Question

Consider the reaction ${{A + B}} \to {{Products}}$. If $[A] = [B] = 1M$ the rate of the reaction is ${{2 \times 1}}{{{0}}^{{{ - 2}}}}{{mold}}{{{m}}^{{{ - 1}}}}{{{s}}^{{{ - 1}}}}$ . What is the rate of reaction if 90 $\% $ of each of the reactants are converted into products?

Answer 1

Hint: The rate of a reaction depends on the concentration of the reactants and the products that participate in the reaction process. The disappearance of the reactant and the increase in the concentration of the product formed have a direct relation with the rate of reaction. The rate of reaction can be calculated by knowing the change in the concentration of the reactants or product in unit time.

Complete step by step answer:
Just like the speed of a vehicle is defined by the distance covered by it in a specific amount of time, the rate of reaction can also be viewed as the change in concentration in time.
The rate of reaction is defined as the change in concentration of reactants or products in unit time.
The rate of the reaction can be expressed as the decrease in the concentration of the involved reactants or the rate of increase in the product concentration in the reaction.
Considering the above reaction ${{A + B}} \to {{Products}}$ we can see that the concentration of A and B will be decreased while the product concentration is bound to be increased. The rate of the expression can be expressed in the terms of either of the two.
The representation of the rate of reaction in terms of the concentration of the reactants is known as rate law. Thus we can use the rate law to form a rate expression for the reaction.
The rate expression for the above reaction can be written as
 ${{Rate = k[A][B]}}$
Here $k$ represents the rate constant and square brackets represent the molar concentration.
The initial concentration of the reactants are given to be 1 $M$ which can be expressed as
$[A] = [B] = 1M$, while the rate is given as ${{2 \times 1}}{{{0}}^{{{ - 2}}}}{{mold}}{{{m}}^{{{ - 1}}}}{{{s}}^{{{ - 1}}}}$.
Putting this data in the above rate expression we get,
$20 \times {10^{ - 2}} = k[1][1]$
$ \Rightarrow k = 20 \times {10^{ - 2}}$
When the reactants are consumed by 90 $\% $ the concentration changes as
 $[A] = [B] = 0.9M$
Putting this new data and the calculated rate constant $k$ in the above rate expression
${{Rate = k[A][B]}}$
$ \Rightarrow Rate = 20 \times {10^{ - 2}}[0.9][0.9]$
$ \Rightarrow Rate = 0.162 = 1.62 \times {10^{ - 1}}mold{m^{ - 1}}{s^{ - 1}}$

So, the answer is $1.62 \times {10^{ - 1}}mold{m^{ - 1}}{s^{ - 1}}$.

Note: 1.The rate of the reaction can be represented in terms of the decrease in the concentration of the reactant as $ - \dfrac{{\Delta [A]}}{{\Delta t}} = - \dfrac{{\Delta [B]}}{{\Delta t}}$ Here the negative sign indicates the decrease in the concentration.
2.The rate can also be expressed in terms of product as $Rate = + \dfrac{{\Delta [\ Product]}}{{\Delta t}}$
Here $\Delta t$ represents the change in time and $\Delta [product]$ represents the change in concentration in $\Delta t$ time.
3.This approach can be used for the calculation of rate when the time interval is given in place of the rate constant.