Example 7.23Flowing equilibrium is strudied by taking 1 mole of N, and

1.	Example 7.23Flowing equilibrium is strudied by taking 1 mole of N, and 3 moles of H2 in a ILflask at a given temperaturestudied by taking 1 mole of N2 and 3 moles of H2N2(g) +3H2() 2NH(g)s(e) formed at equilibrium is neutralised by 200 ml of IM HCI. Calculate equilibrium constant.
Answer» N2 + 3H2 ---------> 2NH3 Initial 1 3 0 P1=4 atm ∑n = 4 Eqbm 1-x 3-3x 2x P2=3 atm ∑n = 4-2x Pα n P1/P2 = n1/n2 4/3 = 4/(4-2x) x= 0.5 PN2= 0.5, PH2= 1.5, PNH3= 1 K = (PNH3)2/ (PN2) (PH2)^3 K= 1 / (0.5) (1.5)^3 the equilibrium constant for dissociation of NH3 is K = (0.5) (1.5)^3

Example 7.23Flowing equilibrium is strudied by taking 1 mole of N, and 3 moles of H2 in a ILflask at a given temperaturestudied by taking 1 mole of N2 and 3 moles of H2N2(g) +3H2() 2NH(g)s(e) formed at equilibrium is neutralised by 200 ml of IM HCI. Calculate equilibrium constant.

Answer»

N2 + 3H2 ---------> 2NH3

Initial 1 3 0 P1=4 atm ∑n = 4

Eqbm 1-x 3-3x 2x P2=3 atm ∑n = 4-2x

Pα n

P1/P2 = n1/n2

4/3 = 4/(4-2x)

x= 0.5

PN2= 0.5, PH2= 1.5, PNH3= 1

K = (PNH3)2/ (PN2) (PH2)^3

K= 1 / (0.5) (1.5)^3

the equilibrium constant for dissociation of NH3 is K = (0.5) (1.5)^3