After Integration [ENDORSED]

[nelms6678]

For the formula after we integrate it yields ln[A] = -k x t
Why is the k negative here, and should it be negative for a second order derivation as well?

[Justin Lau 1D]

The k is negative here because it is left over from the unique average rate law equation, which says Rate=-(1/a) x d[A]/dt. Therefore, when you rearrange the equation to integrate, you get -kt on the right. Because the negative denotes a -1 basically, it is a constant and remains. For the second order rate law, the same principle applies. However, because you are doing -(1/a) x d[A]/dt= k[A]^2 for the second order, the integral of d[A]/[A]^2 is -[A]^-1. Therefore, the negative is moved over to the -kt side which nullifies the negative.

[Justin Chang 2K]

Yeah, zero and first order reactions both have a -kt, but second order reactions have +kt.

[Emily Mei 1B]

Also you can think about k as the slope of the graph for (in this case) ln[A] vs t. The slope of the line will be -k.