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A substance having a half-life period of `30` minutes decomposes according to the first order rate law. The fraction decomposed, the balance remaining after `1.5` hours and time for `60%` decomposition on its doubling the initial concentration will be.A. `87.4, 0.126, 39.7` minB. `80.6, 0.135, 40.8` minC. `90.5, 0.144, 2829` minD. `80.2, 0.135, 26.6 min |
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Answer» Correct Answer - A We know that `K = (0.693)/(t_(1//2)) = (0.693)/(30) = 0.023//"min"^(-1)` Let the initial concentration of the substance `= 100` and the substance decomposed in `1.5` hours `(90 min) = x` then, `a-x = 100-x` Substituting these values in `K = (2.303)/(t) "log"((a)/(a-x))` `rArr 0.0231 = (2.303)/(90) "log" (100)/(100-x)` `"log"(100)/(100-x) = (0.0231 xx 90)/(2.303)` `rArr x = 87.4` So the fraction decomposed in `1.5` hours `= (87.4)/(100) = 0.874` Fraction remaining behind after `1.5` hours `= 1 - 0.874 = 0.126` The time required for `60%` of decomposition `K = (2.303)/(t) "log"((a)/(a-x))` `rArr 0.023 = (2.303)/(t) "log"((100)/(40))` `t = (2.303)/(0.0231) "log"((10)/(4)) = 39.7` minutes Since the reaction is of first order the time required to complex specific fraction is independent of initial concentration (or pressure). Hence `60%` of the reaction will decompose in `39.7` mins in this case also. |
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