\ln K_{2}}{\ln K_{1}} = \frac{E a}{R} (\frac{T_{2} - T_{1}}{T_{1} T_{2}})

\log \frac{K_{2}}{K_{1}} = \frac{E a}{2.303 R} (\frac{T_{2} - T_{1}}{T_{1} T_{2}})

-------eq 1
The reaction is 50% complete that means the half life time is given and we can calculate k1 and k2 from it.

K = \frac{0.693}{t_{\frac{1}{2}}}

so K₁=

\frac{0.693}{20}

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

\log \frac{0.1386}{0.03456} = \frac{E a}{2.303} (\frac{320 - 300}{96000})

solving it for E_a we get 6.65 KJ/mol
option d
Regards!

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