Question

Derive the relation between K_p and K_c.

Answer 1

Let us consider the general reaction in which all reactants and products are ideal gases.

xA + yB ⇌ lC + mD

The equilibrium constant, K_c is,

`"K"_"c" = (["Cl"]^"l"["D"]^"m")/(["A"]^"x"["B"]^"y")`   ...(1)

and K_p is,

`"K"_"p" = ("p"_"C"^"l" xx "p"_"D"^"m")/("p"_"A"^"x" xx "p"_"B"^"y")`   ...(2)

The ideal gas equation is,

PV = nRT

P = `"n"/"V"`RT

since, Active mass = molar concentration = n/V

P = active mass × RT

Based on the above expression the partial pressure of the reactants and products can be expresssed as,

P^X_A = [A]^x[RT]^x

P^Y_B = [A]^y[RT]^y

P^l_C = [A]^l[RT]^l

P^m_D = [A]^m[RT]^m

On substituting in Eqn.(2).,

`"K"_"p" = (["C"]^"l"["RT"]^"l" ["D"]^"m" ["RT"]^"m")/(["A"]^"x" ["RT"]^"X" ["B"]^"y" ["RT"]^"y")`   .....(3)

`"K"_"p" = (["C"]^"l"["RT"]^"l" ["D"]^"m" ["RT"]^("l + m"))/(["A"]^"x"["B"]^"y" ["RT"]^("x + y"))`

`"K"_"p" = (["C"]^"l"["D"]^"m")/(["A"]^"x"["B"]^"y") ["RT"]^(("l + m") - (x + y))`   ....(4)

By comparing equation (1) and (4),

we get

K_p∆ng =K_c (RT) …………..(5)

Where,

∆n_g is the difference between the sum of number of moles of products and the sum of number of moles of reactants in the gas phase.
The following relations become immediately obvious.

when ∆n_g = 0.

K_p = K_c (RT)⁰ = K_c

Example:

\[\ce{H2(g) + l_2(g) <=> 2HI(g)}\]

\[\ce{N2 + O2 <=> 2NO(g)}\]

when ∆n_g = +ve

K_p = K_c(RT)^+ve

K_p > K_c

Example:

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

\[\ce{PCl5(g) <=> PCl3(g) + Cl2(g)}\]

When ∆n_g = -ve

K_p = K_c(RT)^-ve

K_p < K_c

Example:

\[\ce{2H2(g) + O2(g) <=> 2H2O(g)}\]

\[\ce{2SO2(g) + O2(g) <=> 2SO3(g)}\]