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Section 7.3—Changes in State

Section 7.3—Changes in State. Change in State. During a change in state:. We are breaking intermolecular forces. . Breaking intermolecular forces requires energy.

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Section 7.3—Changes in State

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  1. Section 7.3—Changes in State

  2. Change in State During a change in state: • We are breaking intermolecular forces. • Breaking intermolecular forces requires energy • During a change in state, the energy being put into the system is used to breaking IMF’s, not increasing motion (temperature). WE CAN’T USE mc∆T TO GET THIS ENERGY!! A sample with solid & liquid will not rise above the melting point until all the solid is gone. The temp stays the same! WE CAN’T USE mc∆T TO GET THIS ENERGY!! During phase changes we need ANOTHER equation!

  3. Melting - at 1 atm of Pressure • Enthalpy of Fusion (Hfus) - 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 Example: All samples of ice will melt at 0°C and 1atm BUT the more ice you melt, the more energy you need!!! Energy needed to melt 1 g Energy needed to melt Mass of the sample

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

  5. 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

  6. Vaporization – at 1 atm of Pressure • Enthalpy of Vaporization (Hvap) - 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 • Example: All samples of water boil at 100°C BUT the more you have the more energy it takes to boil! Energy needed to boil 1 g Energy needed to boil Mass of the sample

  7. 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?

  8. 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

  9. Vaporizing or Evaporating Gas Increasing molecular motion (temperature) Liquid Melting Condensing Freezing Solid Changes in State go in Both Directions Boiling Breaks All IMF’s!!! Melting only breaks some IMF’s!!!

  10. 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 vaporize a sample, then 547 J would be released when the sample condenses. 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

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

  12. 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

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

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

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

  16. 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 OnlyCp = 0.48 cal/g°C Solid OnlyCp = 0.51 cal/g°C

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

  18. 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

  19. Example 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 calories are needed to change 15.0 g of ice at -12.0°C to steam at 137.0°C?

  20. Example 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 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

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