Question #32f3d

We're asked to find the equilibrium pressures of the three gaseous substances in a reaction at a certain temperature, given the #K_p# and the initial #"COCl"_2# pressure.

Let's first write the equilibrium constant expression for this reaction:

#K_p = ((P_"CO")(P_ ("Cl"_2)))/((P_ ("COCl"_2))) = 4.10xx10^3#

We can set up I.C.E. chart (in the form of neat bullet points), starting with the initial quantities:

Initial:

• #"COCl"_2#: #0.124# #"atm"#

• #"CO"#: #0#

• #"Cl"_2#: #0#

because only #"COCl"_2# is present initially.

According to the coefficients of the chemical equation, each species is in a #1:1# molar ratio, which means the change in pressures is

Change:

• #"COCl"_2#: #-x#

• #"CO"#: #+x#

• #"Cl"_2#: #+x#

Which means the final equilibrium pressures are

• #"COCl"_2#: #0.124-x# #"atm"#

• #"CO"#: #x# #"atm"#

• #"Cl"_2#: #x# #"atm"#

And we can plug these into our equilibrium-constant expression:

#K_p = ((x)(x))/((0.124-x)) = 4.10xx10^3#

Solving for #x#:

#x^2 = (4.10xx10^3)(0.124-x)#

#x^2 = 508.4 - 4100x#

#x^2 + 4100x - 508.4 = 0#

#x = (-4100 +-sqrt(4100^2 - 4(1)(508.4)))/(2(1))#

#= -4100.2#

#= 0.123996#

The obvious value to use is the one that is both positive and not extreme (#0.123996#)

The equilibrium partial pressures of each species is thus

• #"COCl"_2#: #0.124-(0.123996) = color(red)(4xx10^-6# #color(red)("atm"#

• #"CO"#: #color(blue)(0.124# #color(blue)("atm"#

• #"Cl"_2#: #color(green)(0.124# #color(green)("atm"#

Our results reflect the pressure equilibrium lying far to the right.