Question

The experimental data for decomposition of N₂O₅

\[\ce{2N2O5 -> 4NO2 + O2}\]

in gas phase at 318 K are given below:

t/s	0	400	800	1200	1600	2000	2400	2800	3200
102 × [N2O5]/mol L−1	1.63	1.36	1.14	0.93	0.78	0.64	0.53	0.43	0.35

1. Plot [N₂O₅] against t.
2. Find the half-life period for the reaction.
3. Draw a graph between log [N₂O₅] and t.
4. What is the rate law?
5. Calculate the rate constant.
6. Calculate the half-life period from k and compare it with (ii).

Answer 1

t/s	[N2O5] × 102/mol L−1	log [N2O5]
0	1.63	−1.79
400	1.36	−1.87
800	1.14	−1.94
1200	0.93	−2.03
1600	0.78	−2.11
2000	0.64	−2.19
2400	0.53	−2.28
2800	0.43	−2.37
3200	0.35	−2.46

i.

ii. Initial concentration of N₂O₅ = 1.63 × 10⁻² M

Half of this concentration = 0.815 × 10⁻² M

The time corresponding to M concentration = 1440 s

Hence, t_1/2 = 1440 s

iii.

iv. The given reaction is of the first order as the plot, log [N₂O₅] v/s t, is a straight line. Hence, it is a reaction of first order, i.e., the rate law is:

Rate =  k[N₂O₅]

v. For first order reaction,

log R = `- k/(2.303 t) + log R_0`

Therefore, the slope of the graph drawn between log R and t will be `(-k)/2.303`.

∴ The slope of the line = `-k/2.303 = (y_2 - y_1)/(t_2 - t_1)`

`-k/2.303 = (-2.46 - (-1.79))/(3200 - 0)`

`-k/2.303 = -0.67/3200`

`k/2.303 = 0.67/3200`

k = `0.67/3200 xx 2.303`

∴ k = 4.82 × 10⁻⁴ s⁻¹

vi. Half-life is given by,

t_1/2 = `0.639/k`

= `0.693/(4.82 xx 10^-4)`

= 1.438 × 10³ s

= 1438 s

This value, 1438 s, is very close to the value that was obtained from the graph.