For a n^th order reaction, the rate of reaction at time t ,

Rate = K [P_t] ⁿ

Here P_t = pressure at time t, k = constant.

Note :

Here instead of concentration of product, pressure of product is given.

When 5% is reacted at a rate 1 Toss S^$$-$$1

Then un-reacted is 95%..

As initial pressure is 363 Torr

then after 5% reaction completed the pressure will be

= 363 $$ \times $$ $${{95} \over {100}}$$ Torr.

$$\therefore\,\,\,$$ 1 = K $${\left[ {363 \times {{95} \over {100}}} \right]^n}$$ . . . . . . . .(1)

When 33% is reacted at a rate 0.5 , Torr S^$$-$$1 then un-reacted is 67%

So, after 33% reaction completion, the pressure is = $$363 \times {{67} \over {100}}$$ Torr.

$$\therefore\,\,\,$$ 0.5 = K $${\left[ {363 \times {{67} \over {100}}} \right]^n}.......$$ (2)

Dividing (1) by (2), we get

$${1 \over {0.5}} = {{{{\left[ {363 \times {{95} \over {100}}} \right]}^n}} \over {{{\left[ {363 \times {{67} \over {100}}} \right]}^n}}}$$

$$ \Rightarrow \,\,2 = {\left[ {{{95} \over {67}}} \right]^n}$$

$$ \Rightarrow \,\,$$ 2 = $${\left[ {1.41} \right]^n}$$

$$ \Rightarrow \,\,$$ 2 = $${\left[ {\sqrt 2 } \right]^n}$$

$$ \Rightarrow \,\,\,\,2 = {2^{{n \over 2}}}$$

$$\therefore\,\,\,$$ $${n \over 2}$$ = 1

$$ \Rightarrow \,\,\,\,$$ n = 2

This is a 2nd order reaction.