Question

The rate of decomposition of $N_2{O}_5$ in $CC{l}_4$ solution has been studied at 318 K and the following results have been obtained :

t (min)	0	135	342	683	1693
c {M}	2.08	1.91	1.67	1.35	0.57

The half-life period of the reaction is:

• A. $t_{1/2}=1.094\times {10}^{-3}mi{n}^{-1}$
• B. $t_{1/2}=1.067\times {10}^{-3}mi{n}^{-1}$
• C. $t_{1/2}=1.018\times {10}^{-3}mi{n}^{-1}$
• D. $Noneofthese$

Answer 1

The correct option is A $t_{1/2}=1.094\times {10}^{-3}mi{n}^{-1}$

(a) It can be shown that the rate of the reaction does not remain constant with time and therefore, it is not a zero order reaction, We now try integrated first order equation i.e.,
k = $\frac{ln\left(c_0/c_t\right)}{t}=\left(\frac{2.3}{t}log\frac{c_0}{c_t}\right)$
It can be seen that the value of k is almost constant for all the experimental results and hence it is a first order reaction with
(b) k = $6.31\times {10}^{-4}mi{n}^{-1}$
(c) $t_{1/2}=\frac{0.69}{6.31\times {10}^{-4}mi{n}^{-1}}=1.094\times {10}^{-3}mi{n}^{-1}$
Alternatively, if we draw a graph between in c against t, we obtained a straight line with slope = -k.