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

Let the rate of reaction be `R` at `10^(@)C` . Since for every `10^(@)C` rise of temperature, the rate is doubled we have
`10^(@)Coverset(2R)(rarr)20^(@)Coverset(2^(2)R)(rarr)30^(@)Coverset(2^(3)R)(rarr)40^(@)Coverset(2^(4)R)(rarr)`
`50^(@)Coverset(2^(5)R)(rarr)60^(@)Coverset(2^(6)R)(rarr)70^(@)Coverset(2^(7)R)(rarr)80^(@)Coverset(2^(8)R)(rarr)`
`90^(@)Coverset(2^(5)R)(rarr)100^(@)C`
Thus,
`(Rate at 100^(@)C)/(Rate at 10^(@)C)=(2^(9)R)/(R)=2^(9)=512`