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### positive slope in second ordre graph

Posted: **Wed Feb 28, 2018 4:28 pm**

by **Chew 2H**

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

### Re: positive slope in second ordre graph

Posted: **Wed Feb 28, 2018 5:11 pm**

by **Lucian1F**

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

### Re: positive slope in second ordre graph

Posted: **Wed Feb 28, 2018 5:18 pm**

by **Tiffany 1B**

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.