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

Options

A. 256 times
B. 512 times
C. 64 times
D. 128 times

Answer

B

Explanation

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`
`rArrSo," "rate=2^(9)=512"times"`