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In lecture, Professor Lavelle stated that a zero order reaction will be the only time the slope on a graph will be straight. I'm confused because for the graphs in the course reader for 1st and 2nd order reactions, the slopes of those graphs are linear as well, or does it have to do with the fact those graphs are plotted against ln[A] and 1/[A], respectively? If so, does that mean for the zero order reaction's graph, it is in comparison to 1st and 2nd order reaction graphs that are plotted solely against concentration [A]?
Yes, you're correct. The graph of a first order reaction and a second order reaction appear to be linear since the data was manipulated to appear that way by either plotting the data as ln [A] or 1/[A] against time. If the concentrations ([A]) were plotted against time, the first order and second order graphs would appear as exponential growth functions. The zero order reaction is the only graph that will appear linear if the concentration is plotted against time.
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