Use Coupon: CART20 and get 20% off on all online Study Material

Total Price: Rs.

There are no items in this cart.
Continue Shopping
Grade: 12th pass
Derive the rate kinetics of any second order reaction
one year ago

Answers : (1)

23382 Points

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


\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)
one year ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies

Course Features

  • 731 Video Lectures
  • Revision Notes
  • Previous Year Papers
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Test paper with Video Solution

Course Features

  • 141 Video Lectures
  • Revision Notes
  • Test paper with Video Solution
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Previous Year Exam Questions

Ask Experts

Have any Question? Ask Experts

Post Question

Answer ‘n’ Earn
Attractive Gift
To Win!!! Click Here for details