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Chapter 11 Thermochemistry - Heat and Chemical Change

Chapter 11 Thermochemistry - Heat and Chemical Change. 11.3 Heat and changes of state. Things you will learn. Classify the changes that occur during melting/freezing and boiling/condensing Calculate heat changes during changes of state

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Chapter 11 Thermochemistry - Heat and Chemical Change

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  1. Chapter 11Thermochemistry-Heat and Chemical Change 11.3 Heat and changes of state

  2. Things you will learn • Classify the changes that occur during melting/freezing and boiling/condensing • Calculate heat changes during changes of state • Lab-graph the temperature of water as it goes from a mixture of ice and water until it is boiling (with a constant source of heat)

  3. A couple of terms • Molar heat of fusion (∆Hfus) and molar heat of solidification (∆Hsolid). • We learned that the temperature at which some substance melts is the same as the temperature at which it freezes; it should be no surprise that the amount of heat absorbed or released is also the same. • ∆Hfus = -∆Hsolid

  4. A couple more terms • Molar heat of vaporization (∆Hvap) and molar heat of condensation (∆Hcond). • We learned that the temperature at which some substance boils is the same as the temperature at which it condenses; it should be no surprise that the amount of heat absorbed or released is also the same. • ∆Hvap = -∆Hcond

  5. How much heat are we talking about? • The melting of 1 mole of water ice requires 6.01 kJ of energy (1440 calories) • The vaporization of 1 mole of water requires 40.7 kJ of energy (9690 calories) • Remember that it only takes 1 calorie to raise the temperature of 1 g of water 1°C, but 540 calories to change 1 g of water at 100°C to 1 g of steam at 100°C!

  6. Endothermic or exothermic? • H2O(s) H2O(l) • H2O(l) H2O(s) • H2O(l) H2O(g) • H2O(g) H2O(l)

  7. Endothermic or exothermic? • H2O(s) H2O(l) ∆H(fus) = 6.01 kJ/mol • H2O(l) H2O(s) ∆H(solid) = -6.01 kJ/mol • H2O(l) H2O(g) ∆H(vap) = 40.7 kJ/mol • H2O(g) H2O(l) ∆H (cond) = -40.7 kJ/mol • (- ∆H means exothermic) • (+∆H means endothermic)

  8. How much heat is absorbed when 24.8 g H2O(l) at 100°C is converted to steam at 100°C?

  9. How much heat is absorbed when 24.8 g H2O(l) at 100°C is converted to steam at 100°C? • Known: • Mass of water = 24.8 g • ∆H water = 40.7 kJ/mol • Molar mass water = 18 g • Unknown : • kJ absorbed

  10. How much heat is absorbed when 24.8 g H2O(l) at 100°C is converted to steam at 100°C? 24.8 g H2O(l) mol H2O(l) 18 g H2O(l) 40.7 kJ mol H2O(l) 56.1 kJ ∆H(vap) = 40.7 kJ/mol

  11. Heats of solution • Different substances can release heat or absorb heat when they are dissolved in a solvent (we used water) • This is called molar heat of solution ∆H(soln) • These heats are generally written as kJ/mol, and do not depend on the amount of water, but only on the amount of substance being dissolved

  12. How much heat is released when 2.50 moles of NaOH is dissolved in water?

  13. How much heat is released when 2.50 moles of NaOH is dissolved in water? • Known: • Moles of NaOH = 2.50 • ∆H(soln)NaOH = -445.1 kJ/mol • Unknown : • kJ released

  14. How much heat is released when 2.50 moles of NaOH is dissolved in water? -445.1 kJ mol NaOH -1113 kJ 2.50 mol NaOH

  15. Our lab experiment • We will heat a mixture of water and ice with a constant source of heat and plot the temperature of the water (y axis) over time (x axis). • Since the source of energy will remain constant, the time we will plot along the x axis might be considered energy input to the system • We will stop plotting temperature once the water boils. Why?

  16. Homework (HW 14) • Chapter 11.3 • Questions 26-29 • Page 306 • Due: April 15 (red) April 16 (gold)

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