Prove that delta G = delta G^o + RTlnQ

[kotreny]

Please show me the derivation for the formula relating Gibbs free energy change and the reaction quotient. I didn't find it on Google, so I decided to turn here. Is it really that hard?

Thanks for any help.

 

[DrDu]

This formula holds by definition (see http://en.wikipedia.org/wiki/Activity_(chemistry)) and is therefore unprovable.
There are only some cases - ideal gases or solutions- where the activities can be replaced by the concentrations.
For an ideal gas or an ideal solution you should find plenty of discussions how the chemical potential depends on concentration.

 

[epenguin]

Please show me the derivation for the formula relating Gibbs free energy change and the reaction quotient. I didn't find it on Google,

It's here http://en.wikipedia.org/wiki/Chemical_equilibrium for example and in practically any physical chemistry or biophysical chemistry textbook.

It seems to me - am I wrong? - that in very many cases of interest, at least of biochemical interest, although the solutions are not ideal, in the most usual situation where the overall ionic strength, or concentration of charged buffer and salts is very much higher than that of the reactant participants, the activity coefficients are constant with the result that forgetting about them and treating activity of the equilibrium participants of interest as concentration is not misleading even for equilibria between charged species.

 

[kotreny]

Thank you for your help, guys. So let me get this straight. If we double the definition of the standard state concentration, then the activity of a given concentration is halved? Then by this formula (taken from DrDu's link):

$$\mu_i = \mu_i^{\ominus} + RT\ln{a_i}$$

and assuming that the chemical potential does not depend on the definition of standard state, we deduce that RTln 2 is added to the standard chemical potential. So there's an equation relating standard chemical potential and standard concentration, and it involves ln(c^o). But it also appears to involve T, which, if I'm not mistaken, is not the standard temperature? How can this be, and what is the complete equation for standard chemical potential? You can make various assumptions for the equation, just list them.

 

[DrDu]

In general, halving of the standard state concentration will not lead to an RT ln 2 term as
a(c/2) is not equal to a(c)/2. This only holds for ideal gasses and solutions.

Also take in mind that activity a is also a function of T in general.

 

[kotreny]

In general, halving of the standard state concentration will not lead to an RT ln 2 term as a(c/2) is not equal to a(c)/2. This only holds for ideal gasses and solutions.

For now, let's focus on ideal solutions please. Let me state my problem more clearly. If we define the standard state concentration as 2 M instead of 1 M, then by

$$a_i = \gamma_{c,i} c_i/c^{\ominus}\,$$

we see that the activity is halved, since $$c^{\ominus}\,$$ has doubled from 1 molar to 2. Now, by

$$\mu_i = \mu_i^{\ominus} + RT\ln{a_i}$$

we see that dividing $$a_i$$ by 2 must result in

$$\mu_i = \mu_i^{\ominus} + RT\ln{(a_i/2)}$$

$$\mu_i = \mu_i^{\ominus} + RT\ln{a_i} - RT\ln{2}$$

However, I presume that $$\mu_i$$ and T are independent of the standard state concentration, while $$\mu_i^{\ominus}$$ takes on a new value. Therefore, to keep $$\mu_i$$ the same for both sets of equations,

$$\mu_i = \mu_i^{\ominus} (new) + RT\ln{a_i} - RT\ln{2} = \mu_i^{\ominus} + RT\ln{a_i}$$

we must have

$$\mu_i^{\ominus} (new) = \mu_i^{\ominus} + RT\ln{2}$$

which means doubling $$c^{\ominus}$$ must have the effect of adding RT ln 2 to $$\mu_i^{\ominus}$$. But this is very strange, since T should represent the temperature of the system in consideration, which shouldn't seem to have any bearing on the value of mu at standard state. What's wrong with my reasoning here?

 

[DrDu]

Mu standard is also a function of temperature. It's dependence on temperature is given by the van't Hoff equation.

 

[kotreny]

Thank you for taking the time to assist me, DrDu. I'm afraid I still don't understand. It would make sense if mu standard were a function of standard temperature, but the symbol $$T$$ in my previous post represents the temperature corresponding to nonstandard conditions. Mu standard may very well depend on the value of standard concentration, standard temperature, standard pressure, etc., but how can it depend on arbitrarily selected, nonstandard temperature? The most likely scenario I can think of is that mu nonstandard depends on the standard state concentration in some way, though this feels bizarre.

 

[DrDu]

The Delta G^0 or mu^0 in the formulas we are talking about refer to standard concentrations but not to some standard temperature but to the actual temperature.
However, they can be calculated from the corresponding values at standard temperature using van't Hoff equation or similar relations

 

[kotreny]

Ah, I get it. My primary misconception was that the standard state conditions had to include a temperature, such as 298K. There is actually no standard temperature here. Mu^o is the chemical potential when the substance is at standard concentration and the selected T.

Say, this appears to be consistent with the formula

$$\mu_i^{\ominus} = RT\ln{c^{\ominus}}$$

if the solution is ideal. Is this correct?

 

[DrDu]

Ah, I get it. My primary misconception was that the standard state conditions had to include a temperature, such as 298K. There is actually no standard temperature here. Mu^o is the chemical potential when the substance is at standard concentration and the selected T.

Say, this appears to be consistent with the formula

$$\mu_i^{\ominus} = RT\ln{c^{\ominus}}$$

if the solution is ideal. Is this correct?

This formula is certainly not correct, as the argument of the logarithm is not dimensionless.

 

[kotreny]

This formula is certainly not correct, as the argument of the logarithm is not dimensionless.

Oh right. We just need the formula mu=mu^o + RT ln a. Besides, if c^o were 1, then my formula would make mu^o zero, which can't always be right. Thanks for everything.