1.

The activation energy of a reaction is 9 kcal/mol. The increase in the rate constant when its temperature is raised from 395 to 300K is approximately

Answer»

0.1
0.5
1
0.25

Solution :`LOG. (k_2)/(k_1)= 9000/ (2.303xx 2)(5/(395 xx 300))= 0.1103`
Hence, `k_2/k_1 = 1.288`
or `k_2 = 1.288 k_1, i.e., ` INCREASE = 38.8 %`~~`25%


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