Equilibrium of a gas mixture from partial pressures

[RB211]

Homework Statement

Suppose we have a mixture of the gases H2, CO2, CO and, H2O at 1200 K, with partial
pressures pH2 = 0.55 bar, pCO2 = 0.2 bar, pCO = 1.25 bar, and pH2O = 0.1 bar. Is the
reaction described by the equation

CO + H2O <=> H2 + CO2

at equilibrium under these conditions? If not, in what direction will the reaction proceed to
reach equilibrium? Justify your answers.

Homework Equations

K_p = \frac{(P_C/P^0)^c(P_D/P^0)^d}{(P_B/P^0)^b(P_A/P^0)^a}

Where Kp is the equilibrium constant of the reaction based on partial pressures, a,b,c,d are the coefficients in the chemical reaction formula and Pa, Pb, Pc and Pd are the partial pressures of species a,b,c and d and P0 is the total pressure (sum of all partial pressures).

And,

Where Kc is the equilibrium constant based on concentrations, Ru is the universal gas constant, a,b,c,d are defined as above, T is temperature and P0 is as above.

[X_i] = P_i/(R_uT)

where [Xi] is is the molar concentration of species i, Pi is the partial pressure of species i, T and Ru are as defined above.

and the rate equation

where r is the rate of reaction, k is the rate constant, the quantities in the square brackets are molar concentrations and the superscripts are the coefficients in the reaction formula.

The Attempt at a Solution

In this case, from the given equation, I have calculated P0 to be 2.1 bar and Kp for this question to be 0.88. Then, using the second equation, I convert that to Kc, which I find to be also 0.88.

Then, I convert the partial pressures to molar concentration using the third formula to yield the following:

CO = 12.52
H2O = 1
H2 = 5.512
CO2 = 2

all in mols per cubic meter.

The rate equation can then give me the forward and reverse rates of the reaction. Dividing the forward rate by the reverse rate, I have:

\frac{k_f(12.52)(1)}{k_r(5.512)(2)}

Since by definition Kc = Kf/Kr, I can substitute and simplify, which gives me that the result of that ratio is 0.9994 (subject to some rounding error)

Which leads me to conclude that the reaction is approximately at equilibrium but barring rounding error, will progress slightly in the reverse direction.

Is my solution/line of reasoning correct? The final answer is suspiciously close to unity, so I suspect that I've been working in circles and would have gotten this answer with any given values, so if somebody could look this over and let me know if I was on the right track, I would really appreciate it.

Thanks.

 

[Chestermiller]

The equation for K_p doesn't seem correct. It should simply be in terms of partial pressures:

K_p=\frac{P_C^cP_D^d}{P_A^aP_B^b}

Your comparison is not correct either. In the above equation, it is assumed that the reactants and products are at equilibrium. If they are not, the product on the right hand side will not match the left hand side. You need to find out what the equilibrium constant for the reaction actually is at 1200 K. There are graphs of equilibrium constants for various reactions as a function of temperature in textbooks, or you can calculate the equilibrium constant from standard free energies of formation of the reactants and products and their heat capacities. So, step 1 is to determine the actual value of the equilibrium constant at 1200 K.

Chet

 

[Borek]

Suppose we have a mixture of the gases H2, CO2, CO and, H2O at 1200 K, with partial
pressures pH2 = 0.55 bar, pCO2 = 0.2 bar, pCO = 1.25 bar, and pH2O = 0.1 bar. Is the
reaction described by the equation

CO + H2O <=> H2 + CO2

at equilibrium under these conditions? If not, in what direction will the reaction proceed to
reach equilibrium? Justify your answers.

If K (be it Kc or Kp, doesn't matter) is not given, question can't be answered.

 

[RB211]

Thanks for the replies.

@Chet,

That's how it's given in my textbook. The numerical result is the same either way, since in the equation I have, all the terms are divided by the same constant anyways.

That aside, I see your point that I'm essentially making the assumption that the given reaction is already at equilibrium.

So then, if I use this equation:

K_p = exp(-\Delta G_T^0/(R_uT))

Where

\Delta G_T^0 is the sum of the gibbs function of formation at 1200k multiplied by the coefficient of each of the product, minus the same for the reactants.

Those values come from a chart in a textbook, and then I believe that would give the correct Kp. Then, I can repeat the rest of the steps I did originally.

Doing so yields \Delta G_T^0 = 3110 kJ/kmol

Then I get Kp = 0.732

and consequently, the forward rate divided by the reverse rate becomes 0.83, which indicates that to reach equilibrium, the reaction must progress backwards.

Would that be the correct way to do it?

 

[Chestermiller]

Thanks for the replies.

@Chet,

That's how it's given in my textbook. The numerical result is the same either way, since in the equation I have, all the terms are divided by the same constant anyways.

That aside, I see your point that I'm essentially making the assumption that the given reaction is already at equilibrium.

So then, if I use this equation:

K_p = exp(-\Delta G_T^0/(R_uT))

Where

\Delta G_T^0 is the sum of the gibbs function of formation at 1200k multiplied by the coefficient of each of the product, minus the same for the reactants.

Those values come from a chart in a textbook, and then I believe that would give the correct Kp. Then, I can repeat the rest of the steps I did originally.

Doing so yields \Delta G_T^0 = 3110 kJ/kmol

Then I get Kp = 0.732

and consequently, the forward rate divided by the reverse rate becomes 0.83, which indicates that to reach equilibrium, the reaction must progress backwards.

Would that be the correct way to do it?

Yes. Your value of Kp at 1200 is close to the value shown on the graph in my book (as best I can read it from the graph). As far as your interpretation is concerned, I think you meant to say that the reverse reaction is proceeding faster than the forward reaction, so your system is approaching equilibrium from the high side, and the reaction is proceeding in reverse.