Second-order reactions | Kinetics | AP Chemistry | Khan Academy

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The integrated rate law for the second-order reaction A → products is 1/[A]_t = kt + 1/[A]_0. Because this equation has the form y = mx + b, a plot of the inverse of [A] as a function of time yields a straight line. The rate constant for the reaction can be determined from the slope of the line, which is equal to k. View more lessons or practice this subject at www.khanacademy.org/science/a...
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[@namesake5344]

The efforts from sal khan brought us here after a long gap ...
Hope all's are good 😃😃

[@lt.ifeplumz1315]

Can you show the derivation of a second order reaction with respect to two different reactants? As in, R = k[A][B].

[@pboyzzadventures3185]

In Second order reaction the rate of the reaction is equal to the negative of the change of the concentration of A vs time. So the rate of the reaction is the K value raised to the second power. 1/[A] = kt + 1/[A]. This makes the slope a straight increasing line.

[@monpetitgarcon]

We need Sal's version of this! This explaination is suboptimal compared to Sal's

[@daviddavis8502]

Im not getting why Y is 1/At but on the graph it is only t

[@Starnoch]

I forgot I subscribed to khan academy haha...

[@jotarokujo806]

Nice name! I love that word ❤️

[@urfavouriteedits1618]

I am also first here 👍

[@medico-monazir]

First view me