Question

Compound A and B react to form C and D in a reaction that was found to be second-order overall and second-order in A. The rate constant at 30$^{\circ }C$is 0.622 $L\,mo{{l}^{-1}}{{\min }^{-1}}$. What is the half-life of A when $4.10\times {{10}^{-2}}$M of A is mixed with excess B?
A.40 min
B.39.21 min
C.28.59 min
D.None of these

Answer 1

Hint: Before solving this question, We should first know the formula of Half-life.
$Half\,life=\,{{t}_{\dfrac{1}{2}}}$. Now we have to put the given values in the formula to find its half-life.

Complete answer:
The reaction in which the rate is proportional to the square of the concentration of one reactant is the second-order reaction. Their general form is $2A\to Products$.
There is another type of second-order reaction that has a reaction rate that is proportional to the products of concentration of two of the reactants. Their general form is $A+B\to Products$.
The dimerization reaction is the reaction in which two molecules come together from a larger molecule.
The differential rate law of second-order reaction in which $2A\to Products$is :
$rate=\,-(\dfrac{\Delta (A)}{2\Delta T})=k{{A}^{2}}$
If we double the concentration of A, the reaction rate quadruples. The unit of it is moles per liter per second (M/s) and the unit of rate constant of second-order is inverse ${{M}^{-1}}{{s}^{-1}}$.
The molarity can be expressed as mol/L, Rate constant unit is L(mol.s).
Compound A and B react to form C and D :
$A+B\to C+D$
Now, Applying the formula
$Rate=k{{A}^{2}}$
In this k is 0.622
And A is $4.10\times {{10}^{-2}}$
$Rate=0.622\times {{(4.10\times {{10}^{-2}})}^{2}}$
$Half\,life=\,{{t}_{\dfrac{1}{2}}}=\dfrac{1}{0.622\times {{(4.10\times {{10}^{-2}})}^{2}}}$
                        = 39.12 min

So, Option (B) 39.12 min is correct.

Note:
Half-life is the time taken by an isotope of nuclei to be reduced by $\dfrac{1}{2}$ in a sample. The radioactive elements get decayed and become new after each half-life.