Chapter 6. Thermochemistry: Energy Flow and Chemical Change

Transcription

1 Chapter 6 Thermochemistry: Energy Flow and Chemical Change

2 Thermochemistry involves the monitoring of energy transformations that occur with a chemical reaction. CH4(g) + 2 O2 ==> CO2 + 2 H2O + HEAT Reaction gives off heat with rise in temperature in the flask Reactions that evolve heat = EXOTHERMIC Heat is written as a product. NH4NO3 + H2O + HEAT ==> NH4 + + NO3 - Reaction absorbs heat with decrease in temperature Reactions that evolve heat = ENDOTHERMIC Heat is written as a reactant.

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4 Energy is the capacity to do work or produce heat. It takes many faces: Radiant energy comes from the sun and is earth s primary energy source (nuclear fusion) Thermal energy is the energy associated with the vibration and rotational motion of atoms and molecules. Chemical energy is the energy stored within the chemical bonds of substances. Nuclear energy is the energy stored within the neutrons and protons in the atom (E = mc 2 ) Kinetic energy: energy associated with moving mass Potential energy: energy available by virtue of an object s position or height above a reference height.

5 The SI Unit of energy is the Joule (kg m 2 /s 2 ). KE = 1/2 m v 2 = 1/2 (kg)(m/s) 2 = kg m 2 /s 2 = 1 joule Work = F x d = (ma) x d = (kg)(m/s 2 ) x (m) = kg m 2 /s 2 = 1 joule PE = mgh = (kg)(m/s 2 )(m) = kg m 2 /s 2 = 1 joule g = acceleration gravity (9.8 m/s 2 ), and h is object s height. 1 calorie (cal) = 4.18J 1000 calorie (cal) = 1 Cal (food calorie) 1 British Thermal Unit (BTU) = 1055 Joules

6 A single Fritos chip snack has a energy content of 5.0 Cal. How many joules is this? E = 5 Cal 1000 cal 1 Cal 4.18J 1 cal = J What is the kinetic energy (in kilojoules) of a 2300 lb car moving at 55 mi/h? KE = 1 / 2 mv 2 =? 1 Joule = 1 kg m 2 /s 2 1 mile = km = 1609 m 1 lb = kg

7 Energy can be transformed from one form to another but it can not be destroyed. (1st Law) High PE Kinetic Energy = 1/2 mv 2 = 0 Potential Energy = mgh PE and KE exist here No PE Kinetic Energy = 1/2 mv 2 Potential Energy = 0 KE + PE = constant

8 We can describe behavior of physical and chemical phenomena as a tendency to seek lower potentials: High PE High Temp H2(g) + O2(g) Acid + Base No PE Low Temp 2H2O Water + salt

9 Energy is also transformed in chemical reactions. CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g) H 2 O(l) H 2 O(g) CH 4 + 2O 2 H initial H 2 O(g) H final Enthalpy, H ΔH < 0 heat out Enthalpy, H ΔH > 0 heat in CO 2 + 2H 2 O H 2 O(l) H final H initial A Exothermic process B Endothermic process

10 In 1843, James Joule finds that heat and work can be interconverted and are manifestations of the same thing, energy. Joule found 1 cal heat = J work amount of heat required to raise the temperature of 1g of water by 1 C W = Force x distance

11 Thermodynamics uses a specialized vocabulary that we need to get used to. Energy is the capacity to do work and exchange heat-- it can take on many many forms. Work = Energy = Force x Distance. Thermal Energy is the kinetic energy of molecules in motion (translational, rotational, and vibrational). Heat is thermal energy transferred between two objects at different temperatures. Chemical Energy is energy stored in a chemical bond.

12 Thermodynamic Formalism: A system has a bulk property of matter that we call internal energy. Reactants and products are the system; everything else is the surroundings. surroundings We track the energy flow into and out of the system. Loss of energy from the system to the surroundings has a negative sign. energy out - energy in + Increase of the system energy from the surroundings has a positive sign. system

13 The internal energy, E, of a system is a bulk property of matter that on a microscopic level represents all energy within the system. On a microscopic scale the internal energy is: Translational kinetic energy. Molecular rotation. Bond vibration. Intermolecular attractions. Chemical bonds. Electrons. Nuclear Electrochemical

14 The internal energy of a system can only change by exchange of heat (q) and work (w) with the surroundings. Internal energy ΔESys = q + w Surroundings Heat Work +q = heat added to the system +w = work done on the system Internal Energy System Rotation, Vibrations, Electronic, translational energy of molecules Surroundings -q = heat lost from the system -w = work done by the system

15 There are 3-Types of Thermodynamic Systems 1.Open System Matter 2. Closed System 3. Isolated System Matter exchanged Heat or work exchanged Matter not exchanged Heat or work exchanged No matter exchanged No heat or work exchanged

16 Mechanical work can be only be done by a system or on a system by the system being mechanically linked to the environment through P-V work. 2KClO3(s) ==> 2KCl(s) + 3 O2(g) When pistons are attached there could be work done by the system or on the system. We call this: pressure-volume work or PV work.

17 A system can do work on the surroundings or the surroundings can do mechanical work on the system. Note PV has units of energy! J = 1 L atm w = P V height w = F orce Distance F orce Area w = height Area w = P (Area height) w = P V

18 Reworking our Internal Energy Equation, some points of clarification on confusing stuff. Definition of internal energy, E Heat Work ΔESys = q + w Definition of Work work = w = -PΔV Substituting ΔEsys = q + w = q - PΔV = q - P(Vf - Vi)

19 Calculate the work (in kilojoules) done during a reaction in which the volume expands from 12.0 L to 14.5 L against an external pressure of 5.0 atm. Watch the conversion of PV units to joules. P = 5.0 atm ΔV = ( ) L = 2.5 L w = PΔV w = -(5.0 atm)(2.5 L) = L atm (-12.5 L atm)(101 J/L atm) = J = -1.3 kj

20 A sample of nitrogen gas expands in volume from 1.6 L to 5.4 L at constant temperature. What is the work done in joules if the gas expands (a) against a vacuum and (b) against a constant pressure of 3.7 atm? w = -P ΔV (a) ΔV = 5.4 L 1.6 L = 3.8 LP = 0 atm W = -0 atm x 3.8 L = 0 L atm = 0 joules (b) ΔV = 5.4 L 1.6 L = 3.8 LP = 3.7 atm w = -3.7 atm x 3.8 L = L atm w = L atm x J 1L atm = J

21 The 1st Law of Thermodynamics that energy is neither created nor destroyed in any process, it is just transformed between the two boundaries. ΔE system + ΔE surroundings = 0 ΔE system = -ΔE surroundings The total energy of a system and its surroundings remains constant.

22 We can represent the change in internal energy, ΔE of a chemical reaction in a Energy Reaction Diagram. A system does work on its surroundings. The energy is lost by the system. Energy, E E Intial Zn(s) + 2H + (aq) + 2Cl - (aq) ΔE<0 PΔV work done on surroundings E Final H 2 (g) + Zn 2+ (aq) + 2Cl - (aq)

23 We can not measure the absolute internal energy of a system...we can only measure changes in internal energy---we call this a change in state. Change in Internal Energy Heat ΔESYS = q + w Work Measurements are relative to something. What s the tallest mountain in the world? 1. Mount Everest is king if measured from sea level (8,848 meters) 2. Mauna Kea is when measured from the depths of the Pacific Ocean floor. It rises 10,203 meters. 3. Mt. Chimborazo in the Andes triumphs. Although it stands but 6,267 meters above sea level, its peak is the farthest from the earth's core.

24 A state function is any math function that depends only on the initial and final states and not the path taken to between the two states. All are state functions! h PE = mgh Final State ΔE = E final - E initial ΔP = P final - P initial ΔV = V final - V initial ΔT = T final - T initial ΔH = H final - H initial 0 Initial State The potential energy of hiker 1 and hiker 2 is the same even though they took very different paths to get to the mountain top. PE only depends on the initial and final heights only.

25 Why Do We Care About State Functions? A change in a state function tells us that the function has one unique value between any two arbitrary states. This fact is why thermodynamics is so powerful and useful. Final State mgh2 Arbitray State Δh Initial State h = 1

26 A chemical system in equilibrium posses a state function called internal energy. A chemical reaction causes a change in internal energy equal to the difference in heat transfer into or from the system and the work done by or on the system. +/-w +/-q Initial State Final State

27 Determining A Change in Internal Energy When gasoline burns in a car engine, the heat released causes the products CO 2 and H 2 O to expand, which pushes the pistons outward. Excess heat is removed by the car s cooling system. If the expanding gases do 451 J of work on the pistons and the system loses 325 J to the surroundings as heat, calculate the change in energy (ΔE) in J, kj, and kcal. SOLUTION: q = J and w = J ΔE = q + w = -325 J + (-451 J) = -776 J kj -776J 103 J kcal = kJ k 4.18k = kcal

28 Chemists don t use internal energy to monitor energy changes in the lab. Instead, we define and use a related parameter called enthalpy. ΔEsys = q + w = q - PΔV CASE 1 At constant pressure: ΔE = q P - PΔV q P = ΔE + PΔV CASE 2 At constant volume: ΔV=0 ΔE = q V - PΔV ΔE = q V

29 Chemists measure the enthalpies of nearly all chemical reactions and give special names to them.

30 An Exothermic reaction releases heat to the surroundings, ΔH < 0. 2H 2 (g) + O 2 (g) H 2 O (g) 2H 2 O (l) + heat H 2 O (l) + heat ΔH = <0 heat written as a product (given off) An Endothermic reaction absorbs heat from the surroundings into the system, ΔH > 0. heat + 2HgO (s) heat + H 2 O (s) 2Hg (l) + O 2 (g) H 2 O (l) ΔH = >0 heat written as a reactant (consumed)

31 There are 3-Cases When ΔH = ΔE ΔH = ΔE + PΔV I. In systems containing solids and liquids; there is no volume change (or it is small); PΔV = 0 and ΔH = ΔE 2. In reactions where the number of moles of gas does not change from reactants to products; no work is done. N2(g) + O2(g) ==> 2NO(g) PΔV = 0 and ΔH = ΔE 3. Reactions in which the number of moles of gas does change but q is >>> PΔV.

32 A thermochemical equation shows the enthalpy change (ΔH) of a chemical reaction and also the reactions stoichimetric factors. C3H8(g) + 5O 2 (g) 3CO 2 (g) + 4H 2 O(l) ΔH = kj 1) ΔH <0 = REACTION IS EXOTHERMIC--gives off heat 2) 2043 kj are released and can be written as product C3H8(g) + 5O 2 (g) 3CO 2 (g) + 4H 2 O(l) kj 3) more useful stoichiometric factors for calculations 1 mol C3H8(g) = kj 5O 2 (g) = kj Stoichiometeric factors 3CO 2 (g) = kj 4H 2 O(l) = kj

33 Using the Heat of Reaction (ΔH rxn ) to Find Amounts The major source of aluminum in the world is bauxite which is mostly aluminum oxide. Its thermal decomposition can be represented by: Al 2 O 3 (s) 2Al(s) + 3/2O 2 (g) ΔH rxn = 1676 kj If aluminum is produced this way, how many grams of aluminum can form when x 10 3 kj of heat is is used to decompose the oxide? SOLUTION: g Al = kj 2 mol Al 1676 kj g Al 1 mol Al = 32.2 g Al

34 Using the thermochemical equation calculate how much heat is evolved (or what is the enthalpy) when 266 g of white phosphorus (P 4 ) is combusted in air? The molar mass of P4 is g/mol. P 4 (s) + 5O 2 (g) P 4 O 10 (s) ΔHrxn = kj/mol ΔH = H (products) H (reactants) H rxn = 266. g P 4 1 mol P g P kj 1 mol P 4 = 6470 kj

35 In order to apply Hess s Law we must learn to interpret and manipulate thermochemical equations. The rules are: 1. The stoichiometric coefficients always refer to the number of moles of a substance---it s extensive! H 2 O (s) H 2 O (l) ΔH = 6.01 kj 2. If you reverse a reaction, the sign of ΔH changes H 2 O (l) H 2 O (s) ΔH = kj 3. If you multiply both sides of the equation by a factor n, then ΔH must change by the same factor n. 2H 2 O (s) 2H 2 O (l) ΔH = 2 x 6.01 = 12.0 kj 4. The physical states of all reactants and products must be specified in thermochemical equations. H 2 O (s) H 2 O (l) ΔH = 6.01 kj H 2 O (l) H 2 O (g) ΔH = 44.0 kj

36 An energy level diagram summarizes the enthalpy changes from an initial state to a final state in a chemical reaction. CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g) H 2 O(l) H 2 O(g) CH 4 + 2O 2 H initial H 2 O(g) H final Enthalpy, H ΔH < 0 heat out Enthalpy, H ΔH > 0 heat in CO 2 + 2H 2 O H final H 2 O(l) H initial A Exothermic process B Endothermic process

37 Enthalpies (ΔH) of chemical reactions can be determined using the physics of heat transfer a lab method called calorimetry. 1. Law of Conservation of Energy n i=1 q i = 0 2. Physics of heat transfer: the heat absorbed by a substance depends on the material s specific heat capacity, mass and temperature change. q = C ΔT = s m ΔT

38 Principle 1: The amount of heat, q, transferred from an object at higher temperature to an object at lower temperature is proportional to the difference in temperature of the two objects! Thermometer T1 T1 q ΔT q = C ΔT C = heat capacity Thermometer T2 Object 1 Object 2 Heat Q

39 The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a quantity of the substance by one degree Celsius or Kelvin (units of J/ C or J/K or cal/ C). Heat transferred = q (T final - T initial ) q = C ΔT material dependent constant heat capacity temperature difference between objects C has units of Energy per degree T (J/ C, J/K, cal/ C, cal/k) C = q ΔT

40 The heat capacity, C, of any material is found to be proportional to mass of a substance under question. C m C m s mass specific heat Units of s: J/g C, J/g K s is called the specific heat (s) of a substance. It is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius.

41 The specific heat (s) of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius. (How thermally sensitive a substance is to the addition of energy!) q = C ΔT C m s (1) (2) Substituting (2) into 1 gives: Heat (q) absorbed or released: q = s m ΔT

42 The molar heat capacity of a substance is the amount of heat (q) required to raise the temperature of one mole of the substance by one degree Celsius. Heat (q) absorbed or released: q = s m ΔT Units: J/mol C or J/mol K Units: C or K Units: moles

43 How much heat, q, is required to raise the temperature of kg of iron and 1000 kg of water from 25 to 75 C? Note that the specific heat capacity of iron is 0.45 J/g C Heat (q) absorbed or released: q = s m ΔT q Fe = kg 1000 g 1 kg (75 25 ) 0.450J C g = J q H2 O = kg 1000 g 1 kg (75 25 ) 4.18J C g = J

44 Determine the specific heat capacity of lead when grams of lead pellets at C are added to 50.0 ml of water held at 22.0 C. The final temperature of the insulated calorimeter is 28.8 C. Assume the container and thermometer do not absorb heat g C 2. Add pellets to 50.0 ml 22.0 C 3. What s Up? 28.8 C sp. heat?

45 1. Use the law of conservation of energy pertains to ALL OF THE PARTICIPANTS! n i=1 qpb + qwater + qcontainer = 0 q i = 0 1. Conservation of energy is the key fundamental physics! qpb = - qwater mpb cpb ΔTPb = -mwat cwat ΔTwat clead = mwat cwat ΔTwat / mlead ΔTlead q water = mcδt = (50.0 g)(4.184 J/g C)( ) C = 1.40 x10 3 J q lead = mcδt = (150.0 g)(c)( ) C = x10 3 J c lead = 0.13 Jg -1 C -1

46 Enthalpies (ΔH) of chemical reactions can be determined using a lab method called calorimetry. Understand two concepts and you can nail this to the wall! 1. Law of Conservation of Energy q system + q surroundings = 0 n i=1 q i = 0 2. Specific Heat Capacity of Substances q = s m ΔT = C ΔT Heat (q) absorbed depends on specific heat capacity, mass and temperature change.

47 There are two ways we can do calorimetry, (1) at constant pressure calorimetry; 2) at constant volume and Constant Pressure Calorimetry: A constant pressure calorimeter measures the heat change at constant pressure such that qp = ΔH = ΔE - P ΔV Constant Volume Bomb Calorimetry: A bomb calorimeter measures the heat change at constant volume such that qv = ΔE. (SKIP THIS IN BOOK) Skip the constant volume calorimetery in the textbook!

48 Coffee Cup Calorimeter A simple cheap calorimeter---pressure is constant-- so we are measuring: ΔH = q P = ΔE + PΔV Well insulated and therefore isolated. Calorimeter and other parts are assumed to not absorb heat (not true but assumed) heat will come from a chemical reaction, and is denoted: qrxt Energy Conservation is key: Sum the heats n i=1 q i = 0 q cal + q therm + q solution + q rxn = 0 q i = C i ΔT = mi si ΔT No heat enters or leaves!

49 Chemistry in Action Fuel Values of Foods and Other Substances C 6 H 12 O 6 (s) + 6O 2 (g) 6CO 2 (g) + 6H 2 O (l) ΔH = kj/mol All determined using a bomb calorimeter! Little calorie = chemistry 1 cal = J Big Calorie = Food 1 Cal = 1000 cal = 4184 J

50 Coffee-Cup Calorimetry A student mixes 50.0 ml of 1.00 M HCl and 50.0 ml of 1 M NaOH in a coffee-cup calorimeter and finds the mixture goes from C to C. Calculate the Heat of Neutralization for the reaction, assuming that the calorimeter and thermometer absorbs a negligible quantity of heat and that the total volume of the solution is additive; that the water density is g/ml and its specific heat capacity is that of water = 4.18 J/g K. HCl(aq) + NaOH(aq) H 2 O(l ) + NaCl(aq) ΔH =? kj

51 A student mixes 50.0 ml of 1.00 M HCl and 50.0 ml of 1 M NaOH in a coffeecup calorimeter and finds the mixture goes from C to C. Calculate the change Heat of Neutralization per mole acid for the reaction, assuming that the calorimeter loses (or absorbs) only a negligible quantity of heat, that the total volume of the solution is ml, that its density is g/ml and its specific heat is that of water = 4.18 J/g K. HCl(aq) + NaOH(aq) H 2 O(l ) + NaCl(aq) ΔH =? kj/mol n q i = q0 rxt + q soln = 0 and each q i = mi si ΔT i=1 qrxn qrxn qrxn -qcal Δ Δ

52 Hess s Law: The enthalpy change for any reaction is equal to the sum of the enthalpy changes for any individual step in the reaction. The heat of reaction for any reaction is a constant and depends only on the initial and final states (state function). From A Table CH4 (g) + 2O2 ==> CO + 2H2O + 1/2 O2 CO(g) + 2H2O + 1/2 O2(g) ==> CO2 + 2H2O CH4 (g) + 2O2 ==> CO2 + 2H2O ΔH2 = kj ΔH3 = kj ΔH1 =? kj

53 CH4 (g) + 2O2 ==> CO + 2H2O + 1/2 O2 ΔH1 = kj CO(g) + 2H2O + 1/2 O2(g) ==> CO2 + 2H2O ΔH2 = kj CH4 (g) + 2O2 ==> CO2 + 2H2O ΔH3 =? kj Because enthalpy is a state function, its value for any reaction only depends on the final and initial state. This gives us a powerful way to predict enthalpy of many chemical reactions. CH4 (g) + 2O2 H 1 = kj H 3 =? CO(g) + 2H2O + 1/2 O2(g) CO2 + 2H2O H 2 = kj

54 Energy level diagrams are useful pictorials of the thermodynamic transitions that take place during a chemical reaction. CH4 (g) + 2O2 Enthalpy, H H 3 = -890 kj H 1 = kj CO(g) + 2H2O + 1/2 O2(g) H 2 = kj CO2 + 2H2O CH4 (g) + 2O2 ==> CO + 2H2O + 1/2 O2 ΔH1 = kj CO(g) + 2H2O + 1/2 O2(g) ==> CO2 + 2H2O ΔH2 = kj CH4 (g) + 2O2 ==> CO2 + 2H2O ΔH3 = -890 kj

55 Calculate the enthalpy of the following reaction and draw and energy level diagram that summarize the thermodynamic enthalpies involved in this reaction. NO 2 (g) NO(g) + 1/2O 2 (g) ΔH = kj 1/2N 2 (g) + O 2 (g) NO2(g) ΔH = kj ½N 2 (g) + O 2 (g) NO + 1/2O 2 (g) ΔH = kj

56 Two gaseous pollutants that form in auto exhaust are CO and NO. An environmental chemist is studying ways to convert them to less harmful gases through the following equation: CO(g) + NO(g) CO 2 (g) + 1/2N 2 (g) ΔH =? Given the following information, calculate the unknown ΔH: CO(g) + 1/2O 2 (g) N 2 (g) + O 2 (g) CO 2 (g) ΔH A = kj 2NO(g) ΔH B = kj Multiply Equation B by 1/2 and reverse it. CO(g) + 1/2O 2 (g) CO 2 (g) ΔH A = kj NO(g) 1/2N 2 (g) + 1/2O 2 (g) ΔH B = kj CO(g) + NO(g) CO 2 (g) + 1/2N 2 (g) ΔH rxn = kj

57 Physicists and chemists define a standard thermodynamic standard state and denote with a superscript degree sign, ΔH Thermodynamic Standard State Conditions 1 atmosphere pressure 25 C = K 1 Molar concentration for solutions containing solutes ΔHrxn ΔH rxn Conditions of P, T, [] are given in problem and stated The superscript indicates thermodynamic standard state: P = 1 atm, T = 298, [1M]

58 What has changed, not much other than the conditions of pressure, temperature and concentration are implicit when you see the superscript. Example of Reaction N2 (g) + 3H2 (g) > 2NH3 (g) ΔH = kj 1 mol of N2 reacts with 3 mol H2 to produce 3NH3 and evolves kj of heat at 1 atm and 25 C 2N2 (g) + 6H2 (g) > 4NH3 (g) ΔH = 2 X kj 1/3N2 (g) + H2 (g) ----> 2/3NH3 (g) ΔH = 1/3 X kj

59 Types of Heats of Reaction, ΔH rxn ΔH

60 The Standard Enthalpy of Formation, ΔH f, is the enthalpy change associated with the formation of 1 mole of product from its naturally occurring elements under standard state conditions. Examples: Note: 1 mol product! H 2 (g) + O 2 (g) H 2 O(l ) ΔH f = kj/mol 3C(s) + 4H 2 (g) C 3 H 8 (g) ΔH f = kj/mol Ag(s) + 1/2Cl 2 (g) AgCl(s) ΔH f = kj/mol Note: fractional coefficients allowed Oxygen exists as O 2 gas at 25 C Carbon exists as solid graphite (C) at 25 C. Sulfur exits as S8 as a solid at 25 C Water is H 2 O(l ) in its standard state (not ice or water vapor).

61 It is also not possible to measure the absolute enthalpy. We define an arbitrary scale with the standard enthalpy of formation (ΔH f ) as a reference point. 1) The standard enthalpy of formation, ΔHf of any element in its most stable form is zero. ΔH 0 (O 2 ) = 0 f ΔH 0 (O 3 ) = 142 kj/mol f ΔH 0 (C, graphite) = 0 f ΔH 0 (C, diamond) = 1.90 kj/mol f ΔH 0 (S8, rhombic) = 0 kj/mol f The standard enthalpy of formation, ΔHf of many compounds is tabulated in many handbooks. This gives us predictive capability and is someone useful in the real world.

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63 Writing Formation Equations Write balanced equations for the formation of 1 mol of the following compounds from their elements in their standard states and include ΔH 0 f. (a) Silver chloride, AgCl, a solid at standard conditions. (b) Calcium carbonate, CaCO 3, a solid at standard conditions. (c) Hydrogen sulfide, H2S, a gas at standard conditions. PLAN: Use the table of heats of formation for values. Remember 1 mole of product is formed and fractions are allowed as coefficients.

64 Writing Formation Equations Write balanced equations for the formation of 1 mol of the following compounds from their elements in their standard states and include ΔH 0 f. (a) Silver chloride, AgCl, a solid at standard conditions. (b) Calcium carbonate, CaCO 3, a solid at standard conditions. (c) Hydrogen cyanide, HCN, a gas at standard conditions. PLAN: Use the table of heats of formation for values. SOLUTION: (a) Ag(s) + 1/2Cl 2 (g) AgCl(s) ΔH 0 f = kj (b) Ca(s) + C(graphite) + 3/2O 2 (g) CaCO 3 (s) ΔH 0 f = kj (c) O 2 (g) + 1/2N2(g) NO2(g) ΔH 0 f = 33.9 kj

65 We can use heats of formation ΔH f, to determine enthapies of reactions, ΔH rxn. 2-Problem Types To Expect For Enthalpy Calculations Equations where each enthalpy data is provided and you use Hess s Law to calculate enthalpy of another reaction. Use ΔH f data from a table and use the equation: ΔH rxn = Σ ni ΔHi f (products) - Σ mi ΔHj f (reactants)

66 Energy level diagrams are useful pictorials of the thermodynamic transitions that take place during a chemical reaction. Elements Enthalpy, H Reactants decomposition -ΔH 0 f formation ΔH 0 f Products ΔH 0 rxn ΔH rxn = Σ mδh f(products) - Σ nδh f(reactants)

67 The enthalpy of any reaction, ΔH rxn can be obtained by using standard heats of formations and the following equation: ΔH rxn = Σ ni ΔHif (products) - Σ mi ΔHjf (reactants) where Σ means the sum of ni is the respective stoich coefficient for i th product mi is the respective stoich coefficient for each i th reactant aa + bb cc + dd Example: Suppose ΔH rx =? ΔH rx = ΔH f (Products) ΔH f (Reactants) ΔH rx = [cδh f (C) + dδh f (D)] [aδh f (A) + bδh f (B)]

68 Calculating ΔH f from tabulated data Calculate the ΔH comb for the combustion of 2 moles of C 6 H 6 (l ) from enthalpies of formation data table. ΔH f Data From Table In Back of the Textbook (or Handbook of Physics and Chemistry C(s) + O 2 (g) CO 2 (g) ΔH f = H 2 (g) + O 2 (g) H 2 O(l ) 6C(s) + 3H 2 (g) C 6 H 6 (l ) ΔH f = kj ΔH f = kj

69 Calculating ΔH f from tabulated data 6C(s) + 6O 2 (g) 6CO 2 (g) 3H 2 (g) + O 2 (g) 3H 2 O(l ) C 6 H 6 (l ) 6C(s) + 3H 2 (g) ΔH f = 6( kj) ΔH f = 3( kj) ΔH f = ( kj) C 6 H 6 (l ) + O 2 (g) 6CO 2 (g) + 3H 2 O(l) ΔH comb =? ΔH comb = [6( kj) + 3( kj)] - [( kj)] = -3,267.4 kj ΔH rxn = Σ ni ΔHi f (products) - Σ mi ΔHj f (reactants) that s for one mole of benzene only! 2 C 6 H 6 (l ) +15 O 2 (g) 12 CO 2 (g) + 6 H 2 O(l ) ΔH comb = 2 X kj/mol

70 Calculate the heat of decomposition for the following process using standard enthalpies of formation found in Appendix. What kind of reaction is this? Is energy given off or needed to make this reaction proceed? CaCO 3 (s) CaO (s) + CO 2 (g) ΔH rxn = Σ ni ΔHif (products) - Σ mi ΔHif (reactants) ΔH 0 rxn = [ ΔH 0 (CaO) + ΔH 0 (CO 2 )] - [ ΔH 0 (CaCO 3 )] f ΔH 0 rxn = [ ] [ ] = 179 kj f f Reaction is decomposition and endothermic. Energy must be absorbed to proceed.

71 Calculate the standard enthalpy of formation of CS 2 (l) given that: C(graphite) + O 2 (g) CO 2 (g) ΔH 0 = kj rxn S(rhombic) + O 2 (g) SO 2 (g) ΔH 0 = kj rxn CS 2 (l) + 3O 2 (g) CO 2 (g) + 2SO 2 (g) ΔH 0 = kj rxn 1. Write the enthalpy of formation reaction for CS 2 C(graphite) + 2S(rhombic) CS 2 (l) 2. Add the given rxns so that the result is the desired rxn. + C(graphite) + O 2 (g) CO 2 (g) ΔH 0 rxn= kj 2S(rhombic) + 2O 2 (g) 2SO 2 (g) ΔH 0 rxn= x2 kj CO 2 (g) + 2SO 2 (g) CS 2 (l) + 3O 2 (g) ΔH 0 = kj rxn C(graphite) + 2S(rhombic) CS 2 (l) ΔH 0 = (2x-296.1) = 86.3 kj rxn

72 Benzene (C 6 H 6 ) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is kj/mol. 2C 6 H 6 (l) + 15O 2 (g) 12CO 2 (g) + 6H 2 O (l) ΔH 0 rxn = Σ nδh 0 f (products) - ΣmΔH 0 f (reactants) ΔH 0 rxn = [ 12ΔH 0 (CO 6ΔH 0 2 ) + (H 2 O) ] - [ 2ΔH 0 (C 6 H 6 )] f f f ΔH 0 rxn = [ 12x x ] [ 2x49.04 ] = kj kj 2 mol = kj/mol C 6 H 6

73 Nitric acid, whose worldwide annual production is about 8 billion kilograms, is used to make many products, including fertilizer, dyes, and explosives. The first step in the industrial production process is the oxidation of ammonia: 4NH 3 (g) + 5O 2 (g) 4NO(g) + 6H 2 O(g) Calculate ΔH 0 rxn from ΔH 0 f values. PLAN: Look up the ΔH 0 f values in Appendix and use Hess s Law to find ΔH rxn. ΔH 0 f NH 3 (g) = kj/mol ΔH 0 f O 2 (g) = 0 ΔH 0 f H 2 O(g) = kj/mol ΔH 0 f NO(g) = 90.4 kj/mol

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75 Nitric acid, whose worldwide annual production is about 8 billion kilograms, is used to make many products, including fertilizer, dyes, and explosives. The first step in the industrial production process is the oxidation of ammonia: 4NH 3 (g) + 5O 2 (g) 4NO(g) + 6H 2 O(g) Calculate ΔH 0 rxn from ΔH 0 f values. SOLUTION: ΔH rxn = Σ mδh 0 f (products) - Σ nδh 0 f (reactants) ΔH rxn = [4(ΔH 0 f NO(g) + 6(ΔH 0 f H 2 O(g)] - [4(ΔH 0 f NH 3 (g) + 5(ΔH 0 f O 2 (g)] = (4 mol)(90.4 kj/mol) + (6 mol)( kj/mol) - [(4 mol)(-46.3 kj/mol) + (5 mol)(0 kj/mol)] ΔH rxn = -904 kj

76 Heats of Reaction From ΔH f