A certain reaction has the following general reaction aA=bB At a particular temp

Question

A certain reaction has the following general reaction aA=bB At a particular temperature and [A]0=2.00x10^-2 M,concentration versus time data were collected for this reaction,and a plot of ln[A] versus time resulted in a straight line with aslope value of -2.97x10^-2min^-1. a. Determine the rate, the integrated rate law, and the valueof the rate constant for this reaction. b.Calculate the half-life for this reaction. c.How much time is requirede for concentration of A todecrease to 2.50x10^-3M? aA=bB At a particular temperature and [A]0=2.00x10^-2 M,concentration versus time data were collected for this reaction,and a plot of ln[A] versus time resulted in a straight line with aslope value of -2.97x10^-2min^-1. a. Determine the rate, the integrated rate law, and the valueof the rate constant for this reaction. b.Calculate the half-life for this reaction. c.How much time is requirede for concentration of A todecrease to 2.50x10^-3M?

Explanation / Answer

Reaction:     aA --> bB (a) Since the plot of ln[A] vs time is linear, the reaction must befirst order in A, and the graph has the form:     ln[A] = ln[A]0 - kt This is because the graph is linear, i.e. of the form y = mx + c.By comparison, we get: k = slope = 2.97x10-2 min-1 Rearrage to get to the integrated form:       ln[A] = ln[A]0 - kt exp(ln[A]) = exp(ln[A]0 - kt)           [A] =exp(ln[A]0) * exp(-kt)                = [A]0 * exp(-kt)                = 2.00x10-2 M * exp(-2.97x10-2min-1 * t) (remember that ln(exp(#)) = #, and exp(ln(#)) = #) (b) At the half-life, the concentration of A will be0.5*A0.     0.5*[A]0 = [A]0 *exp(-2.97x10-2 min-1 * t)             0.5 = exp(-2.97x10-2 min-1 * t)         ln(0.5) =ln(exp(-2.97x10-2 min-1 * t))         ln(0.5) =-2.97x10-2 min-1 * t                 t = 23.3 min (c) Follow the same process as before:     2.5x10-3 M = 2.00x10-2 M *exp(-2.97x10-2 min-1 * t)             0.125 = exp(-2.97x10-2 min-1 * t)         ln(0.125) =ln(exp(-2.97x10-2 min-1 * t))         ln(0.125) =-2.97x10-2 min-1 * t                     t = ln(0.125) / -2.97x10-2 min-1           = 70.0 min This answer makes sense: since the concentration of A is less thanhalf of the initial concentration, the time taken to reach it willbe greater than the half-life of the reaction.

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