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# Why an elementary reaction becomes second order reaction at low pressure?

SAGAR SINGH - IIT DELHI
879 Points
10 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:

$\ -\frac{d[A]}{dt} = 2k[A]^2$ 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