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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.
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