Integrated Rate Law Concept 3

by Jules Bruno
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for reactions that with second order processes, we use the following equation and it is 1/80 equals positive K. T plus one over a. Oh so 80 here is the final reacting concentration. A. O. Is the initial reaction concentration Now Ks our rate constant in units of remember K equals M two. The negative end plus one times time inverse. And equals the order of our process here. Since it's second order, it's negative two for n Plus one times time inverse. This would mean that when it comes to second order processes, the units for K would be in polarities inverse times time, inverse. So always be on the lookout for that when looking to see if a reaction is second order or not. T is of course time. Now the second order integrated rate law equation is related to the equation of a straight line here, 1/80 is connected to Y que here is equal to our slope. M. T. Is X and one over A O. Is B. Now, if we took a plot of one over A versus time, that's a dead giveaway. That's the second order process. Remember, plots are of Y versus X. So one over reacting concentration matches with this for why that's why it's on the y axis and then the X axis has time because T. Is equal to X. Here. Now here our initial starting is down here and notice that the line is increasing over time. That's because for second order processes, K is positive here it's a positive Katie. We're customizing nada que when it deals with zero order and first order processes. Second order is unique. It's the only one with the increasing slope over time, Right? So just keep these little things in mind that are giving you clues on second order processes.