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    solutions manual chapter 17

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    CHAPTER 17 SPONTANEITY‚ ENTROPY‚ AND FREE ENERGY Questions 11. Living organisms need an external source of energy to carry out these processes. Green plants use the energy from sunlight to produce glucose from carbon dioxide and water by photosynthesis. In the human body‚ the energy released from the metabolism of glucose helps drive the synthesis of proteins. For all processes combined‚ ∆Suniv must be greater than zero (the second law). 12. Dispersion increases the entropy of the universe because

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    Physics

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    CaCl2 = - 795 kJ/ mol Std. enthalpy of atomisation of calcium = + 177 kJ/mol Std. enthalpy of atomisation of Chlorine = +121 kJ/mol 1st ionisation energy of calcium = +590 kJ/mol 2nd ionisation energy of calcium = +1100 kJ/mol Electron Affinity of chlorine = -364 kJ/mol 2. Draw Born-Haber cycles and calculate the L.E. of Cu2O and CuO given: ∆Hf (Cu2O) = -166.7 kJ/mol ∆Hf(CuO) = -155.2 kJ/mol 1st I.E. (Cu) = +750 kJ/mol 2nd I.E. (Cu) = +2000 kJ/mol ∆Hat (Cu) = +339.2 kJ/mol 1st E.A. (O)

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    Hess' Law

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    67)°C= 2699.52 J =2.70 kJ 0.2g Mg x1 mole Mg/24.31 Mg=0.0082 mol Mg -2.70 kJ/0.0082 mol = -329.27 kJ/mol = ΔH Trial 2: q=mCΔT q=10.11g x 4.184 J/g°C x (84.51-24.71)°C=2529.55 J =2.53 kJ 0.2g Mg x1 mole Mg/24.31 Mg=0.0082 mol Mg -2.53 kJ/0.0082 mol = -308.54 kJ/mol = ΔH *Average of the ΔH for Mg= -318.9 kJ/mol MgO: Trial 1: q=mCΔT q= 9.79g x 4.184 J/g°C x (40.27-22.97)°C= 708.63 J = 0.71 kJ 0.2 MgO x 1 mole MgO/40.31 MgO=0.005 mol MgO -0.71 kJ/0.005 mol = -142.0 kJ/mol = ΔH Trial 2: q=mCΔT

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    thermo

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    Chapter 15 (not much on E) Thermodynamics: Enthalpy‚ Entropy & Gibbs Free Energy Thermo 2 Thermodynamics: thermo = heat (energy) dynamics = movement‚ motion Some thermodynamic terms chemists use: System: the portion of the universe that we are considering open system: energy & matter can transfer closed system: energy transfers only isolated system: no transfers Surroundings: everything else besides the system Isothermal: a system that is kept at a constant temperature

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    Single Sex School

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     a  compound  from  its  elements        with  all  substances  in  their  standard  states.   1/2N2(g)  +  O2(g)    NO2(g)      ΔHf°  =  34  kJ/mol    ΔHf°  =  -­‐239  kJ/mol C(s)  +  2H2(g)  +  1/2O2(g)    CH3OH(l) In  Appendix  4   32 STANDARD ENTHALPIES OF FORMATION Conven1onal  Defini1ons  of  Standard  States     For  a  Compound

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    Themochemistry Notes

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    Chemistry 101  Chapter 6  THERMOCHEMISTRY ·  Thermochemistry is the study of the quantity of heat energy released or absorbed in a  chemical reaction.  Example: the burning of fuel:  is a heat­evolving reaction ·  Heat :  ·  Energy:  is a form of energy ­the potential to do work (to move matter)  ­exists in many different forms:  ­ Electrical energy  ­ Kinetic Energy (energy of motion)  ­ Light energy  ­ Heat energy  ­ Chemical energy (energy of substances) ·  Different forms of energy can be interconverted 

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    (O3) in the atmosphere can react with nitric oxide (NO):  O3(g) + NO(g) ( NO2(g) + O2(g)‚ with (H( = -199 kJ/mol‚ (S( = -4.1 J/K·mol. Calculate the (G( for this reaction at 25(C.  A.  1020 kJ/mol B.  -1.22 ( 103 kJ/mol C.  2.00 ( 103 kJ/mol D.  -1.42 ( 103 kJ/mol E.  -198 kJ/mol 2. For the reaction H2(g) + S(s) ( H2S(g)‚ (H( = -20.2 kJ/mol and (S( = +43.1 J/K·mol. Which of these statements is true?  A. The reaction is only spontaneous at low

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    point 107.50C. Heat load on preheater‚ Q = 26017.91 x 1.497(380.5 – 288) = 3.603 x 106 KJ The heating medium is used is dry saturated steam at 420K Steam requirement = 3.603 x 106 2123.4 = 1696.8 kg/hr 2-BUTANOL VAPORIZER:2-butanol feed is vaporized at 107.50C using reaction products. Heat load on vaporizer is ‚ Q = M x λ = 26017.91 x 557.43 kj = 14.5032 x 106 kj/hr Average Cp of reaction products = 2.3 KJ / Kg 0k 14.5032 x 106 = 26017.91 x 2.304 (642 – T) 22 T = 400 K FIRST SUPER HEATER:2-butanol

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    Nothing

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    4. Calculate the heat of vaporization of carbon tetrachloride given that the standard molar enthalpy of formation of liquid carbon tetrachloride is -135.4 kJ/mol and the standard molar enthalpy of formation of gaseous carbon tetrachloride is -102.9 kJ/mol. A. +102.9 kJ/mol B. -32.5 kJ/mol C. +238.2 kJ/mol D. +32.5 kJ/mol E. -238.2 kJ/mol 5. The complete combustion of 1 mole of propane (C3H8) results in the liberation of 488.7

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    Heat of Combustion of Magnesium Background: The students were given full instructions on how to experimentally determine the enthalpy of reaction (ΔHrxn) for the combustion of magnesium ribbon‚ using Hess’s Law. Data Collection: | |Reaction 1 |Reaction 2 | | |(MgO) |(Mg)

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