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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