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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 becomeA. 256 timesB. 512 timesC. 64 timesD. 128 times |
Answer» Correct Answer - B 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"` |
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