Why an elementary reaction becomes second order reaction at low pressure?
Chaitanya Borah , 13 Years ago
Grade 12
Physical Chemistry> kinetics...
1 Answers
SAGAR SINGH - IIT DELHI
Last Activity: 13 Years ago
Dear student,
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:
![\ -\frac{d[A]}{dt} = 2k[A]^2](http://upload.wikimedia.org/math/8/7/3/873a3b7d216517de98cb8ee3a86ceb44.png)
or ![\ -\frac{d[A]}{dt} = k[A][B]](http://upload.wikimedia.org/math/c/f/d/cfd5938b3763344a202d6975a0d68ed9.png)
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 ![-\frac{d[A]}{dt} = k[A]^2](http://upload.wikimedia.org/math/5/8/c/58cd77b68b6d8db6df381fc176b25a1d.png)
. Conflating the 2 inside the constant for the first, derivative, form will only make it required in the second, integrated form.
Please feel free to ask your queries here. We are all IITians and here to help you in your IIT JEE preparation.
All the best.
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Sagar Singh
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