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Why are Equilibrium Constants temperature dependent?
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dbassily
#1 Posted : Friday, August 09, 2013 12:13:56 AM
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Can someone please explain why equilibrium constants are temperature dependent? By increasing the temperature you add kinetic energy into a reaction and so both the forward and reverse reactions should increase, and since equilibrium constants are ratios of products and reactants shouldn't the ratios still be the same? I thought that increasing temperature would only make a reaction reach equilibrium quicker, not affect the quantities of products and reactants at equilibrium.

Could someone let me know where my thinking is incorrect?
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pavelsedach
#2 Posted : Friday, August 09, 2013 4:49:56 PM
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Hello Bassily,

The explanation involves looking at Delta G:

DeltaG = -RTln(K)

Because Gibbs free energy (spontaneity of a reaction) is dependent on temperature, so is K. Essentially the spontaneity of a reaction in either the forward or the reverse reaction increases depending on the relative energies (Delta Hs) and entropies (Delta Ss) of products and reactions.

Where Delta G = Delta H - TDeltaS

Remember that even the boiling of water is a temperature dependent equilibrium. At higher temperatures it favors H2O(g) and at lower temperatures it favors H2O(l)
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pavelsedach
#3 Posted : Friday, August 09, 2013 4:55:55 PM
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Essentially where your thinking is not correct is that

Kinetics (dealing with reaction rate) should be treated separately from Thermodynamics (dealing with equilibrium constants).

Where the two overlap is in reversible reactions where

rate forward = rate reverse at equilibrium

so

kforward x reactants = kreverse x products

However the rate constants k don't change equally with temperature. This is because

k = Ae(-Ea/RT) and the forward activation energy and arrhenius factor can be different from the reverse activation energy and arrhenius factor.

Kinetics is related to reactivity. Thermodynamics is related to stability.
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