States of Matter

1. States of Matter

2. Liquids and Solids • Why are we studying liquids and solids together when we studied gases by themselves? Particles in liquids and solids are closer together than those of gases. • How much energy does it take to go from a solid to a liquid? Not Much • How much energy does it take to go from a liquid to a gas? More

4. Water Water can exist as a solid (ice), a liquid and a gas (water vapor or steam). NOTE: The term __GAS__ is used for substances that normally exist as a gas (at room temperature) while the term __Vapor__ is used for substances that are normally solid or liquid (at room temperature) and have been vaporized. It takes seven times more energy to convert water from a liquid to a gas than from a solid to a liquid. Temperatures: _100ºC_ boiling point __0ºC__ freezing point The solid and liquid states are more alike to one another than to the gaseous state. For example, their densities are much more similar than that of water vapor. Water is colorless and tasteless Water covers approximately 70% of the earth’s surface

5. UNUSUAL PROPETIES OF WATER: • High Boiling Point • High Specific Heat – Amount of energy required to change the temp of 1 gram by 1°C • High Heat of Vaporization – Amount of energy required to change a liquid to a vapor • Density of the Solid is less than the Density of the Liquid • High Surface Tension – imbalance of forces (downward & sideways) at the surface of a liquid; acts like a tight film across it • Universal Solvent – many things dissolve in water allows many reactions to occur

6. Heating Curve of Water

7. Heating Curve of Water Heating curve observations (from left to right as heat is added): • Below 0C water is a solid • At 0C there is no change in temperature. Only a phase change is occurring (solid to liquid) • From 0C up to 100C water is a liquid • At 100C there is no change in temperature. Only a phase change is occurring (liquid to gas) • Above 100C water is a vapor (steam)

8. Heating Curve of Water The opposite of a heating curve is a cooling curve. The curve is the same but the point of view is coming from a cooling perspective. You would read a cooling curve from right to left. Answer the following for a cooling curve: At 100C, what is happening? _Condensation _ At 0C, what is happening? _Freezing_ Why does it take more energy to vaporize water than to melt it? • Because if takes more energy to completely separate individual molecules from one another as they are in a gas • Going from a solid to a liquid only requires molecules to be separated slightly

9. Generic Heating Curve

10. Molar Heats Molar heat of fusion ( DHf )- energy required to melt or freeze 1 mole of a substance (segment B) Ex. water’s molar heat of fusion = 6.02 kJ/mol Molar heat of vaporization ( DHv)- energy required to vaporize or condense 1 mole of a substance (segment D) Ex. water’s molar heat of vaporization = 40.6 kJ/mol

11. Molar Heats As you can see, the molar heat of vaporization of water is greater than the molar heat of fusion. That’s why the “Water and steam” line on the heating curve of water is longer than the “Ice and water” line! • NOTE: You do NOT need to memorize the values for the molar heats of fusion and vaporization for water!

12. Q = mcT How much energy is required to go from one phase to another (the diagonal parts of the curve- segments A, C and E)? Since those sections of the curve are experiencing a temperature change we must use the following equation: Q = mcT Q = heat energy m = mass c = specific heat T = Tfinal – Tinitial

13. Q = mcTNote: The specific heat (c) is different for the solid, liquid and gaseous states of matter. The specific heat value you use is based on what state of matter you are dealing with.

14. Q = nΔH The flat (horizontal segments B and D) regions of the graph are not experiencing any temperature change. We use a different equation here. We must use the molar heats of fusion and vaporization for these regions of the curve: Q = nΔH ΔH depends on which part of the graph you are in (melting/ freezing or vaporizing/ condensing)

15. ΔH and c The following are common constants you will use for water: ΔH fusion = 6.02 kJc liquid = 4.18 J molg°C ΔH vaporization = 40.6 kJc vapor = 2.03 J molg°C c solid = 2.06 J g°C • (You do NOT need to memorize any of these values!)

16. Example • The following is an example of a problem involving a heating curve: How much energy is needed to melt 10. g of water at 0C?

17. Sketch

18. Step 1 Determine what part(s) of the heating curve you are dealing with. • Since it’s a flat area, you must use Q = nΔH. Since you are melting the water, you must use the molar heat of fusion: ΔH f = 6.02 kJmol

19. Step 2: Plug in your known values into the equation Q = nΔHf. • Since ΔHf is per mole (n), your amount must be in moles. 10. g H2O x 1 mol H2O = 0.55 mol H2O 1 18.02 g H2O Q = nΔHf Q = 0.55 mol H2O x 6.02 kJ mol

20. Step 3: • Solve problem. Cancel out like units in the numerators and denominators. Label and round your final answers to the correct number of sig. figs. Q = 0.55 mol H2O x 6.02 kJ = 3.3 kJ mol

21. Example How much energy is needed to raise 10. g of water from 0C to 20. C?

22. Sketch

23. Step 1 • Determine what part(s) of the heating curve you are dealing with. Since it’s a diagonal area, you must use Q = mcT because a temperature change is occurring. In addition, since you are in the liquid range, you must use the specific heat for liquid water.

24. Step 2 Plug in your known values into the equation Q = mcT. Q = mcT T = Tfinal – Tinitial = 20. C - 0C = 20. C Q = 10. g x 4.18 J x 20. C gC

25. Step 3 Solve problem. Cancel out like units in the numerators and denominators. Label and round your final answers to the correct number of sig. figs. Q = 10. g x 4.18 J x 20. C = 836 J = 840 J gC

26. Harder Example How much energy is needed to completely melt 10. g of water at 0C and then raise the water to 20. C?

27. Sketch

28. Step 1 Determine what part(s) of the heating curve you are dealing with. This problem involves both a diagonal and flat region. Therefore, you must use both Q = nΔH and Q = mcT. You solve for both parts separately and then add them together! We already solved both of these parts individually. All that remains is to add them together!

29. Step 2: Plug in your known values into the equation Q = nΔHf. • Since ΔHf is per mole (n), your amount must be in moles. 10. g H2O x 1 mol H2O = 0.55 mol H2O 1 18.02 g H2O Q = nΔHf Q = 0.55 mol H2O x 6.02 kJ mol

30. Step 2 Plug in your known values into the equation Q = mcT. Q = mcT T = Tfinal – Tinitial = 20. C - 0C = 20. C Q = 10. g x 4.18 J x 20. C gC

31. Step 3: • Solve problem. Cancel out like units in the numerators and denominators. Label and round your final answers to the correct number of sig. figs. Q = 0.55 mol H2O x 6.02 kJ = 3.3 kJ mol

32. Step 3 Solve problem. Cancel out like units in the numerators and denominators. Label and round your final answers to the correct number of sig. figs. Q = 10. g x 4.18 J x 20. C = 836 J = 840 J gC

33. Step 4 Add together the final answers to each problem. You are doing this because the question asked for the TOTAL amount of energy needed to perform this change. Remember, you are dealing with a flat and a diagonal region. Be aware of unlike terms!

34. Flat region: Q= 3.3 kJ Diagonal region: Q = 840 J Since both values are in different units, you must convert one of them! 840 J x 1kJ = 0.84 kJ 1 1000 J 0.84 kJ + 3.3 kJ = 4.14 kJ = 4.1 kJ (Remember, round to decimal places in addition and subtraction!)

35. Try it How much energy needs to be removed to condense 10.g of steam at 100. C and cool to 40. C? Realize this is a two-part question. You have a flat and diagonal region on the graph involved. We’ll break this into two parts. Page 6 of your packet

36. Sketch

37. Step 1 Determine what part(s) of the heating curve you are dealing with. This problem involves both a diagonal and flat region. Therefore, you must use both Q = nΔH and Q = mcT. Since you are condensing the water, you must use the molar heat of vaporization: ΔHv= 40.6 kJ/mol. In addition, since you are in the liquid range, you must use the specific heat for liquid water.

38. Flat region: Step 2a: Plug in your known values into the equations: Q = nΔHv. Remember, since ΔH is per mole (n), your amount must be in moles. NOTE: Since we're cooling, energy is being removed. Therefore ΔH must be negative! It’s the same value, just put a negative sign in front of it. We’ve already converted 10. g H2O into moles. It is equal to: ___0.55 mol___ Q = nΔHv Q = (_0.55mol_)(_-40.6KJ_) (Negative because it is 1 mol condensing)

39. Flat region: Step 3a: Solve the problem. Cancel out like units in the numerators and denominators. Label and round your final answers to the correct number of sig. figs. Q = ___-22KJ____

40. Diagonal region: Step 2b: Plug in your known values into the equation Q = mcT. T = (__40ºC___) – (___100ºC__) = ___-60ºC__ NOTE: The temp change is negative! Q = (_10.g__)(__4.18J__)(__-60ºC__) 1 g ºC

41. Diagonal region: Solve each problem. Cancel out like units in the numerators and denominators. Label and round your final answers to the correct number of sig. figs. Q = ___-2500J___

42. Step 4: Sum the values for Q. Watch units and signs! Q= (__-22KJ__) + (__-2.5KJ__) = __-24.5KJ___

43. Inter vsIntramolecular Intramolecular forces- bonding forces that hold the atoms of a molecule together • Ex. covalent and ionic Intermolecular forces- forces among molecules that cause them to aggregate to form a solid or a liquid

44. Helpful Hint: Think INTERstate – a road that goes between two states Think INTRAmurals - a game played with teams from the same school

45. Helpful Hint: • These are the forces that are broken during phase changes • As energy is added, molecules start to vibrate. The molecules will eventually achieve a greater movement and disorder. • As more energy is added, molecules move farther and farther apart until the intermolecular forces are overcome and the substance turns into a gas, where the distance between molecules is too great to have any intermolecular forces.

46. Dipole-dipole attraction • Dipole-dipole attraction- polar molecules have dipole moments (partial positive and partial negative ends); molecules line up so that the partial positive end of one molecule is next to the partial negative end of another molecule NOTE: What two factors determine if a molecule has a dipole (is polar)? • Bond Type – Need Polar Bonds • Molecular Shape – Need Polar Molecule

47. NOTE: • Dipole-dipole attractions are only 1% as strong as a covalent or ionic bond. They are only effective when molecules are close together, such as in solids and liquids. They are not a factor in gases because the molecules are too far apart.

48. Dipole – Dipole Interactions

49. Ex. HCl Dipole- dipole attraction from one HCl molecule to another HClmolecule H─Cl······ H─Cl δ+ δ-δ+ δ- Polar covalent bonds present in the HClmolecule

50. Hydrogen bonding • Hydrogen bonding- special dipole-dipole attraction that occurs among molecules in which hydrogen is bonded directly to a highly electronegative element (ex. N, O, F); the great polarity of the bond and the small size of the hydrogen atom makes this an unusually strong force. Although called Hydrogen “bonding”- they are not a true bond like ionic or covalent!