Question

For the reaction, the values of the initial rate at different reactant concentrations are given in the table below. The rate law for the reaction is:

${\text{[A](mol }}{{\text{L}}^{{\text{ - 1}}}}{\text{)}}$	${\text{[B](mol }}{{\text{L}}^{{\text{ - 1}}}}{\text{)}}$	Initial Rate ${\text{(mol }}{{\text{L}}^{{\text{ - 1}}}}{{\text{s}}^{{\text{ - 1}}}}{\text{)}}$
0.05	0.05	0.045
0.10	0.05	0.090
0.20	0.10	0.71

A.${\text{Rate = }}k{\text{ [A] [B]}}$
B.\[{\text{Rate = }}k{\text{ [A}}{{\text{]}}^{\text{2}}}{\text{ [B}}{{\text{]}}^{\text{2}}}\]
C.\[{\text{Rate = }}k{\text{ [A] [B}}{{\text{]}}^{\text{2}}}\]
D.\[{\text{Rate = }}k{\text{ [A}}{{\text{]}}^{\text{2}}}{\text{ [B]}}\]

Answer 1

Hint: The rate of reaction is the change in concentration of reactant or product in unit time. Change in concentration of reactants affects the rate of reaction. The mathematical relation between the rate of reaction and the concentration of the reaction component is known as the rate law expression. To determine the rate law of reaction it is necessary to calculate the order of the reaction.

Complete Step by step answer: Given reaction to us is:

${\text{2A + B }} \to {\text{ C}}$

We have given data for the initial rate of reaction at various concentrations of reactant.

${\text{[A](mol }}{{\text{L}}^{{\text{ - 1}}}}{\text{)}}$	${\text{[B](mol }}{{\text{L}}^{{\text{ - 1}}}}{\text{)}}$	Initial Rate ${\text{(mol }}{{\text{L}}^{{\text{ - 1}}}}{{\text{s}}^{{\text{ - 1}}}}{\text{)}}$
0.05	0.05	0.045
0.10	0.05	0.090
0.20	0.10	0.71

The mathematical expression for the rate law of given reaction is:

${\text{Rate = }}k{\text{ [A}}{{\text{]}}^{\text{x}}}{\text{ [B}}{{\text{]}}^{\text{y}}}$
Here, x and y are the order of a reaction. The order of the reaction is purely experimental quantity. It may be a whole number or zero or even a fraction.
$k$= rate constant
To determine the order of anyone reactant we have to select any two experiments where the concentration of another reactant is constant.
In experiments 1 and 2 concentration reactant B is constant so using the data of these two experiments calculate the order of reactant A.
So, the rate law for experiment 1 as follows:
\[{\text{Rat}}{{\text{e}}_{\text{1}}}{\text{ = }}k{\text{ [A}}{{\text{]}}_{\text{1}}}^{\text{x}}{\text{ [B}}{{\text{]}}_1}^{\text{y}}\]
Substitute 0.045 ${\text{mol }}{{\text{L}}^{{\text{ - 1}}}}{{\text{s}}^{{\text{ - 1}}}}$ for rate and 0.05 ${\text{mol }}{{\text{L}}^{{\text{ - 1}}}}$ for concentration of A and 0.05 ${\text{mol }}{{\text{L}}^{{\text{ - 1}}}}$ for concentration of B.
\[{\text{0}}{\text{.045 mol }}{{\text{L}}^{{\text{ - 1}}}}{{\text{s}}^{{\text{ - 1}}}}{\text{ = }}k{\text{ (0}}{\text{.05 mol }}{{\text{L}}^{{\text{ - 1}}}}{{\text{)}}^{\text{x}}}{\text{ (0}}{\text{.05 mol }}{{\text{L}}^{{\text{ - 1}}}}{{\text{)}}^{\text{y}}}\]
Similarly, the rate law for experiment 2 as follows:
\[{\text{Rat}}{{\text{e}}_2}{\text{ = }}k{\text{ [A}}{{\text{]}}_2}^{\text{x}}{\text{ [B}}{{\text{]}}_2}^{\text{y}}\]
Substitute 0.090 ${\text{mol }}{{\text{L}}^{{\text{ - 1}}}}{{\text{s}}^{{\text{ - 1}}}}$ for rate and 0.10${\text{mol }}{{\text{L}}^{{\text{ - 1}}}}$ for concentration of A and 0.05 ${\text{mol }}{{\text{L}}^{{\text{ - 1}}}}$ for concentration of B.
\[{\text{0}}{\text{.090 mol }}{{\text{L}}^{{\text{ - 1}}}}{{\text{s}}^{{\text{ - 1}}}}{\text{ = }}k{\text{ (0}}{\text{.1 mol }}{{\text{L}}^{{\text{ - 1}}}}{{\text{)}}^{\text{x}}}{\text{ (0}}{\text{.05 mol }}{{\text{L}}^{{\text{ - 1}}}}{{\text{)}}^{\text{y}}}\]
Now, to calculate the order of reactant A take the ratio of \[{\text{Rat}}{{\text{e}}_2}\] and \[{\text{Rat}}{{\text{e}}_1}\]
\[\dfrac{{{\text{0}}{\text{.090 mol }}{{\text{L}}^{{\text{ - 1}}}}{{\text{s}}^{{\text{ - 1}}}}}}{{0.045{\text{ mol }}{{\text{L}}^{{\text{ - 1}}}}{{\text{s}}^{{\text{ - 1}}}}}}{\text{ = }}\dfrac{{k{\text{ (0}}{\text{.1 mol }}{{\text{L}}^{{\text{ - 1}}}}{{\text{)}}^{\text{x}}}{\text{ (0}}{\text{.05 mol }}{{\text{L}}^{{\text{ - 1}}}}{{\text{)}}^{\text{y}}}}}{{k{\text{ (0}}{\text{.05 mol }}{{\text{L}}^{{\text{ - 1}}}}{{\text{)}}^{\text{x}}}{\text{ (0}}{\text{.05 mol }}{{\text{L}}^{{\text{ - 1}}}}{{\text{)}}^{\text{y}}}}}\]
$2 = {(2)^x}$
$\Rightarrow x = 1$
So, the order of reaction with respect to reactant A is 1.
Now, calculate the order of with respect to reactant B as follows:

\[\dfrac{{{\text{0}}{\text{.090 mol }}{{\text{L}}^{{\text{ - 1}}}}{{\text{s}}^{{\text{ - 1}}}}}}{{0.045{\text{ mol }}{{\text{L}}^{{\text{ - 1}}}}{{\text{s}}^{{\text{ - 1}}}}}}{\text{ = }}\dfrac{{k{\text{ (0}}{\text{.1 mol }}{{\text{L}}^{{\text{ - 1}}}}{{\text{)}}^{\text{x}}}{\text{ (0}}{\text{.05 mol }}{{\text{L}}^{{\text{ - 1}}}}{{\text{)}}^{\text{y}}}}}{{k{\text{ (0}}{\text{.05 mol }}{{\text{L}}^{{\text{ - 1}}}}{{\text{)}}^{\text{x}}}{\text{ (0}}{\text{.05 mol }}{{\text{L}}^{{\text{ - 1}}}}{{\text{)}}^{\text{y}}}}}\]
Substitute 1 for $x$ and solve the equation for$y$.
\[\dfrac{{{\text{0}}{\text{.090 mol }}{{\text{L}}^{{\text{ - 1}}}}{{\text{s}}^{{\text{ - 1}}}}}}{{0.045{\text{ mol }}{{\text{L}}^{{\text{ - 1}}}}{{\text{s}}^{{\text{ - 1}}}}}}{\text{ = }}\dfrac{{k{\text{ (0}}{\text{.1 mol }}{{\text{L}}^{{\text{ - 1}}}}{{\text{)}}^{\text{1}}}{\text{ (0}}{\text{.05 mol }}{{\text{L}}^{{\text{ - 1}}}}{{\text{)}}^{\text{y}}}}}{{k{\text{ (0}}{\text{.05 mol }}{{\text{L}}^{{\text{ - 1}}}}{{\text{)}}^{\text{1}}}s{\text{ (0}}{\text{.05 mol }}{{\text{L}}^{{\text{ - 1}}}}{{\text{)}}^{\text{y}}}}}\]

$2 = 2{(1)^y}$
$\Rightarrow y = 1$

So, the order of reaction with respect to reactant B is 1.
Thus, the rate law equation for the given reaction is
${\text{Rate = }}k{\text{ [A] [B]}}$

Hence, the correct option is (A) ${\text{Rate = }}k{\text{ [A] [B]}}$.


Note: The rate constant value for the given reaction will be always constant. It is independent of the change in concentration of reaction components. Order of reaction is experimental property so cannot be determined from the reaction coefficient.