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Equilibrium constants Ka, Kc, Kx, Kp and rate constants in reversible reactions |
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| Aug15-12, 01:54 AM | #1 |
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Equilibrium constants Ka, Kc, Kx, Kp and rate constants in reversible reactions
Is the statement Ke=k+/k- valid for all equilibrium constants like Ka, Kx, Kp, Kc? All of the expressions I have found for this statement relate Kc, k+ and k- only.
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| Aug15-12, 08:43 PM | #2 |
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| Aug15-12, 09:35 PM | #3 |
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| Aug16-12, 06:48 AM | #4 |
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Equilibrium constants Ka, Kc, Kx, Kp and rate constants in reversible reactions |
| Aug16-12, 09:19 PM | #5 |
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Kc is the most used. It is based on concentration or molarities. Kc=∏(cj^Sj) where Sj is the stoichiometric coefficient of j chemical compound and Cj is its concentration. Concentration is often written as [j]. It has [mol/volume]^Sj units. Kp is used in reactions in vapor phase so its defined through partial pressures of j in the vapor phase. Kp=∏(pj^Sj). It has [pressure]^Sj units. Kx is in terms of mole fractions. It used for liquid phase reactions. Kx=∏(xj^Sj). It's dimensionless. Ka is the Thermodynamic Equilibrium Constant. By definition Ka=∏(aj^Sj). It's dimensionless. All of them are different and I don't know if you can use the concept Ke=k+/k- for all of them. If so, what would be the differences? |
| Aug17-12, 09:16 AM | #6 |
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Yes, they are all related to one another. Start out by writing KE in terms of activities or fugacities. Then express the activities in terms of concentration times activity coefficient, or fugacities in terms of pressure times fugacity coefficient. This will let you see how the different K's are related, and where the k+ and k- can come in.
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| Aug22-12, 10:48 AM | #7 |
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| Aug24-12, 10:32 PM | #8 |
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The easiest way to study this is to assume ideal gas behavior and ideal solutions (no heat of mixing). Start out simple, by relating Kp and Kc. The partial pressure of a species is related to the concentration by p = c RT. Substitute this into the expression for Kp. This should allow you to determine the relationship between Kc and Kp. Starting simple like this is very helpful. Why? If you can't solve for the simplest case, you certainly won't be able to do more complicated cases.
Chet |
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