Derivations
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Derivations
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?
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Zane Mills 1E
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Re: Derivations
Because k must be positive (positive slope in the graph), answering your second question (yes).
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Michaela Capps 1l
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Re: Derivations
For the first question, in the integrated rate law, kt is positive, so that is why we use pos slope.
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Michaela Capps 1l
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Christina Cen 2J
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Re: Derivations
the graph of 1/[reactant] over time has a positive slope which is equal to k for 2nd order reactions
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