# 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?

##### 1 Answer

#K_P = 0.203#

(#"atm"# )

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

(#"M"# )

The first thing you can do is write out

#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

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

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

#" "# (#"atm"# )

Converting to

#PV = nRT#

Given that

#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((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

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

For this expression, be ** sure** to use

#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