# From following data of decomposition of ammonium nitrate in aqueous solution, Determine the order of reaction:

 Time(minutes) 10 15 20 15 $\infty$ Volume of $N_2$(cc) 6.25 9 11.4 13.65 35.05

$\begin{array}{1 1}(a)\;1&(b)\;2\\(c)\;0&(d)\;None\end{array}$

For reaction let it be $0^{th}$ order
So $x=kt,V_t=kt$
At $10sec,\large\frac{V_t}{t}=\frac{6.25}{10}$$=0.625 At 15sec,\large\frac{V_t}{t}=\frac{9}{15}$$=0.6$
At $20sec,\large\frac{V_t}{t}=\frac{11.40}{20}$$=0.57 At 15sec,\large\frac{V_t}{t}=\frac{13.65}{15}$$=0.91$
These are not equals.So it is not $0^{th}$ order.
Let it be first order
So $k_1=\large\frac{1}{t}$$ln\large\frac{V_{\infty}}{V_{\infty}-V_t} V_{\infty}=35.05 The values of k_1 at different times are obtained as follows: At time t\Rightarrow k_1= \large\frac{1}{t}$$ln\large\frac{V_{\infty}}{V_{\infty}-V_t}$
At time $10sec\Rightarrow k_1= \large\frac{1}{10}$$ln\large\frac{35.05}{28.60}$$=0.0197min^{-1}$
At time $15sec\Rightarrow k_1= \large\frac{1}{15}$$ln\large\frac{35.05}{26.05}$$=0.1976min^{-1}$
At time $20sec\Rightarrow k_1= \large\frac{1}{10}$$ln\large\frac{35.05}{23.65}$$=0.01964min^{-1}$
At time $25sec\Rightarrow k_1= \large\frac{1}{25}$$ln\large\frac{35.05}{21.40}$$=0.01971min^{-1}$
A constant value of $k_1$ shows reaction is first order.
Hence (a) is the correct answer.
answered Dec 19, 2013