Chem 112 Class Guide: GIBBS FREE ENERGY
Chapter 19, Sections 5, 6 and 7 Learning Goals:
Upon completion of Chapter 19, Sections 1-4, you should be able to determine the following:
Calculate ΔG°rxn given appropriate variables
Calculate ΔG°rxn given ΔH° and ΔS°
Predict whether a reaction is spontaneous or not based on the sign of ΔG°
Calculate the temperature at which a reaction becomes spontaneous
Calculate ΔG from ΔG°
Calculate K from ΔG°
Chapter Reading Guide: Chapter 19, Sections 5, 6 and 7
Section 5: Gibbs Free Energy
Read Chapter 19.5
American scientist J. Willard Gibbs proposed that the spontaneity of a reaction can be predicted
if the enthalpy AND entropy changes of a chemical reaction are known. He proposed that:
G o  H o  T S o
Where ΔG° is Gibbs Free energy, ΔH° is the enthalpy change in a reaction and ΔS° is the
entropy change for a reaction.
The sign of ΔG° then tells us if a reaction is spontaneous or non-spontaneous
ΔG° > 0 means that the reaction is non-spontaneous as written
ΔG° < 0 means that the reaction is spontaneous as written
ΔG° = 0 means that the reaction is at equilibrium
ΔG° is always less than 0 for a spontaneous process carried out at constant temperature and
pressure.
Given the following general reaction:
aA + bB  cC + dD
The change in entropy can be calculated as follows:
o
Grxn
 (cGC  d GD )  (aGA  bGB )
ΔG° values can be found in Appendix C. Pay attention to phases when choosing the correct
value!
Example:
Calculate ΔGrxn for the following reaction, using ΔG values found in Appendix C:
C2H4 (g) + 3 O2 (g)  2 CO2 (g) + 2 H2O (g)
Grxn   2 mol  G CO2   2 mol  G H 2O   1 mol  G C2 H 4   3 mol  G O2 

kJ 

Grxn   2 mol  394.4
  2 mol
mol 


kJ   
kJ 


 kJ  
 228.57
   1 mol  68.11
  3 mol  0

mol   
mol 


 mol  
Grxn   788.8 kJ  457.14 kJ   68.11 kJ   1314.1 kJ
Example:
Calculate ΔGrxn for the following reaction at 25 °C, if ΔH° = -1111.5 kJ and ΔS° = -29.5 J/K.
C2H4 (g) + 3 O2 (g)  2 CO2 (g) + 2 H2O (g)
Grxn  H  T S
kJ

Grxn  1111.5 kJ   298 K   0.0295
K


  1111.5 kJ + 8.79 kJ = -1102.7 kJ

Note that the ΔGrxn values we calculated for the same reaction using these two methods are
close, but not the same. We would have to run an experiment to determine the actual
value!
Try Practice exercise 19.6
Try Practice exercise 19.7
Try Practice exercise 19.8
Section 6: Free Energy and Temperature
Read Chapter 19.6
You don’t need to know the value of ΔH° or ΔS° to predict if a reaction is spontaneous or
not; merely knowing the signs of the two values is plenty of information. Look at the
following table:
ΔH ΔS ΔG
Spontaneous or No?
+
+
+ or -, depends Spontaneous at High Temperatures
Non-Spontaneous at Low Temperatures
+
+
Non-Spontaneous at ALL Temperatures
+
Spontaneous at ALL Temperatures
+ or -, depends Spontaneous at Low Temperatures
Non-Spontaneous at High Temperatures
Given the values of ΔH° and ΔS° and ΔG°, you can predict at what temperature a reaction
becomes spontaneous. Since you know that when ΔG° is 0, the reaction is at equilibrium, it
makes sense that when ΔG° is 0, the reaction will “flip” spontaneity (ie, go from
spontaneous to not or vice versa). However, when performing these calculations, be wary –
G and H are generally given in kJ while S is given in J!
Try Practice exercise 19.9
Section 7: Free Energy and K
Read Chapter 19.7
ΔG° vs. ΔG
ΔG° means that the reaction is performed under STANDARD CONDITIONS – 298 K and 1 atm.
However, most chemical reactions do not occur at exactly that temperature, so we use the
following conversion:
G o  G  RT ln Q
Where R is 8.314 J/mol·K, T is the temperature and Q is the reaction quotient (recall section
15.6)
Example:
Calculate ΔG at 30 °C for the following reaction, if there are 2.00 atm of C2H4, 3.00 atm of O2,
and 4.00 atm each of CO2 and H2O, and ΔG° = 1102.7 kJ.
C2H4 (g) + 3 O2 (g)  2 CO2 (g) + 2 H2O (g)
 P   P    4.00  4.00
Q
 P  P   2.00 3.00
2
CO2
2
3
C2 H 4
2
2
H 2O
3
 4.74
O2
G o  G  RT ln Q So G  G  RT ln Q
J


G  1.1x106   8.314
  303 K  ln  4.74 
mol
*
K


G  1.1x106 J  3919.9 J  1.096 x106 J  1096 kJ
ΔG° and K
Given the above equation, when a reaction is at equilibrium, ΔG° is 0 and Q = K. That
equation then rearranges to give:
G o   RT ln K
Example:
Calculate ΔG° at 45 °C for the following reaction, if there are 1.67 atm of C2H4, 1.00 atm of
O2, and 2.67 atm each of CO2 and H2O in a reaction at equilibrium.
C2H4 (g) + 3 O2 (g)  2 CO2 (g) + 2 H2O (g)
 P   P    2.67   2.67 
K
 P  P  1.67 1.00
2
CO2
2
3
C2 H 4
2
H 2O
3
2
 30.4
O2
G o   RT ln K
J
J
kJ


G    8.314
 9.03
  318 K  ln 30.4   9027
mol * K 
mol
mol

Try Practice exercise 19.10
Try Practice exercise 19.11
Try Practice exercise 19.12
Learning Resources
Chapter Learning Goals
Chapter 19, Sections 5 through 7 Learning Goals
Pre Class Assignment: This assignment must be completed prior to the next class. Check your syllabus
for the exact due date and time.
Complete to the pre class assignment
(http://berks.psu.edu/clt/chem112/GibbsFreeEnergy_HW.docx)
Submit a copy to the dropbox located in ANGEL called “Pre Class Assignment Submission: Gibbs
Free Energy”
End of Chapter Problems:
Practice with these problems if you are having difficulty with any of the concepts covered in this
class guide AFTER we have met in class. If you cannot easily complete these problems, seek
help from your instructor, your mentor or the learning center
Chapter 19: 57, 59, 61, 63, 69, 79, 81