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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

`64` times

B

`512` times

C

`256` times

D

`128` times

Text Solution

Verified by Experts

The correct Answer is:
B
`(r + (r +100))/(r_(t)) = 2` for each `10^(@)` rise in temperature
`:. (r_(100))/(r_(10)) = (2)^(9) = 512`
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