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Today in Chem104: What determines reaction spontaneity? Entropy The 2 nd Law & No Free Lunch What chemists really use- free energy.

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Presentation on theme: "Today in Chem104: What determines reaction spontaneity? Entropy The 2 nd Law & No Free Lunch What chemists really use- free energy."— Presentation transcript:

1 Today in Chem104: What determines reaction spontaneity? Entropy The 2 nd Law & No Free Lunch What chemists really use- free energy

2 Hypothesis 1: Spontaneous reactions are exothermic HCl + NaOHNa + + Cl- + H 2 O got warm so  H rxn <0  YES!!! Ba(OH) 2. 8H 2 O + 2NH 4 NO 3 Ba(NO 3 ) 2 + 2NH 3 + 10H 2 O got COLD so  H rxn > 0  NO!!! Conclusion 1: Spontaneous reactions can be exothermic or endothemic

3 Next, we encounter ENTROPY ENTROPY: the concept An increase in disorder An increase in energy dispersal symbol S ENTROPY: its magnitude Depends on state of matter, solid< liquid<gas Depends on temperature Depends on complexity of molecule/matter calculated like enthalpy:  S rxn =  S prdt -  S rgt

4 Hypothesis 2: Spontaneous reactions have increased entropy HCl (g) + NH 3 (g)NH 4 Cl(s) Entropy reagent gases >> Entropy solid prdt,  S rxn < 0  NO!!! Ba(OH) 2. 8H 2 O + 2NH 4 NO 3 Ba(NO 3 ) 2 + 2NH 3 + 10H 2 O  S rxn > 0  YES!!! Conclusion 2: Spontaneous reactions can have a decrease in entropy!  S rxn =  S prdt -  S rgt  S rxn = [150 + 2(70) + 10(192)] - [500 + 2(151)]  S rxn = +432 J/K mol S o, J/K mol = 500 2(151) 150 2(70) 10(192)

5 Hypothesis 3: Reaction Spontaneity depends on entropy AND enthalpy HCl (g) + NH 3 (g)NH 4 Cl(s) Heat released goes to surroundings?!!  S rxn =  S prdt -  S rgt  S rxn = [94.6] - [187 + 193]  S rxn = - 285 J/K mol S o, J/K mol = 187 193 94.6  H rxn =  H prdt -  H rgt  H rxn = [-315] - [-92.3 + -46.3]  H rxn = - 176 kJ/mol  H o f,k J/mol = -92.3 -46.3 -315 ENTROPY DECREASED ENTHALPY DECREASED

6 HCl (g) + NH 3 (g)NH 4 Cl(s) Heat released to surroundings should increase  S SURR And the net entropy change is:  S net =  S SURR +  S SYS =  S UNIVERSE  S UNIVERSE = 591 - 285 J/K mol = 206 J/K mol THE ENTROPY of UNIVERSE INCREASED How much? Use this relationship of enthalpy and entropy:  S SURR = -  H sys / T So the addition of 176 kJ/mol heat to surroundings corresponds to:  S SURR = - (-176 kJ/mol) / 298 = 0.591 kJ/K mol  S SURR = 591 J/K mol

7 This is the fundamental requirement long version : a spontaneous change is accompanied by an increase in the total entropy of the system and the surroundings  S TOTAL =  S SURR +  S SYS THE ENTROPY of UNIVERSE INCREASED This is the 2 nd Law of Thermodynamics shorter version : total entropy change must be positive for a spontaneous reactions shortest version : the entropy of the universe is constantly increasing

8 This is the fundamental requirement Paul’s version : No Free Lunch This means, a spontaneous exothermic reaction is not just creating heat ( or energy). The price is Entropy—more disorder in the Universe.… or a greater distribution of energy THE ENTROPY of UNIVERSE INCREASED This is the 2 nd Law of Thermodynamics Paul Grobstein’s version : The 1 st Law: You can’t win. The 2nd Law: You can’t break even. The 3rd Law: You can’t leave the game.


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