[KungKemi]

I noticed that this question has been posted on this forum previously, however, I couldn't see that an actual numerical answer was given. I was just wondering if the following process seemed right, and, as such, if the solution is correct (I have no way of checking). I haven't covered this concept yet in my Chemistry class, but I thought that I'd give it a go anyway:

"Consider the reaction A_(g) + D_(g) C_(g) for which K = 130. Assume that 0.406 mol C(g) is placed in a container fitted with a piston. The temperature is 300.0 K, and the barometric pressure on the piston (which is assumed to be massless and frictionless) is constant at 1.00 atm. The original volume (before the 0.406 mol C(g) begins to decompose) is 10.00 L. What is the volume in the cylinder at equilibrium?

My working was as such:

V_i = 10.00 L
 P = 101.325 kPa
 T = 300.0 K
n_i = 0.406 mol C_(g)

C_C(g) = 0.406/10 = 0.0406 mol/L

   K = [C]/[A][D]
130 = 0.0406/x²
    x = 0.0177 mol/L

        x = y/V_f
0.0177 = x/V_f

n_f = 0.406 - y + y + y = 0.406 + y

          PV_f = n_fRT
101.325V_f = (0.406 + y) × 8.31 × 300.0
            V_f = 24.604y + 9.989

0.0177 = y/(24.604y + 9.989)
  0.177 = 0.565y
     ..  y = 0.312 mol

0.0177 = y/V_f
       V_f = y/0.0177 = 0.312/0.0177 = 17.674 L

If anyone could let me know if this process is right, that would be greatly appreciated. I'm not certain if the volume is allowed to change in the equilibrium expression (as I have assumed).

Thanks,
KungKemi

[Borek]

   K = [C]/[A][D]
130 = 0.0406/x²
    x = 0.0177 mol/L

Some of C decomposed, didn't it? And if so, there were no longer 0.0406 moles of it at equilibrium.

Besides, 0.0406 is number of moles, not concentration nor partial pressure, so you can't be sure it can be plugged directly into the equation this way. Sadly, problem doesn't state whether the K given is Kc or Kp, and they are not interchangeable, especially in this case. I would assume it is Kp, makes more sense for the reaction taking place in the gas phase, then using number of moles makes sense - but is still incorrect, see above.

[KungKemi]

   K = [C]/[A][D]
130 = 0.0406/x²
    x = 0.0177 mol/L

Some of C decomposed, didn't it? And if so, there were no longer 0.0406 moles of it at equilibrium.

Besides, 0.0406 is number of moles, not concentration nor partial pressure, so you can't be sure it can be plugged directly into the equation this way. Sadly, problem doesn't state whether the K given is Kc or Kp, and they are not interchangeable, especially in this case. I would assume it is Kp, makes more sense for the reaction taking place in the gas phase, then using number of moles makes sense - but is still incorrect, see above.

In the textbook that I obtained the question from they just state the standard equilibrium constant as K, and K_p for a gaseous equilibrium shift. However, as for the working, I stated that the concentration of C_(g) is 0.0406 mol/L, since the initial amount is 0.406 mol in a volume of 10.00 L.

As for the equilibrium expression, I think that I was under the impression that you could just plug [C] directly in as you do typically with acid-base equilibria problems. Of course, some C does decompose, but I think that you can only determine this amount by subtracting the amount of the produced products (as can be found from further working of the determined concentration values from the equilibrium expression) from the initial amount (as can be seen in step 7 of my working). Is this correct?

KungKemi

[Borek]

In the textbook that I obtained the question from they just state the standard equilibrium constant as K, and K_p for a gaseous equilibrium shift. However, as for the working, I stated that the concentration of C_(g) is 0.0406 mol/L, since the initial amount is 0.406 mol in a volume of 10.00 L.

In general K is expressed using "activity", whatever it means in a particular case (typically in a reference to the standard state of 1 M and/or 1 bar). However, when the value is given it should be clarified what it means, just in case. Chances are it is defined somewhere and the K in the book is always meant to be the same, but it is not clear when the question is taken out of the context.

As for the equilibrium expression, I think that I was under the impression that you could just plug [C] directly in as you do typically with acid-base equilibria problems.

Even in the acid/base equilibria you don't use total concentrations as you suggest. You follow the stoichiometry to express equilibrium concentrations in terms of the initial concentration and the change (which is where ICE table comes in handy), but from what you wrote it looks like you used initial [C] to calculate concentrations of A and D - which doesn't make sense, as when [C] = 0.0406, [A]=[D]=0.

[AWK]

In Zumdahl Challenge Problem Kc is used.

[KungKemi]

Even in the acid/base equilibria you don't use total concentrations as you suggest. You follow the stoichiometry to express equilibrium concentrations in terms of the initial concentration and the change (which is where ICE table comes in handy), but from what you wrote it looks like you used initial [C] to calculate concentrations of A and D - which doesn't make sense, as when [C] = 0.0406, [A]=[D]=0.

Okay, that does make sense Borek. Let me look at this question a second time.

Thanks, KungKemi

[KungKemi]

Okay, I think that I got it now:

A _(g) + D _(g) C _(g)   K = 130

n_iC = 0.406 mol
T = 300.0 K
P = 101.325 kPa
n_f = 0.406 + x

K = [C]/[A][D]

         130 = (0.406 - x)V_f^-1/(xV_f^-1 × xV_f^-1)
130x²/V_f² = (0.406 - x)/V_f         
 130x²/V_f = 0.406 - x
           V_f = 130x²/(0.406 - x)

PV_f = n_fRT
PV_f = RTn_f/P
PV_f = (8.31 × 300)(0.406 + x)/101.325
PV_f = 24.604(0.406 + x)

24.604(0.406 + x) = 130x²/(0.406 - x)
130.000x² = (9.989 + 24.604x)(0.406 - x)
130.000x² = 4.056 - 9.989x + 9.989x - 24.604x²
130.000x² = 4.056 - 24.604x²
150.604x² = 4.056
             x = 0.162 mol

V_f = 130(0.162)²/(0.406 - 0.162) = 13.974 L

If anything still seems folly please let me know, but I think that this makes more sense now.

Thanks,
KungKemi

[Borek]

At first sight logic looks OK. Can you try to repeat the calculations assuming K given is Kp, not Kc?

[KungKemi]

At first sight logic looks OK. Can you try to repeat the calculations assuming K given is Kp, not Kc?

Well, I know that K, in this case, is K_c (as this is the nomenclature of the text), but I can certainly consider it as K_p as well...

Δn = 1 - (1 + 1) = -1
  R = 0.0821 atm · L/mol · K
  T = 300.0 K

  K_p = KRT^Δn
130 = K/(0.0821 × 300)
    K = 3201.9

n_iC = 0.406 mol
T = 300.0 K
P = 1.00 atm
n_f = 0.406 + x

                 K = [C]/[A][D]
         3201.9 = (0.406 - x)V_f^-1/(xV_f^-1 × xV_f^-1)
3201.9x²/V_f² = (0.406 - x)/V_f         
 3201.9x²/V_f = 0.406 - x
               V_f = 3201.9x²/(0.406 - x)

PV_f = n_fRT
PV_f = RTn_f/P
PV_f = (0.0821 × 300)(0.406 + x)/1
PV_f = 24.630(0.406 + x)

24.630(0.406 + x) = 3201.9x²/(0.406 - x)
3201.90x² = (10.000 + 24.630x)(0.406 - x)
3201.90x² = 4.060 - 10.000x + 10.000x - 24.630x²
3201.90x² = 4.060 - 24.630x²
3226.53x² = 4.060
             x = 0.0355 mol

V_f = 3201.90(0.0355)²/(0.406 - 0.0355) = 10.873 L

I assume that both answers are equally correct (depending on what the value for K is), however, I believe that the first one ought to be the more correct answer (as this is the nomenclature used by the text to express K_c. Nonetheless, I've never done a calculation using the K_p value before so this was very interesting.

KungKemi