By Sunil Bhardwaj

6555 Views


The compound, nitrogen pentoxide, is a volatile solid which decomposes in the gaseous state as well as in the form of its solution in an inert solvent like carbon tetrachloride, chloroform etc. according to the equation $$ N_2O_5(g) \longrightarrow N_2O_4(g) + \frac12{ O }_{ 2 }(g) \qquad \text{ (First Order reaction) } $$

Decomposition of Nitrogen Pentoxide

It can be observed that, volume of \({ O }_{ 2 }\) collected at time t is depends on the amount of \({ N }_{ 2 }{ O }_{ 5 }\) decomposed.

[Amount of \(N_2O_5\) decomposed (x)] \( \propto \) [Volume of \(O_2\) gas collected at any time (\(V_t\))]

i.e. $$ x \propto { V }_{ t } \qquad ...(1)$$ At infinite time the total \({ O }_{ 2 }\) gas collected will depend on the amount of \({ N }_{ 2 }{ O }_{ 5 } \) taken initially.

[Amount of \({ N }_{ 2 }{ O }_{ 5 }\) initially taken (a)] \( \propto \) [Volume of \({ O }_{ 2 }\) gas collected at infinite time \( { V }_{ \infty }\) ]

i.e. $$ a \propto { V }_{ \infty } \qquad ...(2)$$ Substituting these values in the first order equation. $$ k = \frac { 2.303 }{ t } \log { \frac { a }{ a-x } } $$ we get, $$ \boxed { k = \frac { 2.303 }{ t } \log { \frac { { V }_{ \infty } }{ { V }_{ \infty }-{ V }_{ t } } } } $$

Numerical: From the following data for the decomposition of \({ N }_{ 2 }{ O }_{ 5 }\) in carbon tetrachloride solution at 48C, show that the reaction is of the first order and calculate the rate constant.

Time (minutes):10152025
Vol. of \({ O }_{ 2 }\) evolved (ml):6.308.9511.4013.50
Solution: If the reaction is of the first order, it must obey the equation,$$ k = \frac { 2.303 }{ t } \log { \frac { a }{ a-x } } $$ or $$ \boxed { k = \frac { 2.303 }{ t } \log { \frac { { V }_{ \infty } }{ { V }_{ \infty }-{ V }_{ t } } } } $$ In the present case, $${ V }_{ \infty } = 34.75ml$$ The value of k at each time can be calculated as follows:

at time t = 10 min volume of \({ O }_{ 2 }\) evolved \({ V }_{ t } = 6.30\) $$ \therefore k = \frac { 2.303 }{ 10 } \log { \frac { 34.75 }{ 34.75 - 6.30 } } = 0.2303\times \log { 1.221 } = 0.02{ min }^{ -1 }$$

at time t = 15 min volume of \({ O }_{ 2 }\) evolved \({ V }_{ t } = 8.95\) $$ \therefore k = \frac { 2.303 }{ 15 } \log { \frac { 34.75 }{ 34.75 - 8.95 } } = 0.154\times \log { 1.347 } = 0.019{ min }^{ -1 }$$

at time t = 20 min volume of \({ O }_{ 2 }\) evolved \({ V }_{ t } = 11.40\) $$ \therefore k = \frac { 2.303 }{ 20 } \log { \frac { 34.75 }{ 34.75 - 11.40 } } = 0.115\times \log { 1.488 } = 0.019{ min }^{ -1 }$$ Since the value of k comes out to be nearly constant, given reaction is of the first order. The average value of the rate constant = 0.019 \({ min }^{ -1 }\).