Class 12 Chemistry - Chapter Chemical Kinetics NCERT Solutions | The experimental data for decomposition
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).
(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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