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Derive the rate kinetics of any second order reaction

TIYASA DAS , 6 Years ago

Grade 12th pass

1 Answers

Arun

Last Activity: 6 Years ago

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:

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

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

$2\mbox{NO}_2(g) \rightarrow \; 2\mbox{NO}(g) + \mbox{O}_2(g)$

Regards

Arun

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