InterviewSolution
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InterviewSolution
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`75%` of a first-order reaction was completed in `32` minutes. When was `50%` of the reaction completedA. `24` minB. `4` minC. `16` minD. `8` min |
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Answer» Correct Answer - C Let original amount, `N_(0) = 100` Since `75%` completed, so final amount `N = 100 - 75 = 25` As we know `(N_(0))/(N) = 2^(n)`, where `n =` no. of half lives or, `(100)/(25) = 2^(n)` or, `4 = 2^(n)` or, `2^(2)` or, `2^(2) = 2^(n)` `:. n=2` Since total time `= n xx t_(1//2)` `32` minutes `= 2 xx t_(1//2)` `:. t_(1//2) = 16` minutes |
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