Section 7.3—Changes in State What’s happening when a frozen ice pack melts?

Change in State The energy being put into the system is used for breaking IMF’s, not increasing motion (temperature) Breaking intermolecular forces requires energy To melt or boil, intermolecular forces must be broken A sample with solid & liquid will not rise above the melting point until all the solid is gone. The same is true for a sample of liquid & gas

Melting When going from a solid to a liquid, some of the intermolecular forces are broken The Enthalpy of Fusion (Hfus) is the amount of energy needed to melt 1 gram of a substance The enthalpy of fusion of water is 80.87 cal/g or 334 J/g All samples of a substance melt at the same temperature, but the more you have the longer it takes to melt (requires more energy). Energy needed to melt 1 g Energy needed to melt Mass of the sample

Example Example: Find the enthalpy of fusion of a substance if it takes 5175 J to melt 10.5 g of the substance.

Example Example: Find the enthalpy of fusion of a substance if it takes 5175 J to melt 10.5 g of the substance. H = enthalpy (energy) m = mass of sample Hfus = enthalpy of fusion Hfus = 493 J/g

Vaporization When going from a liquid to a gas, all of the rest of the intermolecular forces are broken The Enthalpy of Vaporization (Hvap) is the amount of energy needed to boil 1 gram of a substance The Hvap of water is 547.2 cal/g or 2287 J/g All samples of a substance boil at the same temperature, but the more you have the longer it takes to boil (requires more energy). Energy needed to boil 1 g Energy needed to boil Mass of the sample

Example Example: If the enthalpy of vaporization of water is 547.2 cal/g, how many calories are needed to boil 25.0 g of water?

Example Example: If the enthalpy of vaporization of water is 547.2 cal/g, how many calories are needed to boil 25.0 g of water? H = enthalpy (energy) m = mass of sample Hfus = enthalpy of fusion DH = 1.37×104 cal

Changes in State go in Both Directions Increasing molecular motion (temperature) Changes in State go in Both Directions Gas Vaporizing or Evaporating Liquid Melting Condensing Solid Freezing

Going the other way The energy needed to melt 1 gram (Hfus) is the same as the energy released when 1 gram freezes. If it takes 547 J to melt a sample, then 547 J would be released when the sample freezes. DH will = -547 J The energy needed to boil 1 gram (Hvap) is the same as the energy released when 1 gram is condensed. If it takes 2798 J to boil a sample, then 2798 J will be released when a sample is condensed. DH will = -2798 J

Example Example: How much energy is released with 157.5 g of water is condensed? Hvap water = 547.2 cal/g

Example Example: How much energy is released with 157.5 g of water is condensed? Hvap water = 547.2 cal/g H = enthalpy (energy) m = mass of sample Hfus = enthalpy of fusion Since we’re condensing, we need to “release” energy…DH will be negative! DH = - 8.6×104 cal

Heating Curves Heating curves show how the temperature changes as energy is added to the sample Boiling & Condensing Point Melting & Freezing Point

Going Up & Down Moving up the curve requires energy, while moving down releases energy +DH -DH

States of Matter on the Curve Liquid & gas Energy added breaks remaining IMF’s Liquid Only Energy added increases temp Gas Only Energy added increases temp Solid Only Energy added increases temp Solid & Liquid Energy added breaks IMF’s

Different Heat Capacities The solid, liquid and gas states absorb water differently—use the correct Cp! Liquid Only Cp = 1.00 cal/g°C Gas Only Cp = 0.48 cal/g°C Solid Only Cp = 0.51 cal/g°C

Changing States Liquid & gas Hvap = 547.2 cal/g Solid & Liquid Hfus = 80.87 cal/g

Adding steps together If you want to heat ice at -25°C to water at 75°C, you’d have to first warm the ice to zero before it could melt. Then you’d melt the ice Then you’d warm that water from 0°C to your final 75° You can calculate the enthalpy needed for each step and then add them together

Example Useful information: Cp ice = 0.51 cal/g°C Example: Cp water = 1.00 cal/g°C Cp steam = 0.48 cal/g°C Hfus = 80.87 cal/g Hvap = 547.2 cal/g Example: How many calories are needed to change 15.0 g of ice at -12.0°C to steam at 137.0°C?

Example Useful information: Cp ice = 0.51 cal/g°C Example: Cp water = 1.00 cal/g°C Cp steam = 0.48 cal/g°C Hfus = 80.87 cal/g Hvap = 547.2 cal/g Example: How many calories are needed to change 15.0 g of ice at -12.0°C to steam at 137.0°C? Warm ice from -12.0°C to 0°C 91.8 cal Melt ice 1213 cal Warm water from 0°C to 100°C 1500 cal Boil water 8208 cal Warm steam from 100°C to 137°C 266 cal Total energy = 11279 cal

How many needed to change 40.5 g of water at 25°C to steam at 142°C? Let’s Practice Useful information: Cp ice = 0.51 cal/g°C Cp water = 1.00 cal/g°C Cp steam = 0.48 cal/g°C Hfus = 80.87 cal/g Hvap = 547.2 cal/g Example: How many needed to change 40.5 g of water at 25°C to steam at 142°C?

How many needed to change 40.5 g of water at 25°C to steam at 142°C? Let’s Practice Useful information: Cp ice = 0.51 cal/g°C Cp water = 1.00 cal/g°C Cp steam = 0.48 cal/g°C Hfus = 80.87 cal/g Hvap = 547.2 cal/g Example: How many needed to change 40.5 g of water at 25°C to steam at 142°C? Warm water from 25°C to 100°C 3038 cal Boil water 22162 cal Warm steam from 100°C to 142°C 816 cal Total energy = 26016 cal