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The time required for the decompoistion of `99.9%` fraction of a first order reaction is………….to that of its half-life time.A. 5B. 10C. 15D. 20 |
Answer» Correct Answer - B For a first order reaction, time required to complete any particular fraction is given as `t_(f)=(2.303)/(K)log((1)/(1-f))` Substituting the value of `f=(99.9)/(100)` , we get `t_(99.9%)=(2.303)/(K)log(1)/(1-0.999)` `=(2.303)/(K)log((1)/(0.001))` `=(2.303)/(K)log((10^(3))` `=(2.303)/(K)3log(10)` `=(6.903)/(K)(log(10=1))` Half life of reaction is given as `t_(1//2)=(0.693)/(K)` Dividing the two result we get `(t_(99.9%))/(t_(1//2))=(6.909//K)/(0.693//K)` `=(6.909)/(0.693)` `=9.97~~10` Alternatively, we can calculate `yt_(99.9%)` as follows |
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