# 16.4 Free energy  (Page 4/12)

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$\text{Δ}G=\text{Δ}H-T\text{Δ}S$
$y=b+mx$

Such a plot is shown in [link] . A process whose enthalpy and entropy changes are of the same arithmetic sign will exhibit a temperature-dependent spontaneity as depicted by the two yellow lines in the plot. Each line crosses from one spontaneity domain (positive or negative Δ G ) to the other at a temperature that is characteristic of the process in question. This temperature is represented by the x -intercept of the line, that is, the value of T for which Δ G is zero:

$\text{Δ}G=0=\text{Δ}H-T\text{Δ}S$
$T=\phantom{\rule{0.2em}{0ex}}\frac{\text{Δ}H}{\text{Δ}S}$

And so, saying a process is spontaneous at “high” or “low” temperatures means the temperature is above or below, respectively, that temperature at which Δ G for the process is zero. As noted earlier, this condition describes a system at equilibrium.

## Equilibrium temperature for a phase transition

As defined in the chapter on liquids and solids, the boiling point of a liquid is the temperature at which its solid and liquid phases are in equilibrium (that is, when vaporization and condensation occur at equal rates). Use the information in Appendix G to estimate the boiling point of water.

## Solution

The process of interest is the following phase change:

${\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}\text{O}\left(g\right)$

When this process is at equilibrium, Δ G = 0, so the following is true:

$0=\text{Δ}H\text{°}-T\text{Δ}S\text{°}\phantom{\rule{5em}{0ex}}\text{or}\phantom{\rule{5em}{0ex}}T=\phantom{\rule{0.2em}{0ex}}\frac{\text{Δ}H\text{°}}{\text{Δ}S\text{°}}$

Using the standard thermodynamic data from Appendix G ,

$\begin{array}{ccc}\hfill \text{Δ}H\text{°}& =& \text{Δ}{H}_{\text{f}}^{°}\left({\text{H}}_{2}\text{O}\left(g\right)\right)\phantom{\rule{0.2em}{0ex}}-\text{Δ}{H}_{\text{f}}^{°}\left({\text{H}}_{2}\text{O}\left(l\right)\right)\hfill \\ & =& -\text{241.82 kJ/mol}-\left(-\text{285.83 kJ/mol}\right)=\text{44.01 kJ/mol}\end{array}$
$\begin{array}{ccc}\hfill \text{Δ}S°& =& \text{Δ}{S}_{298}^{°}\left({\text{H}}_{2}\text{O}\left(g\right)\right)\phantom{\rule{0.2em}{0ex}}-\text{Δ}{S}_{298}^{°}\left({\text{H}}_{2}\text{O}\left(l\right)\right)\hfill \\ & =& \text{188.8 J/K·mol}-\text{70.0 J/K·mol}=\text{118.8 J/K·mol}\hfill \end{array}$
$T=\phantom{\rule{0.2em}{0ex}}\frac{\text{Δ}H\text{°}}{\text{Δ}S\text{°}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{44.01\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{J/mol}}{118.8\phantom{\rule{0.2em}{0ex}}\text{J/K·mol}}\phantom{\rule{0.2em}{0ex}}=370.5\phantom{\rule{0.2em}{0ex}}\text{K}=97.3\phantom{\rule{0.2em}{0ex}}\text{°C}$

The accepted value for water’s normal boiling point is 373.2 K (100.0 °C), and so this calculation is in reasonable agreement. Note that the values for enthalpy and entropy changes data used were derived from standard data at 298 K ( Appendix G ). If desired, you could obtain more accurate results by using enthalpy and entropy changes determined at (or at least closer to) the actual boiling point.

## Check your learning

Use the information in Appendix G to estimate the boiling point of CS 2 .

## Answer:

313 K (accepted value 319 K)

## Free energy and equilibrium

The free energy change for a process may be viewed as a measure of its driving force. A negative value for Δ G represents a finite driving force for the process in the forward direction, while a positive value represents a driving force for the process in the reverse direction. When Δ G is zero, the forward and reverse driving forces are equal, and so the process occurs in both directions at the same rate (the system is at equilibrium).

In the chapter on equilibrium the reaction quotient , Q , was introduced as a convenient measure of the status of an equilibrium system. Recall that Q is the numerical value of the mass action expression for the system, and that you may use its value to identify the direction in which a reaction will proceed in order to achieve equilibrium. When Q is lesser than the equilibrium constant, K , the reaction will proceed in the forward direction until equilibrium is reached and Q = K . Conversely, if Q < K , the process will proceed in the reverse direction until equilibrium is achieved.

#### Questions & Answers

what is a balanced equation 4 trioxonitrate (V)acid and sodium hydroxide?
Marcel Reply
proved ur Worth: If A is a of trioxonitrate(V)acid,HNO3' of unknown concentration .B is a standard solution of sodium hydroxide containing 4.00g per dm cube of solution.25cm cube portions solution B required an average of 24.00cm cube of solution A for neutralization,using 2 drops of methyl orange.
Marcel
calculate the concentration of solution B in moles per dm cube
Marcel
calculate the concentration of solution A and B in moles per DM cube
Marcel
finally calculate the concentration in g/dm cube of HNO3 in solution A (H=1,N=14,O=16,Na=23)
Marcel
wat is electrolysis?
Mgbachi Reply
it is the chemical decomposition of a substance when electric current is passed through it either in molten form or aqueous solution
Nuru
list the side effect of chemical industries
Chelsea Reply
how do you ionise an atom
Rabeka Reply
many ways ,but one of them is when the atom becomes heated to a certain temperature the surface electron becomes too energetic and leaves the atom because the attraction between the nucleus and the electron becomes overpowered by the energetic eletron
sunday
also hitting of two atoms can cause transfer of surface electrons
sunday
and when this transfers occur the atom becomes ionised
sunday
who is doing Cape chemistry tomorrow?
caramel Reply
What is hybridization
edmondnti Reply
the mix between different breeds of species in one
Jared
it is the blending of orbitals.
stanley
the mixing of orbital
caramel
are covalent bonds influenced by factors such as temperature and pressure?
patrick Reply
what is catalyst used for mirror test
Sanjay Reply
when an atom looses electron, what does it become?
Abdullahi Reply
it's oxidized and called an ion
Anora
thanks
Abdullahi
Now, I get it
Abdullahi
cation
Anora
can you give an example please, if you don't mind
Abdullahi
a positive ion,become positively charged/a cation.
Janis
sodium plus one is simple cation is exmpl
ajmal
Taking Sodium as example..... it carries a positive charge which means it is positively charged.....when it gains an electron, it is reduced cuz an electron is negatively charged.....also when an atom looses an electron, it becomes positively charged and when it gains, it becomes negatively charged.
Nuru
typically, ionization is the process where an atom looses or gains electron(s) to form ion(s) either a positively or negatively
Nuru
what is copper
Bryan Reply
just an element
Power
Cu
daniel
Why is water a single covalent bond?
Mohamed Reply
nitrogen is a gas whereas phosphorus is solid .Explain.
Jacky Reply
can you explain what you are needing it now better than maybe I'm just not interpreting it what you're needing to know
Alex
cool nitrogen down to around negative 270 °F and it will be solid. now they are both solid
daniel
they are different elements and dats how they are pal.....check the periodic table
Nuru
Nitrogen is a diatomic molecule with relatively weak van de waals forces between the molecules. These forces are overcome when the solid melts or liquid evaporates. Phosphorus forms larger molecules consisting of four phosphorus atoms in a tetradedral shape. The intermolecular forces are stronger
Paul
whats a base
Daksalma Reply
A base is a substance which will neutralize an acid to yield salt and water only
Zainab
base is a substance that produces OH(aq) ions in aqueous solution. Strong soluable bases are in water and are completely dislocated. Therefore weak base ionize slightly...
Roy
a base is a substance that neutralise and acid to form salt and water
Daksalma
write electrolysis of bright solution using either carbon or platinum and write the reaction at the anode or at the cathode
Abdullah Reply
what is the H3O of a solution with the pH of 2.5
Sgt.Elliott_98 Reply
pH<7, therefore there are only H3O+HX3OX+particles in the solution. [H3O+]=10−pH=10−6.99=1.02⋅10−7[HX3OX+]=10−pH=10−6.99=1.02⋅10−7 When the pH is smaller than 6 or greater than 8, one will not notice the difference, but here it is logarithmically speaking  and I'll give you another one if this is ki
Alex
if I'm answering and interpreting what you're asking correctly
Alex
When the pH is smaller than 6 or greater than 8, one will not notice the difference, but here it is logarithmically speaking
Alex
sorry I don't know why that sent again
Alex
We have [H3O+]=10−pH=10−6.99=1.02⋅10−7[HX3OX+]=10−pH=10−6.99=1.02⋅10−7 and [OH−]=10−pOH=10−7.01=9.77⋅10−8[OHX−]=10−pOH=10−7.01=9.77⋅10−8.  Because of H3O++OH−⟶2H2OHX3OX++OHX−⟶2HX2O we are left with [H3O+]=1.02⋅10−7−9.77⋅10−8=4.6⋅10−9
Alex

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Source:  OpenStax, Chemistry. OpenStax CNX. May 20, 2015 Download for free at http://legacy.cnx.org/content/col11760/1.9
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