Question

The reaction, ${\text{2}}{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}} \to {\text{4N}}{{\text{O}}_{\text{2}}} + {{\text{O}}_{\text{2}}}$, was studied and the following data were collected:

Expt. No.	$\left[ {{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}} \right]$ $\left( {{\text{mol }}{{\text{L}}^{ - 1}}} \right)$	Rate of disappearance of${{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}$ $\left( {{\text{mol }}{{\text{L}}^{ - 1}}{\text{ mi}}{{\text{n}}^{ - 1}}} \right)$
1.	$1.13 \times {10^{ - 2}}$	$34 \times {10^{ - 5}}$
2.	$0.84 \times {10^{ - 2}}$	$24 \times {10^{ - 5}}$
3.	$0.62 \times {10^{ - 2}}$	$18 \times {10^{ - 5}}$

Determine (i) order of the reaction, (ii) the rate law, (iii) rate constant for the reaction.

Answer 1

To solve this first write the expression for the rate law for the reaction. From the rate law determine the order of the reaction. From the rate law determine the rate constant for the reaction. The rate law is the expression which shows the change in the concentration of reactant per unit time.

Complete step-by-step answer:
We are given a reaction as follows:
${\text{2}}{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}} \to {\text{4N}}{{\text{O}}_{\text{2}}} + {{\text{O}}_{\text{2}}}$
The change in concentration of a reactant or product per unit time is known as the reaction rate. The expression for the rate of the reaction is known as the rate law.
Thus, for the given reaction, the rate law is,
${\text{Rate}} = k{\left[ {{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}} \right]^x}$ …… (1)
Where $k$ is the rate constant for the reaction.
$x$ is the order of the reaction.
We are given the rate of the reaction and the concentration of ${{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}$ for different experiments. Thus,
For experiment no. 1: $34 \times {10^{ - 5}} = k{\left[ {1.13 \times {{10}^{ - 2}}} \right]^x}$ …… (2)
For experiment no. 2: $24 \times {10^{ - 5}} = k{\left[ {0.84 \times {{10}^{ - 2}}} \right]^x}$ …… (3)
For experiment no. 3: $18 \times {10^{ - 5}} = k{\left[ {0.62 \times {{10}^{ - 2}}} \right]^x}$ …… (4)
The exponent to which the concentration term of a specific reactant is raised in the rate law is known as the order of the reaction with respect to that reactant. This is the partial order.
The sum of the powers of concentration of the reactants in the rate equation of the chemical equation is known as the overall order of the reaction.
Now, calculate the value of x i.e. the order of the reaction. Thus, from equation (2) and equation (3),
$\dfrac{{34 \times {{10}^{ - 5}}}}{{24 \times {{10}^{ - 5}}}} = \dfrac{{k{{\left[ {1.13 \times {{10}^{ - 2}}} \right]}^x}}}{{k{{\left[ {0.84 \times {{10}^{ - 2}}} \right]}^x}}}$
$1.36 = {\left[ {1.345} \right]^x}$
$x = 1$
Thus, the order of the reaction is 1.
Now, in the rate law i.e. equation (1) substitute the value of the order of reaction. Thus, the rate law for the reaction is,
${\text{Rate}} = k\left[ {{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}} \right]$
Now, rearrange the rate law for the rate constant for the reaction. Thus,
$k = \dfrac{{{\text{Rate}}}}{{\left[ {{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}} \right]}}$
Substitute $34 \times {10^{ - 5}}$ for the rate of the reaction, $1.13 \times {10^{ - 2}}$ for the concentration of ${{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}$. Thus,
$k = \dfrac{{34 \times {{10}^{ - 5}}{\text{ mol }}{{\text{L}}^{ - 1}}{\text{ mi}}{{\text{n}}^{ - 1}}}}{{1.13 \times {{10}^{ - 2}}{\text{ mol }}{{\text{L}}^{ - 1}}}}$
$k = 30.08 \times {10^{ - 3}}{\text{ mi}}{{\text{n}}^{ - 1}}$
Thus, the rate constant for the reaction is $30.08 \times {10^{ - 3}}{\text{ mi}}{{\text{n}}^{ - 1}}$.

Note: The unit of the rate constant for the reaction is ${\text{mi}}{{\text{n}}^{ - 1}}$. From the unit of the rate constant we can say that the rate constant for the reaction is independent of the concentration of the reactant. This is because the unit of rate constant does not contain the concentration term.