Chemical Kinetics

Chemical Kinetics

Question

The rate of a reaction becomes four times when the temperature changes from 293 K to 313 K. Calculate the energy of activation (Ea) of the reaction assuming that it does not change with temperature.

[R = 8·314 J K-1 mol-1, log 4 = 0·6021]

Answer

Given: T1 = 293 k

 T2 = 313 k

 R = 8.314 J k-1mol-1

 k2 = 4k1

 Ea =?

The formula used is;
l o g space k subscript 2 over k subscript 1 space equals fraction numerator E subscript a over denominator 2.303 R end fraction open parentheses 1 over T subscript 1 minus 1 over T subscript 2 close parentheses

S u b s t i t u t i n g space v a l u e s comma space w e space g e t space

L o g space fraction numerator 4 k subscript 2 over denominator k subscript 1 end fraction space equals fraction numerator E subscript a over denominator 2.303 space x space left parenthesis 8.314 space J k to the power of negative 1 end exponent m o l to the power of negative 1 end exponent right parenthesis end fraction open parentheses fraction numerator 1 over denominator 293 space k end fraction minus fraction numerator 1 over denominator 313 k end fraction close parentheses

l o g space 4 space equals space fraction numerator E subscript a over denominator 2.303 space x space left parenthesis 8.314 space J k to the power of negative 1 end exponent space m o l to the power of negative 1 end exponent right parenthesis end fraction open parentheses fraction numerator 313 minus 293 k over denominator 293 k space x 313 space k end fraction close parentheses

fraction numerator 0.6021 space x space 2.303 space x space 8.314 J k to the power of negative 1 end exponent space x space 293 space k space x space 313 k over denominator 20 space k end fraction space equals E subscript a

E subscript a space equals fraction numerator 1 comma 057 comma 266.674 over denominator 20 end fraction space equals 52863.33 J space m o l to the power of negative 1 end exponent

equals 52.86 space k J space m o l to the power of negative 1 end exponent

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