# Purpose of natural log in gibbs free energy equation

1. Feb 25, 2013

### thedy

Hi,Im just beginner and I m trying to learn integrals.I m just in starting phase,but still in few tences,not details...How or why we get logarithm in gibbs free energy equation?Because of integration of this equation or due to probability and statistics laws?
Thanks

2. Feb 25, 2013

### DrDu

Maybe you could provide the equation you are referring too?

3. Feb 25, 2013

### thedy

deltaG=-RTlnK
But I mean in general,why there is a ln.And how we get this.By integrating(for now,I don t need why we use integrating)or due to probability(Boltzmann and etc....)
Thanks

4. Feb 25, 2013

### DrDu

This term can be traced back to entropy.
In its easiest form the entropy for a system is some function of $\Omega$ where $\Omega$ is the numer of states available to a system. Now if you are combining two sub systems, with possible number of states $\Omega_1$ and $\Omega_2$, as any state of one system can be combined with any state of the other, $\Omega=\Omega_1\Omega_2$.
But we know that entropy is an extensive quantity, i.e. doubling a system should double the entropy. To achieve this, entropy has to depend on the logarithm of $\Omega$ as the logarithm of a product becomes the sum of the logarithms. This carries over to $\Delta G=\Delta H-T\Delta S$ due to the second term.

5. Feb 25, 2013

### thedy

Thanks,so it has nothing to do with integration...Right?

6. Feb 25, 2013

### Staff: Mentor

In your equation is ΔG the standard change in gibbs free energy for a reaction and K the equilibrium constant for the reaction? Is that what your symbols refer to?

7. Feb 26, 2013

### thedy

Yes,so am I right?Nothing to do with integration?

8. Feb 26, 2013

### DrDu

Depends on what you are integrating.

9. Feb 26, 2013

### thedy

Thanks,but what do you mean with this?

10. Feb 26, 2013

### DrDu

In the easiest case of ideal gasses $\Delta G=\sum_i \nu_i \mu_i=\sum_i \nu_i \mu_i^0+RT \sum_i \ln (p_i/p_i^0)=\Delta G_0+RT (\ln \Pi_i p_i /p_i^0)$. Here $\mu_i$ is the chemical potential of compound i, $\nu_i$ the stochiometric coefficient of compound i in the reaction considered and $p_i$ its pressure ($p_i^0$ the normal pressure, 1 Atm).
Now this relation can be obtained by integrating e.g. $d\mu_i=V_i dp_i$ from $p_i^0$ to $p_i$ with $V_i=RT/p_i$ being the molar volume. So you can derive this formula using steps which include integration if you want. However I would not say that integration is the "reason" for the appearance of the logarithm.

Last edited: Feb 26, 2013
11. Feb 26, 2013

### thedy

OK,so what is the actual reason of this logarithm be in this equation?What inspired physicists?

12. Feb 26, 2013

### ModusPwnd

I always thought the log was used because it has the property: ln(x*y) = ln(x) + ln(y)

13. Feb 26, 2013

### Staff: Mentor

There are different ways of looking at this. One is from the standpoint of statistical thermodynamics, and the other is from the standpoint of classical thermodynamics. From the standpoint of classical thermodynamics, if G0 is the free energy of formation of a substance in a standard state p0, T0, and G is the free energy of the substance at the same temperature and pressure p, then
$$G(p,T_0)=G_0+\int_{p_0}^pVdp$$
This comes directly from the thermodynamic relationship
dG=-SdT+VdP

If the gas is an ideal gas,
$$G(p,T_0)=G_0+RT_0\ln{\frac{p}{p_0}}$$
If the pressure is expressed in atmospheres, and, if the reference state is p0= 1atm, then
$$G(p,T_0)=G_0+RT_0\ln p$$
If you are dealing with a mixture of ideal gases, such that G in the above equation is the partial molar free energy (aka the chemical potential), then the pressure in the above equation is the partial pressure of a species.

In obtaining the equation for the equilibrium constant in terms of the molar free energies of formation of the reactants and products, you add the free energies stoichiometrically. But at equalibrium, the overall change in the free energy is zero. So this leaves you with the equation:

$$RT_0\ln K_p=-\Delta G_0$$

From all this you can see that an integration step was indeed involved in obtaining the final equation for an ideal gas (at least in the classical development). Incidentally, in your equation, you left out a minus sign in front of the change in standard free energy for the reaction.

14. Mar 3, 2013

### thedy

OK,thanks to all for answers....:)