Question

In a reaction between A and B, the initial rate of reaction (${{\text{r}}_{\text{0}}}$) was measured for different initial concentrations of A and B as given below:

A/molL$^{-1}$	0.20	0.20	0.40
B/molL$^{-1}$	0.30	0.10	0.05
$r_0$/molL$^{-1}$s$^{-1}$	$5.07 \times 10^{-5}$	$5.07 \times 10^{-5}$	$1.43 \times 10^{-4}$

What is the order of the reaction with respect to A and B?

Answer 1

Hint: If the order of a reaction with respect to A be x and with respect to B be y. So, the formula for the rate of the reaction is ${{\text{r}}_{\text{0}}}{\text{ = k}}{\left[ {\text{A}} \right]^{\text{x}}}{\left[ {\text{B}} \right]^{\text{y}}}$.

Complete step by step answer:
Let us consider the order of reaction with respect to A be x and with respect to B be y.
Therefore,
${{\text{r}}_{\text{0}}}{\text{ = k}}{\left[ {\text{A}} \right]^{\text{x}}}{\left[ {\text{B}} \right]^{\text{y}}}$

By substituting the three set values from the table, we get:
$5.07 \times 10^{-5}{\text{ = k}}{\left[ {{\text{0}}{\text{.20}}} \right]^{\text{x}}}{\left[ {{\text{0}}{\text{.30}}} \right]^{\text{y}}}$ (1)
$5.07 \times 10^{-5}{\text{ = k}}{\left[ {{\text{0}}{\text{.20}}} \right]^{\text{x}}}{\left[ {{\text{0}}{\text{.10}}} \right]^{\text{y}}}$ (2)
$1.43 \times 10^{-4}{\text{ = k}}{\left[ {{\text{0}}{\text{.40}}} \right]^{\text{x}}}{\left[ {{\text{0}}{\text{.05}}} \right]^{\text{y}}}$ (3)

By dividing the equation (1) by (2), we get:
\[ \Rightarrow \dfrac{{5.07 \times {{10}^{ - 5}}}}{{5.07 \times {{10}^{ - 5}}}} = \dfrac{{k{{\left[ {0.20} \right]}^x}{{\left[ {0.30} \right]}^y}}}{{k{{\left[ {0.20} \right]}^x}{{\left[ {0.10} \right]}^y}}}\]

$ \Rightarrow 1 = \dfrac{{{{\left[ {0.30} \right]}^y}}}{{{{\left[ {0.10} \right]}^y}}}$

$ \Rightarrow {\left( {\dfrac{{0.30}}{{0.10}}} \right)^0} = {\left( {\dfrac{{0.30}}{{0.10}}} \right)^y}$
$ \Rightarrow y = 0$

Dividing the equation (3) by (2), we get:
$\dfrac{{1.43 \times {{10}^{ - 4}}}}{{5.07 \times {{10}^{ - 5}}}} = \dfrac{{k{{\left[ {0.40} \right]}^x}{{\left[ {0.05} \right]}^y}}}{{k{{\left[ {0.20} \right]}^x}{{\left[ {0.30} \right]}^y}}}$

$ \Rightarrow \dfrac{{1.43 \times {{10}^{ - 4}}}}{{5.07 \times {{10}^{ - 5}}}} = \dfrac{{{{\left[ {0.40} \right]}^x}}}{{{{\left[ {0.20} \right]}^x}}}$

$ \Rightarrow \dfrac{{1.43 \times {{10}^{ - 4}}}}{{5.07 \times {{10}^{ - 5}}}} = \dfrac{{{{\left[ {0.40} \right]}^x}}}{{{{\left[ {0.20} \right]}^x}}}$

$ \Rightarrow 2.821 = {2^x}$

$ \Rightarrow \log 2.821 = x\log 2$ (Taking log on both sides)

$ \Rightarrow x = \dfrac{{\log 2.821}}{{\log 2}}$
= 1.496
$ \simeq 1.5$
Therefore, the order of the reaction with respect to A is 1.5 and with respect to be is zero.

Additional Information:
The order of reaction refers to the relationship between the rate of a chemical reaction and the concentration of the species in that reaction. In order to get the order of the reaction, the rate expression of the reaction in the question must be obtained.
The order of reaction can be defined as the power dependence of rate on the concentration of all reactants. For example, the rate of a first-order reaction is dependent only on the concentration of one reactant species in the reaction.

Note: Please note that, in order to get the order of the reaction, the rate expression of the reaction in the question must be obtained. Once the rate equation is obtained, the entire composition of the mixture of all the species in the reaction can be understood.