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the experimental data for decomposition of n2o5...

Question:
The experimental data for decomposition of N₂O₅

[2N₂O₅ → 4NO₂ + O₂]

in gas phase at 318K 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

(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:

(i)

(ii) Time corresponding to the concentration, 1630x10² / 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)

t/s	10² × [N₂O₅] 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

(iv) The given reaction is of the first order as the plot, log[N₂O₅] v/s t, is a straight line. Therefore, the rate law of the reaction is

Rate = k [N₂O₅]

(v) From the plot, log[N₂O₅]

v/s t, we obtain

Slope = -2.46 -(1.79) / 3200-0

= -0.67 / 3200

Again, slope of the line of the plot log[N₂O₅] v/s t is given by

- k / 2.303

.Therefore, we obtain,

- k / 2.303 = - 0.67 / 3200

⇒ k = 4.82 x 10^-4 s^-1

(vi) half life is given by,

t_½ = 0.693 / k

= 0.639 / 4.82x10^-4s

=1.438 x 10³s

Or we can say

1438 S

Which is very near to what we obtain from graph.

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