After Integration  [ENDORSED]

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After Integration

Postby nelms6678 » Wed Feb 28, 2018 10:20 am

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
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Re: After Integration

Postby Justin Lau 1D » Wed Feb 28, 2018 10:41 am

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
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Re: After Integration

Postby Justin Chang 2K » Fri Mar 02, 2018 1:36 am

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

Emily Mei 1B
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Re: After Integration  [ENDORSED]

Postby Emily Mei 1B » Fri Mar 02, 2018 1:17 pm

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.

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