The following results were obtained in the decomposition of `N_(2)O_(5)` in carbon tetrachloride at `40^(@)`C.

Where x denotes volume of oxygen evolved. Show that the reaction is of first order. How long will it take half of original material to decomposes?

Explanation

The decomposition of `N_(2)O_(5)` may be shown as:
`2N_(2)O_(5) to 4NO_(2) + O_(2)`
From the equation, it is evident that the volume of `O_(2)` evolved corresponds to amount of `N_(2)O_(5)` decomposed.
`therefore` Volume of `O_(2)` at time `t(V_(t)) propto` amount of `N_(2)O_(5)` decomposed (X)
Volume of `O_(2)` at time `infty(V_(infty)) propto` amount of `N_(2)O_(5)` initially present (a).
`therefore a/(a-x)=V_(infty)/(V_(infty)-V_(t))`
For first order reaction: `k=(2.303)/t log (a)/(a-x) log (V_(infty))/(V_(infty)-V_(t))`
when t=1200s `k=(2.303)/(1200s) xx log ((34.75)mL)/(34.75 -15.53mL)=(2.303)/(1800s) log (34.75)/(19.22)`
`=(2.303 xx 0.2572)/(1800s) = 3.30 xx 10^(-4)s^(-1)`.
When t=2400s `k=(2.303)/(2400s) log (34.75 mL)/(15.85)=(2.303 xx 0.3409)/(2400s) = 3.28 xx 10^(-4)s^(-1)`
As the value of k comes out to be almost same, the reaction is of first order.
Calculation of half life period `(t_(1//2))`
Average value of k =`((3.31 + 3.30 + 3.28) xx 10^(-4))/(3)=3.30 xx 10^(-4)s^(-1)`
`therefore t_(1//2)=(0.693)/k = 0.693/(3.30 xx 10^(-4)s^(-1)) = 2100s`