QJEEM18Cha41d: JEE Main 2018 Chemistry - Class 12 JEEM18Cha41d Solution: At 518 o C the rate of decomposition of a sample of gaseous…

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.