The reaction, `2N_(2)O_(5(g)) rarr 4NO_(2(g)) + O_(2(g))` follows first-order kinetics. The pressure of a vessel containing only `N_(2)O_(5)` was found to increase from 50 mn hg to 87.5 mm Hg in 30 min. The pressure exerted by the gas after 60 min will be (assume temperature remains constant) :

Options

A. 106.25 mm Hg
B. 116.25 mm Hg
C. 125 mm Hg
D. 150 mm Hg

Answer

A

Explanation

`{:(,2N_(2)O_(5(g)) rarr,4NO_(2(g)),+,O_(2(g)),),(t = 0,50 mm,0,,0,),("t = 30 min",50-2x,4x,,x,),("t = 60 min",50-2y,4y,,y,):}`
At t = 30 min, net pressure = 87.5, therefore `50 - 2x + 4x + x = 87.5, 3x = 37.5 rArr x = 12.5`
The pressure of `N_(2)O_(5)` after 30 min is `50 - 2x rArr 50 - 12.5 xx 2 = 25`
it indicates that after 30 min net pressure of `N_(2)O_(5)` reduces to half, therefore, for next 30 min the pressure of `N_(2)O_(5)` would further reduce to half
`50 - 2y = 12.5 rArr y = 18.75 mm " "` Net pressure t t = 60 min
`P_("total") = 50 - 2y + 4y + y = 50 + 3y = 50 + 3 xx 18.75 = 106.25 mm`