<< Chapter < Page Chapter >> Page >
By the end of this section, you will be able to:
  • Define Gibbs free energy, and describe its relation to spontaneity
  • Calculate free energy change for a process using free energies of formation for its reactants and products
  • Calculate free energy change for a process using enthalpies of formation and the entropies for its reactants and products
  • Explain how temperature affects the spontaneity of some processes
  • Relate standard free energy changes to equilibrium constants

One of the challenges of using the second law of thermodynamics to determine if a process is spontaneous is that we must determine the entropy change for the system and the entropy change for the surroundings. An alternative approach involving a new thermodynamic property defined in terms of system properties only was introduced in the late nineteenth century by American mathematician Josiah Willard Gibbs . This new property is called the Gibbs free energy change ( G )    (or simply the free energy ), and it is defined in terms of a system’s enthalpy and entropy as the following:

G = H T S

Free energy is a state function, and at constant temperature and pressure, the standard free energy change (Δ G °)    may be expressed as the following:

Δ G = Δ H T Δ S

(For simplicity’s sake, the subscript “sys” will be omitted henceforth.)

We can understand the relationship between this system property and the spontaneity of a process by recalling the previously derived second law expression:

Δ S univ = Δ S + q surr T

The first law requires that q surr = − q sys , and at constant pressure q sys = Δ H , and so this expression may be rewritten as the following:

Δ S univ = Δ S Δ H T

Δ H is the enthalpy change of the system . Multiplying both sides of this equation by − T , and rearranging yields the following:

T Δ S univ = Δ H T Δ S

Comparing this equation to the previous one for free energy change shows the following relation:

Δ G = T Δ S univ

The free energy change is therefore a reliable indicator of the spontaneity of a process, being directly related to the previously identified spontaneity indicator, Δ S univ . [link] summarizes the relation between the spontaneity of a process and the arithmetic signs of these indicators.

Relation between Process Spontaneity and Signs of Thermodynamic Properties
Δ S univ >0 Δ G <0 spontaneous
Δ S univ <0 Δ G >0 nonspontaneous
Δ S univ = 0 Δ G = 0 reversible (at equilibrium)

Calculating free energy change

Free energy is a state function, so its value depends only on the conditions of the initial and final states of the system that have undergone some change. A convenient and common approach to the calculation of free energy changes for physical and chemical reactions is by use of widely available compilations of standard state thermodynamic data. One method involves the use of standard enthalpies and entropies to compute standard free energy changes according to the following relation as demonstrated in [link] .

Δ G ° = Δ H ° T Δ S °

Evaluation of δ G ° change from δ H ° and δ S °

Use standard enthalpy and entropy data from Appendix G to calculate the standard free energy change for the vaporization of water at room temperature (298 K). What does the computed value for Δ G ° say about the spontaneity of this process?

Solution

The process of interest is the following:

H 2 O ( l ) H 2 O ( g )

The standard change in free energy may be calculated using the following equation:

Δ G 298 ° = Δ H ° T Δ S °

From Appendix G , here is the data:

Substance Δ H f ° (kJ/mol) S 298 ° (J/K·mol)
H 2 O( l ) −286.83 70.0
H 2 O( g ) −241.82 188.8

Combining at 298 K:

Δ H ° = Δ H 298 ° = Δ H f ° ( H 2 O ( g ) ) Δ H f ° ( H 2 O ( l ) ) = [ −241.82 kJ ( −285.83 ) ] kJ/mol = 44.01 kJ/mol
Δ S ° = Δ S 298 ° = S 298 ° ( H 2 O ( g ) ) S 298 ° ( H 2 O ( l ) ) = 188.8 J/mol·K 70.0 J/K = 118.8 J/mol·K
Δ G ° = Δ H ° T Δ S °

Converting everything into kJ and combining at 298 K:

Δ G 298 ° = Δ H ° T Δ S ° = 44.01 kJ/mol ( 298 K × 118.8 J/mol·K ) × 1 kJ 1000 J
44.01 kJ/mol 35.4 kJ/mol = 8.6 kJ/mol

At 298 K (25 °C) Δ G 298 ° > 0 , and so boiling is nonspontaneous ( not spontaneous).

Check your learning

Use standard enthalpy and entropy data from Appendix G to calculate the standard free energy change for the reaction shown here (298 K). What does the computed value for Δ G ° say about the spontaneity of this process?

C 2 H 6 ( g ) H 2 ( g ) + C 2 H 4 ( g )

Answer:

Δ G 298 ° = 102.0 kJ/mol ; the reaction is nonspontaneous ( not spontaneous) at 25 °C.

Got questions? Get instant answers now!

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
Aislinn Reply
cm
tijani
what is titration
John Reply
what is physics
Siyaka Reply
A mouse of mass 200 g falls 100 m down a vertical mine shaft and lands at the bottom with a speed of 8.0 m/s. During its fall, how much work is done on the mouse by air resistance
Jude Reply
Can you compute that for me. Ty
Jude
what is the dimension formula of energy?
David Reply
what is viscosity?
David
what is inorganic
emma Reply
what is chemistry
Youesf Reply
what is inorganic
emma
Chemistry is a branch of science that deals with the study of matter,it composition,it structure and the changes it undergoes
Adjei
please, I'm a physics student and I need help in physics
Adjanou
chemistry could also be understood like the sexual attraction/repulsion of the male and female elements. the reaction varies depending on the energy differences of each given gender. + masculine -female.
Pedro
A ball is thrown straight up.it passes a 2.0m high window 7.50 m off the ground on it path up and takes 1.30 s to go past the window.what was the ball initial velocity
Krampah Reply
2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.
Sahid Reply
you have been hired as an espert witness in a court case involving an automobile accident. the accident involved car A of mass 1500kg which crashed into stationary car B of mass 1100kg. the driver of car A applied his brakes 15 m before he skidded and crashed into car B. after the collision, car A s
Samuel Reply
can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
Joseph Reply
Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
Joseph
Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
Joseph
"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true
Ryan
what's motion
Maurice Reply
what are the types of wave
Maurice
answer
Magreth
progressive wave
Magreth
hello friend how are you
Muhammad Reply
fine, how about you?
Mohammed
hi
Mujahid
A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
yasuo Reply
Who can show me the full solution in this problem?
Reofrir Reply
Got questions? Join the online conversation and get instant answers!
Jobilize.com Reply
Practice Key Terms 3

Get Jobilize Job Search Mobile App in your pocket Now!

Get it on Google Play Download on the App Store Now




Source:  OpenStax, Chemistry. OpenStax CNX. May 20, 2015 Download for free at http://legacy.cnx.org/content/col11760/1.9
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Chemistry' conversation and receive update notifications?

Ask