By Sunil Bhardwaj

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$$NH_3NO_2 \longrightarrow 2H_2O + N_2 \qquad \text{ (First Order reaction) }$$ 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 $${ min }^{ -1 }$$.

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