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Number of natural life times `(T_(av))` required for a first order reaction to achieve `99.9%` level of comletion isA. `6.9`B. `1.5`C. `0.105`D. `9.2` |
Answer» Correct Answer - A Time required for `99.9%` reaction is `10` times of half-life period. `T_(99.9) = 10T_(50)` But `T_(50) = (0.693)/(k)` `:. T_(av) = (1)/(k)` and `T_(50) = 0.693 T_(av)` `:. T_(99.9) = 10 xx 0.693 T_(av) = 6.93 T_(av)` |
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