The experimental data for decomposition of N2O5 in gas phase at 318K are given below: [2N2O5 → 4NO2 + O2] (i) Plot [N2O5] against t. (ii) Find the half-life period for the reaction. (iii) Draw a graph between log[N2O5] 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)

Question 4.15: The experimental data for decomposition of N2O5 in gas phase at 318K are given below:
[2N₂O₅ → 4NO₂ + O₂]

(i) Plot [N2O5] against t.
(ii) Find the half-life period for the reaction.
(iii) Draw a graph between log[N2O5] 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) Plot [N2O5] against t.

Draw the graph

(ii) Find the half-life period for the reaction.
Half concentration ½(1.63) x 10 ^-2 =0.815 x 10 ^-2 mol L^-1
The time is 1520 sec
So half life is 1520 sec
(iii) Draw a graph between log[N2O5] and t.

Draw the graph now we get

(iv)What is the rate law?
We are getting a straight line when we draw a graph between log[N2O5] and t
Hence it is first order reaction
Rate law for this reaction
Rate = k [N₂O₅]
(v) Calculate the rate constant.
Use the formula

Slope

Use first and last value from the above table, we get

M = - 2.09 x 10 ^{- 4}
Plug the value in above formula

k = 2.303(- m)
k = 2.303 x –(- 2.09 x 10 ^{- 4}) = 4.82 x 10^{– 4} s^-1
(vi) Calculate the half-life period from k and compare it with (ii).
Use the formula of half life
k = 0.693 / t_1/2
t_1/2 = 0.693/ k = 0.693/ 4.82 x 10^{– 4} s^-1 = 1438 s
This value is very close to the value obtained from the graph