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Solids, Liquids, Energy & Heat

Solids, Liquids, Energy & Heat. Intermolecular Forces “Inter” => between “molecular” => molecules. Intermolecular forces apply only to covalent bonds. Intermolecular Forces vs. Intramolecular Forces. Intra molecular Forces. Inter molecular Forces.

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Solids, Liquids, Energy & Heat

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  1. Solids, Liquids, Energy & Heat

  2. Intermolecular Forces“Inter” => between “molecular” => molecules Intermolecular forces apply only to covalent bonds

  3. Intermolecular Forces vs. Intramolecular Forces Intramolecular Forces Intermolecular Forces • Intermolecular forces are between neighboring molecules • Weaker than intramolecular forces • Types: • London dispersion forces • dipole-dipole forces • hydrogen bonding • Intramolecular forces are within one molecule • a.k.a. Bonding • Stronger than intermolecular forces • Types: • Ionic Bonding • Covalent Bonding

  4. Intermolecular Forces • Intermolecular forces vary in strength and properties: bond strength incr. Hydrogen bonds Dipole-dipole London dispersion properties incr. Boiling point, melting point, surface tension, and viscosity and

  5. London Dispersion Forces(van der Waals Forces) • Very weak forces • Occurs when an electron cloud temporarily gets distorted and forms a temporary positive and negative charge. • The ONLY forces of attraction that exist between nonpolar molecules (e.g. Ar, C8H18 )

  6. London Dispersion Forces

  7. Dipole-Dipole Attraction • Polar molecules that orient themselves so that + and – ends of the dipoles are close to each other.

  8. Hydrogen Bonding • Strongest dipole-dipole attraction when hydrogen is bound with either: N, O, F. • Molecule would contain –NH, -OH, or -FH

  9. Hydrogen Bonding in DNA

  10. Steps to Determine Intermolecular Forces 1) Draw & determine the VSEPR Shape for each molecule: I2 H2Se NH3 • • • • • • • • • Se H • H N H • • I I • • • • • • H H bent linear pyramid

  11. 2) Determine if each molecule is a polar or non-polar structure: Options: (a) There are no polar bonds  non-polar structure (b) Every polar bond has an opposite (no lone pairs on center)  non-polar structure (c) Not every polar bond has an opposite (lone pairs on center)  polar structure I2 H2Se NH3 • • • • • Se H • H N H • • • • • • I I H • • H • • • • polar polar non-polar

  12. 3) Determine the type of intermolecular force for each molecule: Options: (a) non-polar structure  London dispersion forces (b) polar structure with H-O, H-N, or H-F bond(s)  hydrogen bonding (c) polar structure without H-O, H-N, or H-F bond(s) dipole-dipole forces I2 H2Se NH3 dipole-dipole forces hydrogen bonding London dispersion forces

  13. Energy * + & - indicate whether the heat is flowing into or out of the system* Energy is the capacity to do work. Energy allows us to “do things”, such as drive, cook eggs, and read the words on this page. Often energy is used to change the temperature of a substance. When you change the temperature of a substance, you are reflecting the random motion of the molecules. There are two common units of energy – the calorie and the joule.

  14. A calorie (cal) is defined as the amount of energy required to raise the temperature of one gram of water by 1C. The joule (J) is the unit of energy in the SI System. 1 calorie = 4.184 J Example – Express 22.4 cal of energy in joules. 4.184 J 22.4 cal X _________________ = 93.7 J 1 cal 1 kcal = 1000 cal 1 kJ = 1000 J

  15. The amount of energy required to raise the temperature of a substance depends on three things:1. amount of substance (mass)2. the amount of the temperature change3. type or identity of the substanceDifferent substances respond to heat differently – 4.184 J raises 1 g of water - 1C BUT 4.184 J raises gold 32CSpecific Heat Capacity – the amount of energy required to change the temperature of one gram of a substance by 1C

  16. Formula to calculate energy requirements- Q = (s)(m)(T) Q = energy required, J s = specific heat capacity, J/gC m = mass of substance, g T = change in temperature, C T = Tfinal - Tinitial Include: Given, formula, setup, answer, SF, units

  17. Specific Heat, Heat of Fusion, and Heat of Vaporization of Some Selected Substances

  18. Ex. – A 5.63 g sample of solid copper is heated from 21C to 32C. How much energy in joules is required? m = 5.63 g T = Tf – Ti T = 32C - 21C = 11C Q = ? s = 0.385 J/gC Q = s m T Q = (0.385 J/gC) (5.63 g) (11 C ) Q = 24 J

  19. Example – A 2.8 g sample of pure metal requires 10.1 J of energy to change its temperature from 21C to 36C. What is the metal? s = __Q__ m T m = 2.8 g Q = 10.1 J T = Tf – Ti T = 36C - 21C = 15C s = ? Look on chart Silver s = ___10.1 J___ (2.8 g)(15C) s = 0.24 J/gC

  20. Heating Curve for Water

  21. When a substance melts or boils, there is no change in temperature (no ∆T). Heat of Fusion (∆Hfus) is the amount of energy required to melt a substance Heat of Vaporization (∆Hfus) is the amount of energy required to boil a substance The formulas to use are: For melting or freezing, Q=m∆Hfus For boiling or condensing, Q=m∆Hvap

  22. Molar Heat of Fusion and Molar Heat of Vaporization have units of J/mol Heat of Fusion, Heat of Vaporization, and Specific Heat capacity are characteristic properties and can be used to identify a substance. Ex. How much energy is required to melt 8.93 moles of gold with no change in temperature? Ex. How much energy is absorbed when 5.34 g of chloroform condenses from a gas? q=m∆Hfus g Au 196.97 m = 8.93 mol Au Hfus Au = x _________ = 1760 g Au q = (1760 g)(63.0 J/g) 1 mol Au 63.0 J/g q = 111000 J q=m∆Hvap m = 5.34 g Hvap chloroform = q = (5.34 g)(264 J/g) 264 J/g q = 1410 J

  23. Enthalpy (H) – Heat of reaction • H = exothermic reaction - heat flows out of the system + H = endothermic reaction - heat flows into the system feels hot or gives off light feels cold

  24. Entropy(S) – Disorder Second law of thermodynamics - the entropy of the universe is always increasing Solid  liquid  gas Ordered  disordered Also: # of moles and number of bonds

  25. Ex: Which of the following has more entropy? • solid or gaseous phosphorus • KBr (s) or KBr (aq) • CH4(g) or C3H8(g)

  26. Ex: Does entropy increase or decrease for each of the following reactions? • (NH4)2Cr2O7(s) Cr2O3(s) + 4 H2O (l) + N2(g) b. Mg(OH)2(s)MgO(s) + H2O (g) c. PCl3 (g) + Cl2(g) PCl5(g) INCREASE INCREASE DECREASE

  27. Phase Diagram • Triple pointis the temperature and pressure at which all three phases can exist in equilibrium.Above the critical point, molecules are unable to liquefy.

  28. Normal Boiling Point:  Normal Freezing Point the temperature at which a liquid boils at 1 atmosphere of pressure. the temperature at which a liquid freezes at 1 atmosphere of pressure. Phase Changes: Melting Freezing Boiling Condensing Sublimation Deposition solid liquid liquid  solid liquid  gas gas liquid solid gas gas  solid

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