11 moles `N_(2)` and 12 moles of `H_(2)` mixture reacted in 20 litre vessel at 800 K. After equilibrium was reached, 6 mole of `H_(2)` was present. `3.58` litre of liquid water is injected in equilibrium mixture and resultant gaseous mixture suddenly cooled to 300K. What is the final pressure of gaseous mixture? Neglect vapour pressure of liquid solution. Assume (i) all `NH_(3)` dissolved in water (ii) no change in volume of liquid (iii) no reaction of `N_(2)` and `H_(2)` at 300 K.

Options

A. 18.47 atm
B. 60 atm
C. 22.5 atm
D. 45 atm

Answer

C

Explanation

`{:(, N_(2) , + , 3 H_(2) , hArr , 2 NH_(3)) , ("Initial moles " , 11 , , 12 , , 0) , ("At equilibrium" , 9 ,, 6 ,, 4):}`
Moles of `N_(2)` and `H_(2)` present at equilibriun = 15
After addition of water : `NH_(3) (g) + H_(2) O (l) to NH_(4) OH (l)`
Volume of vessel available for gaseous mixture of `N_(2)` and `H_(2) = 20 - 3.58 = 16.42` L
Pressure exerted by gaseous mixture at 300 K = `(15 xx 0.0821 xx 300)/(16.42) implies 22.5` atm