Zero vs. Second Order slopes

$\frac{d[R]}{dt}=-k[R]^{2}; \frac{1}{[R]}=kt + \frac{1}{[R]_{0}}; t_{\frac{1}{2}}=\frac{1}{k[R]_{0}}$

Henry Krasner 1C
Posts: 64
Joined: Fri Sep 28, 2018 12:15 am

Zero vs. Second Order slopes

Can someone explain to me why the second order rxn's slope is k whereas the zero order rxn's slope is -k? What does this signify conceptually?

Saman Andalib 1H
Posts: 73
Joined: Fri Sep 28, 2018 12:16 am

Re: Zero vs. Second Order slopes

You have to be careful when comparing what the slopes signify when comparing the slopes of different order reactions. Remember that for a 0 order reaction, the axes are time and [Reactant] whereas the slopes of a 2nd order reaction graph are time and 1/[reactant]. Direct graphical comparisons are therefore risky to do because of the inherent differences in their dimensions. Instead, it could be useful to examine the derived equations

For 0 order [A]=-kt+[A]initial
For 2nd order 1/[A]=kt+1/[A]initial

Think about the differences in how the 'kt' term affects the concentration of the reactant. Since there is a direct relationship in the 0 order, a negative kt is used to denote that the reactant concentration decreases over time. Since the relationship in the 2nd order reaction is the inverse, any increase to the 'kt+1/[A]initial' side of the equation would require the 1/[A] side to increase as well. How does this happen? With a subsequent decrease in the current concentration of the reactant [A], which is what we hypothetically are looking for. Therefore, the +kt term in the 2nd order equation is simply a nuance to make sure the concentration value of the reactant continues to decrease over time. Hope this helped :)