## positive slope in second ordre graph

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

Chew 2H
Posts: 34
Joined: Fri Sep 29, 2017 7:05 am

### positive slope in second ordre graph

Could someone explain why there is a positive linear slope for the second order reaction while the slopes are both negative for zero order and first order? Wouldn't the positive slope mean that the concentration in increasing? I'm alittle confused

Lucian1F
Posts: 87
Joined: Fri Sep 29, 2017 7:07 am
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### Re: positive slope in second ordre graph

The graph is plotting 1/concentration against time and since as time increases, the concentration of a reactant will decrease, but since the y axis is 1/concentration, the y value will increase as time goes on since the y value will increase as concentration decreases because it is in the denominator

Tiffany 1B
Posts: 32
Joined: Fri Sep 29, 2017 7:05 am

### Re: positive slope in second ordre graph

Hi,
So I think the difference in these has to do with the way it is integrated, using the second order differential rate law (-1/a* d[A]/dt = k[A]^2) as starting point then separating variables and integrating using the power rule (which is in his integration rules tab on the website) you get -[A]^-1/-1=kt +C in which the negative signs cancel so then everything in the equation is positive and 1/[A]=kt+C. In regards to the concentration, the answer is no the slope is positive but the concentration isn't increasing because the graph is 1/[A] (and not just[A]) plotted against time which means that 1/[A] increases as [A] decreases, think 1/.5= 2 and 1/.25=4.
Hope that helps.