Question

Thermal decomposition of ${{N}_{2}}{{O}_{5}}$ follows first order kinetics according to the reaction shown below:
${{N}_{2}}{{O}_{5}}\to 2N{{O}_{2}}+\dfrac{1}{2}{{O}_{2}}$
If the initial pressure of ${{N}_{2}}{{O}_{5}}$ is 100 mm and after 10 minutes the pressure developed is 130 mm, then find out the rate constant of the reaction.

Answer 1

Hint: In first order kinetics at any point rate of reaction is directly proportional to the concentration of reactants and concentration of products formed totally depends on how many reactants have reacted and what is the stoichiometry of the reaction. Now as we know pressure of a gaseous substance is directly proportional to its pressure at constant temperature, therefore the ratio of change in concentrations will always equal to the ratio of change in their partial pressures.

Complete step by step answer:
 ${{N}_{2}}{{O}_{5}}\to N{{O}_{2}}+\dfrac{1}{2}{{O}_{2}}$
According to reaction stoichiometry, when 1 mole of ${{N}_{2}}{{O}_{5}}$ decomposes, it forms 1 mole $N{{O}_{2}}$ and half mole${{O}_{2}}$.
We also know that at a given temperature for constant volume, pressure will always be proportional to the number of moles of the gas; therefore change in pressure will also be proportional to the change in the number of moles.
- So, we can say that according to reaction stoichiometry, if pressure of ${{N}_{2}}{{O}_{5}}$ decreases by x unit then pressure of $N{{O}_{2}}$ will be x unit and pressure of oxygen gas will be $\dfrac{x}{2}$ units.
 In the system,
Partial pressure of ${{N}_{2}}{{O}_{5}}$ = $(100-x)mm$
$\rightarrow$ Partial pressure of $N{{O}_{2}}$ = $x$ mm
$\rightarrow$ Partial pressure of ${{O}_{2}}$= $\dfrac{x}{2}$ mm
- According to dalton’s partial pressure law,
Total pressure of the system will be equal to the sum of the partial pressure of all gases.
$P={{p}_{{{N}_{2}}{{O}_{5}}}} + {{p}_{N{{O}_{2}}}}+{{p}_{{{O}_{2}}}}$
$130=100 - x + x + \dfrac{x}{2}$
$\rightarrow$ $130=100 + \dfrac{x}{2}$
$\rightarrow$ $130-100 = \dfrac{x}{2}$
$\rightarrow$ $\dfrac{x}{2} = 30$
$x = 30\times 2$
$x = 60mm$

- After 10 minutes
$\rightarrow$ Pressure of ${{N}_{2}}{{O}_{5}}$ = 100 - 60
$\rightarrow$ Pressure of ${{N}_{2}}{{O}_{5}}$ = 40 mm
- Now let’s use first order integrated rate equation to find out the value of rate constant
$k=\dfrac{2.303}{{{t}_{{}}}}\log \dfrac{p}{{{p}_{0}}}$
Here, $p$ is the pressure of reactant at time t and ${{p}_{0}}$ is the pressure of the reactant at the start of the reaction.
${{p}_{0}} = 100mm$
$p = 40mm$
$t = 10\min $

- Now put all these values into the integrated rate equation
$\rightarrow$ $k = \dfrac{2.303}{100}\log \dfrac{100}{40}$
$\rightarrow$ $k = \dfrac{2.303}{100}\log 2.5$
$\rightarrow$ $k=\dfrac{2.303}{100}\times 0.4$
$\rightarrow$ $k = \dfrac{0.9212}{100}$
$\rightarrow$ $k = 9.212\times {{10}^{-3}}{{\min }^{-1}}$
Therefore value of rate constant will be $9.212\times {{10}^{-3}}{{\min }^{-1}}$

Additional Information :All the radioactive reactions and all the decomposition reactions on a catalyst surface at low pressure are always first order reactions.

Note: Rate constant of a first order reaction doesn’t depend on the concentration of reactant. It is inversely proportional to the half life of the reaction and its value is equal to $\dfrac{0.693}{{{t}_{1/2}}}$ and unit will always be $tim{{e}^{-1}}$. Half life of radioactive decay is very important as it tells about the stability of radioactive substances and some methods like carbon dating become an important criterion to calculate the life of the fossil.