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# the derivation of second order reaction rate constant expression when there are two different reactants are reacting.

Ashwin Sinha
520 Points
9 years ago

Dear Neeta Gupta,

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 (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

$\frac{1}{[A]} = \frac{1}{[A]_0} + kt$

or

$\frac{[A]}{[B]} = \frac{[A]_0}{[B]_0} e^{([A]_0 - [B]_0)kt}$

[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 $\ t_ \frac{1}{2} = \frac{1}{k[A]_0}$. 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: $\ln{}r = \ln{}k + 2\ln\left[A\right]$

Examples of a Second-order reaction
• $2\mbox{NO}_2(g) \rightarrow \; 2\mbox{NO}(g) + \mbox{O}_2(g)$

Plz. apptrove my answer by clicking ''Yes'' given below, if u loved it... Plz.. Plz... Plz...

E.Nandhini ketha
115 Points
4 years ago
Chemical kinetics, also known as reaction kinetics, Chemical kinetics is the study of rates of chemical processes.