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 :

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answer is B.

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Detailed Solution

$\frac{^{r} 100 ° C}{^{r} 10 ° C} = 2^{(\frac{T_{2} - T_{1}}{10})} = 2^{(\frac{100 - 10}{10})} = 2^{9} = 512$ (where 2 is temperature coefficient of reaction)

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