Question

For the decomposition, ${{\text{N}}_{\text{2}}}{{\text{O}}_5}{\text{(g)}}\, \to {{\text{N}}_{\text{2}}}{{\text{O}}_4}{\text{(g)}}\,\, + \,1/2\,{{\text{O}}_2}({\text{g)}}$ , the initial pressure of ${{\text{N}}_{\text{2}}}{{\text{O}}_5}$is $114$mm and after $20$s, the pressure of reaction mixture becomes $133$mm. calculate the rate of reaction in terms of
(a) Change in atm ${{\text{s}}^{ - 1}}$
(b)Change in molarity ${{\text{s}}^{ - 1}}$ Given that reaction is carried out at ${\text{127}}{\,^{\text{o}}}{\text{C}}$

Answer 1

Hint: To determine the rate of reaction we will determine the pressure of the product. By dividing the pressure of the product with time we can determine the rate of reaction in the atm ${{\text{s}}^{ - 1}}$unit. By using the ideal gas equation we can determine the molar concentration. By dividing the molar concentration with time we will determine the rate of reaction in the molarity${{\text{s}}^{ - 1}}$ unit.

Complete answer:
The decomposition of nitrogen pentoxide is as follows:
${{\text{N}}_{\text{2}}}{{\text{O}}_5}{\text{(g)}}\, \to {{\text{N}}_{\text{2}}}{{\text{O}}_4}{\text{(g)}}\,\, + \,1/2\,{{\text{O}}_2}({\text{g)}}$
We will write the initial pressure and change in pressure as follows:
                         $\, \,\,\,\,\,\,\,\,\,\,\, {{\text{N}}_{\text{2}}}{{\text{O}}_5}{\text{(g)}}\to {{\text{N}}_{\text{2}}}{{\text{O}}_4}{\text{(g)}} + 1/2\,{{\text{O}}_2}({\text{g)}}$

Initial pressure	$114$	$0$	$0$
After $20$s	$114 - x$	$x$	$x/2$

After 20 s the total pressure is 13mm
$\Rightarrow 114\, +x\,+ \dfrac{x}{2}\, = \,133\,{\text{mm}}$
$\Rightarrow 114\, + \dfrac{x}{2}\, = \,133\,{\text{mm}}$
$\Rightarrow \dfrac{x}{2}\, = \,133\, - 114$
$\,x\, = \,9.5\,\,{\text{mm}}$

Divide the pressure by $760$ mmHg to convert the pressure into atm.
$760\,{\text{mm}}\,{\text{ = }}\,{\text{1}}\,{\text{atm}}$
$9.5\,{\text{mm = 0}}{\text{.0125}}\,{\text{atm}}$

(a) Change in atm ${{\text{s}}^{ - 1}}$
The average rate is defined as the change in pressure of reactant or product concerning per unit time over a specified period.
The formula of average rate is as follows:
   ${\text{r}}\,{\text{ = }}\,\dfrac{{{\Delta p}}}{{{\Delta t}}}$
Where,
 ${\text{r}}\,$is the average rate
 ${{\Delta p}}$is the change in pressure
 ${{\Delta t}}$is the change in time

Change in pressure of product is${\text{0}}{\text{.0125}}$atm, initial time is zero second and final time is $20$s.
So, change in time is,
$\Rightarrow {{\Delta t}}\,{\text{ = }}\,{\text{20}}\,{\text{s}}\, - 0\,{\text{s}}$
$\Rightarrow {{\Delta t}}\,{\text{ = }}\,{\text{20}}\,{\text{s}}$

Substitute $20\,{\text{s}} $for$ {{\Delta t}} $and $ {\text{0}}{\text{.0125} }$atm for$ {{\Delta p}}$.
\[{\text{r}}\,{\text{ = }}\,\dfrac{{0.0125\,{\text{atm}}}}{{20\,{\text{s}}}}\]
\[{\text{r}}\,{\text{ = }}\,0.000625\,{\text{atm}}\,.{{\text{s}}^{ - 1}}\]


(b) Change in molarity ${{\text{s}}^{ - 1}}$
The ideal gas equation is as follows:
${\text{pV}}\,{\text{ = }}\,{\text{nRT}}$

Molarity is defined as the mole per liter.
So, on rearranging the ideal gas equation for molarity,
$\Rightarrow\dfrac{{\text{n}}}{{\text{V}}} = \,\dfrac{{\text{p}}}{{{\text{RT}}}}$
$\Rightarrow{\text{M}} = \,\dfrac{{\text{p}}}{{{\text{RT}}}}$
Where, M is the molarity.
The formula of average rate in term of molarity is as follows:
${\text{r}}\,{\text{ = }}\,\dfrac{{{\Delta M}}}{{{\Delta t}}} $
Convert temperature from ${\text{127}}{\,^{\text{o}}}{\text{C}}$to K as follows:
$\Rightarrow{\text{273}}\,{\text{ + }}\,{\text{127}}{\,^{\text{o}}}{\text{C}}\,{\text{ = }}\,{\text{400}}\,{\text{K}}$

Substitute \[0.000625\,{\text{atm}}\,.{{\text{s}}^{ - 1}}\]for pressure, $0.0831\,{\text{L}}\,{\text{atm}}\,\,{\text{mo}}{{\text{l}}^{ - 1}}{{\text{k}}^{ - 1}}$ for R and ${\text{400}}$for T.
$\Rightarrow{\text{M}} = \,\dfrac{{{\text{0}}{\text{.0125}}\,{\text{atm}}}}{{0.0831\,{\text{L}}\,{\text{atm}}\,\,{\text{mo}}{{\text{l}}^{ - 1}}{{\text{k}}^{ - 1}}\, \times 400\,{\text{K}}}}$
$\Rightarrow {\text{M}} = \,0.00038\,{\text{mol}}\,\,{{\text{L}}^{ - 1}}$

Substitute $0.00038\,{\text{mol}}\,\,{{\text{L}}^{ - 1}}$for molarity and $20\,{\text{s}}$for${{\Delta t}}$.
$\Rightarrow{\text{r}}\, = \,\dfrac{{0.00038\,{\text{mol}}\,\,{{\text{L}}^{ - 1}}}}{{20\,{\text{s}}}}$
$\Rightarrow{\text{r}}\, = \,1.9\, \times {10^{ - 5}}{\text{mol}}\,\,{{\text{L}}^{ - 1}}{{\text{s}}^{ - 1}}$

Therefore, (a) Change in term of atm ${{\text{s}}^{ - 1}}$is \[0.000625\,{\text{atm}}\,.{{\text{s}}^{ - 1}}\] (b) Change in term of molarity ${{\text{s}}^{ - 1}}$is $1.9\, \times {10^{ - 5}}{\text{mol}}\,\,{{\text{L}}^{ - 1}}{{\text{s}}^{ - 1}}$.

Note:The average rate is defined as the change in molar concentration of reactant or product concerning per unit time over a specified period. In case of gaseous species, the average rate is defined as the change in pressure. Molarity is defined as the mole of solute dissolved in a liter of solution. Stoichiometry of reactant and product is important to determine the changes. So, a balanced reaction is necessary.