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### Derivations

Posted: **Sun Mar 18, 2018 8:51 am**

by **Ardo 2K**

When deriving the integrated law for a second order reaction why isn't the kt negative, and also does a positive kt value mean that the slope of the line of 1 over the concentration of reactant plotted against time will be positive as well?

### Re: Derivations

Posted: **Sun Mar 18, 2018 10:01 am**

by **Zane Mills 1E**

Because k must be positive (positive slope in the graph), answering your second question (yes).

### Re: Derivations

Posted: **Sun Mar 18, 2018 1:58 pm**

by **Michaela Capps 1l**

For the first question, in the integrated rate law, kt is positive, so that is why we use pos slope.

### Re: Derivations

Posted: **Sun Mar 18, 2018 1:58 pm**

by **Michaela Capps 1l**

It is postitive because y=mx+b so the slope is positive.

### Re: Derivations

Posted: **Sun Mar 18, 2018 4:18 pm**

by **Christina Cen 2J**

the graph of 1/[reactant] over time has a positive slope which is equal to k for 2nd order reactions