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 X [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).
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(i)
(ii) Time corresponding to the concentration:
1630x102 / 2 mol L-1 = 81.5 mol L-1 is the Half Life.
As per the Graph, the Half Life is Obtained as 1450 s.
(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 get
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
Hence, we get:
- 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 get from Graph.
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