20.3. Free Energy and Equilibrium http://www.ck12.org
Lesson Summary
- The value of the free energy change (∆G) is zero when a reaction is at equilibrium. The Gibbs free energy
equation can be used to solve for the temperature, which equals∆H/∆S. - Two states of a substance are in equilibrium with one another during a change of state.
- The standard free energy change for a reversible reaction is related to the equilibrium constant by the equation
∆G° =−RT ln(Keq). A large Keqvalue corresponds to a negative∆G°, while a small Keqvalue corresponds to
a positive∆G°.
Lesson Review Questions
Reviewing Concepts
- What is the general value of∆G° for a reaction that strongly favors the formation of products? That strongly
favors the reactants? - What is the value of∆G when a reaction is at equilibrium?
- A certain reaction has a free energy change of∆G = +10 kJ/mol at a certain temperature. Does this mean that
no products are formed in this reaction? Explain. - A reaction has a negative∆H°, a negative∆S°, and a positive∆G° at a given temperature. In order to make
formation of the products more favorable, should the temperature be raised or lowered? Explain.
Problems
- For a certain reaction,∆H° =−29.2 kJ/mol and∆S° =−141 J/K•mol. Is the reaction spontaneous at 25°C? If
not, at what temperatures would it become spontaneous? - Find the Celsius temperatures at which reactions with the following∆H° and∆S° values would be sponta-
neous.
a.∆H° = +76.5 kJ/mol;∆S° = +231 J/K•mol
b.∆H° =−50.2 kJ/mol;∆S° = +97.7 J/K•mol
c.∆H° = +19.1 kJ/mol;∆S° =−73.0 J/K•mol - For the nearly insoluble compound silver iodide, Ks p= 8.3× 10 −^17. Calculate∆G° (in kJ/mol) for the
following reaction at 25°C: AgI(s)⇀↽Ag+(aq)+I−(aq). - Calculate the equilibrium constant at 25°C for the following reaction in which hydrogen and bromine combine
to form hydrogen bromide. H 2 (g)+Br 2 (g)⇀↽2HBr(g) ∆G◦=− 106 .4 kJ - Aqueous ammonia combines with water in an equilibrium reaction to form the ammonium ion and the
hydroxide ion according to the following equation: NH 3 (aq) +H 2 O(l)⇀↽NH+ 4 (aq) +OH−(aq). For this
reaction at 25°C,∆H° = 3.36 kJ/mol and∆S° =−78.9 J/K•mol.
a. Calculate∆G° at 25°C.
b. Calculate Keqat 25°C. - Carbon in the form of diamond can form graphite according to the following equation: C(diamond)⇀↽
C(graphite). Use the thermodynamic data below (Table20.4).
a. Calculate the∆G° for the reaction in which diamond changes to graphite. (Hint: first calculate∆H° and
∆S° from the data table.)
b. Calculate the equilibrium constant, Keq.
c. Is the reaction spontaneous at 25°C? If so, do we need to be concerned about diamonds turning into
graphite? Explain.