InterviewSolution
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InterviewSolution
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For the reaction A + B ` to` products, the following initial rates were obtained at various given initial concentration `{:("S.No.",,"[A]mol/L",,"[B] mol/L",,"Initial rate M/s"),(1.,," "0.1,," "0.1,," "0.05),(2.,," "0.2,," "0.1,," "0.10),(3.,," 0.1",," 0.2",," 0.05"):}` Determine the half-life period. |
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Answer» `"rate" = k [A]^(x)[B]^(y)` ` 0.05 = k[0.1]^(x)[0.1]^(y) `......(i) ` 0.10 = k[0.2]^(x)[0.1]^(y)` .......(ii) ` 0.05 = k[0.1]^(x) [0.2]^(y)`......(iii) `(ii) div (i)` ` (0.10)/(0.05) = (2)^(x)` ` = x = 1` ` (iii) div (i) ` ` (0.05)/(0.05) = (2)^(y)` ` y = 0` `"rate" = k[A]^(1)[B]^(0)` It is a first order reaction. `k = ("rate")/([A]) = 0.5 s^(-1)` ` t _(½) = (0.693)/k = (0.693)/(0.5)` ` t _(½) = 1.386 s ` Or ` t _(½) = (0.693)/k ` ` k_(2) = (0.693)/(25) " " 350 K` ` k_(1) = (0.693)/(50) " " 300 K ` ` k_(2)/k_(1) = 2 ` `log. k_(2)/k_(1) = E_(a)/(2.303 R) [1/T_(1)-1/T_(2)]` ` log 2 = E_(a)/(2.303 x 8.314) [(350-300)/(350 x 300)]` ` Ea = 12.104" kJ"//mol`. |
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