11 moles N2 and 12 moles of H2 mixture reacted in 20 L vessel at 800 K. After equilibrium was reached, 6 mole of H2 was present. 3.58 L of liquid water is injected in equilibrium mixture and resultant gaseous mixture suddenly cooled to 300 K. What is the final pressure of gaseous mixture/? Neglect vapour pressure of liquid solution. Assume (i) all NH3 dissolved in water (ii) no change in volume of liquid (iii) no reaction of N2 and H2 at 300 K (Given R=0.821 L atm mol−1 K−1)

11 moles N2 and 12 moles of H2 mixture reacted in 20 L vessel at 800 K

1.	11 moles N2 and 12 moles of H2 mixture reacted in 20 L vessel at 800 K. After equilibrium was reached, 6 mole of H2 was present. 3.58 L of liquid water is injected in equilibrium mixture and resultant gaseous mixture suddenly cooled to 300 K. What is the final pressure of gaseous mixture/? Neglect vapour pressure of liquid solution. Assume (i) all NH3 dissolved in water (ii) no change in volume of liquid (iii) no reaction of N2 and H2 at 300 K (Given R=0.821 L atm mol−1 K−1)
Answer» $11 m o l e s$ $N_{2}$ and $12 m o l e s$ of $H_{2}$ mixture reacted in $20 L$ vessel at $800 K$ . After equilibrium was reached, $6 m o l e$ of $H_{2}$ was present. $3.58 L$ of liquid water is injected in equilibrium mixture and resultant gaseous mixture suddenly cooled to $300 K$ . What is the final pressure of gaseous mixture/? Neglect vapour pressure of liquid solution. Assume (i) all $N H_{3}$ dissolved in water (ii) no change in volume of liquid (iii) no reaction of $N_{2}$ and $H_{2}$ at $300 K$ (Given $R = 0.821 L a t m m o l^{- 1} K^{- 1}$ )