# 0.23 Free energy and thermodynamic equilibrium  (Page 2/6)

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## Foundation

We have come a long way to reach this point in our studies, so we have a substantial foundation to build on. We know all the elements of the Atomic Molecular Theory, including the models for molecular structure and bonding, and we have developed the postulates of the Kinetic Molecular Theory. We have observed and defined phase transitions and phase equilibrium, and we have also observed equilibrium in a variety of reaction systems, including acids and bases. We will assume an understanding of the energetics of chemical reactions, including the idea of a “state function” and the concept of Hess’ Law.

In the previous study, we developed the expression ∆S-∆H/T>0 as a predictor of whether a process will be spontaneous. Chemists generally rewrite this inequality in a different and somewhat more convenient form by multiplying both sides by –T :

-T∆S+∆H<0

This is the same inequality, but take notice of the “less than” sign which replaced the “greater than” sign as a result of multiplying by T.

Both terms in this inequality have units of energy, so ∆H - T∆S is often referred to as an energy. However, it is important to remember that this combination of terms arises from calculating the changes in entropy of the system and its surroundings, and the “less than” sign arises from the Second Law of Thermodynamics which deals with entropy and not energy. As such, chemists refer to this combination of terms as the “free energy” rather than the energy. Specifically, we define a new term as G = H – TS, which is also called the “Gibbs free energy.” If the temperature of a process is constant, then for that process ∆G = ∆H - T∆S. This means that

∆G<0 for any spontaneous process at constant T and P.

This might look and sound new and complicated, but this is just the Second Law of Thermodynamics in a succinct and convenient form. To see this, let’s recall that

∆S universe = ∆S system + ∆S surroundings = ∆S - ∆H/T

If we multiply this equation by –T , we get

-T∆S universe = -T∆S + ∆H = ∆G

These equations show that the change in the free energy is just the change in the entropy of the universe multiplied by –T . Since S universe >0, then ∆G<0. As such, when you see G for “free energy,” you might find it convenient to think of it as the “negative entropy of the universe.” For that reason, during a spontaneous process, the free energy of the system always decreases. We will use this notation alternately with the entropy notation.

## Observation 1: thermodynamic description of phase equilibrium

In the previous study, we carefully analyzed the freezing of water as an interesting example of a spontaneous process for which ∆Sº <0 but for which ∆S° - ∆H°/ T>0. The discussion above tells us that, when the freezing of water is at equilibrium, then ∆S° - ∆H°/T = 0. We should be able to solve this equation for T to predict the temperature at 1 atm when liquid water and solid water are in equilibrium. We can calculate ∆ = liquid - solid = 22.0 J/mol·K, and ∆Hº = ∆Hº f (liquid) - ∆Hº f (solid) = 6.01 kJ/mol. When we solve for T we get T = 273 K, which is of course the melting point temperature of water. This calculation shows the power of our predictions from the Second Law of Thermodynamics. Not only can we determine what processes are spontaneous, but we also determine the conditions necessary for a process to come to equilibrium!

#### Questions & Answers

find the 15th term of the geometric sequince whose first is 18 and last term of 387
Jerwin Reply
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
virgelyn Reply
hmm well what is the answer
Abhi
how do they get the third part x = (32)5/4
kinnecy Reply
can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
hmm
Abhi
is it a question of log
Abhi
🤔.
Abhi
I rally confuse this number And equations too I need exactly help
salma
But this is not salma it's Faiza live in lousvile Ky I garbage this so I am going collage with JCTC that the of the collage thank you my friends
salma
Commplementary angles
Idrissa Reply
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
hii
Uday
hi
salma
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
Kevin Reply
a perfect square v²+2v+_
Dearan Reply
kkk nice
Abdirahman Reply
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
Kim Reply
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
Embra Reply
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
Nancy Reply
rolling four fair dice and getting an even number an all four dice
ramon Reply
Kristine 2*2*2=8
Bridget Reply
Differences Between Laspeyres and Paasche Indices
Emedobi Reply
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
Mary Reply
how do you translate this in Algebraic Expressions
linda Reply
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
Crystal Reply
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
Chris Reply
what's the easiest and fastest way to the synthesize AgNP?
Damian Reply
China
Cied
types of nano material
abeetha Reply
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials​ and their applications of sensors.
Ramkumar Reply
what is nano technology
Sravani Reply
what is system testing?
AMJAD
preparation of nanomaterial
Victor Reply
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
good afternoon madam
AMJAD
what is system testing
AMJAD
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
can nanotechnology change the direction of the face of the world
Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
bamidele Reply
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
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Source:  OpenStax, Concept development studies in chemistry 2013. OpenStax CNX. Oct 07, 2013 Download for free at http://legacy.cnx.org/content/col11579/1.1
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 By By Mistry Bhavesh