56 g of N2 reacts 9 g of H2 in a closed vessel to form NH3. how much N

1.	56 g of N2 reacts 9 g of H2 in a closed vessel to form NH3. how much N2 is left?
Answer» ANSWERGiven: The reaction is N 2 +3H 2 ⇄2NH 3 56g6g27.5kg INITIAL no. of moles of H 2 = 26 =3molN 2 = 2.856 =2molInitial number of moles 2N 2 + 33H 2 ⇄02NH 3 At EQUILIBRIUM, (2−x)(3−3x) 2x∴ Number of moles of NH 3 = 1727.54 =1.62i.e. 2x=1.62x=0.81∴ At equilibrium,No. of moles of N 2 =2−x=2−0.81=1.19No. of moles in H 2 =3(1−x)=3(1−0.81)=0.57∴ Equilibrium concentration of ,[N 2 ]= 11.19 =1.19M[H 2 ]= 10.57 =0.57M[NH 3 ]= 11.62 =1.62M∴ Equilibrium constant,K C = [N 2 ][H 2 ] 3 [NH 3 ] 2 = 1.19×0.57 3 1.62 2 =11.9≈12mol 2 lit 2 Answered By

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