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The reaction `Ararr B` follows first order kinetics. The time taken for `0.8 mol` of `A` to produce `0.6 mol` of `B` is `1 hr`. What is the time taken for the conversion of `9.0 mol` of `A` to Product `0.675 mol` of `B` ?A. `0.25h`B. `2h`C. `1H`D. `0.5h` |
Answer» Correct Answer - C Rate constant of first of this order of first order reaction `K=(2.303)/(t) Log""((A)_(0))/((A)_(t))` `or k=(2.303)/(t)xxlog_(10)"" (0.8)/(0.2)` because 0.6 mole of B is formed suppose `t_(1) ` hour are reguireed for changing the concentration of A from 0.9 mole to 0.675 mole of B remaining mole of A =0.9-0.675=0.225 `therefore k=(2.303)/(t_(1))log_(10)(0.9)/(0.225)` From Eqs. (i) and(ii) `(2.303)/(1) log_(10)(0.8)/(0.2)=(2.303)/(t_(1)) log_(10)(0.9)/(0.225)` `2.303 log _(10 ) 4=(2.303)/(t_(1))log _(10)4` `t_(1) =1h` |
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