A certain gas mixture is held at #395^@ "C"# has the following initial partial pressures: #P_(Cl_2) = 351.4#, #P_(CO) = 342.0#, #P_(COCl_2) = 0#, all in #"torr"#. Find #K_P^@#?

Since Michael and Stefan managed to help me to figure this out, I'll put an answer here.

We got #K_P^@ = 22.17#. The back of my book gives #K_P^@ = 22.2#.

The main idea is that for ideal gases, at fixed #V# and at the same/fixed #T# (the actual value doesn't matter), a change in #mols# of gas influences the change in the total pressure, which follows Dalton's Law of partial pressures:

#sum_i^N P_i = P_1 + P_2 + . . . + P_N = P_"tot"#

which we will use for #P_(i,eq)# and #P_("tot",eq)#.

First, we can construct an ICE table to determine the expression for each equilibrium partial pressure.

#"CO"(g) " "+" " "Cl"_2(g) " "rightleftharpoons" " "COCl"_2(g)#

#"I"" "342.0" "" "" "351.4" "" "" "" "" "0#
#"C"" "-x" "" "" "-x" "" "" "" "" "+x#
#"E"" "342.0-x" "351.4-x" "" "" "x#

The equilibrium partial pressures can be expressed as Dalton's Law of partial pressures:

#P_"CO" + P_("Cl"_2) + P_("COCl"_2) = P_("tot",eq) = "439.5 torr"#

#342.0 - x + 351.4 cancel(- x + x) = 439.5#

#693.4 - x = 439.5#

#color(green)(x) = 693.4 - 439.5 = color(green)("253.9 torr")#

From this, we can construct the expression to calculate #K_P^@# (where #P^@ = "750.062 torr"# is the standard pressure, which is also equal to #"1 bar"#):

#color(blue)(K_P^@) = (P_(COCl_2)/P^@)/((P_(Cl_2)/P^@)(P_(CO)/P^@))#

#= ((253.9)/(750.062))/(((351.4 - 253.9)/(750.062))((342.0 - 253.9)/(750.062))#

#= color(blue)(22.17)#