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 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

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 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

5. Endothermic vs. Exothermic Chemistry, Raymond Chang, 11 th edition5

6. Mullis6 Sample Heat of Fusion Problem q = mcΔT If the temperature change includes a phase change, you must split the problem into steps to separate the phase change. Use the heat of fusion or heat of vaporization instead of specific heat for the phase change. Ex. Calculate the heat that must be absorbed by 10.0 grams of ice (H 2 O) at -9.00 ºC to convert it to water at 25.0 ºC. 1. Solid warming: -9.0 ºC to 0 ºC: (10.0 g) x (2.09 J/gºC) x (0.0 – (-9.00 ºC) ) = 188 J 2. Phase Change at 0 ºC: (10.0 g) x (334 J/g) = 3340 J 3. Liquid warming: 0 ºC to 25.0 ºC: (10.0 g) x (4.18 J/gºC) x (25.0 - 0 ºC ) = 1045 J Total amount of heat absorbed = 4573 J ~ 4570 J

7. Mullis7 A Bigger Heat of Fusion Problem q = mcΔT Calculate the heat needed to convert 25.0 grams of ice (H 2 O) from -30.0 ºC to steam at 150. ºC. 1. Solid warming: -9.0 ºC to 0 ºC: (25.0 g) x (2.09 J/gºC) x (0.0 – (-30.0 ºC)) = 1568 J 2. Phase Change at 0 ºC: (25.0 g) x (334 J/g) = 8350 J 3. Liquid warming: 0 ºC to 100. ºC: (25.0 g) x (4.18 J/gºC) x (100. - 0 ºC ) = 10450 J 4. Phase Change at 0 ºC: (25.0 g) x (2260 J/g) = 56500 J 5. Vapor warming: 100. ºC to 100.0 ºC: (25.0 g) x ( 2.03 J/gºC) x (150. – 100. ºC ) = 2537 J Total amount of heat absorbed = 79405 J ~ 79.4 kJ

8. Mullis8 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

9. Mullis9 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

10. Mullis10 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

11. Mullis11 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.

12. Mullis12 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.

13. ΔH : Enthalpy ΔH f is heat of formation- the energy required to form the substance ΔH f is heat of formation- the energy required to form the substance An element in its standard state (g,l or s) under standard conditions has ΔH f value of zero. An element in its standard state (g,l or s) under standard conditions has ΔH f value of zero. Examples: ΔH f for Fe(s) = 0 Examples: ΔH f for Fe(s) = 0 ΔH f for Cl 2 (g) =0 …BUT ΔH f for Cl(g) ≠0 -Atomic Cl is at a higher potential energy than the covalently bonded diatomic molecule. Mullis13

14. ΔH rxn : Enthalpy (Heat of reaction) ΔH rxn is a state function and is calculated by finding the sum of ΔH f values for the products and subtracting the sum of the reactants. ΔH rxn = Σ ΔH f products - Σ ΔH f reactants Ex: C 2 H 5 OH + 3O 2  2CO 2 + 3H 2 O Mullis14

15. Ex: C 2 H 5 OH + 3O 2  2CO 2 + 3H 2 O(l) ΔH rxn = Σ ΔH f products - Σ ΔH f reactants ΔH rxn =[2(ΔH f CO2 ) + 3(ΔH f H2O )] – [ΔH f C2H5OH + 3(ΔH f O2 )] From standard ΔH f table of values: ΔH rxn =[2(-393.5) + 3(-285.8)] – [-277.7+ 3(0)] = -1367 kJ Recall that ΔH f for an element in its standard state is zero. This shows that the combustion of ethanol by this equation has a change in enthalpy of -1367 kJ. Since the coefficient of ethanol in the equation is 1, it is also true that one mole of ethanol produces 1367kJ of energy. This shows that the combustion of ethanol by this equation has a change in enthalpy of -1367 kJ. Since the coefficient of ethanol in the equation is 1, it is also true that one mole of ethanol produces 1367kJ of energy. The reaction is exothermic: The sign of ΔH rxn is negative. There is more energy in the reactants than the products. The reaction is exothermic: The sign of ΔH rxn is negative. There is more energy in the reactants than the products. 15

16. SubstanceFormulaΔH f ° (kJ/mol) SubstanceFormulaΔH f ° (kJ/mol) Acetylene C 2 H 2 (g) 226.7Hydrogen chloride HCl (g) -92.30 Ammonia NH 3 (g) -46.19Hydrogen fluoride HF (g) -268.6 Benzene C 6 H 6 (l) 49.0Hydrogen iodide HI (g) 25.9 Calcium carbonate CaCO 3 (s) -1207.1Methane CH 4 (g) -74.8 Calcium oxide CaO (s) -635.5Methanol CH 3 OH (l) -238.6 Carbon dioxide CO 2 (g) -393.5Propane C 3 H 8 (g) -103.85 Carbon monoxide CO (g) -110.5Silver chloride AgCl (s) -127.0 Diamond C (s) 1.88 Sodium bicarbonateNaHCO 3 (s) -947.7 Ethane C 2 H 6 (g) -84.68Sodium carbonate Na 2 CO 3 (s) -1130.9 Ethanol C 2 H 5 OH (l) -277.7Sodium chloride NaCl (s) -410.9 Ethylene C 2 H 4 (g) 52.30Sucrose C 12 H 22 O 11 (s ) -2221 Glucose C 6 H 12 O 6 (s ) -1273Water H 2 O (l) -285.8 Hydrogen bromide HBr (g) -36.23Water vapor H 2 O (g) -241.8

17. Mullis17 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

18. Mullis18 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.

19. Mullis19 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

20. Mullis20 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