Question

For a second order reaction in which both the reactants have equal initial concentration, the time taken for $20\%$ completion of reaction is 500 seconds. What will be the time taken for $60\%$ of the reaction?
A. 500 sec
B. 1000 sec
C. 3000 sec
D. 1500 sec

Answer 1

In this question, we will first we will see the definition of second order reactions kinetic and then derive the expression for the rate of reaction which is $k = \dfrac{1}{t}.\dfrac{x}{{a(a - x)}}$ . We will use the given information and this formula to get the answer.

Complete answer:
Let us first see what a second order reaction is?
These are the chemical reactions which depend on concentrations of two first order reactants.
The rate of these reactions can be given as:
r= k[A][B]. Where,
k is constant for second order.
[A] and [B] are concentrations of reactant A and B.
If A and B has equal initial concentration = a mol/L
Let ‘x’ mol/L of reactant is used in reaction to form the product then at time = t, the left amount of reactant is [a-x]. Therefore the rate of reaction is given as:
$\dfrac{{dx}}{{dt}} = k{(a - x)^2}$.
$\Rightarrow \dfrac{{dx}}{{{{(a - x)}^2}}} = kdt$
On integrating both sides, we have:
$\Rightarrow \int {\dfrac{{dx}}{{{{(a - x)}^2}}}} = \int {kdt}$
$\Rightarrow \dfrac{1}{{a - x}} = kt + c$ --------- (1)
At t=0, x=0.
$\therefore c = \dfrac{1}{a}$
Putting this value in equation 1, we get:
$\Rightarrow \dfrac{1}{{a - x}} = kt + \dfrac{1}{a}$
On rearranging the above equation, we have:
$k = \dfrac{1}{{t \times a}}.\left( {\dfrac{x}{{a - x}}} \right)$ ------ (2)
It is given that at t= 500 seconds, x = 20%. So, (a-x) = 80%.
Putting these values in equation 2, we have:
$k = \dfrac{1}{{500}}.\dfrac{{20}}{{100 \times 80}} = 5 \times {10^{ - 6}}{\sec ^{ - 1}}$
Equation 2 can be written as:
$t = \dfrac{1}{{k \times a}}.\left( {\dfrac{x}{{a - x}}} \right)$
For x =60% , (a-x) =40%. Putting these values in above equation, we get:
$t = \dfrac{1}{{5 \times {{10}^{ - 6}}}}.\dfrac{{60}}{{100 \times 40}} = 3 \times {10^3}\sec$ =3000 seconds.

Therefore, option, C is correct.

Note:
In this type of question, you should know how to write the expression for rate of reaction for first or second order reaction. The half life for a second order reaction is ${t_{1/2}} = \dfrac{1}{{ka}}$ , where ‘a’ is the initial concentration of reactants. Therefore, half life for a second order reaction is inversely proportional to concentration of the reactants.