Energy Heat Work Heat Capacity Enthalpy

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1 Energy Heat Work Heat Capacity Enthalpy 1 Prof. Zvi C. Koren

2 Thermodynamics vs. Kinetics Thermodynamics Thermo = Thermo + Dynamics E (Note: Absolute E can never be determined by humans!) Can a reaction occur??? (Is it spontaneous)??? Kinetics Diamond vs. Graphite Thermodynamically unstable Kinetically stable Thermodynamically stable C(di)? C(gr) 2 Prof. Zvi C. Koren

3 Energy Kinetic Energy = K.E. = energy of motion = ½ mv 2 Potential Energy = P.E. = energy of position = m g h, g = 9.81 m/s 2, h = height Energy Units SI unit: Joule (J) 1 J = 1 kg m 2 /s 2 ( kms ) [Note units of mv 2 ] = 1 V C = 1 Pa m 3 = 10 7 erg, 1 erg = 1 g cm 2 /s 2 ( cgs ) ev 1 cal J (exactly) 1 BTU = J 1 kw hr = 3.6 MJ 1 L atm J (exactly) 1 erg = ev R = Gas Constant = energy/mol K = L atm/mol K = 1.99 cal/mol K = 8.31 J/mol K 3 Prof. Zvi C. Koren

4 The First Law of Thermodynamics Popular Expression: Law of Conservation of Energy Mathematical Expression (and more general): E = q heat + w work w: For example, compression or expansion Note the equation is NOT any of the following: E = q + w or E = q + w E is a state function q and w are path functions w (P 1,V 1,T 1 ) (P 2,V 2,T 2 ) q E 1 E 2 E = E 2 - E 1 4 Prof. Zvi C. Koren

5 Work and Heat Work: Energy transferred when an object is moved against an opposing force. Note: Two conditions need to be met: movement and resistance. Heat: Energy transferred from a hot body (at T high ) to a cold body (at T low ) WORK: w = f opposing d = f opposing x needed for direction Surroundings (piston) System p opposing f opposing m gas expansion compression m w = f opposing x = pa x = p V w = P V f = force d = distance f opposing = weight = mg w = f d = mg h Also: p = f/a f = p A For piston of constant weight or if open to atmosphere: p opposing = P internal 5 Prof. Zvi C. Koren

6 Qualitative Differences Between q and w w causes ordered motion Work that the SYSTEM performs causes the molecules (or atoms) in an object in the SURROUNDINGS to all move in the same direction. q causes random motion Heat flowing from SYSTEM to SURROUNDINGS causes the molecules in the surroundings to move in random directions. 6 Prof. Zvi C. Koren

7 Heat Capacity, C Amount of heat (q) needed to raise a quantity of substance by 1 degree For: 1 gram specific heat capacity Units of C cal (or J)/g deg For: 1 mole molar heat capacity Units of C cal (or J)/mol deg For example, for water: C specific = 1 cal/g deg, C molar =? 7 Prof. Zvi C. Koren

8 Solids Liquids Gases Specific Heat Capacities of Some Common Substances Substance Al Fe Cu Au H 2 O(s) NH 3 (l) C 2 H 5 OH(l) H 2 O(l) H 2 O(g) O 2 (g) N 2 (g) Name Aluminum Iron Copper Gold Ice Wood Concrete Glass Granite Ammonia Ethanol Water (liquid) Steam Oxygen Nitrogen C (J/g deg) Law of Dulong and Petit: Molar Heat Capacities of Elemental Metals 3R cal/mol deg Note: C(liq) > C(solid) Why? (one of the highest) 8 Prof. Zvi C. Koren

9 Heating Substances Let s heat water or Cu from one temperature to another. What are the factors that affect how much heat is needed in each case? q bathtub q cup Why? Start with C. Recall its units: q = C m t C n t (Note the overall units) If: q = + (endothermic), q = (exothermic) 9 Prof. Zvi C. Koren

10 Heat Transfer For example: A hot iron rod is placed in cold water. Eventually, everything comes to thermal equilibrium Recall: q = C m t מאזן חום בתהליכי העברת חום Heat Balance Σq i = 0 (assuming no heat loss to surroundings) q hot + q cold = 0 q hot = q cold Problems involving 7 parameters in 1 equation 10 Prof. Zvi C. Koren

11 Temp. Heating Involving Phase Changes (Physical processes) Heating Curve (Cooling Curve) T bp l Vaporization: l v H vap (H 2 O: 2260 J/g) Condensation: v l, H cond =? g T mp s Fusion: s l H fus (H 2 O: 333 J/g) For H 2 O: C (s) = 2.06 J/g deg C (l) = J/g deg = 1.0 cal/ g deg = 1.88 J/g deg C (g) Solidification (or Crystallization): l s, H solid (or cryst) = time For example: Calculate the heat ( thermal energy ) required for the following process: 10.0 g, (ice, -10 o C) (steam, 120 o C) 11 Prof. Zvi C. Koren Prof. Zvi C. Koren

12 Temp. ( 0 C) Heating/Cooling Curve for Water. 1 mol water is heated from 10 0 C to C. A constant heating rate of 100 J/min is assumed. 100 vaporization [Constant Pressure: P = f/a; f = w = mg] Piston m s s l l l v g 0 fusion Time (min) Prof. Zvi C. Koren

13 Temp. ( 0 C) Calculation of the Heats Involved With Each Step in the Heating/Cooling Curve 100 vaporization Symbol Name Value for H 2 O C(l) specific heat capacity (of liquid) 1.00 cal/g deg 4.18 J/g deg C( l ) C-bar molar heat capacity (of liquid) cal/mol deg 75.2 J/mol deg ΔH fus heat of fusion 333 J/g ΔH vap heat of vaporization 2250 J/g 0 fusion Find the values of C(s) and C(g) of H 2 O Time (min) Prof. Zvi C. Koren

14 Enthalpy H E + PV (a convenient definition for H) Enthalpy (Note: Absolute H can never be determined. Why? H = E + (PV) = q + w + (PV) = q - P V + P V, [P] H = q P q P open to atmosphere: P internal = p external = constant p external P internal Rxn V is constant (not P) Note: For constant volume processes, V = 0: w = P V = 0 And E = q V q V Rxn. 14 Prof. Zvi C. Koren

15 Calculating Heat of Reaction, H rxn : Recall, H E + PV H rxn = E rxn + (PV) rxn Energy of Reaction, E rxn H rxn E rxn + RT n g For example, consider the following rxn.: 2A(s) + B(l) + 4C(g) 2D(g) + E(g) H rxn E rxn + RT n g H rxn E rxn + RT (3-4) H rxn E rxn - RT (PV) rxn = (PV) products - (PV) reactants = (PV) products(s,l,g) - (PV) reactants(s,l,g) But, for a given quantity, V gas >> V solid,liquid PV gas >> PV solid,liquid (PV) products(g) - (PV) reactants(g) Assume ideal gases: PV = nrt, (nrt) products(g) - (nrt) reactants(g) = RT n g H = q P E = q V 15 Prof. Zvi C. Koren

16 Calculating Heat of Reaction, H rxn : Hess s Law of Heat Summation If a rxn is made up of other rxns, then the heats are summed. Why? Because H (like E) is a state function. For example: H A Reactants Products Problem: H B HC H D H E [Recall Born-Haber Cycle] H A = H B + H C + H D + H E (1) (2) (3) Find H for rxn (1), (1) A + 2D C, H 1 =? From the following data: (2) A + 2B 5C, H 2 = 50 kj (3) B 2C + D, H 3 = -75 kj Solution: Because rxn(1) = rxn(2) 2 rxn(3), H 1 = H 2 2 H 3 = 50 2 (-75) = 200 kj. By the way, of course it s the same for E: E 1 = E 2-2 E 3 But recall: K 1 = K 2 / K Prof. Zvi C. Koren

17 Calculating Heat of Reaction, H rxn : Heats (or Enthalpies) of Formation The Standard State standard state of a substance (at a specific T) = most stable state of the substance at 1 atm (or 1 bar) at that T. For example, the standard state for nitrogen: At 25 o C: N 2 (diatomic) and a gas; At 2,000,000 o C: N (monatomic) and a gas, probably even N + ; At -270 o C: crystalline (solid) At other temps., between T mp and T bp, the liquid is most stable Another example, the standard state for carbon: At 25 o C: graphite (solid); At 1 atm, graphite is more stable than diamond. At 2,000,000 o C: C (monatomic) and a gas; (continued) 17 Prof. Zvi C. Koren

18 Formation (continued) Formation is a rxn where: 1 mole of a compound is formed from its elements in their standard state (most stable form at 1 atm) For example, formation of CH 4 : o C(s,gr) + 2H 2 (g) CH 4 (g), H f (CH 4 ) (measured from q P of the rxn) o H f = standard heat (or enthalpy) of formation Note: (any property) final initial H rxn H final H initial = ΣH products ΣH reactants So, in effect: o H f (CH 4 ) = H(CH 4 ) H(C) 2 H(H 2 ) So, the enthalpy of formation of a compound is in effect the relative enthalpy of that compound, that is, its enthalpy relative to the enthalpies of the elements from which it is composed. (continued) 18 Prof. Zvi C. Koren

19 (continued) Consider the following combustion reaction: C 6 H 6 (l) + 7½ O 2 (g) 6 CO 2 (g) + 3H 2 O(l) Recall: (any property) final initial H rxn H final H initial = ΣH products ΣH reactants H comb rxn = 6 H(CO 2 ) + 3 H(H 2 O) H(C 6 H 6 ) 7½ H(O 2 ) (This last equation is correct but not useful, because we can never know absolute H. So, we must use relative enthalpies, that is enthalpies of formation, H f.) So, build the overall reaction from a series of formation reactions (in the Hess-way): 6 (C + O 2 CO 2 ), 6 H f (CO 2 ) 3 (H 2 + ½ O 2 H 2 O), 3 H f (H 2 O) (6 C + 3 H 2 C 6 H 6 ), H f (C 6 H 6 ) C 6 H 6 (l) + 7½ O 2 (g) 6 CO 2 (g) + 3H 2 O(l), H comb rxn =? o o o o H comb = 6 H f (CO 2 ) + 3 H f (H 2 O) H f (C 6 H 6 ) We can generalize that for any reaction: H o rxn = H o f - H o f P R H o f (element) 0 19 Prof. Zvi C. Koren

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21 thermometer stirrer Calorimetry Measuring heats of combustion reactions water O 2 ignition wires Heat Capacity of Calorimeter (or Calorimeter Constant ), C calorimeter, in energy /deg: Bomb Calorimeter (constant V) For every experiment use the same overall calorimeter mass. Calibrate Calorimeter: Use a weighed mass of substance with known ΔH comb. (continued) 21 Prof. Zvi C. Koren

22 (continued) Measuring Heats of Reaction (q V ) with a Bomb Calorimeter : 3 parts: Reaction, Bomb Apparatus, Water Heat Balance Equation: Σq i = 0 q rxn + q bomb + q water = 0 q rxn = q bomb + q water = C bomb t bomb + (C m t) water [Recall: units of C bomb = energy /deg] For example, consider the following combustion of octane: C 8 H 18 (l) O 2 (g) 8 CO 2 (g) + 9 H 2 O(l) 1.0 g of octane burns in a constant-volume calorimeter (C bomb = 837 J/deg) containing 1.20 kg of water. The temperature rises from o C to o C. Calculate: (a) the heat of comb., q V, for this quantity of octane, (b) H comb for 1 mole of octane. Answer to (a): q rxn = q bomb + q water = C bomb t bomb + (C m t) water = [C bomb + (C m) water ] t; t t f t i = [(837 J/deg) + (4.184 J/g deg) (1200 g)] (8.20 o C) = 48.1 kj q V = 48.1 kj Answer to (b): First calculate # of moles: 1.0 g (1 mol/114 g) = mol q V = 48.1 kj / mol E comb = q V = 5,500 kj/mol q P = H comb = E comb + RT( n g ) = -5.5x10 6 J/mol + (8.31 J/mol K)(298 K)( 4.5) = 5.5x10 6 J/mol 22 Prof. Zvi C. Koren