Chemistry Week 12

Chemistry Week 12 - (4) Thermodynamic Influences - Entropy...

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(4) Thermodynamic Influences - Entropy TEXTBOOK READING : BLB-10 , Chapter 19 Practice PROBLEMS: (Ch 19) 12, 32, 40, 48, 54, 56, 65, 76, 87 Demonstration : The cryophorus using Br 2 (l) (melting point is 7 ° C) or H 2 O(l) (melting point = 0 ° C.) Br 2 (l) T 1 T 2 Heat The process Br 2 (l) ⎯→ Br 2 (s) is exothermic, i.e., H < 0 and releases heat to the surroundings. Therefore, in the glass apparatus, the second law of thermodynamics explains why heat moves from high temperature to low temperature, so that the temperature throughout the cryophorus can be the same value. With the liquid bromine in the warmer side, heat is flowing from its flask to the other flask. As heat is removed, the liquid bromine will then condense to the solid. (c) 3rd Law of Thermodynamics : “There is a temperature so low that it cannot be reached (absolute zero).” The temperature of the colder reservoir in the engine we have discussed can never reach 0 K (the efficiency of the engine would be perfect, or there would be no heat losses!). The third law states that the entropy of any pure substance at T = 0 K is zero . As temperature increases, the absolute entropy of a substance increases. Solids (well-ordered, rigid) --- Liquids (somewhat ordered, less rigid) --- Gases (disordered, moving) ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯→ Entropy increases.
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Therefore, the zero of entropy occurs at 0 K for every pure substance! At 298.15 K, the entropy of a substance in its standard state is some value greater than zero, in units of J/mol K. (The only exception to this is 1 M H + (aq), whose entropy is defined as 0 J/mol K). Entropy describes the extent of "randomness" or "chaos" in a system. The greater the disorder, the greater the entropy. We can talk about absolute entropy at any temperature, due to the third law. Compare the structures of two different isomers of C 5 H 12 : neopentane and n-pentane. S ° (neopentane) = 280 J/mol K and S ° (n-pentane) = 320 J/mol K. The higher absolute entropy occurs with the molecule that is more flexible. There are many more different geometrical arrangements possible for the n-pentane molecule than for the neopentane molecule. Other examples include: S ° (F 2 ,g) = 202.7 J/mol K (MW = 38.0 g/mol) S ° (Cl 2 ,g) = 233.0 J/mol K (MW = 71.0 g/mol) S ° (Br 2 ,g) = 245.3 J/mol K (MW = 159.9 g/mol) S ° (I 2 ,g) = 260.6 J/mol K (MW = 253.8 g/mol) (Entropy increases with mass – more translational degrees of freedom.) S ° (N 2 ,g) = 191.5 J/mol K (MW = 28.0 g/mol) Singlet ground state S ° (O 2 ,g) = 205.0 J/mol K (MW = 32.0 g/mol) Triplet ground state S ° (F 2 ,g) = 202.7 J/mol K (MW = 38.0 g/mol) Singlet ground state (Entropy depends on having unpaired electrons: Singlet = 0 unpaired electrons; triplet = 2 unpaired electrons) S ° (CO 2 ,g) = 213.6 J/mol K (MW = 44.0 g/mol) Linear; Singlet (O=C=O) Entropy; S 0 (J/mol K) Temperature (K) 0 0 T f T b SOLID Ordered, Rigid LIQUID GAS Disordered, Nonrigid MELTING FREEZING fus fus f H S T ∆= BOILING CONDENSATION vap vap b H S T
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S ° (N 2 O,g) = 220.0 J/mol K (MW = 44.0 g/mol) Linear; Singlet (N=N=O) S ° (NO 2 ,g) = 240.5 J/mol K (MW = 46.0 g/mol) Bent; 1 unpaired electron S ° (O 3 ,g) = 237.6 J/mol K (MW = 48.0 g/mol) Bent; Singlet
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Chemistry Week 12 - (4) Thermodynamic Influences - Entropy...

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