By Sunil Bhardwaj

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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 $$N_2O_5(g) \longrightarrow N_2O_4(g) + \frac12{ O }_{ 2 }(g) \qquad \text{ (First Order reaction) }$$

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 $${ min }^{ -1 }$$.

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