1.
En un envase de 2.0 litros se añaden 1.50 moles de N ?2 , 2.40 moles de H 2 ?. Al llegar a
equilibrio se forma 0.84 moles de NH 3 ?. Calcule el valor para la constante de equilibrio de la
reacción: N ?2(g) + 3 H ?2(g) ? ??? 2 NH 3(g)
Solution
initial concentration of N2 = mol of N2 / volume
= 1.50 moles / 2.0 L
= 0.75 M
initial concentration of H2 = mol of H2 / volume
= 2.40 moles / 2.0 L
= 1.20 M
final concentration of NH3 = mol of NH3 / volume
= 0.84 moles / 2.0 L
= 0.42 M
ICE Table:
[N2]Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â [H2]Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â [NH3]
initial            0.75              Â
1.2Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0
change             -1x                -
3x                +2x
equilibrium        0.75-1x            1.2-
3x             +2x
Given at equilibrium,
[NH3] = 0.42
+2x = 0.42

2.
x = 0.21
Equilibrium constant expression is
Kc = [NH3]^2/[N2][H2]^3
Kc = (+2x)^2/(0.75-1x)(1.2-3x)^3
Kc = (+2*0.21)^2/(0.75-1*0.21)(1.2-3*0.21)^3
Kc = 1.76
Answer:Â Â 1.76