Chemistry 1A. Chapter 6

Transcription

1 Chemistry 1A Chapter 6

2 Some Chemical Changes Release Energy Combustion of Methane CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(l) +

3 Some Chemical Changes Absorb Energy

4 Energy Terms Energy = the capacity to do work Work, in this context, may be defined as what is done to move an object against some sort of resistance.

5 Two Types of Energy Kinetic Energy = the energy of motion = 1/2 mμ 2 Potential Energy = energy by virtue of position or state

6 Law of Conservation of Energy

7 Endergonic Change more stable + energy less stable system lesser capacity + energy greater capacity to do work to do work lower PE + energy higher PE coin in hand + energy coin in air above hand

8 Coin and Potential Energy

9 Bond Breaking and Potential Energy

10 Exergonic Change less stable system more stable + energy greater capacity lesser capacity + energy to do work to do work higher PE lower PE + energy coin in air above hand coin on ground + energy

11 Bond Making and Potential Energy

12 Which higher energy? Is it kinetic or potential? 428 m/s Ar atoms or 456 m/s Ar atoms? 428 m/s Ar atoms or 428 m/s Kr atoms? Na + close to Cl or Na + and Cl far apart? ROOR or 2 RO H(g) and O 2 (g) or HO 2 (g) Solid CO 2 or gaseous CO 2

13 Units of Energy Joule (J) = kg m 2 s J = 1 cal kj = 1 kcal 4184 J = 1 Cal (dietary calorie) kj = 1 Cal

14 Approximate Energy of Various Events

15 More Terms External Kinetic Energy = Kinetic energy associated with the overall movement of a body Internal Kinetic Energy = Kinetic energy associated with the random motion of the particles within a body

16 External and Internal Kinetic Energy

17 Heat Heat = Energy transfer from a region of higher temperature to a region of lower temperature due to collisions of particles.

18 Heat Transfer

19 Radiant Energy Radiant Energy is electromagnetic energy that behaves like a stream of particles. It has a dual Nature Particle photons = tiny packets of radiant energy photons/second from a flashlight bulb Wave oscillating electric and magnetic fields describes effect on space, not true nature of radiant energy

20 A Light Wave s Electric and Magnetic Fields

21 Radiant Energy Spectrum

22 Endergonic Change more stable + energy less stable system lesser capacity + energy greater capacity to do work to do work lower PE + energy higher PE

23 Exergonic Change less stable system more stable + energy greater capacity lesser capacity + energy to do work to do work higher PE lower PE + energy

24 Bond Breaking and Potential Energy

25 Bond Making and Potential Energy

26 Exergonic (Exothermic) Reaction weaker bonds stronger bonds + energy less stable more stable + energy higher PE lower PE + energy

27 Exothermic Reaction

28 Endothermic Reaction stronger bonds + energy weaker bonds more stable + energy less stable lower PE + energy higher PE NH 4 NO 3 (s) + energy NH 4+ (aq) + NO 3 (aq)

29 Energy and Chemical Reactions

30 Factors that Affect Heats of Chemical Reactions Nature of the reaction, including the states of the reactants and products Amount of reactant Conditions for the reaction Constant pressure or constant volume Temperature and pressure 2C 2 H 6 (g) + 7O 2 (g) 4CO 2 (g) + 6H 2 O(l) ΔH = 3.08 x 10 3 kj

31 Conversion Factors from ΔH 2C 2 H 6 (g) + 7O 2 (g) 4CO 2 (g) + 6H 2 O(l) ΔH = 3.08 x 10 3 kj

32 Relationship Between ΔH and the Chemical Equation Whatever you do to the equation, do the same to the ΔH. Reverse the equation change the sign of the ΔH. Multiply the coefficients by some number multiply the ΔH by the same number.

33 Gas

34 Gas Model Gases are composed of tiny, widely-spaced particles. For a typical gas, the average distance between particles is about ten times their diameter.

35 Gas Model (cont.) Because of the large distance between the particles, the volume occupied by the particles themselves is negligible (approximately zero). For a typical gas at room temperature and pressure, the gas particles themselves occupy about 0.1% of the total volume. The other 99.9% of the total volume is empty space. This is very different than for a liquid for which about 70% of the volume is occupied by particles.

36 Gas Model (cont.) The particles have rapid and continuous motion. For example, the average velocity of a helium atom, He, at room temperature is over 1000 m/s (or over 2000 mi/hr). The average velocity of the more massive nitrogen molecules, N 2, at room temperature is about 500 m/s. Increased temperature means increased average velocity of the particles.

37 Gas Model (cont.) The particles are constantly colliding with the walls of the container and with each other. Because of these collisions, the gas particles are constantly changing their direction of motion and their velocity. In a typical situation, a gas particle moves a very short distance between collisions. Oxygen, O 2, molecules at normal temperatures and pressures move an average of 10 7 m between collisions.

38 Gas Model (cont.) There is no net loss of energy in the collisions. A collision between two particles may lead to each particle changing its velocity and thus its energy, but the increase in energy by one particle is balanced by an equal decrease in energy by the other particle.

39 Ideal Gas The particles are assumed to be point-masses, that is, particles that have a mass but occupy no volume. There are no attractive or repulsive forces at all between the particles.

40 Gas Properties and their Units Pressure (P) = Force/Area units 1 atm = kpa = 760 mmhg = 760 torr 1 bar = 100 kpa = atm = mmhg Volume (V) unit usually liters (L) Temperature (T)? K = --- C Number of gas particles expressed in moles (n)

41 Decreased Volume Leads to Increased Pressure P α 1/V if n and T are constant

42 Relationship between P and V

43 Boyle s Law The pressure of an ideal gas is inversely proportional to the volume it occupies if the moles of gas and the temperature are constant.

44 Increased Temperature Leads to Increased Pressure P α T if n and V are constant

45 Relationship between P and T

46 Gay-Lussac s Law The pressure of an ideal gas is directly proportional to the Kelvin temperature of the gas if the volume and moles of gas are constant.

47 Increased Moles of Gas Leads to Increased Pressure P α n if T and V are constant

48 Relationship between n and P

49 Relationship Between Moles of Gas and Pressure If the temperature and the volume of an ideal gas are held constant, the moles of gas in a container and the gas pressure are directly proportional.

50 Ideal Gas Equation

51 Constant Pressure or Volume Heat at Constant Pressure = ΔH = change in enthalpy Heat at Constant Volume = ΔE = change in the total energy E = KE + PE

52 Standard Heat Changes ΔH at K and 1 atm = ΔH ΔE at K and 1 atm = ΔE

53 Standard Changes in Enthalpy (ΔH ) and Total Internal Energy (ΔE ) 2C 2 H 6 (g) + 7O 2 (g) 4CO 2 (g) + 6H 2 O(l) ΔH = 3.08 x 10 3 kj ΔE = 3.09 x 10 3 kj ΔH = Heat at constant pressure at 25 C and 1 atm ΔE = Heat at constant volume at 25 C and 1 atm

54 Heat at Constant Pressure and Volume (Example 1)

55 Sign Conventions Heat evolved is negative. Heat absorbed is positive. Work done by the system leads to heat lost by the system, so it is negative. Work done on the system adds heat to the system, so it is positive.

56 Heat at Constant Pressure and Volume (Example 2)

57 Heat at Constant Pressure and Volume (Example 3)

58 Goal: to Convert from ΔE to ΔH ΔE is easier to determine accurately in the laboratory. Because most chemical reactions are run at constant pressure, ΔH values are usually used to describe heats of reaction.

59 Conversion from ΔE to ΔH We saw for the KClO 3 and NH 3 reactions, that q V = q rxn So q P = q rxn + q work q P = q V + q work ΔH = ΔE + w For chemical changes at constant pressure and temperature, the only work is work done by a gas as it expands (negative) or on a gas as it is compressed (positive).

60 Relationship between Heat at Constant and Constant Pressure ΔH = ΔE + w ΔH = ΔE + FΔd ΔH = ΔE + P ΔV

61 Relationship between Heat at Constant and Constant Pressure (2) ΔH = ΔE + PΔV PV = nrt (ideal gas equation) If pressure and temperature are kept constant, the only way to increase the volume is to increase moles of gas. PΔV = (Δn)RT ΔH = ΔE + (Δn)RT Δn = moles (g) products moles (g) reactants Δn = coef. (g) products coef. (g) reactants

62 Heat at Constant Pressure Conventions ΔH describes kj per mole of primary reactant (or kj per the number of moles equal to the coefficient in the balanced equation for each reactant and product). q P describes kj per amount of substance in a change.

63 Heat at Constant Volume Conventions ΔE describes kj per mole of primary reactant (or kj per the number of moles equal to the coefficient in the balanced equation for each reactant and product). q V describes kj per amount of substance in a change.

64 Ways to get ΔH From Tables (ΔH for combustion or ΔH of formation) From calorimeter data (bomb or open) From other ΔH s General Law of Hess Heat of formation problem

65 Bomb Calorimeter

66 Exercise

67 Derivation of Calorimeter Equation q lost = q gained q rxn = [q products + q calorimeter + q water + q surr ] q rxn [q calorimeter + q water ] q rxn = [q cal + q w ]

68 Heat Capacity and Specific Heat (Capacity) Heat capacity, C = the heat energy (kj) necessary to raise the temperature of an object (such as a calorimeter) by 1 ºC (or 1K). Units of kj/ºc or kj/k it s the same number for each unit Specific heat (capacity), c = the heat energy (kj) necessary to raise the temperature of 1 g of a substance (such as water) by 1 ºC (or 1K). Units of kj/g ºC or kj/g K it s the same number for each unit

69 Bomb Calorimeter Problems Calculating ΔH - Part 1 Given mass of reactant, mass of H 2 O, T initial, T final, and C cal Steps Assign variables Calculate q V ΔT = T 2 T 1 Watch signs.

70 Bomb Calorimeter Problems Calculating ΔH - Part 2 Calculate ΔE Calculate ΔH

71 Bomb Calorimeter Problems Calculating C cal - Part 1 Given mass of reactant, mass of H 2 O, T initial, T final, and ΔH Steps Assign variables Calculate ΔE

72 Calorimeter Problems Calculating C Cal - Part 2 Calculate q V Calculate C cal

73 Law of Hess If a reaction can be viewed as a sum of two or more equations, the ΔH for the net reaction is equal to the sum of the ΔH s for the intermediate reactions.

74 Law of Hess Example

75 General Law of Hess Problems ΔH net = ΣΔH intermediate Write intermediate equations and their ΔH s. Rearrange intermediate equations so they add to yield the desired net equation. Whatever you did to the intermediate equations, so the same to their ΔH s. Add the new intermediate ΔH s.

76 Heat of Formation ΔH f = Heat at constant pressure for the formation of one mole of substance from its elements in their standard states (at K and 1 atm).

77 Standard States of Elements Metals Hg(l) or Symbol(s) e.g. Zn(s) Noble gases Symbol(g) e.g. Ne(g) Diatomic elements H 2 (g), N 2 (g), O 2 (g), F 2 (g), Cl 2 (g), Br 2 (l), I 2 (s) Carbon C(graphite) Other elements S 8 (s), Se 8 (s), P 4 (s), As 4 (s), Sb 4 (s)

78 Heat of Formation Equation Chemical reactions could take place in two steps. Conversion from reactants to elements in their standard states. ΔH 1 = - ΣΔH f (reactants) Conversion of elements in their standard states to products. ΔH 2 = ΣΔH f (products) The overall ΔH rxn would be equal to the sum of the ΔH s for the two steps. ΔH rxn = ΔH 1 + ΔH 2 = ΔH 2 + ΔH 1 ΔH rxn = ΣΔH f (products) + ( ΣΔH f (reactants)) ΔH rxn = ΣΔH f (products) ΣΔH f (reactants)