K in Rate Laws [ENDORSED]

[Jared Smith 1E]

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?

[Nisarg Shah 1C]

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.

[Justin Folk 3I]

Because for second order reactions the y-axis is 1/[A]. As time progresses the concentration of [A] decreases (assuming A is reactant) and thus 1/A increases, therefore slope of graph >0