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H 2 O (s) H 2 O (  ) H 2 O (g). Heat & Changes of State.

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Presentation on theme: "H 2 O (s) H 2 O (  ) H 2 O (g). Heat & Changes of State."— Presentation transcript:

1 H 2 O (s) H 2 O (  ) H 2 O (g)


3 Heat & Changes of State

4 vaporization boiling melting Heat & Changes of State condensation freezing sublimation deposition sublimation

5 Heat & Changes of State Fusion 0  C Vaporization 100  C Molar Heat of: Latent Heat of: 80 cal/g 540 cal/g Heat involved per gram Heat involved per mole

6 Energy Requirements for changing state: In ice the water molecules are held together by strong intermolecular forces. The energy required to melt 1 gram of a substance is called the latent heat of fusion For ice it is 80.0 cal/g The energy required to change 1 gram of a liquid to its vapor is called the latent heat of vaporization For water it is 540 cal/g

7 It takes more energy to vaporize water than to melt it. This is because in melting you weaken the intermolecular forces. Here about 1/6 of the hydrogen bonds are broken. In vaporization you totally break them. All the hydrogen bonds are broken Fusion is when a solid melts to form a liquid Vaporization is when a liquid evaporates to form a gas.

8 A-B = Solid ice, temperature is increasing. Particles gain kinetic energy, vibration of particles increases. Heating and cooling curve for water heated at a constant rates. Ice

9 B-C = Solid starts to change state from solid to liquid. Temperature remains constant as energy is used to break inter- molecular bonds. H 2 O (s)  H 2 O (  ) energy required  80 cal/g 0ºC

10 C-D = temperature starts to rise once all the solid has melted. Particles gain kinetic energy. Liquid water

11 D-E = Liquid starts to vaporize, turning from liquid to gas. The temperature remains constant as energy is used to break inter- molecular forces. H 2 O (  )  H 2 O (g) energy required  540 calg 100ºC

12 E-F = temperature starts to rise once all liquid is vaporized. Gas particles gain kinetic energy. steam

13 Calculating Energy Changes TemperatureTemperature Heat Added ( o C)  H f = 80 cal/g  H v = 540 cal/g 3. c l = 1.00 cal/g o C 5. c v = 0.50 cal/g o C 1. c s = 0.50 cal/g o C Heating Diagram

14 Calculating Energy Changes Q (gained or lost) = m x L. H. (fusion/vaporization) Phase change: Temperature change: heat = mass specific (T f - T i ) heat Q (gained or lost) = m c  T

15 Problem How much energy is required to heat 25 g of liquid water from 25  C to 100  C and change it to steam?

16 Step 1: Calculate the energy needed to heat the water from 25  C to 100  C Q = 25g  1.0 cal g -1  C –1  75  C = Q = m  c   T

17 Step 2: Vaporization: Use the Latent Heat to calculate the energy required to vaporize 25g of water at 100  C.25g  1mol H 2 O / 18g mol -1 H 2 O = 1.4 mol H 2 O  vap H (H 2 O) = 1.4 mol H 2 O  40.6kJ/mol = 57 kJ Q = 25.0 g  540 cal/g =

18 Total energy change is:

19 18.0 g x 0.5 cal/g o C x 7 o C 18.0 g x 80.0 cal/g 18.0 g x 1.0 cal/g o C x 100 o C 18.0 g x 540.0 cal/g 18.0 g x 0.5 cal/g o C x 25 o C = 63 cal = 1440 cal = 1800 cal = 9720 cal = 225 cal = 13248 cal Calculating Energy Changes Calculate the total amount of heat needed to change 1 mole of ice at -7 o C to steam at 125 o C. TT  phase TT TT

20 Intermolecular Forces Intra-molecular forces are (within the molecule) while inter-molecular forces are (between molecules) Types of inter-molecular forces dipole-dipole (1% as strong as covalent bonds) POLAR MOLECULES A special type of dipole-dipole force is the hydrogen bond. These form between molecules that contain a hydrogen atom bonded to a very electronegative element like N, O or F. Hydrogen bonds are very strong compared to an ordinary dipole-dipole bond. E.g HF, NH 3, H 2 O all form hydrogen bonds Hydrogen bonding 10% as strong as covalent bonds

21 London dispersion forces (instantaneous and induced dipoles) NON-POLAR MOLECULES

22 Non-polar molecule Movement of electrons causes an instantaneous dipole This induces a dipole in a nearby molecule This instantaneous dipole will effect any nearby molecules

23 Water molecules are polar molecules. The  - oxygen forms intermolecular bonds with the  + hydrogen of another water molecules. Water has a special type of intermolecular bond called a hydrogen bond. Inter-molecular forces

24 Ice molecules are locked in fixed positions, held by intermolecular-bonds. Ice is less dense than liquid water because the molecules are further apart than in liquid water.

25 Other properties of Liquids: Many properties are due to the forces between the particles. Why do some liquids exhibit capillary action? Why are some liquids more viscous than others? Why do liquids on a surface form droplets? Hg H 2 O

26 The inward force or pull which tends to minimize the surface area of any liquid is surface tension. This allows insects to walk on water!

27 Surface tension is caused by hydrogen bonding between water molecules. The more polar a liquid the stronger its surface tension. The smallest surface area a liquid can form is a sphere. Hgpure H 2 OH 2 O with detergent Surfactants are chemicals that decrease the surface tension of water, detergents and soaps are examples.

28 Viscosity is the resistance to motion of a liquid. Maple syrup is more viscous than water. But water is much more viscous than gasoline or alcohol. The stronger the attraction between molecules of a liquid, the greater its resistance to flow and so the more viscous it is.

29 Capillary action is the spontaneous rising of a liquid in a narrow tube. Two forces are responsible for this action: Cohesive forces,the intermolecular forces between molecules of the liquid Adhesive forces, between the liquid molecules and their container Hg H 2 O If the container is made of a substance that has polar bonds then a polar liquid will be attracted to the container. This is why water forms a concave meniscus while mercury forms convex meniscus

30 The fact that water has both strong cohesive (intermolecular) forces and strong adhesive forces to glass, it pulls itself up a glass capillary tube. This also allows it to be drawn up high into trees like giant redwoods.

31 If you place a liquid in a container, then some of the particles will have enough kinetic energy to evaporate. You will notice the amount of liquid decreasing. Dynamic - at the same time some of these gaseous molecules condense to reform liquid. In an open container all the liquid will eventually evaporate out if they have enough kinetic energy. Volatility – the higher the vapor pressure, the more volatile the liquid. Temperature – vapor pressure increases in a non-linear fashion with increasing temperature.

32 Closed system - In a sealed container, molecules will start to evaporate and the liquids volume will decrease. No, both evaporation and condensation continue. But an equilibrium has been reached. The rate of evaporation = the rate of condensation But some of these molecules will then condense and after a short time the volume of the liquid will not change. Has evaporation and condensation stopped?

33 When water is heated bubbles of vapor form within it. The vapor pressure in the bubble is the same as the vapor pressure of the water at that temperature. As long as this vapor pressure is less than atmospheric pressure the bubbles collapse. When the temperature of the water reaches a point that the vapor pressure of the bubble equals atmospheric pressure, the bubbles don’t collapse, they get larger and more form and escape as steam. The water begins to boil. The normal boiling point of a liquid occurs at 1 atm.

34 Calculating Energy Changes: Solid to liquid How much energy is required to melt 8.5 g of ice at 0  C? The molar heat of fusion for ice is 6.02 kJmol -1 Step 1: How many moles of ice do we have? n = m/Mn = 8.5g / 18gmol -1 = 0.47 mol H 2 O Step 2: Use the equivalence statement to work the energy (6.02 kJ is required for 1 mol H 2 O) kJ = 0.47 mol H 2 O  6.02 kJ / mol H 2 O = 2.8kJ

35 What is specific heat capacity? The amount of energy required to change the temperature of one gram of a substance by 1  C. Another name for specific heat is a calorie (1 calorie = 4.184 Joules) Specific heat capacity of liquid water (H 2 O (L) ) is 4.18 J g -1  C –1. Water (s) = 2.03 J g -1  C –1  0.5 cal/g to break up ice Water (g) = 2.0 J g -1  C –1 10  C11  C

36 Calculating the energy to increase the temperature of liquid water. Calculating specific heat using the equation: Q = ms (t f  t i ) orQ = energy (heat) required Q = ms  T ors = specific heat capacity Heat (H) = ms (t f  t i )m = mass of the sample  T = change in temperature in  C EXAMPLE: How much energy does it take to heat 10g of water from 50 to 100  C ? Specific heat capacity of water = 4.184 J g -1  C –1 Q = m  s   T Q = (10g)  (4.184 J g -1  C -1)  (50  C) = 2.1  10 3 J

37  vap H (H 2 O) = 1.4 mol  40.6kJ/mol = 57 kJ


39 Explain these trends in Boiling points Boiling point is effected by the strength of the inter-molecular forces between liquid molecules. The general trend is an increase in B.P. due the greater size of the molecules and hence the greater intermolecular forces The anomalous B.P. for H 2 O, HF, and NH 3 are explained by the fact that they exhibit hydrogen bonding.

40 Heating Diagram

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