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confused about Keq and temperature

Discussion in 'MCAT: Medical College Admissions Test' started by ssh18, Feb 22, 2007.

  1. ssh18

    ssh18 Member
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    I was hoping someone could help me out with this. I was reading through the chapter on equilibrium and it says that the Keq can change only if temperature changes - whereas changes in reactant/product concentration or pressure doesn't change Keq. I can't seem to figure out why temperature would change Keq. I mean, once equilibrium is reached, shouldn't the value of Keq remain constant?
     
  2. Funky

    Funky This space is for sale
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    Dont forget that K is a constant that is only used in standard state conditions. K much like the rate constant will most definitely change based on temperature. How it changes is hard to measure because you can't use the law of mass action and you must find out through experimentation.
    Changing the reactant/product concentration or pressure will knock it out of equilibrium until the reaction fixes itself (le chatelier's principle). You use the reaction quotient to find out which direction equilibrium is shifted.
    You can say that when you add a reactant/product or change the pressure, the system fixes itself towards Keq.
     
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  3. Beau Geste

    Beau Geste yah mo b there
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    Tee hee - everytime I see the title of this thread it looks like "Confused about Keg and temperature" - I can't help but think "well, cold is best!"

    :)
     
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  4. Richter915

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    hmm...assuming ur deltaG is a non zero number, the equation:

    deltaG = -RT ln(Keq)

    should explain why T is the only thing to effect Keq
     
  5. OP
    OP
    ssh18

    ssh18 Member
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    thanks guys! that does make sense. Here's another thing that's been really bothering me all evening and I've been trying to find/rationalize the answer to this but it doesn't seem to make sense.
    I read somewhere that in hot areas (like a desert) the atmospheric pressure is low vs. pressure near an icy area.
    I'm really confused. I understand that with an increase in temperature, hot air rises (it is less dense) but I don't get the pressure relation here. I also understand that with an increase in altitude, pressure decreases as you have less air pushing on you basically.
    Someone please correct me. Thanks!
     
  6. Funky

    Funky This space is for sale
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    the only thing i can really think of that would answer your question is since pressure is F/A, the air molecules will be moving around faster in a desert because it is in a higher temperature area and will disperse somewhere else. Maybe this overcomes the fact that air molecules go from high to low pressure.
     
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  7. Richter915

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    hmm that may be a possibility but I was thinking more along the lines of...in warm areas the gas molecules move faster but are spread farther apart so there're less gas particles per area than in a colder region. So force is related to mass as we all know and with less mass per area there ought to be less force per area (assuming the same a for the two areas...which is true on earth...g)...and therefore less pressure. i'm just throwing ideas out there.
     
  8. punkindrublic

    punkindrublic calls shenanigans
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    My thought would be that because they are at a lower temperature, and thus moving less quickly, it would allow for more intermolecular attraction between molecules, which could allow for more mass and thus force over a given area...
     
  9. killinsound

    Physician 10+ Year Member

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    that is not 100% correct...

    that equation is used to calculate the delta G given the equilibrium constant at some temperature.

    K is affected by temperature through the arrhenius equation. k = Ae^(-Ea/RT)

    Remember that the arrhenius constant is the RATE constant 'k' ...

    however, 'k' foward / 'k' backward = equilibrium constant K.

    Since the rate constants are related to the temperature on a log scale, it will not be proportional in change between the foward and the backward, and you will get a deviation from the standard K.

    in addition to temperature, through the arrhenius equation, we can also see that activation energy affects the equilibrium constant and rate constant.
     
  10. RAD11

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    :laugh: I was thinking the same thing :thumbup:
     

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