Slope of a Second Order Reaction Plot

[vincesmetona_3I]

Why is it that the slope of a second order plot is negative? It seems to me that it should be negative because the concentration of reactants decrease with time.

[Charlene Tang 1B]

The positivity/negativity of the slope has nothing to do with concentration, but rather the order of the reaction.
We can use the 2nd order differential rate law to obtain (-1/a)(d[A]/dt)=k[A]^2 --> (-d[A]/[A]^2) = kdt
When you integrate that you'll get 1/[A] = kt + 1/[A]_o, and when you plot 1/[A] against t you'll get a positive slope because k is positive.

Hope this helps!