By Sunil Bhardwaj

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$$ NH_3NO_2 \longrightarrow 2H_2O + N_2 \qquad \text{ (First Order reaction) } $$

Decomposition of Ammonium Nitrite

It can be observed that, volume of \({ N }_{ 2 }\) collected at time t is depends on the amount of \({ NH }_{ 4 }{ NO }_{ 2 }\) decomposed.

[Amount of \(NH_4NO_2\) decomposed (x)] \( \propto \) [Volume of \(N_2\) gas collected at any time (\(V_t\))]

i.e. $$ x \propto { V }_{ t } \qquad ...(1)$$ At infinite time the total \({ N }_{ 2 }\) gas collected will depend on the amount of \({ NH }_{ 4 }{ NO }_{ 2 } \) taken initially.

[Amount of \({ NH }_{ 4 }{ NO }_{ 2 }\) initially taken (a)] \( \propto \) [Volume of \({ N }_{ 2 }\) 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: The decomposition of an aqueous solution of ammonium nitrite was studied by placing the apparatus in a thermostat maintained at a particular temperature. The volume of nitrogen gas collected at different intervals of time was as follows:

Time (minutes):10152025\(\infty\)
Vol. of \({ O }_{ 2 }\) evolved (ml):6.259.0011.4013.6535.05
From the above data prove that the reaction is of the first order.

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, $${ V }_{ \infty } = 35.05ml$$ The value of k at each time can be calculated as follows:

at time t = 10 min volume of \({ N }_{ 2 }\) evolved \({ V }_{ t } = 6.25\) $$ \therefore k = \frac { 2.303 }{ 10 } \log { \frac { 35.05 }{ 35.05 - 6.25 } } = 0.2303\times \log { \left( 1.217 \right) } = 0.0196{ min }^{ -1 }$$

at time t = 15 min volume of \({ N }_{ 2 }\) evolved \({ V }_{ t } = 9.00\) $$ \therefore k = \frac { 2.303 }{ 15 } \log { \frac { 35.05 }{ 35.05 - 9.00 } } = 0.154\times \log { \left( 1.345 \right) } = 0.0198{ min }^{ -1 }$$

at time t = 20 min volume of \({ N }_{ 2 }\) evolved \({ V }_{ t } = 11.40\) $$ \therefore k = \frac { 2.303 }{ 20 } \log { \frac { 35.05 }{ 35.05 - 11.40 } } = 0.115\times \log { \left( 1.482 \right) } = 0.0196{ min }^{ -1 }$$

at time t = 25 min volume of \({ N }_{ 2 }\) evolved \({ V }_{ t } = 13.65\) $$ \therefore k = \frac { 2.303 }{ 25 } \log { \frac { 35.05 }{ 35.05 - 13.65 } } = 0.092\times \log { \left( 1.638 \right) } = 0.0197{ min }^{ -1 }$$ 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 = 0.0197 \({ min }^{ -1 }\).