Question

For the non-equilibrium process, $A+B\to ~products$ , the rate is first-order w.r.t A and second-order w.r.t B. If 1.0 mole each of A and B is introduced into a 1.0 L vessel and the initial rate was$1.0\times {{10}^{-2}}mol{{L}^{-1}}{{s}^{-1}}$, rate when half reactants have been turned into products is:
(A) $1.25\times {{10}^{-3}}mol{{L}^{-1}}{{s}^{-1}}$
(B) $1.0\times {{10}^{-2}}mol{{L}^{-1}}{{s}^{-1}}$
(C) $2.50\times {{10}^{-3}}mol{{L}^{-1}}{{s}^{-1}}$
(D) $2.0\times {{10}^{-2}}mol{{L}^{-1}}{{s}^{-1}}$

Answer 1

Hint: Attempt this question by using the definitions of rate of reaction and order of reaction. Rate of reaction is the speed at which a chemical reaction takes place and order of reaction describes the relationship between the rate of a chemical reaction and the concentration of the species taking part in it.
Formula used:
Molarity (M) = $\dfrac{n}{V}$
where,
n = number of moles
V = volume in litres

Complete answer:
-Let us first understand the concept of rate of reaction and its order.
-The rate of reaction (or reaction rate) is the speed at which a chemical reaction takes place and is defined as proportional to the increase in the concentration of a product and to the decrease in the concentration of a reactant per unit time.
-The order of reaction can be referred to the power dependence of rate on the concentration of all reactants which means it tells us that, which species’ concentration affects the rate of the reaction.
Now we will use the information provided in the question as follows:-
-It is given that rate is first-order w.r.t A and second-order w.r.t B, so the overall rate of reaction (r) is:-
$r=k{{[A]}^{1}}{{[B]}^{2}}$
where,
k = rate constant
-The values provided in the question are:-
r = $1.0\times {{10}^{-2}}mol{{L}^{-1}}{{s}^{-1}}$
Number of moles of A = 1 mole
Number of moles of B = 1 mole
-Calculation of concentration terms:-
[A] = $\dfrac{1 mole}{1 L}$= 1M
[B] =$\dfrac{1 mole}{1 L}$= 1M
On putting all these values in $r=k{{[A]}^{1}}{{[B]}^{2}}$, we get:-
$1.0\times {{10}^{-2}}mol{{L}^{-1}}{{s}^{-1}}$= $k{{[1mole/L]}^{1}}{{[1mole/L]}^{2}}$
k= $1.0\times {{10}^{-2}}mo{{l}^{-2}}{{L}^{2}}{{s}^{-1}}$
-Now, calculation of concentration terms when half reactants have been turned into products:-
[A] = $\dfrac{\dfrac{1}{2} mole}{1 L}$= 0.5M
[B] =$\dfrac{\dfrac{1}{2} mole}{1 L}$= 0.5M
On putting all these values in $r=k{{[A]}^{1}}{{[B]}^{2}}$, we get:-
r= $1.0\times {{10}^{-2}}mo{{l}^{-2}}{{L}^{2}}{{s}^{-1}}$.${{[0.5mole/L]}^{1}}{{[0.5mole/L]}^{2}}$
r= $1.25\times {{10}^{-3}}mol{{L}^{-1}}{{s}^{-1}}$

Hence, the rate of the reaction when half reactants have been turned into products is (A) $1.25\times {{10}^{-3}}mol{{L}^{-1}}{{s}^{-1}}$.

Note:
The misunderstandings we have while dealing with kinetics questions are as follows:
(A)The order of a reaction is equal to the molecularity of the reaction. This is true but only when it is an elementary reaction, not for complex reactions as molecularity is a positive integral value whereas order can be zero, negative, positive or fractional.
(B)Also we should not forget to consider stoichiometry while writing or using the rate expression or our obtained results will be wrong.