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        Why an elementary reaction becomes second order reaction at low pressure?
6 years ago

SAGAR SINGH - IIT DELHI
879 Points
										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:
$\ -\frac{d[A]}{dt} = 2k[A]^2$ or $\ -\frac{d[A]}{dt} = k[A][B]$ or $\ -\frac{d[A]}{dt} = 2k[B]^2$
In several popular kinetics books, the definition of the rate law for second-order reactions is $-\frac{d[A]}{dt} = k[A]^2$.  Conflating the 2 inside the constant for the first, derivative, form  will only make it required in the second, integrated form.

All the best.
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Sagar Singh
B.Tech, IIT Delhi


6 years ago
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