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Thermochemistry Chapter 5 BLB 12th.

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Presentation on theme: "Thermochemistry Chapter 5 BLB 12th."— Presentation transcript:

1 Thermochemistry Chapter 5 BLB 12th

2 Expectations Heat & enthalpy – same or different? Heat calculations:
Temp. change Phase change (11.4, p. 438) Reactions Enthalpy calculations Read the chapter, study, and apply!

3 5.1 The Nature of Energy Chemistry ⇐ ? ⇒ Energy
Energy – capacity to do work or transfer heat Potential – stored energy; chemical Kinetic – released energy; energy of motion; thermal Electrostatic potential – interaction between charged particles

4 Examples of Kinetic Energy

5 Energy, cont. Units of energy: Joule (J) – SI unit of energy;
calorie (cal) amount of energy required to raise the temperature of exactly 1 gram of pure water by 1°C (from 14.5°C to 15.5°C) 1 cal = J (exactly) Calorie (dietary calorie),Cal 1 Cal = 1000 cal = 1 kcal

6 Energy, cont. System and Surroundings
System – component(s) of interest Open – matter and energy can be exchanged between system and surroundings Closed – can exchange energy but not matter Isolated – neither energy nor matter can be exchanged Surroundings – everything outside of the system

7 Energy, cont. Transferring Energy
Work (w) – energy used to move an object against a force; w = F x d Heat (q) – energy transferred from a hotter object to a cooler one Energy – capacity to do work or transfer heat; ΔE = q + w

8 Combustion heat & work

9 5.2 The First Law of Thermodynamics
Energy can be neither created nor destroyed. Energy is conserved. Internal energy, E – sum of all the kinetic and potential energy of the system’s components What kinds of energy are in here? What changes could occur?

10 5.2 The First Law of Thermodynamics
More interested in the change in energy: ΔE = Efinal – Einitial Need to give number, units, and sign for all thermodynamic quantities. ΔE > 0 - Efinal > Einitial, system has gained energy; endergonic ΔE < 0 - Efinal < Einitial, system has lost energy; exergonic Note: Opposite change occurs with respect to the surroundings.

11 Figure 5.5 Changes in internal energy.

12 Energy, heat & work ΔE = q + w
Sign of ΔE depends upon sign and magnitude of q and w.

13 Figure 5.7 Sign conventions for heat and work.

14 Exothermic Efinal < Einitial

15 Sample Exercise 5.2 A(g) + B(g) → C(s)
System loses 1150 J of heat to the surroundings. The piston move downwards doing 480 J of work on the system. ΔE = ?

16 Calculate ΔE (in J); exothermic or endothermic?
Balloon heating by adding 900 J of heat and expands doing 422 J of work on atmosphere. 50 g of H2O cooled from 30°C to 15°C losing 3140 J of heat. Reaction releases 8.65 kJ of heat, no work done.

17 Heat or Thermal Energy (q)
Exothermic: system → surroundings Heat energy released to surroundings q < 0 e.g. combustion reaction, crystallization Surroundings get warmer Endothermic: system ← surroundings Heat energy flows into the system q > 0 e.g. melting, boiling, dissolution of NH4NO3 Surroundings get colder

18 Heat, cont. Evidenced by a change in temperature
Spontaneously transferred from the hotter to the cooler object Atoms or molecules with more energy move faster Temperature-dependent Extensive property (depends on amount) Total energy of system is the sum of the individual energies of all the atoms and molecules of the system.

19 Work (w) Force acting over a distance w = F x d = −PΔV
Compression: work ← surroundings Work is done on the system. ΔV < 0 w > 0 Expansion: work → surroundings Work is done on the surroundings. ΔV > 0 w < 0

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21 Heat & Work, cont. Work and heat are pathways by which energy can be transferred. State function – depends only on the system’s present state; independent of the pathway; internal energy, P, V, ΔE, ΔH, ΔS are state functions Energy is a state function, as is enthalpy.

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23 5.3 Enthalpy Enthalpy – heat flow at constant pressure; from Gr. enthalpien – to warm Enthalpy change (ΔH) – energy transferred as heat at constant pressure; ΔH = Hproducts – Hreactants H = E + PV For a constant pressure: ΔH = ΔE + PΔV ΔH = ΔE − w = qP

24 5.3 Enthalpy ΔH < 0 exothermic: reactants → products + heat
ΔH > 0 endothermic: reactants + heat → products ΔH = q/mol Enthalpy (or heat) of reaction, ΔHrxn – enthalpy change that accompanies a reaction

25 5.4 Enthalpies of Reaction, ΔHrxn
ΔH is an extensive property; value depends upon the BALANCED equation. 2 H2(g) + O2(g) → 2 H2O(g) ΔH = kJ per 2 moles of H2O H2(g) + ½ O2(g) → H2O(g) ΔH = −241.8 kJ per mole of H2O

26 Figure 5.14 Exothermic reaction of hydrogen with oxygen.

27 5.4 Enthalpies of Reaction
For reverse reactions: ΔH values are equal in magnitude, but opposite in sign. For water: ΔHvap = kJ/mol ΔHcond = −44.0 kJ/mol

28 For the combustion of methane:

29 5.4 Enthalpies of Reaction
ΔH is dependent upon physical state. ΔHf values: C6H6(g) = 82.9 kJ/mol C6H6(l) = 49.0 kJ/mol H2O(l) → H2O(g) ΔH = +44 kJ

30 CH3OH(g) → CO(g) + 2 H2(g) ΔHrxn = +90.7 kJ
Exothermic or endothermic? Heat transferred for 1.60 kg CH3OH? If 64.7 kJ of heat were used, how many grams of H2 would be produced?

31 CH3OH(g) → CO(g) + 2 H2(g) ΔHrxn = +90.7 kJ
ΔH of reverse reaction? Heat (in kJ) released when 32.0 g of CO(g) reacts completely?

32 5.5 Calorimetry Calorimetry – science of measuring heat flow
Calorimeter – a device used to measure heat flow Coffee-cup calorimeter ⇒

33 Heat Capacity and Specific Heat
Heat capacity (C) - amount of heat required for a 1°C temperature change: J/°C = J/K extensive property ΔT in K = ΔT in °C

34 Heat Capacity and Specific Heat
Specific heat capacity (Cs) – heat capacity for 1 g; J/g·°C or J/g·K Molar heat capacity – heat capacity for 1 mole; J/mol·°C or J/mol·K

35 Heat Capacity and Specific Heat
Specific heat values (more on p. 176): Fe 0.45 J/g·K glass 0.84 J/g·K water 4.18 J/g·K (highest of all liquids and solids except ammonia)

36 Calculating heat (q) q = Cs x m x ΔT
To calculate the quantity of heat transferred: q = Cs x m x ΔT q – heat (J) Cs – specific heat (J/g∙K) m – mass (g) ΔT – change in temp. (K or °C)

37 Calculate the heat (in J) required to raise the temperature of 62
Calculate the heat (in J) required to raise the temperature of 62.0 g toluene from 16.3°C to 38.8°C. The specific heat of toluene is 1.13 J/g·K.

38 Calculate the specific heat of lead if 78
Calculate the specific heat of lead if 78.2 J of heat were required to raise the temperature of a 45.6-g block of lead by 13.3°C.

39 Constant-Pressure Calorimetry
Constant-pressure, ΔH = qP and ΔE = qP + w Assume no heat is lost to surroundings. Usually exothermic (qrxn < 0) Applications: Heat transfer between objects Reactions in aqueous solutions Use specific heat of water (4.18 J/g·K). Use mass (or moles) of solution.

40 Solution Calorimetry heat lost by reaction = heat gained by solution
−qrxn = qsoln qrxn = −(Cs,soln x msoln x ΔT) Enthalpy of reaction (ΔHrxn) per mole ΔHrxn = qrxn/mol of specified reactant

41 A 19. 6-g piece of metal was heated to 61. 67°C
A 19.6-g piece of metal was heated to 61.67°C. When the metal was placed into 26.7 g water, the temperature of the water increased from to 35.00°C. Calculate specific heat of the metal.

42 A 15.0-g piece of nickel at 100.0°C is dropped into a coffee-cup calorimeter containing 55.0 g H2O at 23.0°C. What is the final temperature of the water and nickel after reaching thermal equilibrium? The specific heat capacity of nickel is J/g·K and of water is 4.18 J/g·K.

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44 In a coffee-cup calorimeter, 2
In a coffee-cup calorimeter, 2.50 g of MgO was reacted with 125 mL of 1.0 M HCl. The temperature increased by 9.6°C. Calculate the enthalpy of reaction per mole of MgO for the following reaction Mg2+(aq) + H2O(l) → MgO(s) + 2 H+(aq)

45 Constant-Volume Calorimetry (p. 178)
Bomb calorimetry No work is done (ΔV = 0), so ΔE = qV Used for combustion reactions The bomb components absorb the heat lost by the reaction. Heat capacity of the bomb (Ccal) needed to calculate the heat of combustion (reaction) qrxn = −(Ccal x ΔT)

46 Bomb Calorimeter

47 A g sample of a new organic substance is combusted in a bomb calorimeter (Ccal = 8.74 kJ/K). The temperature of the bomb increased from 22.14°C to 26.82°C. What is the heat of combustion per gram of the substance?

48 Phase Changes (Fig , p. 439) (or crystallization)

49 Phase Changes (p. 440) Endothermic → Cs = 1.84 J/g·K
ΔHvap = kJ/mol ← Exothermic Cs = 4.18 J/g·K ΔHfus = 6.01 kJ/mol Cs = 2.03 J/g·K

50 Heat transfer – phase changes
To calculate the quantity of heat transferred during a change of state: q = ΔHprocess x m or q = ΔHprocess x mol No change in temperature, so no ΔT. For a complete process, add together the heat transferred for each segment. See Sample Exercise 11.3, p. 441.

51 11. 46 Calculate the heat transferred for the conversion of 35
11.46 Calculate the heat transferred for the conversion of 35.0 g of the fluorocarbon, C2Cl3F3, from a liquid at 10.00°C to a gas at °C. Data: b.p. 47.6°C, ΔHvap= kJ/mol, Cs(liquid) = 0.91 J/g·K, Cs(gas) = 0.67 J/g·K

52 5.6 Hess’s Law If a reaction is carried out in a series of steps, ΔHrxn will equal the sum of the ΔH values of the individual steps. Hess’s Law works because ΔH is a state function, i.e. it only depends upon the initial reactant and the final product states.

53 Hess’s Law Example

54 Hess’s Law Example

55 Calculation of H C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
Imagine this as occurring in three steps: C3H8 (g)  3 C (graphite) + 4 H2 (g)

56 Calculation of H C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
Imagine this as occurring in three steps: C3H8 (g)  3 C (graphite) + 4 H2 (g) 3 C (graphite) + 3 O2 (g)  3 CO2 (g)

57 Calculation of H C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
Imagine this as occurring in three steps: C3H8 (g)  3 C (graphite) + 4 H2 (g) 3 C (graphite) + 3 O2 (g)  3 CO2 (g) 4 H2 (g) + 2 O2 (g)  4 H2O (l)

58 Calculation of H C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
The sum of these equations is: C3H8 (g)  3 C (graphite) + 4 H2 (g) 3 C (graphite) + 3 O2 (g)  3 CO2 (g) 4 H2 (g) + 2 O2 (g)  4 H2O (l) C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)

59 Based on the following reactions: ΔH, kJ N2(g) + O2(g) → 2 NO(g) 180
Based on the following reactions: ΔH, kJ N2(g) + O2(g) → 2 NO(g) NO(g) + O2(g) → 2 NO2(g) − N2O(g) → 2 N2(g) + O2(g) − Calculate the ΔHrxn for the following reaction: N2O(g) + NO2(g) → 3 NO(g)

60 Based on the following reactions: ΔH, kJ C2H2(g) + 5/2 O2(g) → 2 CO2(g) + H2O(l) −1300. C(s) + O2(g) → CO2(g) −394 H2(g) + ½ O2(g) → H2O(l) −286 Calculate the ΔHrxn for the following reaction: 2 C(s) + H2(g) → C2H2(g)

61 5.7 Enthalpies of Formation
Standard state of a substance – pure form at atmospheric pressure (1 atm) and temperature of interest (usually 25°C). Standard enthalpy - ΔH° Standard enthalpy of formation, ΔHf° - the change in enthalpy for the formation of 1 mole of a substance from its elements in their standard states, kJ/mol of product. ΔHf° of an element in its most stable form = 0 kJ/mol

62 Standard Enthalpies of Formation (see Appendix C, p. 1059)

63 Reactions for ΔHf°

64 Calculating Enthalpies of Reaction
ΔH°rxn = Σn ΔHf°(products) − Σ n ΔHf°(reactants)

65 5.73 (c) Calculate ΔHrxn for the following reaction: N2O4(g) + 4 H2(g) → N2(g) + 4 H2O(g)

66 Calculate ΔHrxn for the following reaction: 2 KOH(s) + CO2(g) → K2CO3(s) + H2O(g)

67 5.8 Foods and Fuels Glucose is our body’s fuel source.
Carbs and fats are metabolized into glucose. Excess fat is stored. What’s the big deal? Take a look at the “fuel value”.

68 Figure: 05-T04 Title: Table 5.4 Caption: Compositions and Fuel Values of Some Common Foods

69 Nonbiological Fuel

70


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