CHEM 1112: Week 4
CHEM 1112: Week 4 CHEM 1112
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This 4 page Class Notes was uploaded by Hayley Seal on Sunday February 7, 2016. The Class Notes belongs to CHEM 1112 at George Washington University taught by Martin Zysmilich in Spring 2016. Since its upload, it has received 44 views. For similar materials see General Chemistry 2 in Chemistry at George Washington University.
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Date Created: 02/07/16
CHEM 1112 Dr. Martín Zysmilich Class Notes for February 2-4 Entropy (February 2) Units of enthalpy: kJ/mol Units of entropy: J/K∙mol 3 Law of Thermodynamics: the entropy of a perfect crystal of any substance at 0 K is equal to 0 o All molecules in the system have a fixed position; only one possible arrangement o Also, all molecules are in the lowest energy state possible: no translational or rotational energy (no kinetic energy) and lowest possible vibrational energy o Increased temperature increases entropy; therefore, entropy has an absolute value because it can be compared to the absolute value of 0 at 0 K Enthalpy does not have to be expressed as ΔS like enthalpy must be expressed as ΔH Entropy at 1 atm = standard entropy (S°) o Entropy is a state function, so Hess’s Law can be applied to entropy as well as enthalpy o Molecular complexity increases entropy (more ways of movement and more ways to store energy in more complex molecules) Trends in Entropy o S°gas> S°liquidS°solid o S° increases with molar mass o S° increases with molecular complexity ΔS° (change in entropy) for a chemical reaction: aA + bB → cC + dD ΔS° = enthalpy of products – enthalpy of reactants = (cS°C+ dS°D) – (aS°A+ bS° B SAMPLE PROBLEM: Given: Calculate the enthalpy change for the reaction of formation of water a) as a gas. S°H2 (g) 130.6 J/K∙mol H 2 (g) ½O 2 (g) H O2 (g) S°O2 (g) 205.0 J/K∙mol S°H2O (g) 188.8 J/K∙mol ΔS° = S° productsS°reactants S°H2O (l)69.9 J/K∙mol = S°H2O (g)(S°H2 (g) ½S°O2 (g) = 188.8 J/K∙mol – (130.6 J/K∙mol + ½(205.0 J/K∙mol)) = -44.3 J/K∙mol b) as a liquid. H2 (g) ½O 2 (g)→ H 2 (l) ΔS° = S°products S°reactants = S°H2O (l)(S°H2 (g) ½S° O2 (g) = 69.9 J/K∙mol – (130.6 J/K∙mol + ½(205.0 J/K∙mol)) = -163.2 J/K∙mol Predicting the Sign of ΔS° sys o Processes that lead to increased entropy: Melting Vaporization Sublimation Temperature increase Reaction resulting in a greater number of gas molecules SAMPLE PROBLEM: Determine the sign of ΔS° : sys a) decomposition of KClO 3 (s)o give K2O (s)l 2 (g)and O 2 (g) KClO 3 (s)K O2+(s) 2 (g) O2 (g) creation of gas molecules: ΔS° sys> 0 b) condensation of water vapor on a cold surface H 2 →(g)O 2 (l) condensation goes to a more ordered system: ΔS° sys< 0 c) reaction of NH 3 (g)nd HCl (g)give NH Cl 4 (s) NH 3 (g)HCl (g)H Cl 4 (s) less gas molecules in products: ΔS° sys< 0 Change in entropy of a system is not enough to predict spontaneity of a process: condensation of water has negative entropy but we know it is spontaneous o Spontaneity depends on ΔS of the universe, which in total increases in entropy o Universe = system + surroundings o Therefore, the entropy of both the system and the surroundings are needed to determine the spontaneity of a process Calculating ΔS surr Δ???????????????? o Δ???? ???????????????? = − ???? o If the system is losing heat (exothermic), the surroundings absorb that heat (endothermic) and vice versa, which is the reason for the negative sign o At low temperatures, increase in temperature greatly increases the energy of the system o At higher temperature, adding energy doesn’t change the energy of the system quite as much because the distribution of energy among molecules is more broad at higher temperatures nd 2 Law of Thermodynamics: ΔS univ 0 for spontaneous processes ΔSuniv= ΔSsys+ ΔSsurr Δ???????????????? ΔSuniv= ΔSsys– ???? –TΔS univ –TΔS sys+ ΔHsys ΔG ΔG = ΔH – TΔS ΔG = Gibbs Free Energy (state function) o ΔG < 0 for a spontaneous process o ΔG > 0 for a nonspontaneous process o ΔG = 0 for an equilibrium process Under standard conditions: ΔG° = ΔH° – TΔS° How temperature affects the sign of ΔG: ΔH ΔS Spontaneity + + Spontaneous at high temperatures - - Spontaneous at low temperatures + - Never spontaneous at any temperature - + Always spontaneous at any temperature Calculating ΔG° rxn ΔG° rxn= Σ(ΔG°products– Σ(ΔG°reactants SAMPLE PROBLEM: Is this reaction spontaneous at 200°C under standard conditions? If yes, at what temperature does it become nonspontaneous? If no, at what temperature does it become spontaneous? CaCO → CaO + CO 3 (s) (s) 2 (g) Given: CO2 (g) CaO (s) CaCO3 (s) ΔHf°(kJ/mol) -393.5 -635.6 -1206.9 S° (J/K-mol) 213.6 39.8 92.9 200°C = 473 K ΔG° rxn= (ΔH° – TΔS° products– (ΔH° – TΔS° reactants = [(-393.5 kJ/mol – 635.6 kJ/mol) – 473 K (213.6 J/K-mol + 39.8 J/K-mol)] – [-1206.9 kJ/mol – 473 K (92.9 J/K-mol)] = (-1029 kJ/mol – 119.9 kJ/mol) – (-1207 kJ/mol – 43.94 kJ/mol) = -1149 kJ/mol + 1251 kJ/mol = 101.9 kJ/mol ΔG° rxn> 0, therefore the process is nonspontaneous at 200°C It changes to spontaneous at ΔG° rxn= ΔH° – TΔS° = 0 ΔH° = TΔS° ΔH° T = ΔS° T = (−393.5 kJ/mol – 635.6 kJ/mol) – (−1206.9 kJ/mo= 1108 K = 835°C (39.8J/K−mol +213.6J/K−mol −92.9J/K−mol The class on February 4 was a TA-led review session for the exam.
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