The reactant concentration in a zero-order reaction was 9.00 Solution The reacta

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The reactant concentration in a zero-order reaction was 9.00×10-2 after 130 and 2.50×10-2 after 300 . What is the rate constant for this reaction? i got answer for 3.82x10^-4 b.What was the initial reactant concentration for the reaction described in Part A? c.The reactant concentration in a first-order reaction was 7.70×10-2 after 15.0 and 5.60×10-3 after 95.0 . What is the rate constant for this reaction? d. The reactant concentration in a second-order reaction was 0.370 after 140 and 1.40×10-2 after 715 . What is the rate constant for this reaction? a. integrated zero order rate law [A] = [A]0 - k·t You the concentration at two different times, i.e. [A]1 = [A]0 - k·t1 [A]2 = [A]0 - k·t2 eliminate unknown by taking the difference: [A]1 - [A]2 = -k·t1 + k·t2 = k·(t2 - t1) Hence: k = ([A]1 - [A]2) / (t2 - t1) = (9.0×10?²M - 2.5×10?²M) / (300s - 130s) = 3.82×10?4Ms?¹ b. [A]1 = [A]0 - k·t1 => [A]0 = [A]1 + k·t1 = 9.0×10?²M + 3.82×10?4Ms?¹·130s = 13.97×10?²M c. integrated first order rate law: ln[A] = ln[A]0 - k·t Again you know two concentrations at their time: ln[A]1 = ln[A]0 - k·t1 ln[A]2 = ln[A]0 - k·t2 => ln[A]1 - ln[A]2 = k·(t2 - t1) => k = (ln[A]1 - ln[A]2)/(t2 - t1) = ln( [A]1/[A]2 )/(t2 - t1) = ln( 7.7×10?²M/5.6×10?²M )/(95s - 15s) = 3.98×10?³s?¹ d. integrated second order rate law: 1/[A] = 1/[A]0 - k·t same procedure as in Part a) and c) 1/[A]1 = 1/[A]0 + k·t1 1/[A]2 = 1/[A]0 + k·t2 => 1/[A]2 - 1/[A]1 = k·(t2 - t1) => k = (1/[A]2 - 1/[A]1 )/(t2 - t1) = ln( 1/1.4×10?²M - 1/0.37M)/(715s - 140s) = 1.195×10?¹M?¹s?¹

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