In a zero order reaction for every 10∘C rice 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
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B
512 times
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C
64 times
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D
128 times
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Solution
The correct option is B512 times 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.so, rate =29=512 times
alternate method with every 10∘ rise in temperature, rate becomes double.
So r′r=2(100−1010)=29=512 times