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Question: For the decomposition, \({{\text{N}}_{\text{2}}}{{\text{O}}_5}{\text{(g)}}\, \to {{\text{N}}_{\text{...

For the decomposition, N2O5(g) →N2O4(g)  + 1/2 O2(g){{\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 N2O5{{\text{N}}_{\text{2}}}{{\text{O}}_5}is 114114mm and after 2020s, the pressure of reaction mixture becomes 133133mm. calculate the rate of reaction in terms of
(a) Change in atm s−1{{\text{s}}^{ - 1}}
(b)Change in molarity s−1{{\text{s}}^{ - 1}} Given that reaction is carried out at 127 oC{\text{127}}{\,^{\text{o}}}{\text{C}}

Explanation

Solution

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 s−1{{\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 molaritys−1{{\text{s}}^{ - 1}} unit.

Complete answer:
The decomposition of nitrogen pentoxide is as follows:
N2O5(g) →N2O4(g)  + 1/2 O2(g){{\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:
            N2O5(g)→N2O4(g)+1/2 O2(g)\, \,\,\,\,\,\,\,\,\,\,\, {{\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 pressure1141140000
After 2020s114−x114 - xxxx/2x/2

After 20 s the total pressure is 13mm
⇒114 +x +x2 = 133 mm\Rightarrow 114\, +x\,+ \dfrac{x}{2}\, = \,133\,{\text{mm}}
⇒114 +x2 = 133 mm\Rightarrow 114\, + \dfrac{x}{2}\, = \,133\,{\text{mm}}
⇒x2 = 133 −114\Rightarrow \dfrac{x}{2}\, = \,133\, - 114
 x = 9.5  mm\,x\, = \,9.5\,\,{\text{mm}}

Divide the pressure by 760760 mmHg to convert the pressure into atm.
760 mm  =  1 atm760\,{\text{mm}}\,{\text{ = }}\,{\text{1}}\,{\text{atm}}
9.5 mm = 0.0125 atm9.5\,{\text{mm = 0}}{\text{.0125}}\,{\text{atm}}

(a) Change in atm s−1{{\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:
r  =  ΔpΔt{\text{r}}\,{\text{ = }}\,\dfrac{{{\Delta p}}}{{{\Delta t}}}
Where,
r {\text{r}}\,is the average rate
Δp{{\Delta p}}is the change in pressure
Δt{{\Delta t}}is the change in time

Change in pressure of product is0.0125{\text{0}}{\text{.0125}}atm, initial time is zero second and final time is 2020s.
So, change in time is,
⇒Δt  =  20 s −0 s\Rightarrow {{\Delta t}}\,{\text{ = }}\,{\text{20}}\,{\text{s}}\, - 0\,{\text{s}}
⇒Δt  =  20 s\Rightarrow {{\Delta t}}\,{\text{ = }}\,{\text{20}}\,{\text{s}}

Substitute 20 s20\,{\text{s}} forΔt{{\Delta t}}and 0.0125 {\text{0}}{\text{.0125} }atm forΔp {{\Delta p}}.
r  =  0.0125 atm20 s{\text{r}}\,{\text{ = }}\,\dfrac{{0.0125\,{\text{atm}}}}{{20\,{\text{s}}}}
r  =  0.000625 atm .s−1{\text{r}}\,{\text{ = }}\,0.000625\,{\text{atm}}\,.{{\text{s}}^{ - 1}}

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

Molarity is defined as the mole per liter.
So, on rearranging the ideal gas equation for molarity,
⇒nV= pRT\Rightarrow\dfrac{{\text{n}}}{{\text{V}}} = \,\dfrac{{\text{p}}}{{{\text{RT}}}}
⇒M= pRT\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:
r  =  ΔMΔt{\text{r}}\,{\text{ = }}\,\dfrac{{{\Delta M}}}{{{\Delta t}}}
Convert temperature from 127 oC{\text{127}}{\,^{\text{o}}}{\text{C}}to K as follows:
⇒273  +  127 oC  =  400 K\Rightarrow{\text{273}}\,{\text{ + }}\,{\text{127}}{\,^{\text{o}}}{\text{C}}\,{\text{ = }}\,{\text{400}}\,{\text{K}}

Substitute 0.000625 atm .s−10.000625\,{\text{atm}}\,.{{\text{s}}^{ - 1}}for pressure, 0.0831 L atm  mol−1k−10.0831\,{\text{L}}\,{\text{atm}}\,\,{\text{mo}}{{\text{l}}^{ - 1}}{{\text{k}}^{ - 1}} for R and 400{\text{400}}for T.
⇒M= 0.0125 atm0.0831 L atm  mol−1k−1 ×400 K\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}}}}
⇒M= 0.00038 mol  L−1\Rightarrow {\text{M}} = \,0.00038\,{\text{mol}}\,\,{{\text{L}}^{ - 1}}

Substitute 0.00038 mol  L−10.00038\,{\text{mol}}\,\,{{\text{L}}^{ - 1}}for molarity and 20 s20\,{\text{s}}forΔt{{\Delta t}}.
⇒r = 0.00038 mol  L−120 s\Rightarrow{\text{r}}\, = \,\dfrac{{0.00038\,{\text{mol}}\,\,{{\text{L}}^{ - 1}}}}{{20\,{\text{s}}}}
⇒r = 1.9 ×10−5mol  L−1s−1\Rightarrow{\text{r}}\, = \,1.9\, \times {10^{ - 5}}{\text{mol}}\,\,{{\text{L}}^{ - 1}}{{\text{s}}^{ - 1}}

Therefore, (a) Change in term of atm s−1{{\text{s}}^{ - 1}}is 0.000625 atm .s−10.000625\,{\text{atm}}\,.{{\text{s}}^{ - 1}} (b) Change in term of molarity s−1{{\text{s}}^{ - 1}}is 1.9 ×10−5mol  L−1s−11.9\, \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.