Consider the following equilibrium at 388.6 K: #NH_4HS(s) rightleftharpoons NH_3(g) + H_2S(g)#. The partial pressure of each gas is 0.203 atm. How do you calculate #K_P# and #K_C# for the reaction?

#K_P = 0.203#
(#"atm"#)

#K_C = 6.37 xx 10^(-3)#
(#"M"#)

The first thing you can do is write out #K_P#.

#K_P = (P_(NH_3)P_(H_2S))/(P_(NH_4HS))#

where #P_i# is the partial pressure of gas #i#.

The stoichiometries are all 1:1:1, so all the exponents are #1#. Thus, as all partial pressures are given, we have a very simple equilibrium expression to evaluate:

#color(blue)(K_P) = (("0.203 atm")("0.203 atm"))/(("0.203 atm"))#

#=# #color(blue)("0.203")#
#" "#(#"atm"#)

Converting to #K_C#, we assume all the gases dealt with are ideal gases, so that we can use the ideal gas law:

#PV = nRT#

Given that #n_i/V_i = [i]# for a given gas #i#, we can substitute all the pressure terms in #K_P# with concentration terms as follows. For a given gas reaction

#aA + bB -> cC + dD#,

we would have written

#K_P = (P_C^cP_D^d)/(P_A^aP_B^b)#.

Thus, by substituting

#P_i^(n_i) = ((n_iRT)/V_i)^(n_i) = ([i]RT)^(n_i)#,

we then have:

#K_P = (([C]RT)^(c)([D]RT)^(d))/(([A]RT)^(a)([B]RT)^(b))#

#= stackrel(K_C)overbrace(([C]^c[D]^d)/([A]^a[B]^b))((RT)^(c)(RT)^(d))/((RT)^a(RT)^b)#

At this point we've separated out the very definition of #K_C#. Now, just use the properties of exponents to condense the #RT# terms together.

#= K_C((RT)^(c+d))/((RT)^(a+b))#

#= K_C(RT)^((c+d)-(a+b))#

But since the stoichiometries of a reaction can be related through the moles of reactants and products, we might as well call the exponent

#Deltan_(gas) = (n_c + n_d) - (n_a + n_b)#,

where #Deltan_(gas)# is the mols of product gases minus the mols of reactant gases.

Therefore, to convert from #K_P# to #K_C#, we simply have:

#color(green)(K_P = K_C(RT)^(Deltan_(gas)))#

For this expression, be sure to use #bb(R = "0.08206 L"cdot"atm/mol"cdot"K")#. To get #K_C# then, we simply have:

#color(blue)(K_C) = (K_P)/((RT)^(Deltan_(gas)))#

#= (0.203 cancel"atm")/[("0.08206 L"cdotcancel"atm""/mol"cdotcancel"K")(388.6 cancel"K")]^((1 + 1) - (1))#

#= color(blue)(6.37 xx 10^(-3))#

Technically, the units are #("mol"/"L")^((1+1) - (1)) = "mol"/"L"#, but generally #K_C# is reported without units.