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Help with Integrated Rate Laws?

Topic: How to write an equation with a slope of 0
June 26, 2019 / By Paise
Question: A certain reaction has the following general form: aA -----> bB At a particular temperature and [A]o = 2.80 x 10^-3 M, concentration vs. time data were collected for this reaction, and a plot of 1/ [A] vs time resulted in a straight line with a slope value of 3.60 x 10^/2 L/mol s. a. Determine the rate law, the integrated rate law, and the value of the rate constant for this reaction. b. Calculate the half-life for this reaction. c. How much time is required for the concentration of A to decrease to 7.00 x 10^-4 M? Thank you!!! Help is much appreciated!!

Best Answers: Help with Integrated Rate Laws?

Maralyn | 3 days ago
let's start here... for the rxn. A --> B + C rate = change in amount of A per unit time.. right? and if the volume is constant.. rate = change in concentration of A per time and in math we write "change in" with "d"... and in chemistry we write concentration with [ ] ... so that rate = -d[A] / dt where.. - is because [A] is decreasing [A] is concentration of A ok so far? ******* also... we know that rate is proportional to concentration... the higher the concentration of A, the faster the rate because the more frequent the collisions. In math we write proportions as rate α [A] and in math we solve proportions by inserting a constant and an = sign rate = k x [A] and since some reaction mechanisms involve 1 molecule of A decomposing and others involve 2 or more A's colliding...there is an exponent on that [A] rate = k x [A]^n where n is the "order" of the reaction n = 0 is zero order n = 1 is first order n = 2 is second order etc **** combining those 2 rate = -d[A] / dt = k x [A]^n **** now we just need to solve.. the idea is.. we have two data points.. point 1 is.. time = zero.. [A] = [Ao].. (ie we start with initial conc = Ao) point 2 is.. time = t.. [A] = [At].. (ie.. after time = t has elapsed, conc = At) and we integrate.. **** for n=0 -d[A] / dt = k x [A]^0 d[A] = -k x dt integrating from (0, [Ao]) to (t, [At]) .. . . [At]... . ..... t [A]...|... = -k x t.. | ... .. .[Ao]... ... .. 0 or.. [At] = -kt + [Ao] ***** if n=1 -d[A] / dt = k x [A]^1 1/[A] x d[A] = -k x dt ln[At] = - kt + ln[Ao] ****** if n=2.. -d[A] / dt = k x [A]² 1/[A]² d[A] = -k x dt integrating.. - 1/[At] - - 1/[Ao] = -kt 1/[At] = +kt +1/[Ao] ***** now we have these equations.. n = 0.. .. [At] = -kt + [Ao] n = 1.. .. ln[At] = -kt + ln[Ao] n = 2.. .. 1/[At] = +kt + 1/[Ao] and if you look just at the 3rd equation..and.. let y = 1/[At] x = t m = +k b = 1/[Ao] you get this equation y = mx + b which is a line.. right? so.. IF the reaction is 2nd order.. a plot of 1/[A] vs t will give a straight line likewise.. if the rxn is 0 order.. [A] vs t gives a line if the rxn is 1st order.. ln[A] vs t gives a line ***** ***** so.. you have a plot of 1/[A] vs t give a straight line.. therefore.. the rxn is 2nd order in [A] rate = k x [A]² and of course.. +k = the slope of the line = 3.60x10^-2 /Mxsec.. so that.. rate = +3.60x10^-2 / Mxsec x [A]² the integrated rate law is then. 1/[A] = +3.60x10^-2 / Mxsec x time + 1/(2.80x10^-3M) **** half life is when [At] = 1/2 [Ao].. 1/[At] = +kt + 1/[Ao] t 1/2 = (1/[At] - 1/[Ao]) / k t 1/2 = (1/(1/2[Ao]) - 1/[Ao]) / k t 1/2 = (2/[Ao] - 1/[Ao]) / k t 1/2 = (1/[Ao]) / k t 1/2 = (1 / 2.80x10^-3M) / (3.60x10^-2 /Mxsec) = 9920 sec *** [At] = 7.00x10^-4M 1/[At] = +kt + 1/[Ao] t = (1/[At] - 1/[Ao]) / k t = (1/7.00x10^-4M - 1/2.80x10^-3M) / (3.60x10^-2 /Mxsec) = ___ sec I'll let you finish.
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Originally Answered: Can an LDS marriage last if you do not like your in-laws?
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Kinborough
undemanding sillyhil who's gonna answer you mate why cant you pop in the library for a sec and verify some books or study your notes and follow the examples. no you may still have given you this question in case you have no longer been taught mate. dont be aggravated ladybird that's the actuality mate
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