[niobium]

I have done many tries on this exercise, but couldn't get to the right answer.
We are given this reaction: [tex] \ce{COCl_2_{(g)}} = \ce{CO_{(g)}} +\ce{Cl_2_{(g)}} [/tex]

Along with its equilibrium constant [tex]K = 8.3 \times 10^{-4} [/tex] (at  360°C)

We are told to calculate the molar concentrations [tex] [\ce{CO}], [\ce{Cl_2}], [\ce{COCl_2}][/tex] at equilibrium when [tex]10,0 \text{ mol} [/tex] of [tex]\ce{COCl_2} [/tex] decompose initially in a volume of [tex]5,00 \text{ L} [/tex]

Here is my try:

First of all I've computed the initial partial pressure of the reactant using the ideal gas law. I got [tex]P_i =1.0525 \times 10^7 \text{ Pa} [/tex] (keeping 3 significant figures).

Then, I've expressed the equilibrium constant in terms of X, which is the partial pressure of [tex] \ce{CO}[/tex] at equilibrium.

We have [tex]K=\frac{X^2}{P_i - X} [/tex]

After solving for X, I got [tex] X= 93.4661 \text{ Pa} [/tex] (keeping again 3 significant figures).

Finally, in order to have molar concentrations, we use the ideal gas law again:
[tex] P_{\ce{CO}}=[\ce{CO}]_{eq} R T[/tex]

This yields [tex] [\ce{CO}]_{eq} = 0.0178 \text{ mol/m^3}[/tex]

Which is different from the numerical answer given in my classnotes for the concentration of CO at equilibrium.


Can anyone guide me or tell me my errors in my reasonning ?

[Borek]

Along with its equilibrium constant [itex]K = 8.3 \times 10^{-4} [/itex] (at  360°C)

K_c or K_p?

[Hunter2]

From the written text it sounds first to calculate the molar concentrations. But the question in deed is  which K was given, what Borek already mentioned.

[niobium]

Thanks for your replies, well it isn't specified if it is Kc or Ks, but as you can see I have done the calculations for partial pressures and it turned out to be an incorrect answer, numerically.

What leaves me doubtful is that, assuming it were a [itex] K_c [/itex], how could we compute the concentrations, given all present species are gases ? Are we assuming (without mentionning it) that there is water involved ? And are we also assuming that the gases are dissolved in water ?

[Hunter2]

That doesnt matter. It's not dissolved in water it's in a 5 l sealed Container. You have a volume given and mole given.
In the beginning you  have 10 mol/5l = 2 mol/l of Phosgen.
Your K does it have a unit given.

I got 0,0403 mol/l for x

[niobium]

Then if I do the calculation for [tex] K = K_c = \frac{x^2}{2,00-x}[/tex]

I get for [itex]x[/itex], which is the molar concentration of [itex]\ce{CO}[/itex] at equilibrium: [tex]x=0.0403 \text{ M}[/tex] which is, if 2 Significant Figures are kept, [itex] x=0.040[/itex] M

This is not quite the same as the answer given, which is [itex] x=0.041[/itex] M

Where did this go wrong ? I took the positive solution to the quadratic (because concentrations are positive). Could this be a typo from my classnotes ?

[Hunter2]

Your calculation is correct. Probably the constant has more digits or a typo.

[niobium]

Probably the constant has more digits.

Got it.

Thanks to all you guys