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I noticed that depending on the reaction order, the slope of the graph can be equal to positive or negative k. For first and zero-order reactions, the slope of the graph is equal to -k, but for second order reactions, the slope is equal to +k. Can someone please explain to me why one is different from the other two?
It depends on the derivation of the integrated rate laws. When you integrate the zero and first-order laws you get a -kt term, and when you integrate the second order law you get a positive kt term. In this case, where we relate these integrated laws to y = mx+b, m (the slope) is equal to positive or negative k.
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