1.

28 g of N_(20 and 6g of H_(2) were keip at 40^(@)C in 1 litre vesscel the equilibrium mixture contained 24.54g of NH_(3). The approximate value of K_(c)for the above reaction can be (in "mole" ^(-2) "litre"^(2)

Answer»

75
50
25
100

Solution :`N_(2)+3H_(2)hArr 2NH_(3)`
`{:("Initial conc.",1,3,0),("at equlibrium",1-0.81,3-2.43,1.62),(,0.19,0.57,):}`
No. of MOLES of `N_(2)=28/28=3` mole
No. of moles of `H_(2)=6/2=3` moles
`K_(c)=([NH_(3)]^(2))/([N_(2)][H_(2)]^(3))=([1.62]^(2))/([0.19][0.57]^(3))=75`


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