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Question 15

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,

t½ = 0.693 / k

= 0.639 / 4.82x10-4 s

=1.438 x 103 s

Or we can say

1438 S

Which is very near to what we obtain from graph.

 

 

 

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