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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
`(r_(100^(@)C))/(r_(10^(@)C))=2^((T_(2)-T_(1))/(10))=2^((100-10)/(10))=2^(9)=512` (where 2 is temperature coefficient of reaction)
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