Question

Derive a general expression for the equilibrium constant Kp and Kc for the reaction, \[\ce{3H2(g) + N2(g) <=> 2NH3(g)}\].

Answer 1

Let us consider the formation of ammonia in which, ‘a’ moles nitrogen and h’ moles hydrogen gas are allowed to react in a container of volume V. Let ‘x’ moles of nitrogen react with 3 x moles of hydrogen to give 2x moles of ammonia.

\[\ce{3H2(g) + N2(g) <=> 2NH3(g)}\]

	N2	H2	NH3
Initial number of moles	a	b	0
number of moles reacted	X	3x	0
Number of moles at equilibrium	a -x	b - 3x	2x
Active mass or molar concentration at equilibrium	`("a -x")/"V"`	`("b" - "3x")/"V"`	`(2x)/"V"`

Applying law of mass action,

K₂ = `(["NH"_3]^2)/(["N"_2]["H"_2]^3)`

= `("2x"/"V")^2/((("a - x")/"V")(("b" - 3"x")/"V")^3)`

= `("4x"^2/"V"^2)/((("a - x")/"V")(("b" - 3"x")/"V")^3)`

= `(4x^2"V"^2)/(("a - x")("b - 3x")^3)`

The equilibrium constant K_P can also be calculated as follows:

`"K"_"P" = "K"_"C" ("RT")^(Delta"ng")`

Δn_g = n_p - n_r = 2 - 4 = - 2

`"K"_"P" = (4x^2"V"^2)/(("a - x")("b - 3x")^3) xx ["RT"]^-2`

Total number of moles at equilibrium,

n = a - x + b - 3x + 2x = a + b - 2x

`"K"_"P" = (4x^2"V"^2)/(("a - x")("b - 3x")^3) xx ["PV"/"n"]^-2`

`"K"_"P" = (4x^2"V"^2)/(("a - x")("b - 3x")^3) xx ["n"/"PV"]^2`

`"K"_"P" = (4x^2"V"^2)/(("a - x")("b - 3x")^3) xx [("a + b - 2x")/"PV"]^2`

`"K"_"P" = (4x^2("a + b - 2x")^2)/("P"^2("a - x")("b - 3x")^3)`