If you like us, please share us on social media.
The latest UCD Hyperlibrary newsletter is now complete, check it out.

MindTouch
http://mindtouch.com

This file and accompanying files are licensed under the MindTouch Master Subscription Agreement (MSA).

At any time, you shall not, directly or indirectly: (i) sublicense, resell, rent, lease, distribute, market, commercialize or otherwise transfer rights or usage to: (a) the Software, (b) any modified version or derivative work of the Software created by you or for you, or (c) MindTouch Open Source (which includes all non-supported versions of MindTouch-developed software), for any purpose including timesharing or service bureau purposes; (ii) remove or alter any copyright, trademark or proprietary notice in the Software; (iii) transfer, use or export the Software in violation of any applicable laws or regulations of any government or governmental agency; (iv) use or run on any of your hardware, or have deployed for use, any production version of MindTouch Open Source; (v) use any of the Support Services, Error corrections, Updates or Upgrades, for the MindTouch Open Source software or for any Server for which Support Services are not then purchased as provided hereunder; or (vi) reverse engineer, decompile or modify any encrypted or encoded portion of the Software.

A complete copy of the MSA is available at http://www.mindtouch.com/msa

Thermodynamics of Mixing

The molar Gibbs free energy of mixing is

$\Delta G_{m,mix} = RT(x_i) \ln(x_i )$

or for a two component solution

$\Delta G_{m,mix} = nRT(x_A\ln x_A + x_B \ln x_B)$

and for more than two components

$\Delta G_{m,mix} = nRT(x_A \ln x_A + x_B \ln x_B+ x_c \ln x_c+ ...)$

For the first equation, $$m$$ denotes Gibbs free energy per mole of solution, and $$x_i$$ is the mole fraction of component $$i$$ in the liquid state. Note that when calculating the free energy of mixing in an ideal solution is always negative, since each $$\ln x_i$$ term becomes be negative value. Thus showing how ideal solutions are always completely miscible.

How the Equations Calculate Gibbs Free Energy For Ideal Solutions

The origins of the m value comes from the equation of partial molar quantities which is the combination of the first and second laws of thermodynamics's equations for an open system. Note that these equations are expressed in terms of chemical potentials of their individual components.

First law's:

$dU = (dq) - p(dV) = (dq) + w$

Second law's:

$dq_{rev}= T(dS)$

$$dq$$ represents the change in enthalpy or heat, $$w$$ is work, $$dS$$ is the change in entropy, $$T$$ is temperature, and $$dU$$ is the changes of internal energy. By combining these two equations, integrating them, and placing in the number of moles for a solution, the new formula becomes

$dU = TdS = pdV + m_1\,dn_1+m_2\,dn_2$

where $$m$$ came from was the replacement of dU divided by dn (the change of moles). However it can also be $$dG$$ (the change in Gibbs Free energy) divided by $$dn$$. By using the integrated form of $$G$$:

$G = m_1n_1+m_2n_2+...$

one can find the Gibbs free energy of the ideal mixture for multiple components:

$\Delta G_{m,mix} = nRT(x_A \ln x_A + x_B \ln x_B+x_c \ln x_c+...)$

Ideal Vapor Pressure Relationship

Raoult's Law says that $$P_A$$,the experimental partial pressure of the solvent, $$x_A$$ is the mole fraction of the solvent, and $$P^*_A$$ is the vapor pressure of the pure solvent at the given temperature.

$P_A = x_AP^*_A$

For ideal solutions, the entire composition range of temperature and concentration and shows no internal energy change on mixing and no attractive force between components. The equilibrium vapor pressure of the pure component is the mole fraction of the component in solution. It can also be shown that volumes are strictly additive for ideal solutions.

Why is Raoult's law applicable for ideal solutions?

1. mi = m*i+ RT ln Xi                        (formula for chemical potentials only)
2. ml = mg           (the partial gibbs energy in both liquid and gas form is equal*)
3. m*l+ RT ln Xl = m^g + RT ln (P1/1 atm) (for the second, ideal conditions in gas form)
4. In liquid form, the ideal solution is pure, thus its vapor pressure is also pure. Replace m*l with m^g + RT ln (P*1/1 atm)
5. Subtract: [m^g + RT ln (P1/1 atm)] - [m^g + RT ln (P*1/1 atm)]
6. ln Xl = ln (P1/P*1)                    (get rid of natural logs)
7. p1 = XlP*1        (Raoult's law)

References

1. Petrucci, Harwood, Herring, Madura. General Chemistry: Principles & Modern Applications, Ninth Ed. Upper Saddle River, NJ: Pearson Education, Inc., 2007.
2. Salzman, W.R. "Mixtures; Partial Molar Quantities; Ideal Solutions". Chemical Thermodynamics. Last updated: 21 Oct 04. http://www.chem.arizona.edu/~salzmanr/480a/480ants/mixpmqis/mixpmqis.html
3. Petrucci, Herring, Madura, Bissonette. General Chemistry: Principles and Modern Applications, Tenth Ed. Upper Saddle River, NJ: Pearson Education, Inc., 2011.

Contributors

• Ryan Woods (UCD), Shelley Chu (UCD)

22:57, 27 Dec 2013

Classifications

(not set)
(not set)

Textbook Maps

An NSF funded Project