The reactant concentration in a zero-order reaction was 8.00 Solution *** a ***

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*** a *** for the reaction... A ---> B + C... rate = -d[A] / dt = k x [A]^n where.. [ ] means concentration. d[A] / dt means change in concentration of A over time - is because concentration of A is decreasing k is the rate constant. [A] is the concentration of A at any given time n is the order of the reactant and if n = 0, then º means the reaction is zero order in A and zero order overall in this case. And that equation came from this concept. rate = change in concentration of A over time... so.. rate = - d[A] / dt and.. rate is proportional to the concentration of A at any given point in time. The larger the concentration of A, the faster the rate... so.. rate is proportional to [A] and if you put in a proportionality constant rate = k [A] therefore... -d[A] / dt = k x [A]º for a zero order reaction. rearranging.. 1 / [A]º d[A] = -k dt and since.. [A]º = 1 (1/1) d[A] = -k dt integrating from [Ao] to [At].. (initial concentration - which is a constant.. to final concentration which varies by time) [At] - [Ao] = -k x (t - to) and if to = 0...ie.. we start the clock at time = 0.. then... [At] - [Ao] = -k t rearranging... [At] = -kt + [Ao] which is of the form y = mx + b if... y = [At] m = -k x = t b = [Ao] right? and since y = mx + b is a line.. a plot of time on the x-axis and concentration of reactant on the y-axis will yield a straight line of slope = -k and intercept = [Ao].. initial concentration. since you have only 2 data points.. we can just say.. k = - m = - ([A2] - [A1]) / (t2 - t1) = - (4.00x10^-2M - 0.100M) / (305s - 190s) = 5.22x10^-4 M/s *** B *** that's the intercept of that line... [At] = -k x t + [Ao] [Ao] = [At] + (5.22x10^-4 M/s) x t pick either point... [Ao] = (0.100M) + (5.22x10^-4 M/s) x (190s) = 0.199M *** C *** same deal except.. rate = -d[A] / dt = k x [A]¹ rearranging... 1 / [A]¹ d[A] = -k dt integrating.. ln[At] - ln[Ao] = -kt rearranging... ln[At] = -kt + ln[Ao] also of the form.. y = mx + b so a plot of t vs ln[At] gives slope = -k and intercept = ln[Ao] ie... k = - (ln[A2] - ln[A1]) / (t2 - t1) = - ln([A2] / [A1]) / (t2 - t1) k = - ln(7.50x10^-3 / 7.80x10^-2) / (100s - 40.0s) k = 0.0390 / s... notice the units? M/M has cancelled out. *** d *** again.. same deal as before...except 2 instead of 1 or 0 rate = -d[A] / dt = k x [A]² rearranging... 1 / [A]² d[A] = -k dt integrating.. - 1/[At] - -1/[Ao] = -kt rearranging... 1/[At] -1/[Ao] = kt rearranging... 1/[At] = kt + 1/[Ao] also of the form.. y = mx + b so a plot of t vs 1/[At] gives slope = k and intercept = 1/[Ao] ie... k = + (1 / [A2] - 1 / [A1]) / (t2 - t1) k = + (1/8.50x10^-2M - 1/0.810M) / (710s - 265s) k = 0.0237 / (Mxsec). again.. notice the units? M is on the bottom this time. *********** here's another example problem. http://answers.yahoo.com/question/index;…

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