A student has studied the decomposition of a gas $$AB_3$$ at 25°C. He obtained the following data

p (mm Hg)	50	100	200	400
Relative $$t_{1/2}$$ (s)	4	2	1	0.5

The order of the reaction is

For an $$n^{th}$$ order reaction, the half-life is related to the initial concentration or pressure by

$$t_{1/2}\propto \frac{1}{p^{,n-1}}$$

or equivalently,

$$t_{1/2}\propto p^{,1-n}$$

Using two data points from the table,

$$p_1=50,\quad (t_{1/2})_1=4$$

$$p_2=100,\quad (t_{1/2})_2=2$$

Taking the ratio,

$$\frac{(t_{1/2})1}{(t{1/2})_2}=\left(\frac{p_1}{p_2}\right)^{1-n}$$

Substituting the values,

$$\frac{4}{2}=\left(\frac{50}{100}\right)^{1-n}$$

$$2=\left(\frac{1}{2}\right)^{1-n}$$

Using $$\frac{1}{2}=2^{-1}$$,

$$2^1=(2^{-1})^{1-n}$$

$$2^1=2^{-(1-n)}$$

$$2^1=2^{n-1}$$

Equating the exponents,

$$1=n-1$$

$$n=2$$

Thus, the reaction is second order.

This is also evident from the data since doubling the pressure from $$50$$ to $$100$$ reduces the half-life from $$4$$ to $$2$$, indicating

$$t_{1/2}\propto \frac{1}{p}$$

which is the characteristic dependence of a second-order reaction.

Hence, the correct answer is

$$n=2$$