Question

For the second order reaction, if the concentration of reactant changes from 0.08M to 0.04M in 10 minutes. Calculate the time at which concentration of reactant becomes 0.01M.

Answer 1

Hint: In a second order reaction, the rate of reaction is directly proportional to square of concentration of reactant or the sum of exponents of two reactants is two. The integrated rate law equation for a second order reaction is \[\dfrac{1}{{{A}_{0}}}-\dfrac{1}{A}=kt\].

Complete answer:
In a second order reaction, the sum of exponents of the reactant in the rate law equation is two.
-In second order reaction, rate is directly proportional to concentration of a second order reactant or two first order reactant.
-There can be two first order reactants or one second order reactant.
-consider a reaction having one second order reactant, the integrated rate law can be derived as follows:
Rate is given as $\dfrac{-d[A]}{dt}=k{{[A]}^{2}}$
To obtain integrated rate equation, it can be written as
$\dfrac{d[A]}{{{[A]}^{2}}}=-kdt$
Integrating both the sides with respect to concentration change from time 0 to t
$_{{{A}_{O}}}{{\int }^{{{A}_{t}}}}\dfrac{d[A]}{{{[A]}^{2}}}=-{{k}_{0}}{{\int }^{t}}dt$
So by apply law of integration,
\[\dfrac{1}{{{[A]}_{t}}}-\dfrac{1}{{{[A]}_{0}}}=kt\]
Required integrated equation for second order reaction is obtained.
By substituting values given in data, rate constant k can be calculated.
$\dfrac{1}{0.08}-\dfrac{1}{0.04}=k.10$
K=1.25
So, time at which concentration of reactant becomes 0.01M can be determine using same equation,
$\begin{align}
  & \dfrac{1}{0.08}-\dfrac{1}{0.01}=1.25t \\
 & \\
\end{align}$
t=70mins
so, the time at which concentration of the reactant becomes 0.01M is 70minutes.

Note:
For determining the concentration or time, an integrated rate equation can be used. Half-life can be also determined for the reaction. Half-life is the time required for concentration of the reactant to reach to its half value. In order to determine any value, rate constant must be calculated.