A calculation of equilibrium constant of a reaction of phosgene

[zenterix]

Homework Statement

Phosgene ($COCl_2(g)$) is a chemical warfare agent that decomposes by the reaction

$$\mathrm{COCl_2(g)\rightleftharpoons CO(g)+Cl_2(g)}$$

with equilibrium constant $K=8.3\times 10^{-4}$ at $360^\circ\text{C}$#.

Calculate the $\mathrm{[CO],[Cl_2]}$, and $\mathrm{[COCl_2]}$ when 10 mol of phosgene decompose at $360^\circ\text{C}$ and reach equilibrium in a $5.00\text{L}$ flask

Relevant Equations

The way I though about this problem is the following.

We start with $n_i$ moles of phosgene and zero of the products.

The initial total pressure and also partial pressure of phosgene is $\frac{n_iRT}{V}$.

At equilibrium, suppose $x$ moles of phosgene have reacted.

Then we have $n_i-x$ moles of phosgene left and $x$ mole of each of the products.

The equilibrium partial pressures are thus $\frac{(n_i-x)RT}{V}, \frac{xRT}{V}$, and $\frac{xRT}{V}$ for phosgene and the two products, respectively.

Then

$$K=\frac{P_{CO}P_{Cl_2}}{P_{COCl_2}}=8.3\times 10^{-4}$$

Now, we have to be careful with values and units because the more precise equation is

$$K=\frac{(P_{CO}/P^\circ)(P_{Cl_2}/P^\circ)}{(P_{COCl_2}/P^\circ)}=8.3\times 10^{-4}$$

where $P^\circ=1\ \text{bar}$

We have an equation with one unknown, $x$. When I solve for $x$ I reach the result of $0.028\ \text{mol}$.

It is not clear to me exactly what the problem is asking though.

Here is the answer provided by MIT OCW

 

[Borek]

Is this Kc, or Kp?

 

[Mayhem]

The question is asking for the final concentrations of the compounds. Your approach looks right but I didn't verify the numbers myself.

Going from amount of substance to a concentration should be piece of cake for you when you know the vessel volume, considering the work you showed here.

@Borek I'd be surprised if they gave Kc for a problem concerning gases.

 

[Borek]

I'd be surprised if they gave Kc for a problem concerning gases.

Yes, but the answer given is suggesting calculations done with concentrations (2M is what you get after pushing 10 moles into 5L vessel), which is what made me wonder.

 

[zenterix]

Is this Kc, or Kp?

The problem statement gives us $K$. It seems to me this is $K_p$.

But I did indeed forget about the existence of $K_c$.

Suppose the given $K$ is actually $K_c$.

Then, if $x$ moles of phosgene react to generate $x$ moles each of $\mathrm{CO(g)}$ and $\mathrm{Cl_2(g)}$ then at equilibrium we have molar concentrations of $\frac{10-x}{5}$, $\frac{x}{5}$, and $\frac{x}{5}$.

$K_c$ is then $\frac{x^2}{10-x}=8.3\cdot 10^{-4}$ which we solve for $x$ to get $0.09$.

This doesn't alight with the given solution unfortunately.

 

[Chestermiller]

The problem statement gives us $K$. It seems to me this is $K_p$.

But I did indeed forget about the existence of $K_c$.

Suppose the given $K$ is actually $K_c$.

Then, if $x$ moles of phosgene react to generate $x$ moles each of $\mathrm{CO(g)}$ and $\mathrm{Cl_2(g)}$ then at equilibrium we have molar concentrations of $\frac{10-x}{5}$, $\frac{x}{5}$, and $\frac{x}{5}$.

$K_c$ is then $\frac{x^2}{10-x}=8.3\cdot 10^{-4}$ which we solve for $x$ to get $0.09$.

This doesn't alight with the given solution unfortunately.

Shouldn't it be $\frac{x^2}{10-x}=(8.3\cdot 10^{-4})(5)$

 

[zenterix]

Shouldn't it be

Yes, you are right. I made an algebraic mistake.

We have

$$K_c=8.3\cdot 10^{-4}=\frac{\frac{x}{5}\frac{x}{5}}{\frac{10-x}{5}}=\frac{x^2}{5(10-x)}$$

which gives

$$x=0.201$$

and so the equilibrium molar concentrations are

$$\frac{(10-0.201)\text{mol}}{5\text{L}}=1.96\text{M}$$

$$\frac{0.201\text{mol}}{5\text{L}}=0.04\text{M}$$

So indeed the problem gave us $K_c$.

 

[zenterix]

Given that we now know that $K_c=8.3\cdot 10^{-4}$, we can calculate $K$, which comes out to 0.04368.

In the OP, that approach works if we use this value of $K$ in the calculation with partial pressures.