What happens when you double the concentration of a second order reaction?

The rate does not depend on the concentration. Whatever you do to the concentration, the rate will not change.

First Order

#"rate" = k["A"]^1 = k["A"]#

The rate is directly proportional to the concentration.

If you double the concentration, you double the rate.
If you triple the concentration, you triple the rate.
If you halve the concentration, you halve the rate, and so on.

Second Order

#"rate" = k["A"]^2#

The rate is proportional to the square of the concentration.

If you double the concentration, you multiply the rate by four.
If you triple the concentration, you multiply the rate by nine.
If you halve the concentration, you divide the rate by four, and so on.

Since concentration changes during an experiment, we must measure the initial rate of the reaction, before the concentration has had a chance to decrease.

We set up an experiment and measure the rate. Then we do another experiment in which we change only the concentration of component A. Let's say we double the concentration of A.

Second order reactions can be defined as chemical reactions wherein the sum of the exponents in the corresponding rate law of the chemical reaction is equal to two. The rate of such a reaction can be written either as r = k[A]2, or as r = k[A][B].

Table of Contents

What is a Second Order Reaction?

From the rate law equations given above, it can be understood that second order reactions are chemical reactions which depend on either the concentrations of two first-order reactants or the concentration of one-second order reactants.

Since second order reactions can be of the two types described above, the rate of these reactions can be generalized as follows:

r = k[A]x[B]y

Where the sum of x and y (which corresponds to the order of the chemical reaction in question) equals two.

Examples of Second Order Reactions

A few examples of second order reactions are given below:

\(\begin{array}{l}H^{+} + OH^{-} \rightarrow H_{2}O\end{array} \)

\(\begin{array}{l}C + O_{2} \rightarrow CO + O\end{array} \)

The two examples given above are the second order reactions depending on the concentration of two separate first order reactants.

\(\begin{array}{l}2NO_{2} \rightarrow 2NO + O_{2}\end{array} \)

\(\begin{array}{l}2HI \rightarrow I_{2} + H_{2}\end{array} \)

These reactions involve one second order reactant yielding the product.

Differential and Integrated Rate Equation for Second Order Reactions

Considering the scenario where one second order reactant forms a given product in a chemical reaction, the differential rate law equation can be written as follows:

\(\begin{array}{l}\frac{-d[R]}{dt} = k[R]^{2}\end{array} \)

In order to obtain the integrated rate equation, this differential form must be rearranged as follows:

\(\begin{array}{l}\frac{d[R]}{[R]^{2}} = -k dt\end{array} \)

Now, integrating on both sides in consideration of the change in the concentration of reactant between time 0 and time t, the following equation is attained.

\(\begin{array}{l}\int_{[R]_{0}}^{[R]_{t}} \frac{d[R]}{[R]^{2}} = -k \int_{0}^{t} dt\end{array} \)

From the power rule of integration, we have:

\(\begin{array}{l}\int \frac{dx}{x^{2}} = -\frac{1}{x} + C\end{array} \)

Where C is the constant of Integration. Now, using this power rule in the previous equation, the following equation can be attained.

\(\begin{array}{l}\frac{1}{[R]_{t}} – \frac{1}{[R]_{0}} = kt\end{array} \)

Which is the required integrated rate expression of second order reactions.

Graph of a Second Order Reaction

Generalizing [R]t as [R] and rearranging the integrated rate law equation of reactions of the second order, the following reaction is obtained.

\(\begin{array}{l}\frac{1}{[R]} = kt + \frac{1}{[R]_{0}}\end{array} \)

Plotting a straight line (y=mx + c) corresponding to this equation (y = 1/[R] , x = t , m = k , c = 1/[R]0)

It can be observed that the slope of the straight line is equal to the value of the rate constant, k.

Half-Life of Second-Order Reactions

The half-life of a chemical reaction is the time taken for half of the initial amount of reactant to undergo the reaction.

Therefore, while attempting to calculate the half-life of a reaction, the following substitutions must be made:

\(\begin{array}{l}[R] = \frac{[R]_{0}}{2}\end{array} \)

And,

\(\begin{array}{l}t = t_{1/2}\end{array} \)

Now, substituting these values in the integral form of the rate equation of second order reactions, we get:

\(\begin{array}{l}\frac{1}{\frac{[R]_{0}}{2}} – \frac{1}{[R]_{0}} = kt_{1/2}\end{array} \)

Therefore, the required equation for the half life of second order reactions can be written as follows.

\(\begin{array}{l}t_{1/2} = \frac{1}{k[R]_{0}}\end{array} \)

This equation for the half life implies that the half life is inversely proportional to the concentration of the reactants.

What happens if you double concentration?

What happens when you triple the concentration of a second order reaction?

How is the rate of reaction affected if the concentration of A is doubled?

What happens to the rate of first order reaction when the concentration is doubled?