 # Explain the derivative and integral forms of second-order reactions

To understand how the rate of the reaction will change, according to the changes in the concentration of the reactants and the products, the integrated form of the equation is used as well as the differential form of the equation is used. The integrated rate equation shows that how the concentration of the species in the reaction will change over time, while the differential rate law can tell us that how the rate of the reaction will change with the change in the time. The language form is plotted in the form of the graph yield a linear function and therefore it is more convenient to look at the graph. And even both of the differential and integrated form of the equation can be derived with the help of the expression for the reaction rate.

CASE- A+A→P(Second-order reaction with the single reactant)

The rate at which A decrease in the above reaction can be expressed by using the differential form of the rate equation.

−d[A]dt=k[A]2 (11)

The 11th equation can be rearranged as

d[A][A]2=−kdt (12)

Since we are only interested in the changing of concentration of A over a while, we need to integrate between t=0 to t=t, this is the time of interest.

∫[A]t[A]od[A][A]2=−k∫t0dt (13)

As per the rule of integration, we solve the above equation(power rule):

dxx2=−1x+constant (14)

After it, we can obtain the integrated form of the rate equation.

1[A]t−1[A]o=kt (15)

By arranging the integrated form of the rate equation, we can obtain an equation of the line:

1[A]t=kt+1[A]o (16)

The important part of the whole process is not dominantly describing the integrated form of the rate law equation, other than that the thing which is important to notice or learn is that how the equation directly relates to the graph which ultimately provides a linear relationship. In this case and for all the second-order reaction as well, the linear plot of 1/[A]t versus time will yield the following graph.

This above graph is useful in a variety of ways. If we are only aware of the concentration at specific times for a reaction, we can able to create our graph which is similar to the above graph. And if the graph yields a straight line, then the reaction will be a second-order reaction. (1) (0)