Question

If the rate of decomposition of ${{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}$ during a certain time internal is $2.4 \times {10^{ - 4}}\,{\text{mol}}\,\,{{\text{L}}^{ - 1}}{\min ^{ - 1}}$ .
${{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}\, \to \,{\text{2N}}{{\text{O}}_{\text{2}}}\,{\text{ + }}\,\dfrac{{\text{1}}}{{\text{2}}}{{\text{O}}_{\text{2}}}$
What is the rate of formation of ${\text{N}}{{\text{O}}_{\text{2}}}$ and ${{\text{O}}_{\text{2}}}$ ${\text{mol}}\,\,{{\text{L}}^{ - 1}}{\min ^{ - 1}}$.
A. $2.3 \times {10^{ - 5}}$ and $1.2 \times {10^{ - 5}}$ respectively.
B. $3.8 \times {10^{ - 4}}$ and $0.6 \times {10^{ - 4}}$ respectively.
C. $2.4 \times {10^{ - 4}}$ and $1.5 \times {10^{ - 4}}$ respectively.
D. $4.8 \times {10^{ - 4}}$ and $1.2 \times {10^{ - 4}}$ respectively.

Answer 1

Hint: Rate of reaction is the change in concentration of reactant and product with per unit time. The rate of decomposition of the reactant with stoichiometry is equal to the rate of formation of the product with stoichiometry. So, we can put the rate of formation equal to the rate of decomposition.

Formula used: Rate of reaction${\text{ = }}\, \pm \dfrac{{\text{1}}}{{{\text{stoichiometry}}}}\dfrac{{{\text{d}}\left[ {{\text{conc}}{\text{.}}} \right]}}{{{\text{dt}}}}$

Complete step by step answer:
The rate of reaction is defined as the change in concentration of reactant and product with time. The rate of reaction includes the rate of formation of product and rate of decomposition of reactant.
The given reaction is:
$\Rightarrow {{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}\, \to \,{\text{2N}}{{\text{O}}_{\text{2}}}\,{\text{ + }}\,\dfrac{{\text{1}}}{{\text{2}}}{{\text{O}}_{\text{2}}}$

The rate of reaction is written as follows:
Rate of reaction ${\text{ = }}\, \pm \dfrac{{\text{1}}}{{{\text{stoichiometry}}}}\dfrac{{{\text{d}}\left[ {{\text{conc}}{\text{.}}} \right]}}{{{\text{dt}}}}$

The rate of decomposition of ${{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}$ is as follows:
Rate of decomposition of ${{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}{\text{ = }}\, - \dfrac{{{\text{d}}\left[ {{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}} \right]}}{{{\text{dt}}}}$
$\Rightarrow - \dfrac{{{\text{d}}\left[ {{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}} \right]}}{{{\text{dt}}}} = 2.4 \times {10^{ - 4}}$

The rate of formation of products is as follows:
Rate of formation of ${\text{N}}{{\text{O}}_2}$ $ = + \dfrac{1}{2}\dfrac{{{\text{d}}\left[ {{\text{N}}{{\text{O}}_2}} \right]}}{{{\text{dt}}}}$
Rate of formation of ${{\text{O}}_2}$ $ = + 2\dfrac{{{\text{d}}\left[ {{{\text{O}}_2}} \right]}}{{{\text{dt}}}}$
The rate of formation of product is equal to the rate of decomposition of reactant. so we can write,
$\Rightarrow - \dfrac{{{\text{d}}\left[ {{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}} \right]}}{{{\text{dt}}}} = + \dfrac{1}{2}\dfrac{{{\text{d}}\left[ {{\text{N}}{{\text{O}}_2}} \right]}}{{{\text{dt}}}}$ and
$\Rightarrow + \dfrac{{{\text{d}}\left[ {{\text{N}}{{\text{O}}_2}} \right]}}{{{\text{dt}}}} = 2 \times \dfrac{{{\text{d}}\left[ {{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}} \right]}}{{{\text{dt}}}}$
$\Rightarrow + \dfrac{{{\text{d}}\left[ {{\text{N}}{{\text{O}}_2}} \right]}}{{{\text{dt}}}} = 2 \times 2.4 \times {10^{ - 4}}$
$\Rightarrow + \dfrac{{{\text{d}}\left[ {{\text{N}}{{\text{O}}_2}} \right]}}{{{\text{dt}}}} = 4.8 \times {10^{ - 4}}$

For rate of formation of oxygen we can write,
$\Rightarrow - \dfrac{{{\text{d}}\left[ {{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}} \right]}}{{{\text{dt}}}} = + 2\dfrac{{{\text{d}}\left[ {{{\text{O}}_2}} \right]}}{{{\text{dt}}}}$ and
$\Rightarrow + \dfrac{{{\text{d}}\left[ {{{\text{O}}_2}} \right]}}{{{\text{dt}}}} = \dfrac{1}{2} \times \dfrac{{{\text{d}}\left[ {{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}} \right]}}{{{\text{dt}}}}$
$\Rightarrow + \dfrac{{{\text{d}}\left[ {{{\text{O}}_2}} \right]}}{{{\text{dt}}}} = \dfrac{1}{2} \times 2.4 \times {10^{ - 4}}$
$\Rightarrow + \dfrac{{{\text{d}}\left[ {{{\text{O}}_2}} \right]}}{{{\text{dt}}}} = 1.2 \times {10^{ - 4}}$
So, the rate of formation of ${\text{N}}{{\text{O}}_{\text{2}}}$ is $4.8 \times {10^{ - 4}}$${\text{mol}}\,\,{{\text{L}}^{ - 1}}{\min ^{ - 1}}$and rate of formation of ${{\text{O}}_{\text{2}}}$ is $1.2 \times {10^{ - 4}}$ ${\text{mol}}\,\,{{\text{L}}^{ - 1}}{\min ^{ - 1}}$.

Therefore option (D) $4.8 \times {10^{ - 4}}$and $1.2 \times {10^{ - 4}}$ respectively, is correct.

Note: Positive sign in the rate of formation of the product indicates the formation of a species or the concentration of that species is increasing and the negative sign in the rate of decomposition indicates the decomposition of a species or the concentration of that species is decreasing.