InterviewSolution
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InterviewSolution
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Show that in the case of first order reaction, the time required for `99.9%` completion of the reaction is 10 times that required for `50%` completion `(log 2 = 0.3010)` |
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Answer» `t_(1//2)=(0.693)/(k)` For `99.9%to a=100` `a-x =100 -99.9 =0.1` `k=(2.303)/(k)log ""(100)/(0.1)` But ` k =(0.693)/(t_(1//2))=(2.303)/(0.93)xxt_(1//2)log 1000=3.33xx t_(1//2)log 1000=3.33xxt_(1//2) xx3 =9.99xxt_(1//2)` `t_(99.9%)`is 10 times `t_(1//2)` |
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