Jump to a New ChapterIntroduction to the SAT IIIntroduction to the SAT II Chemistry TestStrategies for Taking the SAT II Chemistry TestThe Structure of MatterThe States of MatterReaction TypesStoichiometryEquilibrium and Reaction RatesThermodynamicsDescriptive ChemistryLaboratoryBasic Measurement and Calculation ReviewChemical Formulas Review: Nomenclature and Formula WritingPractice Tests Are Your Best Friends
 9.1 Enthalpy 9.2 Spontaneous Reactions 9.3 Heat Capacity and Specific Heat 9.4 Enthalpies of Reactions 9.5 Hess’s Law

 9.6 More About Entropy 9.7 Gibb’s Free Energy 9.8 Practice Questions 9.9 Explanations
Gibb’s Free Energy
As we said earlier, the two driving forces for chemical reactions are enthalpy and entropy. If a chemical reaction is endothermic, it must result in an increase in entropy, and if a reaction results in a decrease in entropy, it must be exothermic. For the SAT II Chemistry test, you will be expected to know how to use given entropy (S) and enthalpy (H) values to calculate if a reaction will be spontaneous or not, and you can do so by using the Gibb’s free energy (G) equation:
DG = DH - TDS
The Gibb’s free energy equation combines all the information that we have learned thus far. But what does the Gibb’s free energy value tell us about a reaction? It tells us the following:
1. If G is negative, the reaction is spontaneous in the forward direction.
2. If G is equal to zero, the reaction is at equilibrium.
3. If G is positive, then the reaction is nonspontaneous in the forward direction, but the reverse reaction will be spontaneous.
4. for elements at standard state (pure elements at 25ºC and 1 atm are assigned a value of zero).
The Gibb’s free energy equation can be used to calculate the phase change temperature of a substance. During a phase change, equilibrium exists between phases, so if the G is zero, we know that the reaction is in equilibrium.
Example
Find the thermodynamic boiling point of
H2O(l)H2O(g)
given the following information:
Hvap = +44 kJ  Svap = 118.8 J/K
Explanation
You would solve this problem by setting the equation equal to zero since in equilibrium, G has a value of 0.
0 = (44,000 J ) - (T )(118.8 J/K)
Now solve for T: the answer is 370K, the boiling point of water.
Here’s a handy reference table for interpreting what enthalpy and entropy values say about chemical reactions:
DH DS Result
Negative Positive Spontaneous at all temperatures
Positive Positive Spontaneous at high temperatures
Negative Negative Spontaneous at low temperatures
Positive Positive Never spontaneous
Much as is the case with both enthalpy and entropy, you can calculate DG using the following equation:
The units for DG are the same as the units as for enthalpy: J/K.
Now try using the above equation in a problem.
Example
Find the free energy of formation for the oxidation of water to produce hydrogen peroxide.
2H2O(l) + O2(g)2H2O2(l)
given the following information:
 ∆Gf˚ H2O(l) -56.7 kcal/mol O2(g) 0 kcal/mol H2O2(l) -27.2 kcal/mol
Explanation
Plugging all of the values you were given into the equation (remember that elements have a DGf˚ of 0), you get
[2(-27.2)] - [2(-56.7) + 1(0)] = 59.0 kcal/mol
 Jump to a New ChapterIntroduction to the SAT IIIntroduction to the SAT II Chemistry TestStrategies for Taking the SAT II Chemistry TestThe Structure of MatterThe States of MatterReaction TypesStoichiometryEquilibrium and Reaction RatesThermodynamicsDescriptive ChemistryLaboratoryBasic Measurement and Calculation ReviewChemical Formulas Review: Nomenclature and Formula WritingPractice Tests Are Your Best Friends
Test Prep Centers
SparkCollege
 College Admissions Financial Aid College Life