Question

The experimental data for decomposition of `N_(2)O_(5)`
`[2N_(2)O_(5)to 4NO_(2)+O_(2)]`
in gas phase at 318K are given below:

(i) Plot `[N_(2)O_(5)]` against t.
(ii) Find the half-life period for the reaction.
(iii) Draw a graph between log`[N_(2)O_(5)]` and t.
(iv) What is the rate law ?
(v) Calculate the rate constant.
(vi) Calculate the half-life period from k and compare it with (ii).

Answer 1

(ii) Time corresponding to the concentration , `(1.630 xx 10^(2))/(2) "mol" L^(-1) = 81.5 "mol" L^(-1)` is the half life . From the graph , the half life is obtained as 1450 s
(iii)
(iv) The given reaction is of the first order as the plot , log `[N_(2)O_(5)]` v/s t , is a straight line . Therefore , the rate law of the reaction si
Rate = `k[N_(2)O_(5)]`
(v) From the plot , log `[N_(2)O_(5)] ` v/s t , we obtain
Slope = `(-2.46 -(-1.79))/(3200 - 0)`
`= (-0.67)/(3200)`
Again , slope of the line of the plot log`[N_(2)O_(5)]` v/s t is given by `-(k)/(2.303)`
Therefore , we obtain
`-(k)/(2.303) = - (0.67)/( 3200)`
`implies k = 4.82 xx 10^(-4) s^(-1)`
(vi) Half- life is given by
`t_(1//2) = (0.639)/(k)`
= `(0.693)/(4.82 xx 10^(-4))s`
= `1.438 xx 10^(3)`s
= 1438 s
This value , 1438 s is very close to the value that was obtained from the graph .