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Show that in case of a first order reaction, the time required for 99.9% of the reaction to take place is about ten times than that required for half the reaction. |
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Answer» SOLUTION :For REACTION of first order, `t_(1//2)=(2.303)/(k)log""(a)/(a-(a)/(2))=(2.303)/(k)log""2=(2.303)/(k)(0.3010)` `t_(99.9%=(2.303)/(k)log=(a)/(a-0.999a)=(2.303)/(k)LOG10^(-3)=(2.303)/(k)xx3:.(t_(99.9%))/(t_(1//2))=(3)/(0.3010)=10` |
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