Q.
In a zero order reaction for every 10∘C rise of temperature, the rate is doubled. If the temperature is increased from 10∘C to 100∘C, the rate of the reaction will become .
For 10∘ rise in temperature, n = 1
so rate = 2n=21=2
When temperature is increased from 10∘C to 100∘C, change in temperature =100−10=90∘C
i.e. n=9
So, rate = 29 = 512 times
Alternate method with every 10∘ rise in temperature, rate becomes double,
so rr′=2(10100−10)=29= 512 times.