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 ΔE = q + W ΔE = q + W What are the symbols? What are the symbols? Internal Energy: ΔE Internal Energy: ΔE Heat transferred (into or out of system): q Heat transferred (into or out of system): q Work done (on or by the system): w Work done (on or by the system): w Enthalpy = heat gained or lost by system: ΔH Enthalpy = heat gained or lost by system: ΔH ΔH < 0: Exothermic process ΔH < 0: Exothermic process

2. Mullis2 q and ΔH: Signs Sign is determined by the experience of the system: Heat in System (The reaction occurs in here!) ΔH > 0 q > 0 Both q & Δare positive. Both q & Δ H are positive. Endothermic Rxn System (The reaction occurs in here!) ΔH < 0 q < 0 Both q & Δare negative. Both q & Δ H are negative. Exothermic Rxn Heat out

3. Mullis3 Relationships: ΔH, ΔE, q and w SignCondition q, ΔH + Heat transferred to system q, ΔH - Heat transferred from system w+ Work done by surroundings w- Work done by system ΔEΔEΔEΔE+ q > 0 and w > 0 ΔEΔEΔEΔE- q < 0 and w < 0 ΔE = q + w ΔE = ΔH - P ΔV (P ΔV is the amount of work done by expanding gases, but in most reactions, there is a very small volume change, so ΔE ~ ΔH in many cases.)

4. Mullis4 Internal Energy Internal energy = ΔH – PΔV = ΔH – ΔnRT Internal energy = ΔH – PΔV = ΔH – ΔnRT If number of moles of gas change in the reaction, include the # moles term. If number of moles of gas change in the reaction, include the # moles term. The combustion of C 3 H 8 (g) to produce gaseous products has ΔH r = -2044.5 kJ-mol -1 at 298 K. Find the change in internal energy for this reaction. The combustion of C 3 H 8 (g) to produce gaseous products has ΔH r = -2044.5 kJ-mol -1 at 298 K. Find the change in internal energy for this reaction. C 3 H 8 (g) + 5O 2 (g)  3CO 2 (g) + 4H 2 O (g) Δn = 7-6 = 1 ΔnRT = 1 mol (8.314 J mol -1 K -1 )(298K) = 2478J Int. Energy = -2044.5 kJ -2.478 kJ = -2047 kJ

5. Mullis5 Calorimetry Calorimeter = Measures temp change in a process Calorimeter = Measures temp change in a process “Bomb” calorimeter = constant volume “Bomb” calorimeter = constant volume Under constant pressure, heat transferred = enthalpy change ( q = ΔE ) Under constant pressure, heat transferred = enthalpy change ( q = ΔE ) 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

6. Mullis6 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

7. Mullis7 Bomb Calorimeter 1.00 g ethanol, C 2 H 5 OH, is burned in a bomb calorimeter with heat capacity of 2.71 kJ/g ºC. The temp of 3000 g H 2 O rose from 24.28 to 26.22 ºC. Determine the ΔE for the reaction in kJ/g of ethanol and then in kJ/mol ethanol. Specific heat of water is 4.184J/g ºC. q = ΔE = mc Δ T Δ T = 26.22 - 24.28- = 1.94 ºC Heat to warm H 2 O = (3000g)(4.18J)(1.94 ºC) = 24.3 x 10 3 J = 24.3 kJ g ºC g ºC Add this to heat to warm calorimeter = (2.71kJ)(1.94 ºC) = 5.26 kJ g ºC g ºC Total heat absorbed by calorimeter and water = 29.6 kJ = 29.6 x 10 3 J Since this is combustion, heat is liberated so change sign for heat of reaction: - 29.6kJ/g ethanol -29.6kJ | 46.1 g = -1360 kJ ethanol g | mol mol g | mol mol

8. Mullis8 Hess’s Law Law of heat summation Law of heat summation The enthalpy change for a reaction is the same whether it occurs by one step or by any series of steps. The enthalpy change for a reaction is the same whether it occurs by one step or by any series of steps. A state function such as enthalpy does not depend on the steps A state function such as enthalpy does not depend on the steps State functions are analogous to your bank account: Many combinations of deposits and withdrawals result in what you watch: The account balance. State functions are analogous to your bank account: Many combinations of deposits and withdrawals result in what you watch: The account balance. ΔHº rxn = ΔHº a + ΔHº b + ΔHº c +… ΔHº rxn = ΔHº a + ΔHº b + ΔHº c +… Application is an accounting/algebra exercise to get resulting reaction ΔHº rxn by adding reactions of known ΔH. Application is an accounting/algebra exercise to get resulting reaction ΔHº rxn by adding reactions of known ΔH.

9. Mullis9 Hess’s Law Write resulting equation and arrange the step equations to get products in step reactions on same side of products in the final reaction. Write resulting equation and arrange the step equations to get products in step reactions on same side of products in the final reaction. If reaction is reversed, change sign of ΔH. If reaction is reversed, change sign of ΔH. Multiply step equations as needed to get the resulting equation. Fractions are allowed. Multiply step equations as needed to get the resulting equation. Fractions are allowed. Recall that coefficients correspond to moles. Recall that coefficients correspond to moles.

10. Mullis10 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