Chapter 12 Intermolecular Forces of Attraction

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1 Chapter 12 Intermolecular Forces of Attraction

2 Intermolecular Forces Attractive or Repulsive Forces between molecules. Molecule Molecule Intramolecular Forces bonding forces within the molecule. H O H

3 Review: Characteristics of Physical States

4 Review: Phase Changes - Changes in Physical States Solid Liquid Liquid Solid Liquid Gas Gas Liquid Solid Gas Gas Solid Melting (Fusion) Freezing Vaporization Condensation Sublimation Deposition

5 Which of the physical changes are endothermic or exothermic? Endothermic Melting Sublimation Vaporization Exothermic Freezing Deposition Condensation

6 Phase Changes of pure substances require a specific amount of energy per mole ( H) H of a Phase Change: The amount of energy required to produce a phase change for one mole of a substance (kj/mol). Note: The VALUE of H vap. = H cond. However, the sign is different depending on exothermic or endothermic processes.

7 Examples H 2 O (l) H 2 O (g) H vap = 40.7 kj/mol H 2 O (g) H 2 O (l) H cond. = kj/mol H 2 O (s) H 2 O (l) H fus = 6.02 kj/mol H 2 O (l) H 2 O (s) H freez = kj/mol

8 Heats of Vaporization & Fusion for Various Substances

9 Quantitative Aspects of Phase Changes Calculating the amount of energy for a substance to undergo phase changes. If you had grams of water at o C, how much energy would be released when the water cooled to o C. Some important facts to know: Specific Heat of H 2 O (g) : 33.1 J/mol- o C Specific Heat of H 2 O (l) : 75.4 J/mol- o C (4.184 J/g o C) Specific Heat of H 2 O (s) : 37.6 J/mol- o C

10 Temperature (ºC) Stage 1 Stage 2 Stage 3 Stage 4 Stage GAS GAS LIQUID LIQUID 0 H 0 vap LIQUID SOLID H 0 fus 40 Heat removed SOLID

11 Temperature (ºC) Stage GAS Heat removed

12 Calculating the amount of heat absorbed or released when a substance undergoes a temperature change: q = mc T or q = nc T n = (25.00gH 2 O)(1mol H 2 O) (18.02 g H 2 O) n = mol H 2 O q = (1.387 mol)(33.1j/mol o C)(100.0 o C o C) q = J = kj

13 Temperature (ºC) Stage 1 Stage GAS GAS LIQUID 0 H 0 vap 40 Heat removed

14 Calculating the amount of heat gained or lost during a phase change. q = n H? q = mol ( kj/mol) q = kj

15 Temperature (ºC) Stage 1 Stage 2 Stage GAS GAS LIQUID LIQUID 0 H 0 vap 40 Heat removed

16 q = nc T q = 1.387mol (75.4 J/mol o C)(0.0 o C o C) q = J = kj

17 Temperature (ºC) Stage 1 Stage 2 Stage 3 Stage GAS GAS LIQUID LIQUID 0 H 0 vap LIQUID SOLID H 0 fus 40 Heat removed

18 q = n H fus q = 1.387mol(-6.02 kj/mol) q = kj

19 Temperature (ºC) Stage 1 Stage 2 Stage 3 Stage 4 Stage GAS GAS LIQUID LIQUID 0 H 0 vap LIQUID SOLID H 0 fus 40 Heat removed SOLID

20 q = nc T q = 1.387mol (37.6 J/mol o C)(-40.0 o C 0.0 o C) q = J = kj

21 Total Heat Change q = (heat lost by steam + heat of condensation + heat lost by liquid + heat of freezing + heat lost by solid) Remember, you can t add kj with J! 1,000 J = 1 kj

22 kj kj kj kj kj kj Final Answer: kj of heat are released

23 Exam Question(s) You should be able to calculate the amount of heat released or absorbed for any substance given H s, C s, and temps.

24 Phase Changes and Equilibrium Equilibrium The rate of the forward reaction (or process) is equal to the rate of the reverse reaction (or process). Example: At 0 o C and 1 atm H 2 O (s) H 2 O (l) Rate (forward) = Rate Equilibrium

25 Example Open Container (Evaporation) Closed Container (Equilibrium)

26

27 Disturbing Equilibrium (Placing a Stress on the System) 1. Add more vapor to the container 2. Remove some of the vapor 3. Add heat 4. Remove heat Note: If left alone the system will eventually return to equilibrium.

28 The pressure, at a particular temperature, exerted by a vapor (gas) in equilibrium with its liquid in a closed container is called the vapor pressure. The temperature at which the vapor pressure is equal to the external (atmospheric pressure) is the boiling point of a substance.

29 The Effects of Temperature on Equilibrium As the temperature is increased so will the vapor pressure

30 At a particular temperature, the weaker the intermolecular forces of attraction of a substance, the higher the vapor pressure.

31 Clausius-Clapeyron Equation ln P 2 = - H vap P 1 R 1-1 T 2 T 1 R = J/mol-K T must be in Kelvin

32 Problem Solving Calculate the vapor pressure of water at 85 o C. Hint: Remember that water s vapor pressure is equal to 1.00 atm (760. torr) at its b.p.

33 Answer P 1 = 760. torr P 2 =? T 1 = K T 2 = K ln P = J/mol torr J/mol-K K K P 2 = 439 torr or atm

34 Phase Diagrams Regions - Various phases Lines between regions Equilibrium of phases Critical Point Point at which the substance has the density of a gas and solvent capability of a liquid. Triple Point Point at which all three physical states exist in equilibrium

35

36 Types of Intermolecular Forces

37

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