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one mole of n2 and 3 moles of h2 are placed in a closed container at a pressure of 4 atm. The pressure falls to 3 atm at the same temperature when N2 + 3H2 gives 2NH3. the equilibrium constant for dissociation of NH3 is
N2 + 3H2 ---------> 2NH3Initial 1 3 0 P1=4 atm ∑n = 4Eqbm 1-x 3-3x 2x P2=3 atm ∑n = 4-2x P α nP1/P2 = n1/n24/3 = 4/(4-2x)x= 0.5 PN2 = 0.5, PH2 = 1.5, PNH3 = 1 K = ( PNH3)2 / (PN2) (PH2)^3K= 1 / (0.5) (1.5)^3the equilibrium constant for dissociation of NH3 is K = (0.5) (1.5)^3
N2 + 3H2 ---------> 2NH3Initial 1 3 0 P1=4 atm ∑n = 4Eqbm 1-x 3-3x 2x P2=3 atm ∑n = 4-2x P α nP1/P2 = n1/n24/3 = 4/(4-2x)x= 0.5 PN2 = 0.5, PH2 = 1.5, PNH3 = 1 K = ( PNH3)2 / (PN2) (PH2)^3K= 1 / (0.5) (1.5)^3
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