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11 moles of n2 and 12 moles of h2 mixture reacted in 20 litre vessel vessel at 800 k . after equilibrium was reached , 6 moles of h2 was present .6.5 litre of liquid water was injected in equilibrium mixture and resultant gaseous was suddenly cooled at 300k what is the final pressure of gaseous mixture? negative vapour pressure of liquid solution . assume all NH3 dissoled in water, no change in the volume of liquid , no reaction of N2 AND H2 AT 300K ?

11 moles of n2 and 12 moles of h2 mixture reacted in 20 litre vessel vessel at 800 k . after equilibrium was reached , 6 moles of h2 was present .6.5 litre of liquid water was injected in equilibrium mixture and resultant gaseous was suddenly cooled at 300k what is the final pressure of gaseous mixture? negative vapour pressure of liquid solution . assume all NH3 dissoled in water, no change in the volume of liquid , no reaction of N2 AND H2 AT 300K ?

Grade:11

2 Answers

SAGAR SINGH - IIT DELHI
878 Points
13 years ago

Dear student,

Use the concept of equivalents,

equivalent weight=moleculare weight/ n factor

vimalskatiyar
11 Points
7 years ago
@Sagar Singh
 
Kuchh bhi?
Which equivalents you are talking about...

Moles of N2 & H2 are combining
Post reaction, NH3 is dissolved; moles left are 6 for H2 & hence 9 for N2 (since 6 mole H2 must have combined with 2 moles N2)

15 Moles in 16.42 L of volume available for gases (rest occupied by 3.58 L of water)

P = nRT/V
= 22.5 atm

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