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Urmi Roy

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- Thread starter Urmi Roy
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Urmi Roy

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- #2

eigenperson

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Let's look at a simple reaction: A

In aqueous solution, you have a zillion molecules -- a term I will use to refer to ions and atoms as well -- whizzing around and bashing into each other. Let's guess that the average molecule collides with another molecule once every femtosecond (don't assume this is accurate), and suppose there are n molecules (including water and the species we're interested in) in the entire solution.

Look at a particular A

Now look at the number of AB molecules that break up every femtosecond. The total number of AB molecules in the container is n[AB], but in a given femtosecond, each one has a probability of only P

If you assume the solution is at equilibrium, then the concentrations of the various species must not be changing. So in every femtosecond, the number of AB molecules forming should be equal to the number of AB molecules breaking up. Therefore, at equilibrium, we should have nP

So that's why you might expect the rule to be what it is.

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Urmi Roy

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Thanks a lot for that very detailed explanation, eigenperson!

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Borek

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Chestermiller

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Borek

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Urmi Roy

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It turns out that:

Affinity of a reaction (A)=R*T*ln(K/((x1^v1)*(x2^v2)))

Where v1, v2 are stoichiometric coefficients, and x1, x2 are mole fractions of each reactant...

So if the reaction is in equilibrium, A=0 and K= ((x1^v1)*(x2^v2))

How my question is

- #8

Yanick

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The derivation using thermodynamics is the more "fundamental" one because using rate laws and such is only valid when speaking of elementary reactions. Trying to derive an equilibrium constant expression using rate laws which describe a multi-step mechanism will not necessarily give you the correct equilibrium constant expression because the order of each reactant in a rate law need not equal the stoichiometric coefficient of that species in the chemical equation. On the other hand, using the condition of equilibrium (Ʃ

- #9

DrDu

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Basically it is due to the chemical potential of gases and ideal solutions depending on the logarithm of pressure/ concentration. These logarithms are added to calculate Gibbs enthalpy of the reaction, which corresponds to a multiplication of the partial pressures/ concentrations.

That's clearest for an ideal gas. The thermodynamical properties of an ideal gas only depend on its partial pressure p:

##dU=TdS-pdV+\mu dN##

As ##\mu=(\partial U/\partial N)_{S,V}## and ##p=(\partial U/\partial V)_{S,N}##,

so ##(\partial \mu/\partial V)_{S,N}=-(\partial p/\partial N)_{S,V}=-kT/V=-p/N## where in the last steps I used the ideal gas law. From it you also get ## dV=NkT/p^2 dp##,

so, at constant S, ##d\mu=\partial \mu/\partial V)_{S,N}dV=(-p/N)(NkT/p^2) dp=-kTdp/p##.

Upon integration over p you get a nice logarithm.

That's clearest for an ideal gas. The thermodynamical properties of an ideal gas only depend on its partial pressure p:

##dU=TdS-pdV+\mu dN##

As ##\mu=(\partial U/\partial N)_{S,V}## and ##p=(\partial U/\partial V)_{S,N}##,

so ##(\partial \mu/\partial V)_{S,N}=-(\partial p/\partial N)_{S,V}=-kT/V=-p/N## where in the last steps I used the ideal gas law. From it you also get ## dV=NkT/p^2 dp##,

so, at constant S, ##d\mu=\partial \mu/\partial V)_{S,N}dV=(-p/N)(NkT/p^2) dp=-kTdp/p##.

Upon integration over p you get a nice logarithm.

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Borek

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Urmi Roy

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The K is basically the Q if the reaction is not in equilibrium?

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Borek

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What are x1, v1, x2 and v2?

The K is basically the Q if the reaction is not in equilibrium?

No. K is K. If equation uses K, it means K, period.

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