Question 32f3d

Jul 15, 2017

${P}_{{\text{COCl}}_{2}} = 4 \times {10}^{-} 6$ $\text{atm}$ (essentially $0$, but not exactly)

${P}_{\text{CO}} = 0.124$ $\text{atm}$

${P}_{{\text{Cl}}_{2}} = 0.124$ $\text{atm}$

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 ${\text{COCl}}_{2}$ 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:

• ${\text{COCl}}_{2}$: $0.124$ $\text{atm}$

• $\text{CO}$: $0$

• ${\text{Cl}}_{2}$: $0$

because only ${\text{COCl}}_{2}$ 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:

• ${\text{COCl}}_{2}$: $- x$
• $\text{CO}$: $+ x$

• ${\text{Cl}}_{2}$: $+ x$

Which means the final equilibrium pressures are

• ${\text{COCl}}_{2}$: $0.124 - x$ $\text{atm}$

• $\text{CO}$: $x$ $\text{atm}$

• ${\text{Cl}}_{2}$: $x$ $\text{atm}$

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

${K}_{p} = \frac{\left(x\right) \left(x\right)}{\left(0.124 - x\right)} = 4.10 \times {10}^{3}$

Solving for $x$:

${x}^{2} = \left(4.10 \times {10}^{3}\right) \left(0.124 - x\right)$

${x}^{2} = 508.4 - 4100 x$

${x}^{2} + 4100 x - 508.4 = 0$

$x = \frac{- 4100 \pm \sqrt{{4100}^{2} - 4 \left(1\right) \left(508.4\right)}}{2 \left(1\right)}$

$= - 4100.2$

$= 0.123996$

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

• ${\text{COCl}}_{2}$: 0.124-(0.123996) = color(red)(4xx10^-6 color(red)("atm"

• $\text{CO}$: color(blue)(0.124 color(blue)("atm"

• ${\text{Cl}}_{2}$: color(green)(0.124 color(green)("atm"#

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