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Grade 11Physical Chemistry

Two Moles of NH3 gas are introduced into a previously evacuated one litre vessel in which it partially dissociates at high temperature . AT equilibrium one mole of NH3 remain,the value of equilibrium constant

Profile image of Prateek
8 Years agoGrade 11
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4 Answers

Profile image of BIKRAM
8 Years ago
              Reaction: 2NH3 = 3H2 + N2
Initial amount(mol)    2          –   –  
Final amount(mol)  (2-1)       3         1
                              =1
Equlibrium Constan,K=(1 × 33)/1= 9 unit
Profile image of Savita Singh
7 Years ago
Rx             2NH3 = N2 + 3H2
Initial mole= 2            0       0
Final Mole = 1           1/2    3/2 
Equilibrium constant = 3/2*3/2*3/2*1/2= 27/16 ans 
      
 
 
 
 
 
Profile image of AVNI AGRAWAL
7 Years ago
    no of moles                       2NH_{3}\rightleftharpoons 3H_{2} + N_{2}
at t=0                                         2
at t=teq                                     2-x              3x/2        x/2
but as per the question,
2-x =1   ---> x = 1
thus,                                           1                 3/2        ½
 
 
equilibrium constant=     [H2]3 × [N2]/[NH3]2
substituting values.. we get,
equilibrium constant=  33/2 × 1/2=27/16
Profile image of Kushagra Madhukar
5 Years ago
Dear student,
Please find the answer to your question.
 
                   2NH3 = N2 + 3H2
Initial mole= 2            0       0
Final Mole = 1           1/2    3/2
 
Equilibrium constant =     [H2]3 × [N2]/[NH3]2
                               =  (3/2)3  × (1/2) / 12
                               =   27/16
 
Thanks and regards,
Kushagra