1.

The activation energy for a reaction is 9.0 K cal/mol . The increase in the rate constant when its temperature is increased from 298 K to 308 K is

Answer»

0.63
0.5
1
0.1

Solution :2.303 log `(k_(2))/(k_(1)) = (E_(a))/(R) [(T_(2) - T_(1))/(T_(1)T_(2))]`
log `(k_(2))/(k_(1)) = (9.0 XX 10^(3))/(2.303 xx 2) [(308 - 298)/(308xx 298)]`
`(k_(2))/(k_(1)) = 1.63 , K_(2) = 1.63 K_(1) , (1.63 k_(1) - k_(1))/(k_(1))xx 100 = 63.0` %


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