Question

Derive the expression for second order reaction with unequal concentration for a reaction $A + B \to product$ when ( $1$ ) if $a > b$ & ( $2$ ) if $b > a$ , where “a $mol/d{m^3}$ ” and “ b $mol/d{m^3}$ ” are the initial concentrations of A and B respectively?

Answer 1

We should know about the second order reaction. It is a reaction in which the sum of the exponents in the corresponding rate law of the chemical reactions is equal to two. It depends on the concentration of both the reactants.

Complete answer:
The second order reaction is given as:
$A + B\xrightarrow{k}products$
As we know that the rate of second order reaction is
$r = k[A][B]$
We can also write the above equation in other forms such as
Let $x\,\,mol/d{m^3}$ be the amount reacted in time ‘ $t$ ’. Then the equation is written as:
$\dfrac{{dx}}{{dt}} = k[A][B]$
Let ‘ $a$ ’ and ‘ $b$ ’ be the initial concentration of $A$ and $B$ respectively and it is given in the question that both the concentration of $A$ and $B$ are not equal. Mathematically, we can write as $a \ne b$ . Then,
$\dfrac{{dx}}{{dt}} = k(a - x)(b - x)$
Or,
$\int\limits_0^x {\dfrac{{dx}}{{(a - x)(b - x)}} = } \int\limits_0^t {kdt} = kt$
Now, to solve this integration, we use the method of partial fraction to evaluate the first integral.
It gives that,
$\int\limits_0^x {\dfrac{{dx}}{{(a - x)(b - x)}} = \int\limits_0^x {\dfrac{{dx}}{{(b - a)(a - x)}} + \int\limits_0^x {\dfrac{{dx}}{{(a - b)(b - x)}}} } }$
Now, we get
$= \dfrac{1}{{b - a}}\int\limits_0^x {\dfrac{{dx}}{{(a - x)}} + \dfrac{1}{{a - b}}\int\limits_0^x {\dfrac{{dx}}{{(b - x)}}} }$
Then, we use method of u-substitution method to solve the new integrals values, we get
$= \dfrac{1}{{b - a}}\int\limits_0^x {\dfrac{{dx}}{{(a - x)}} + \dfrac{1}{{a - b}}\int\limits_0^x {\dfrac{{dx}}{{(b - x)}}} }$
$= \dfrac{1}{{b - a}}\ln \left( {\dfrac{a}{{a - x}}} \right) + \dfrac{1}{{a - b}}\ln \left( {\dfrac{b}{{b - x}}} \right)$
Taking $\dfrac{1}{{b - a}}$ as common from the above equations,
$= \dfrac{1}{{b - a}}\left( {\ln \,\left( {\dfrac{a}{{a - x}}} \right) - \ln \left( {\dfrac{b}{{b - x}}} \right)} \right)$
$\therefore \dfrac{1}{{b - a}}\ln \left( {\dfrac{{a(b - x)}}{{b(a - x)}}} \right) = kt$
This is the integrated rate law expression.
Arranging the above equation, we get
$\therefore \ln \left( {\dfrac{{a(b - x)}}{{b(a - x)}}} \right) = (b - a)kt$
We can write the rate law, in terms of original symbols.
$\ln \left( {\dfrac{{{{[A]}_0}[B]}}{{{{[B]}_0}[A]}}} \right) = ({[B]_0} - {[A]_0})kt$
This rate law works for all the values of concentration in which both the concentration i.e. ‘ $a$ ’ and ‘ $b$ ’ of $A$ and $B$ respectively are not equal ( $a \ne b$ ) whether it is $a > b$ or $b > a$ .

Note:
It must be remembered that the overall order of reaction is two for the second order reaction. We know that the rate for second order reaction is written as $r = k[A][B]$ . Sometimes it is also expressed as $r = k{[A]^2}$ when one reactant is given.