Answer please

Dear Student,
For a first order reaction

$\mathrm{log}\frac{{K}_{2}}{{K}_{1}}=\frac{Ea}{2.303R}\left(\frac{{T}_{2}-{T}_{1}}{{T}_{1}{T}_{2}}\right)$  -------eq 1
The reaction is 50% complete that means the half life time is given and we can calculate k1 and k2 from it.
$K=\frac{0.693}{{t}_{1}{2}}}$
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
​​$\mathrm{log}\frac{0.1386}{0.03456}=\frac{Ea}{2.303}\left(\frac{320-300}{96000}\right)$
solving it for E we get 6.65 KJ/mol
option d
Regards!

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