1. Mullis1 First Law of Thermodynamics (Law of Conservation of Energy) The combined amount of matter and energy in the universe is constant. The combined amount of matter and energy in the universe is constant. Potential energy Kinetic energy Potential energy Kinetic energy Enthalpy = heat gained or lost by system: ΔH Enthalpy = heat gained or lost by system: ΔH ΔH < 0: Exothermic process ΔH < 0: Exothermic process ΔH > 0: Endothermic Process ΔH > 0: Endothermic Process

2. Endothermic vs. Exothermic Chemistry, Raymond Chang, 11 th edition2

3. Mullis3 q and ΔH: Signs Sign is determined by the experience of the system: Heat in System (The reaction occurs in here!) ΔH > 0 Δis positive. Δ H is positive. Endothermic Rxn System (The reaction occurs in here!) ΔH < 0 Δis negative. Δ H is negative. Exothermic Rxn Heat out

4. Mullis4 Calorimetry Calorimeter = Measures temp change in a process Calorimeter = Measures temp change in a process Heat capacity = amount of heat to raise temp by 1K (or 1º C) Heat capacity = amount of heat to raise temp by 1K (or 1º C) Specific heat = heat capacity for 1 g of a substance Specific heat = heat capacity for 1 g of a substance (Symbol for specific heat is usually C) Amount of heat absorbed by a substance calculated using mass, specific heat and temperature change: Amount of heat absorbed by a substance calculated using mass, specific heat and temperature change: q = ΔE = mc Δ T

5. Mullis5 Heat Measurements Using Calorimeter 50.0 mL of 0.400 M CuSO 4 at 23.35 ºC is mixed with 50.0 mL of 0.600 M NaOH at 23.35 ºC in a coffee-cup calorimeter with heat capacity of 24.0 J/ ºC. After reaction, the temp is 25.23 ºC. The density of the final solution is 1.02 g/mL. Calculate the amount of heat evolved. Specific heat of water is 4.184J/g ºC. q = ΔE = mc Δ T Mass = (50 mL+50 mL)(1.02g/mL) = 102 g Δ T = 25.23-23.35 = 1.88 ºC Heat absorbed by solution = (102g)(4.18J)(1.88 ºC) = 801J g ºC g ºC Add this to heat absorbed by calorimeter = (24.0J)(1.88 ºC) = 45.1J g ºC Total heat liberated by this reaction = 846 J

6. Mullis6 Hess’s Law Example C 2 H 5 OH + 3O 2  2CO 2 + 3H 2 O ΔH = -1367 kJ/mol C 2 H 4 + 3O 2  2 CO 2 + 2H 2 O ΔH = -1411 kJ/mol Find ΔH for C 2 H 4 + H 2 O  C 2 H 5 OH 2CO 2 + 3H 2 O  C 2 H 5 OH + 3O 2 ΔH = 1367 kJ/mol C 2 H 4 + 3O 2  2 CO 2 + 2H 2 O ΔH = -1411 kJ/mol ___________________________________________ C 2 H 4 + H 2 O  C 2 H 5 OHΔH = -44 kJ/mol

7. Mullis7 Heating Curve at Constant Pressure Curve is flat during phase changes. Area A: Temperature remains constant until all the solid has become liquid because melting requires energy. Once the energy is no longer required for phase change, kinetic energy again increases.

8. Mullis8 Heating Curve at Constant Pressure, cont. Length of horizontal line A is proportional to the heat of fusion. The higher the heat of fusion, the longer the line. Line B is longer than A because heat of vaporization is higher than heat of fusion. Ex: For water: Heat of fusion = 334 J/g Heat of vaporization =2260J/g ΔH fus = 6.02 kJ/mol ΔH vap = 40.7 kJ/mol

9. Mullis9 Heating Curve at Constant Pressure, cont. Gas and solid warming slopes are steeper than that for liquid warming. The specific heat of the liquid phase is usually greater than that of the solid or gas phase. Ex: for water: a. Solid = 2.09 J/g-ºC b. Liquid = 4.18 J/g-ºC c. Gas = 2.03 J/g-ºC