Ashwin Sinha
Last Activity: 12 Years ago
Dear Neeta Gupta,
A second-order reaction depends on the concentrations of one second-order reactant, or two first-order reactants.
For a second order reaction, its reaction rate is given by:
or
or ![\ -\frac{d[A]}{dt} = 2k[B]^2](http://upload.wikimedia.org/math/7/8/1/781cda9b25d2e92fb50c2f6a30188f41.png)
In several popular kinetics books, the definition of the rate law for second-order reactions is
. Conflating the 2 inside the constant for the first, derivative, form will only make it required in the second, integrated form (presented below). The option of keeping the 2 out of the constant in the derivative form is considered more correct, as it is almost always used in peer-reviewed literature, tables of rate constants, and simulation software.[8]
The integrated second-order rate laws are respectively
![\frac{1}{[A]} = \frac{1}{[A]_0} + kt](http://upload.wikimedia.org/math/6/6/4/664842f28db997704b7ddbc61f6a7be8.png)
or
![\frac{[A]}{[B]} = \frac{[A]_0}{[B]_0} e^{([A]_0 - [B]_0)kt}](http://upload.wikimedia.org/math/0/2/6/0261d2c287f23e6971bba8e5cdf68b22.png)
[A]0 and [B]0 must be different to obtain that integrated equation.
The half-life equation for a second-order reaction dependent on one second-order reactant is
. For a second-order reaction half-lives progressively double.
Another way to present the above rate laws is to take the log of both sides: ![\ln{}r = \ln{}k + 2\ln\left[A\right]](http://upload.wikimedia.org/math/c/f/0/cf04309d2e7d76fc386ebfcf6877723f.png)
- Examples of a Second-order reaction
Plz. apptrove my answer by clicking ''Yes'' given below, if u loved it... Plz.. Plz... Plz...