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In a zero-order reaction for every 10^(@...

In a zero-order reaction for every `10^(@)` 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

A

256 times

B

512 times

C

64 times

D

128 times

Text Solution

Verified by Experts

The correct Answer is:
B
for `10^(@)` rise in temperature ,n=1
SO rate `=2^(n)=2^(1)=2`
when temperature is increased from `10^(@)C` to `100^(@)C` change in temperature
`=100-10= 90^(@)C`
`i.e., n=9`
SO rate `=2^(9) =512 ` times
Alternate method with every `10^(@)` rise in termperature rate becomes double .
`So,(I')/(I)=2^(((100-10)/(10)))=2^(9)=512 `times.
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