InterviewSolution
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InterviewSolution
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Two reacants `A` and `B` are present such that `[A_(0)] = 4[B_(0)]` and `t_(1//2)` of `A` and `B` are `5` and `15` mintute respectively. If both decay folliwing `I` order, how much time later will concentrations of both of them would be equal?A. `15 min`B. `10 min`C. `5 min`D. `12 min` |
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Answer» Correct Answer - A Amount of `A` left in `n_(1)` halves`= ((1)/(2))^(n_(1)) [A_(0)]` Amount of `B` left in `n_(2)` halves `= ((1)/(2))^(n_(2))[B_(0)]` At the end, according to the question `([A_(0)])/(2^(n_(1))) ([B_(0)])/(2^(n_(2))) rArr (4)/(2^(n_(1))) = (1)/(2^(n_(2))), ([A_(0)] = 4[B_(0)])` `(2^(n_(1)))/(2^(n_(2))) = 4 rArr 2^(n_(1)-n_(2)) = (2)^(2) rArr n_(1) - n_(2) = 2` `:. n_(2) = (n_(1) - 2)` ....(i) Also `t = n_(1) t_(1//2(A)), t = n_(2)t_(1//2(B))` (Let concentration of both become equal after time `t`) `:. (n_(1) xx t_(1//2(A)))/(n_(2) xx t_(1//2(B))) = 1 rArr (n_(1) xx 5)/(n_(2) xx 15) = 1 rArr (n_(1))/(n_(2)) = 3` ...(ii) form Eqs. (i) and (ii), we get `n_(1) = 3, n_(2) = 1` `t = 3 xx 5 = 15 min` |
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