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16.4 Free energy  (Page 4/12)

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Δ G = Δ H − T Δ S
y = b + m x

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:

Δ G = 0 = Δ H − T Δ S
T = Δ H 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.

A graph is shown where the y-axis is labeled, “Free energy,” and the x-axis is labeled, “Increasing temperature ( K ).” The value of zero is written midway up the y-axis with the label, “delta G greater than 0,” written above this line and, “delta G less than 0,” written below it. The bottom half of the graph is labeled on the right as, “Spontaneous,” and the top half is labeled on the right as, “Nonspontaneous.” A green line labeled, “delta H less than 0, delta S greater than 0,” extends from a quarter of the way up the y-axis to the bottom right of the graph. A yellow line labeled, “delta H less than 0, delta S less than 0,” extends from a quarter of the way up the y-axis to the middle right of the graph. A second yellow line labeled, “delta H greater than 0, delta S greater than 0,” extends from three quarters of the way up the y-axis to the middle right of the graph. A red line labeled, “delta H greater than 0, delta S less than 0,” extends from three quarters of the way up the y-axis to the top right of the graph.
These plots show the variation in Δ G with temperature for the four possible combinations of arithmetic sign for Δ H and Δ S .

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:

H 2 O ( l ) ⟶ H 2 O ( g )

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

0 = Δ H ° − T Δ S ° or T = Δ H ° Δ S °

Using the standard thermodynamic data from Appendix G ,

Δ H ° = Δ H f ° ( H 2 O ( g ) ) − Δ H f ° ( H 2 O ( l ) ) = − 241.82 kJ/mol − ( − 285.83 kJ/mol ) = 44.01 kJ/mol
Δ S ° = Δ S 298 ° ( H 2 O ( g ) ) − Δ S 298 ° ( H 2 O ( l ) ) = 188.8 J/K·mol − 70.0 J/K·mol = 118.8 J/K·mol
T = Δ H ° Δ S ° = 44.01 × 10 3 J/mol 118.8 J/K·mol = 370.5 K = 97.3 °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)

Got questions? Get instant answers now!

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

Three charges q_{1}=+3\mu C, q_{2}=+6\mu C and q_{3}=+8\mu C are located at (2,0)m (0,0)m and (0,3) coordinates respectively. Find the magnitude and direction acted upon q_{2} by the two other charges.Draw the correct graphical illustration of the problem above showing the direction of all forces.
Kate Reply
To solve this problem, we need to first find the net force acting on charge q_{2}. The magnitude of the force exerted by q_{1} on q_{2} is given by F=\frac{kq_{1}q_{2}}{r^{2}} where k is the Coulomb constant, q_{1} and q_{2} are the charges of the particles, and r is the distance between them.
Muhammed
What is the direction and net electric force on q_{1}= 5µC located at (0,4)r due to charges q_{2}=7mu located at (0,0)m and q_{3}=3\mu C located at (4,0)m?
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what is the change in momentum of a body?
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please solve
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8m/s²
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What is Thermodynamics
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velocity can be 72 km/h in question. 72 km/h=20 m/s, v^2=2.a.x , 20^2=2.a.50, a=4 m/s^2.
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A boat travels due east at a speed of 40meter per seconds across a river flowing due south at 30meter per seconds. what is the resultant speed of the boat
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50 m/s due south east
Someone
which has a higher temperature, 1cup of boiling water or 1teapot of boiling water which can transfer more heat 1cup of boiling water or 1 teapot of boiling water explain your . answer
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I believe temperature being an intensive property does not change for any amount of boiling water whereas heat being an extensive property changes with amount/size of the system.
Someone
Scratch that
Someone
temperature for any amount of water to boil at ntp is 100⁰C (it is a state function and and intensive property) and it depends both will give same amount of heat because the surface available for heat transfer is greater in case of the kettle as well as the heat stored in it but if you talk.....
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about the amount of heat stored in the system then in that case since the mass of water in the kettle is greater so more energy is required to raise the temperature b/c more molecules of water are present in the kettle
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field is a region of space under the influence of some physical properties
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No. According to Isac Newtons law. this two bodies maybe you and the wall beside you. Attracting depends on the mass och each body and distance between them.
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Are you really asking if two bodies have to be charged to be influenced by Coulombs Law?
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like charges repel while unlike charges atttact
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What is specific heat capacity
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Specific heat capacity is a measure of the amount of energy required to raise the temperature of a substance by one degree Celsius (or Kelvin). It is measured in Joules per kilogram per degree Celsius (J/kg°C).
AI-Robot
specific heat capacity is the amount of energy needed to raise the temperature of a substance by one degree Celsius or kelvin
ROKEEB
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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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