- #1

RB211

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## 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

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

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.

[itex][X_i] = P_i/(R_uT) [/itex]

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:

[itex]\frac{k_f(12.52)(1)}{k_r(5.512)(2)}[/itex]

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.

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