Question

For the reaction,
\[{\text{A + B}} \to {\text{Product}}\]
Rate law is rate \[{\text{ = k}}{\left[ {\text{A}} \right]^{\text{2}}}\left[ {\text{B}} \right]\]
Where \[{\text{k = 5}} \times {\text{1}}{{\text{0}}^{{\text{ - 5}}}}{\left( {{\text{mol/L}}} \right)^{{\text{ - 2}}}}{\text{mi}}{{\text{n}}^{{\text{ - 1}}}}\]
Determine the time (in mins) in which concentration of A becomes half of its initial concentration. The initial concentration of A and B are \[{\text{0}}{\text{.2M}}\]and \[{\text{2}} \times {\text{1}}{{\text{0}}^{\text{3}}}{\text{M}}\]respectively.

Answer 1

Hint: We also need to know that chemical kinetics is one of the topics used to study the kinetics nature of the chemical reaction. It is used to optimize the chemical reaction for industrial purposes. The reactant to the product so many parameters are required. All are optimised by using this chemical kinetics. Chemical kinetics is used as a mechanism of reactant to product in the chemical reaction.
Formula used:
The rate of the reaction is directly proportional to the concentration of reactant and inversely proportional to the time of the reaction.
The rate of the reaction is equal to the product of the concentration of the reactant with respect to that order of the reaction.
\[{\text{rate = k[A}}{{\text{]}}^{\text{m}}}{{\text{[B]}}^{\text{n}}}\]
Here, k is proportionality constant, known as rate constant.
A and B are reactants of the reaction.
m and n are the order of the reaction of A and B respectively.
The time period calculation,
\[{{\text{t}}_{{\text{1/2}}}}{\text{ = }}\dfrac{{{\text{0}}{\text{.6932}}}}{{\text{k}}}\]
Here, the time period is t.

Complete answer:
For the reaction,
\[{\text{A + B}} \to {\text{Product}}\]
Rate law is rate \[{\text{ = k}}{\left[ {\text{A}} \right]^{\text{2}}}\left[ {\text{B}} \right]\]
Where \[{\text{k = 5}} \times {\text{1}}{{\text{0}}^{{\text{ - 5}}}}{\left( {{\text{mol/L}}} \right)^{{\text{ - 2}}}}{\text{mi}}{{\text{n}}^{{\text{ - 1}}}}\]
\[{\text{rate = k[A}}{{\text{]}}^{\text{m}}}{{\text{[B]}}^{\text{n}}}\]
Rate law is rate \[{\text{ = k}}{\left[ {\text{A}} \right]^{\text{2}}}\left[ {\text{B}} \right]\]
Here, k is proportionality constant, known as rate constant.
A and B are reactants of the reaction.
The order of the reactions of A and B are \[2\] and \[{\text{1}}\] respectively.
The overall order of the reaction is \[1 + 2 = 3\] respectively.
Hence, the above reaction is a third order reaction.
The initial concentration of A is\[{\text{0}}{\text{.2M}}\].
The initial concentration of B is \[{\text{2}} \times {\text{1}}{{\text{0}}^{\text{3}}}{\text{M}}\].
Calculate the time (in mins) in which concentration of A becomes half of its initial concentration.
Hence, the time period calculation is reduced to half,
\[{{\text{t}}_{{\text{1/2}}}}{\text{ = }}\dfrac{{{\text{0}}{\text{.6932}}}}{{\text{k}}}\]
\[{{\text{t}}_{{\text{1/2}}}}{\text{ = }}\dfrac{{{\text{0}}{\text{.6932}}}}{{{\text{2k}}}}\]
Now we can substitute the known values we get,
\[ \Rightarrow {{\text{t}}_{{\text{1/2}}}}{\text{ = }}\dfrac{{{\text{0}}{\text{.6932}}}}{{{\text{2}} \times {\text{5}} \times {\text{1}}{{\text{0}}^{{\text{ - 5}}}}{\text{mi}}{{\text{n}}^{{\text{ - 1}}}}}}\]
\[ \Rightarrow {{\text{t}}_{{\text{1/2}}}}{\text{ = }}\dfrac{{{\text{0}}{\text{.6932}}}}{{{\text{1}}{{\text{0}}^{{\text{ - }}4}}{\text{mi}}{{\text{n}}^{{\text{ - 1}}}}}}\]
\[{{\text{t}}_{{\text{1/2}}}}{\text{ = 0}}{\text{.6932}} \times {\text{1}}{{\text{0}}^4}{\text{min}}\]
\[{{\text{t}}_{{\text{1/2}}}}{\text{ = 6}}{\text{.932}} \times {\text{1}}{{\text{0}}^3}{\text{min}}\]
\[ \Rightarrow {{\text{t}}_{{\text{1/2}}}}{\text{ = 7}} \times {\text{1}}{{\text{0}}^3}{\text{min}}\]
For the reaction,
\[{\text{A + B}} \to {\text{Product}}\]
Rate law is rate \[{\text{ = k}}{\left[ {\text{A}} \right]^{\text{2}}}\left[ {\text{B}} \right]\], Where \[{\text{k = 5}} \times {\text{1}}{{\text{0}}^{{\text{ - 5}}}}{\left( {{\text{mol/L}}} \right)^{{\text{ - 2}}}}{\text{mi}}{{\text{n}}^{{\text{ - 1}}}}\]
The time (in mins) in which concentration of A becomes half of its initial concentration is
\[{\text{7}} \times {\text{1}}{{\text{0}}^3}{\text{min}}\].

Note:
We also need to know that the rate of the reaction is an important factor for the study of reaction. The rate of reaction is an important concept for chemical kinetics. Rate of the reaction depends on the concentration of the reactant. The rate of reaction is also calculated by using the concentration of the product in the chemical reaction. Depending on the concentration, the sign of the rate will change.