Iit jee 1997

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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

1

rahul nair ·2010-02-15 22:54:30

k1(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}

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Gone.. ·2010-02-15 23:02:07

thanks rahul.

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