The Temperature The Temperature Dependence of SpontaneityDependence of Spontaneity
Example 15-18: Use thermodynamic data to estimate Example 15-18: Use thermodynamic data to estimate the normal boiling point of water.the normal boiling point of water.
Problem: Potassium chlorate, one of the common oxidizing agents inexplosives, fireworks, and matchheads, undergoes a solid-state redoxreaction when heated, in which the oxidation number of Cl in the reactant is higher in one of the products and lower in the other (disproportionation):
Effect of Temperature on Reaction SpontaneityThe temperature at which a reaction occurs influences the magnitude of the T S term. By scrutinizing the signs of H and S, we can predict the effect of temperature on the sign of G and thus on the spontaneity of a process at any temperature.
1. Reaction is spontaneous at all temperatures: Ho < 0, So > 0
2. Reaction is nonspontaneous at all temperatures: Ho > 0, So < 0
Temperature-independent cases (opposite signs)
2 H2O2 (l) 2 H2O(l) + O2 (g) Ho = -196 kJ and So = 125 J/K
3 O2 (g) 2 O3 (g) Ho = 286 kJ and So = - 137 J/KTemperature-dependent cases (same signs)
3. Reaction is spontaneous at higher temperature: Ho > 0 and So > 0
4. Reaction is spontaneous at lower temperature: Ho < 0 and So < 0
2 N2O(g) + O2 (g) 4 NO(g) Ho = 197.1 kJ and So = 198.2 J/K
2 Na(s) + Cl2 (g) 2 NaCl(s) Ho = - 822.2 kJ and So = - 181.7 J/K
Problem: An important reaction in the production of sulfuric acid is theoxidation of SO2 (g) to SO3 (g):
At 298 K, Go = -141.6 kJ; Ho = -198.4 kJ; and So = -187.9 J/K.(a) Use the data to decide if this reaction is spontaneous at 25oC and how Go will change with increasing T.(b) Assuming Ho and So are constant with T, is the reaction spontaneous at 900.oC?
Plan: (a) We examine the sign of Go to see if the reaction is spontaneous and the signs of Ho and So to see the effect of T.(b) We use Equation 20.6 to calculate Go from the given Ho and So at the higher T (in K).
Solution: (a) Since Go < 0, the reaction is spontaneous at 298 K: a mixture of SO2 (g), O2 (g), and SO3 (g) in their standard states (1 atm) willspontaneously yield more SO3 (g). With So < 0, the term -T So > 0 andbecomes more positive at higher T. Therefore, Go will be less negative, and the reaction less spontaneous, with increasing T.
(b) Calculating Go at 900.oC (T= 273 + 900. = 1173 K):
Go = Ho - T So = - 198.4 kJ - [(1173 K)(-187.9 J/K)(1 kJ/1000 J)] = 22.0 kJ
Since Go > 0, the reaction is nonspontaneous at higher T.
Calculating G at Nonstandard Conditions–IProblem: The oxidation of SO2, which we discussed earlier, is too slow
at 298 K to be useful in the manufacture of sulfuric acid. To overcome this low rate, the process is conducted at an elevated temperature.(a) Calculate K at 298 K and at 973 K. Go
298 = -141.6 kJ/mol ofreaction as written; using Ho and So values at 973 K, Go
973 = -12.12 kJ/mol of reaction as written. (b) In experiments to determine the effect of temperature on reactionspontaneity, sealed containers are filled with 0.500 atm SO2, 0.0100 atm O2, and 0.100 atm SO3 and kept at 25oC and at 700oC. Inwhich direction, if any, will the reaction proceed to reach equilibrium at each temperature?(c) Calculate G for the system in part (b) at each temperature.Plan: (a) We know Go, T, and R, so we can calculate the K’s fromEquation 12.10.(b) Calculate Q, and compare it with each K from part (a).
Calculating G at Nonstandard Conditions–IIPlan : Continued(c) Since these are not standard-state pressures, we calculate G at eachT from Equation 20.11 with the values of Go (given) and Q [found inpart (b)].Solution: (a) Calculating K at the two temperatures:
At 298 K, the exponent isGo = -RT ln K so K = e-( G/RT)
Calculating G at Nonstandard Conditions-III(b) Calculating the value of Q :
Since Q < K at both temperatures, the denominator will decrease and thenumerator increase–more SO3 will form–until Q equals K. However, at 298 K, the reaction will go far to the right before reaching equilibrium, whereas at 973 K, it will move only slightly to the right.
(c) Calculating G, the nonstandard free energy change, at 298 K and 973 K
Q = = = 4.00p2
SO3
p2SO2 x pO2
0.1002
0.5002 x 0.0100
G298 = Go + RT ln Q = -141.6 kJ/mol + (8.31 J/mol K x x 298 K x ln 4.00) = -138.2 kJ/mol
1 kJ1000 J
G973 = Go + RT ln Q = - 12.12 kJ/mol + (8.31 J/mol K x x 973 K x ln 4.00 = - 0.90 kJ/mol
pure Npure N22OO44 has has GGrxnrxn < 0. < 0. Reaction proceeds until Reaction proceeds until
GGrxn rxn = 0 - the minimum in = 0 - the minimum in G(reaction) - see graph.G(reaction) - see graph.
At this point, both NAt this point, both N22OO44 and and NONO22 are present, with more are present, with more NONO22.. This is a This is a reactant-favored reactant-favored
reaction. reaction.
Thermodynamics and KThermodynamics and Keq eq (3)(3)
We make machines that make other machines, machines that mow lawns, move mountains, and go to the moon.
We eat concentrated energy in the form of food, process that energy to synthesize complex biochemicals and run our organism, excreting diffused energy as body heat and less concentrated energy substances.
We use concentrated energy fuels to gather all kinds of materials from all parts of the world and, without any energetic limitation, arrange them in ways that please us.
We affect non-spontaneous reactions (pure metals from ores, synthesizing curative drugs from simple compounds, altering DNA).
All of the biochemical systems that run our bodies are maintained and regulated by feedback subsystems, composed of complex substances that are synthesized internally by thermodynamically nonspontaneous reactions, affected by utilizing energy ultimately transferred from the metabolism of food. When these feedback subsystems fail energy can no longer be processed to carry out the many reactions we need for life that are contrary to the direction predicted by the second law.
Every organic chemical of the 10,000 different kinds in our bodies is metastable, synthesized by a nonspontaneous reaction and kept from instant oxidation in air by activation energies.
Living creatures are energy processing systems that cannot function unless biochemical cycles operate synchronically to use energy to oppose second law predictions.
time Changes in entropy of an isolated system with time. Entropy increases with time until equilibrium is reached
Change in the Gibbs free energy at constant temperature and pressure. Under these conditions G decreases during the course of a spontaneous change until equilibrium is reached.
Comparing the chemical potential of the Comparing the chemical potential of the real gas to the chemical potential of an real gas to the chemical potential of an ideal gas at the same pressure ideal gas at the same pressure
The fugacity coefficients are obtained from The fugacity coefficients are obtained from the compression factors (Z) as shown the compression factors (Z) as shown below below
If system is If system is notnot simple, e.g. open system or simple, e.g. open system or mixtures, then for e.g. substances A and Bmixtures, then for e.g. substances A and B
(see lab manual, the five pages following error analysis)(see lab manual, the five pages following error analysis)
For constant For constant T,P T,P and and nnAA (e.g. 1 kg water in expt (e.g. 1 kg water in expt
2)2)
Find Find VV does not vary does not vary linearly with linearly with nnBB
For a reversible, equilibrium process (we’ll take For a reversible, equilibrium process (we’ll take dwdwsuch as a phase change or chemical reaction (at const. such as a phase change or chemical reaction (at const. T, T, PP))
( ( iceice liquid liquid ))dn dn = 0 = 0
i.e. equilibrium requires i.e. equilibrium requires iceice liquid liquid
Hence Hence criteriacriteria for for equilibriumequilibrium equal equal T T thermal equilthermal equil equal equal PP mechanical equilmechanical equil equal equal chemical equilchemical equil
