The reactant concentration in a second-order reaction was 0.220M after 205s and

Question

The reactant concentration in a second-order reaction was 0.220M after 205s and 7.10

Explanation / Answer

The integrated rate laws for simple nth order reaction, i.e. d[AB]/dt = -k·[AB]n are 1/[AB]n?¹ = 1/[AB]0n?¹ + (n-1)·k·t for n?1 and ln[AB]n?¹ = ln[AB]0n?¹ - k·t for n=1 That means a plot of 1/[AB]n?¹ (or ln[AB]) versus t yields a straight line with slope (n-1)·k (or -k). Therefore a straight line of 1/[AB] versus t plot indicates that you have second order reaction, i.e. n=2. The slope is equal to the rate constant: k = 5.1×10?²M?¹s?¹ Rate law is rate = -d[AB]/dt = k·[AB]² Integrated second order rate law is: 1/[AB] = 1/[AB]0 + k·t Half life t½ is time elapsed until [AB] has dropped to (1/2)·[AB]0. i.e.: 2/[AB]0 = 1/[AB]0 + k·t½ => t½ = 1/(k·[AB]0) = 1/(5.1×10?²M?¹s?¹ · 0.54M) = 363.1s The concentration changes of the components are linked by the reaction equation. One mole of A is produced per mole AB decomposed, hence: - ?[AB] = ?[A] -([AB] - [AB]0) = [A] - [A]0 => [A] = [A]0 + [AB]0 - [AB] analogously for product B [B] = [B]0 + [AB]0 - [AB] Here we got the particular case that no product is present initially. [A]0 = [B]0 = 0 => [A] = [B] = [AB]0 - [AB] Reactant concentration after 80 sec is [AB] = 1/(1/[AB]0 + k·t) = 1/(1/0.220M + 5.1×10?²M?¹s?¹·80s) = 0.116M So the product concentrations are: [A] = [B] = [AB]0 - [AB] = 0.220M - 0.116M = 0.104M

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