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If the graph of time plotted against ln[A] is linear, then this indicates that the reaction with respect to reactant A is 1st order. The integrated rate law for 1st order reactions is ln[A] = -kt + ln[Ao]. If you look at this equation as if it is in the form of y = mx + b, you can see that the slope would be -k, not +k, due to the negative sign in front of kt. For zero order reactions, the slope is also -k for graphs of time plotted against [A]. However, the slope for 1/[A] vs. time for 2nd order reactions is +k.
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