Question

The experimental data for decomposition of \(N_2O_5\)
\([2N_2O_5→4NO_2+O_2]\)
in gas phase at \(318 \ K\) 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

1. Plot \([N_2O_5]\) against t.
2. Find the half-life period for the reaction.
3. Draw a graph between \(log\ [N_2O_5]\) 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

(i) 

(ii) Time corresponding to the concentration, \(\frac {1.630 \times 10^2}{2 }\ mol L^{-1}\) = \(81.5\  mol L^{-1}\) is the half life. From the graph, the half life is obtained as 1450 s.
(iii) 

t/s	102 x [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, \(log [N_2O_5]\) v/s \(t\), is a straight line. Therefore, the rate law of the reaction is
\(Rate = k [N_2O_5]\)

(v) From the plot, log [N₂O₅] v/s t, we obtain

\(Slop = \frac  {-2.46-(-1.79)}{3200-0}\)

\(Slop\) =\(-\frac {0.67}{3200}\)
Again, slope of the line of the plot log [N₂O₅] v/s t is given by

=\(-\frac {k}{2.303}\)

Therefore, we obtain,

\(-\frac {k}{2.303} =-\frac {067}{3200}\)

\(k = 4.82 \times 10^{-4} s^{-1}\)

(vi) Half-life is given by,
 
\(t½ = \frac {0.693}{k}\)

= \(\frac  {0.693 }{4.82\times10^{-4}} s\)

= \(1.438 \times 10^3 \ s\)

= \(1438  \ s\)

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