Time(minutes) | 10 | 15 | 20 | 15 | $\infty$ |

Volume of $N_2$(cc) | 6.25 | 9 | 11.4 | 13.65 | 35.05 |

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

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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.

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