21 where the water is the system & everything else is the surroundings H2O(s) H2O(l)where the water is the system & everything else is the surroundingsentropy of water increases, therefore Ssys > 0the vaporization of water is an endothermic process,therefore heat is flowing from the surroundings to the system & the random motion of the atoms in the surroundings decrease.thus, Ssurr < 0so, Suniv = Ssys(+) + Ssurr(-) ; which predominates?we know that above 1000C, water spontaneously evaporates and that below 1000C, water spontaneously condenses, thus TEMPERATURE must have an effect on relative importance of Ssys and Ssurr .
22 Temperature Dependence S = J/mol (1/T) Ssurr = -H/T introduces central idea that the entropy changes in the surroundings are primarily determined by heat flow.exothermic process increases the entropy of the surroundings, thus is an important driving force for spontaneity. Remember that a system tends to undergo changes that lower its energy.The SIGN on Ssurr depends on the direction of the heat flow.The MAGNITUDE of Ssurr depends on the temperataure.
89 information, determine 3. Using thermodynamicinformation, determinethe boiling point ofbromine.
90 Thermodynamics and Keq FACT: Product-favored systems have Keq > 1.
91 Thermodynamics and Keq Therefore, both ∆G˚rxn and Keq are related to reaction favorability.
92 Thermodynamics and Keq Keq is related to reaction favorability and thus to ∆Gorxn.The larger the value of K the more negative the value of ∆Gorxn
93 Thermodynamics and Keq ∆Gorxn= - RT lnKwhere R = J/K•mol
94 Thermodynamics and Keq ∆Gorxn = - RT lnKCalculate K for the reactionN2O4 2 NO2∆Gorxn = +4.8 kJK = 0.14When ∆G0rxn > 0, then K < 1
95 ∆G, ∆G˚, and Keq∆G is change in free energy at non-standard conditions.∆G is related to ∆G˚∆G = ∆G˚ + RT ln Q where Q = reaction quotient
96 ∆G, ∆G˚, and Keq When Q < K or Q > K, reaction is spontaneous. When Q = K reaction is at equilibriumWhen ∆G = 0 reaction is at equilibriumTherefore, ∆G˚ = - RT ln K
97 Product favored reaction ∆G, ∆G˚, and KeqProduct favored reaction–∆Go and K > 1In this case∆Grxn is < ∆Gorxn , so state with both reactants and products present is MORE STABLE than complete conversion.
98 ∆G, ∆G˚, and Keq Product-favored reaction. 2 NO2 N2O4 ∆Gorxn = – 4.8 kJHere ∆Grxn is less than ∆Gorxn , so the state with both reactants and products present is more stable than complete conversion.
99 Thermodynamics and Keq Overview ∆Gorxn is the change in free energy when reactants convert COMPLETELY to products.
100 Keq is related to reaction favorability. When ∆Gorxn < 0, reaction moves energetically “downhill”
101 4. For the following reaction, calculate the temperature at which the reactants are favored.
103 5. How much useful work can be obtained from an engine fueled with 75 5. How much useful work can be obtained from an engine fueled with 75.0 L of hydrogen at 10 C at 25 atm?
104 reacting silver with water. 6. The reaction to split water into hydrogen and oxygen can be promoted by firstreacting silver with water.2 Ag(s) + H2O(g) Ag2O(s) + H2(g)Ag2O(s) 2 Ag(s) + 1/2 O2(g)