11thanks rahul.
The time required for 10% completion of a first order reaction at 298 K is equal to that required for its 25 % completion at 308 K.If pre exponential factor for the reaction is 3.56 x 109s-1,calculate its rate constant at 318 K and also energy of activation.
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2 Answers
1k1(298K)=ln(100/90)/t
K2(308K)=ln(100/75)/t
k2/k1=ln(100/75)/ln(100/90)
k2/k1=2.73
logk2/k1=Ea\times 1000/(2.3\times 8.314\times 298\times 308)
solving,Ea=76.5kJ approx.
K(318)=Ae^{-Ea/RT}
=9.2\times 10^{-4}