Consider the second-order reaction aA $\rightarrow$ products (which has a first half-life of 24 s). If the concentration of A after 15.4 s is 0.40 M, determine the initial concentration of A.

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The first half-life of a second-order reaction is given by:t1/2 = 1/(kA0)where k is the rate constant and A0 is the initial concentration of the reactant.We are given that t1/2 = 24 s and [A] = 0.40 M after 15.4 s. Substituting these values into the equation above, we get:24 s = 1/(k * 0.40 M)k = 1/(24 s * 0.40 M) = 0.0104 M-1 s-1Now we can use the integrated rate law for a second-order reaction to determine the initial concentration of A:1/[A]t - 1/[A]0 = ktwhere [A]t is the concentration of A at time t and [A]0 is the initial concentration of A.Substituting the values we know into the equation above, we get:1/0.40 M - 1/[A]0 = 0.0104 M-1 s-1 * 15.4 s[A]0 = 0.66 MTherefore, the initial concentration of A is 0.66 M.

Consider the Second-Order Reaction AA →\rightarrow→ Products (Which Has a First Half-Life of 24 S)

Consider the Second-Order Reaction AA $\rightarrow$ Products (Which Has a First Half-Life of 24 S)