Question

Question #32f3d

Answer 1

`P_ ("COCl"_2) = 4xx10^-6` `"atm"` (essentially `0`, but not exactly)

`P_"CO" = 0.124` `"atm"`

`P_ ("Cl"_2) = 0.124` `"atm"`

Explanation:

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.