1. 1 Zumdahl Ch 6 AP Chemistry

2.  From Greek therme (heat); study of energy changes in chemical reactions  Energy: capacity do work or transfer heat  Joules (J) or calories (cal); 1 cal = 4.184 J 1. Kinetic: energy of motion; dependent on mass & velocity  Applies to motion of large objects & molecules  Linked to thermal energy (object’s T above 0 K) 2 James Prescott Joule (1818-1889)

3. 2. Potential: stored in “fields” (gravitational and electrical/magnetic); dependant on position relative to another object  Applies to large objects where gravity is overriding force, but not significantly to molecules where gravity is negligible and electrostatic forces dominate  Associated with chemical energy; stored in arrangement of atoms or subatomic particles (electrostatic & nuclear forces, bonding between atoms) 3

4.  Force: a push or pull on an object  Work: energy transferred to move an object a certain distance against a force: W = (F)(d)  Heat: energy transferred from a hotter object to a colder one  Temperature: property that reflects the random motions (kinetic energies) of particles  State Function/Property: only depends on the present state, does not depend on past or future. Pathway independent  Energy is a state function, work and heat are not. 4

5.  System –the part of the universe on which you focus your attention  Closed system: easiest to study because exchanges energy with surroundings but matter is not exchanged.  Surroundings – include everything else in the universe  Universe – System and the surroundings

6.  “0 th Law”: 2 systems are in thermal equilibrium when they are at the same T.  Thermal equilibrium is achieved when the random molecular motion of two substances has the same intensity (and therefore the same T.)  1 st Law: Energy can be neither created nor destroyed, or, energy is conserved.  energy gained by surroundings = energy lost by system  2 nd and 3 rd Laws: discussed later 6

7.  Includes:  Translational motion  Rotational motion of particles through space  Internal vibrations of particles.  It is difficult to measure all E, so the change in internal energy (  E) is typically measured:  E = E final - E initial  E > 0Increase in energy of system (gained fromsurroundings)  E < 0Decrease in energy of system (lost to surroundings) 7

8.  When a system undergoes a chemical or physical change, the change in internal energy (E) is equal to the heat (q) added or liberated from the system plus the work (w) done on or by the system:  E = q + w 8 First Law of Thermodynamics

9. q > 0 Heat is added to system q < 0 Heat is removed from system (into surroundings) w > 0 Work done to system w < 0 System does work on surroundings Always need a quantity and a sign (+/-) 9

10.  Heat flowing into a system from its surroundings is defined as positive; q has a positive value  The system gains heat as the surroundings cool down

11.  Heat flowing out of a system into its surroundings is defined as negative; q is a negative value  The system loses heat as the surroundings heat up.

12.  Calculate Δ E for a system undergoing an endothermic process in which 15.6 kJ of heat flows and where 1.4kJ of work is done on the system.  E = q + w q = +15.6 kJ (endothermic) w = +1.4 kJ (work done on the system)  E = 15.6 kJ + 1.4 kJ = + 17.0 kJ gained by system 12

13.  P = F/A (F = P x A)  Work = force x distance = F x Δ h  Work = F x Δ h = P x A x Δ h  Δ V = (V f –V i ) = A x Δ h  w = P Δ V  Need to have opposite signs b/c when a gas expands ( Δ V is positive), work flows into the surroundings (w is negative)  w = - P Δ V 13

14.  Octane and oxygen gases combust within a closed cylinder in an engine. The cylinder gives off 1150 J of heat and a piston is pushed down by 480 J during the reaction. What is the change in internal energy of the system?  q is (-) since heat leaves system; w is (-) since work is done by system. Therefore,  E = q + w = (-1150 J) + (-480 J) = - 1630 J  1630 J has been liberated from the system (C 8 H 18 and O 2 ) added to the surroundings (engine, atmosphere, etc.) 14

15.  Calculate the work associated with the expansion of a gas from 46 L to 64 L at a constant external pressure of 15 atm. w = - P Δ V P = 15 atm Δ V = 64 L – 46 L = 18 L w = - 15 atm x 18 L = -270 L  atm Gas expands, so it does work on the surroundings 1 L  atm = 101.3 J 15

16.  http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/activa2.swf http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/activa2.swf 16 E E

17.  Property of a system that is determined by specifying its condition or its “state”  The value of a state function depends only on its present state and not on the history of the sample.  T & E are state functions. Consider 50 g of water at 25°C: E H2O does not depend on how the water got to be 25°C (whether it was ice that melted or steam that condensed or…)  Work (w) and heat (q) are not state functions because the ratio of q and w are dependent on the scenario. Consider the combustion of gasoline in a car engine vs. burning in the open. 17

18.  Measurement of heat flow  Heat capacity, C: amount of heat required to raise T of an object by 1 K (or 1 °C) q = C  T  Specific heat (or specific heat capacity, c ): heat capacity of 1 g of a substance q = m c  T Ex: How much energy is required to heat 40.0 g of iron (c = 0.45 J/(g °C) from 0.0ºC to 100.0ºC? q = m c  T = (40.0 g)(0.45 J/(g °C))(100.0 – 0.0 ºC) = 1800 J 18

19.  Calorimetry is an experimental technique used to measure the heat transferred in a physical or chemical process.  The apparatus used in this procedure is called as a “ Calorimeter ”. There are two types of calorimeters- Constant pressure (coffee cup) and constant volume (bomb calorimeter). We will discuss constant pressure calorimeter in detail.  Constant Pressure Calorimeter : The coffee cup calorimeter is an example of this type of calorimeter. The system in this case is the “contents” of the calorimeter and the surroundings are cup and the immediate surroundings. 19

20. During the rxn: q rxn + q solution = 0  where q rxn is the heat gained or lost in the chemical reaction and q solution is the heat gained or lost by solution. Heat exchange in this system (q rxn ), is equal to enthalpy change. Assuming no heat transfer takes place between the system and surroundings, q rxn + q solution = 0 chemlab.truman.edu 20

21.  Since most reactions occur in containers open to the air, w is often negligible. If a reaction produces a gas, the gas must do work to expand against the atmosphere. This mechanical work of expansion is called PV (pressure-volume) work.  Enthalpy (H): change in the heat content (q p ) of a reaction at constant pressure H = E + PV  H =  E + P  V(at constant P)  H = (q p + w) + (-w)  H = q p 21

22.  Sign conventions  H > 0 Heat is gained from surroundings +  H in endothermic reaction  H < 0 Heat is released to surroundings -  H in exothermic reaction 22

23. 23 chemlab.truman.edu

24. 24 Sample Problem #1: 200. ml of 0.400 M HCl is mixed with the same amount and molarity of NaOH solution, inside a coffee-cup calorimeter. The temperature of the solutions before mixing was 25.10 o C, and 27.78 o C after mixing and letting the reaction occur. Find the molar enthalpy of the neutralization of the acid, assuming the densities of all solutions are 1.00 g/ml and their specific heat capacities are 4.18 J/(g * K).

25. 25 Sample Problem #2: 1. 0.500g of magnesium chips are placed in a coffee-cup calorimeter and 100.0 ml of 1.00 M HCl is added to it. The reaction that occurs is: Mg (s) + 2HCl (aq)  H 2 (g) + MgCl 2 (aq) The temperature of the solution increases from 22.2 o C to 44.8 o C. What’s the enthalpy change for the reaction, per mole of Mg? (Assume specific heat capacity of solution is 4.18 J/(g * °C) and the density of the HCl solution is 1.00 g/ml.) Ans. -7.60 x 10 5 J/mol Mg

26. 26 Sample Problem #3: 1. 5.00 g of a metal is heated in an oven to 100°C. It is then placed in 100. g of H 2 O at 25.0°C. The final temp of the solution is 30.0 °C. Calculate the specific heat of the metal.

27.  Also called heat of reaction: 1. Enthalpy is an extensive property (depends on amounts of reactants involved). Ex: CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O (l)  H rxn = - 890. kJ  Combustion of 1 mol CH 4 produces 890. kJ … of 2 mol CH 4 → (2)(-890. kJ) = -1780 kJ What is the  H of the combustion of 100. g CH 4 ? 27

28. 2.  H reaction = -  H reverse reaction CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O (l)  H = - 890. kJ CO 2 (g) + 2 H 2 O (l)  CH 4 (g) + 2 O 2 (g)  H = +890. kJ 28

29. 88 kJ 802 kJ CO 2 (g) + 2 H 2 O (g) 3.  H rxn depends on states of reactants and products. CO 2 (g) + 2 H 2 O (g)  CH 4 (g) + 2 O 2 (g)  H = 802 kJ 2 H 2 O (l)  2 H 2 O (g)  H = 88 kJ So: CO 2 (g) + 2 H 2 O (l)  CH 4 (g) + 2 O 2 (g)  H = 890. kJ CH 4 (g) + 2 O 2 (g) 2 H 2 O (l) 2 H 2 O (g) CO 2 (g) + 2 H 2 O(l) CH 4 (g) + 2 O 2 (g) 890. kJ

30.  If a rxn is carried out in a series of steps,  H rxn =  (  H steps ) =  H 1 +  H 2 +  H 3 + … 30 Ex. What is  H rxn of the combustion of propane? C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O (l) 3 C (s) + 4 H 2 (g)  C 3 H 8 (g)  H 1 = -103.85 kJ C (s) + O 2 (g)  CO 2 (g)  H 2 = -393.5 kJ H 2 (g) + ½ O 2 (g)  H 2 O (l)  H 3 = -285.8 kJ  H  rxn = 103.85 + 3(- 393.5) + 4(- 285.8) = - 2219.8 kJ 3[ ] 3( ) 4[ ] 4( ) Germain Hess (1802-1850) C 3 H 8 (g)  3 C (s) + 4 H 2 (g)  H 1 = +103.85 kJ

31.  Formation: a reaction that describes a substance formed from its elements NH 4 NO 3 (s)  Standard enthalpy of formation (  H f  ): forms 1 mole of compound from its elements in their standard state (at 298 K) C 2 H 5 OH (l)  H f  = - 277.7 kJ   H f  of the most stable form of any element equals zero. H 2, N 2, O 2, F 2, Cl 2 (g) Br 2 (l), Hg (l) C (graphite), P 4 (s, white), S 8 (s), I 2 (s) 31 Ex: 2 N 2 (g) + 4 H 2 (g) + 3 O 2 (g)  2 2 C (graphite) + 3 H 2 (g) + ½ O 2 (g) 

32. 32 The complete (long) way to calculate the ENTHALPY of any reaction is to take apart the reactant molecules piece by piece to elements, and then build the product molecules from all of the pieces, bond by bond.

33. 33 A GREAT SHORTCUT exists however: To estimate the enthalpy of a reaction Δ R H ° we need only to know the enthalpies of formation of the reactants and the products. (we calculate the large RED arrow by the difference between the known small GREEN arrows)

34. 34 An example: The combustion of liquid benzene : Firstly, we need to define the overall reaction, and balance it to 1 mole of the molecule we’re interested in : 1 C 6 H 6 (l) + 15/2 O 2 (g)  6 CO 2 (g) + 3 H 2 O (l)Δ R H ° = ??? Δ R H ° = Δ f H ° { products } - Δ f H ° { reactants } Secondly, we need to look up the Δ f H ° values corresponding to both the reactants and the products, and multiply by the moles:

35. 35 Δ f H ° { products }= 6 ( - 394 kJ/mol )+ 3 ( - 286 kJ/mol )= - 3222 kJ/mol to form the products Δ f H ° { reactants } = ( + 49 kJ/mol )+ 15/2 ( zero )= + 49 kJ/mol to form the reactants ----------------------------------------------- Overall reaction Δ R H ° = - 3268 kJ/mol

36. 36 Enthalpies of combustion Δ C H ° are also commonly tabulated : ( converted into enthalpies per gram, for convenience ) methaneΔ C H ° = - 55 kJ/g octaneΔ C H ° = - 48 kJ/g dodecaneΔ C H ° = - 51 kJ/g methanolΔ C H ° = - 23 kJ/g glucoseΔ C H ° = - 16 kJ/g carbohydratesΔ C H ° ~ - 18 kJ/g tristearinΔ C H ° = - 38 kJ/g human beings at age 20 need around 10,000 kJ per day of energy from combustions (~12,000 kJ for men, and ~9000 kJ for women).

37. 37 Ex. Combustion of propane: C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O (l) Given:Compound  H  rxn (kJ/mol) C 3 H 8 (g)-103.85 CO 2 (g)-393.5 H 2 O (l)-285.8 H 2 O (g)-241.82  H  rxn = [3(- 393.5) + 4(- 285.8)] – [1(-103.85) + 5(0)] = - 2219.8 kJ

38.   mount of energy required to break a particular bond between two elements in gaseous state. Given in kJ/mol. Remember, breaking a bond always requires energy!  Bond enthalpy indicates the “strength” of a bond.  Bond enthalpies can be used to figure out  H rxn. Ex: CH 4 (g) + Cl 2 (g) → CH 3 Cl (g) + HCl (g)  H rxn = ? 1 C-H & 1 Cl-Cl bond are broken (per mole) 1 C-Cl & 1 H-Cl bond are formed (per mole)  H rxn ≈  (H bonds broken ) -  (H bonds formed ) Note: this is the “opposite” of Hess’ Law where  H rxn =  H products –  h reactants Bond Enthalpy link 38

39. Ex: CH 4 (g) + Cl 2 (g) → CH 3 Cl (g) + HCl (g)  H rxn = ? BondAve  H/molBondAve  H/mol C-H413Cl-Cl242 H-Cl431C-Cl328 C-C348C=C614  H rxn ≈  (H bonds broken ) -  (H bonds formed )  H rxn ≈ [(1(413) + 1(242)] – [1(328) + 1(431)]  H rxn ≈ -104 kJ/mol  H rxn = -99.8 kJ/mol (actual) Note:2 C-C ≠ 1 C=C 2(348) = 696 kJ≠ 614 kJ 39

40. H Absorb E, break 1 C-H and 1 Cl-Cl bond Release E, form 1 C-Cl and 1 H-Cl bond CH 4 (g) + Cl 2 (g) CH 3 Cl (g) + HCl (g)  H rxn *CH 3 (g) + H(g) + 2 Cl(g)  H rxn =  (H bonds broken ) +  (- H bonds formed )  H rxn =  (H bonds broken ) -  (H bonds formed )

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