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For a first order reaction, the ratio of time for the completion of `99.9%` and half of the reaction isA. `8`B. `10`C. `9`D. `12` |
Answer» Correct Answer - B For reaction first order, `t_(1//2) = (2.303)/(k) "log"(a)/(a-(a)/(2))` `= (2.303)/(k) log2 = (2.303)/(k) = 0.3010` `t_(99.9%) = (2.303)/(k)"log"(a)/(a-0.999 a)` `= (2.303)/(k)log10^(-3) = (2.303)/(k) xx 3` `(t_(99.9%))/(t_(1//2)) = (3)/(0.3010) ~~ 10`. |
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