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Topic: Temperature dependence of the equilibrium constant  (Read 3482 times)

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Offline Rutherford

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Temperature dependence of the equilibrium constant
« on: September 13, 2013, 10:25:40 AM »
For the reaction CO+0.5O2 :rarrow: CO2 ΔH=-283kJ/mol abd ΔG=-257.1kJ/mol establish an expression for the temperature dependence of the equilibrium constant Kp.

I know that ΔG=-RTlnKc. ΔG also depends on the temperature, so what to do now?

Offline Babcock_Hall

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Re: Temperature dependence of the equilibrium constant
« Reply #1 on: September 13, 2013, 10:30:12 AM »
Do you mean ΔG or ΔG°?

Offline Rutherford

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Re: Temperature dependence of the equilibrium constant
« Reply #2 on: September 13, 2013, 10:59:33 AM »
ΔG° is given.

Offline Babcock_Hall

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Re: Temperature dependence of the equilibrium constant
« Reply #3 on: September 13, 2013, 11:19:53 AM »
Do you know of an equation that relates ΔG° to temperature?

Offline Rutherford

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Re: Temperature dependence of the equilibrium constant
« Reply #4 on: September 13, 2013, 11:26:20 AM »
Aside the one I wrote I know ΔH-TΔS=ΔG, so ΔH-TΔS=RTlnKc, but then I get the expression for Kc, but it should be Kp, right? Their expression for Kp is the one I got for Kc  ??? Why?

Offline Big-Daddy

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Re: Temperature dependence of the equilibrium constant
« Reply #5 on: September 13, 2013, 11:35:02 AM »
You're getting confused by the Kp = Kc * (RT)^n relation (n is the change in number of moles of gas per mole of the forward reaction equation that occurs) here?

I think the crucial point is that ΔG=-RT*ln(K) actually has K referring not to concentrations or partial pressures but to activities, and if Kp is expressed in terms of partial pressures (fugacities to be exact) then ΔG=-RT*ln(Kp) holds because fugacities are activities for gases (I think), whereas Kc is never exact but in solution is often approximated to be equal to K.

Offline Rutherford

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Re: Temperature dependence of the equilibrium constant
« Reply #6 on: September 13, 2013, 12:07:49 PM »
Yes, that expression was in my head. By the way, how is it derived?

Offline Big-Daddy

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Re: Temperature dependence of the equilibrium constant
« Reply #7 on: September 13, 2013, 01:14:06 PM »
Units. Let's take generalized equilibrium aA + bB -> cC + dD, everything is gaseous. Kp has units Pa^(d+c-a-b), Kc has units M^(d+c-a-b), and to convert from concentration to pressure we have to use P=nRT/V=CRT (C=n/V). So the units of Kp are then those of (C * RT)^(d+c-a-b) which we can break up into C^(d+c-a-b) * (RT)^(d+c-a-b) which is Kc * (RT)^(d+c-a-b). (I used C and M interchangeably because M is the unit and C is the term, but hope there is no confusion :P )

Offline Rutherford

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Re: Temperature dependence of the equilibrium constant
« Reply #8 on: September 13, 2013, 01:47:13 PM »
Understood. Thank you for the explanation ;).

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