Chapter 5 Principles of Chemical Reactivity: Energy and Chemical Reactions

Transcription

1 Chapter 5 Principles of Chemical Reactivity: Energy and Chemical Reactions Jeffrey Mack California State University, Sacramento

2 Energy & Chemistry Questions that need to be addressed: How do we measure and calculate the energy changes that are associated with physical changes and chemical reactions? What is the relationship between energy changes, heat, and work? How can we determine whether a chemical reaction is product-favored or reactant-favored at equilibrium? How can we determine whether a chemical reaction or physical change will occur spontaneously, that is, without outside intervention?

3 Units of Energy 1 calorie = heat required to raise temp. of 1.00 g of H 2 O by 1.0 C cal = 1 kilocalorie = 1 kcal 1 kcal = 1 Calorie (a food calorie ) SI units for energy: joule (J) 1 cal = exactly J James Joule

4 Energy & Chemistry Some Basic Principles Energy as the capacity to do work or transfer heat (q). Heat (q) is NOT temperature, temperature is a measure of kinetic energy. Energy is divided into two basic categories: Kinetic energy (the energy associated with motion) Potential energy (energy that results from an object s position). The law of conservation of energy requires that energy can neither be created nor destroyed. However, energy can be converted from one type into another.

5 Energy & Chemistry Energy can be divided into two forms: Kinetic: Energy of motion: Thermal Mechanical Electrical Potential: Stored energy: Gravitational Electrostatic Chemical & Nuclear

6 Potential & Kinetic Energy Kinetic energy energy of motion Translation

7 Potential & Kinetic Energy Potential energy energy a motionless body has by virtue of its composition and position.

8 Energy & Chemistry Burning peanuts supply sufficient energy to boil a cup of water. Burning sugar (sugar reacts with KClO 3, a strong oxidizing agent)

9 Thermal Equilibrium Energy transfer as heat will occur spontaneously from an object at a higher temperature to an object at a lower temperature. Transfer of energy as heat continues until both objects are at the same temperature and thermal equilibrium is achieved. At thermal equilibrium, the object with a temperature increase has gained thermal energy, the object with a temperature decrease has lost thermal energy.

10 System & Surroundings SYSTEM The object under study SURROUNDINGS Everything outside the system Energy flows between the two

11 Types of Systems Open: energy and matter can be exchanged with the surroundings. Closed: energy can be exchanged with the surroundings, matter cannot. Isolated: neither energy nor matter can be exchanged with the surroundings. A closed system; energy (not matter) can be exchanged. After the lid of the jar is unscrewed, which kind of system is it? 11

12 Directionality of Energy Transfer When energy leaves the system and goes into the surroundings, the process is said to be EXOTHERMIC. In the case of thermal energy, the temperature of the system decreases. (q system < 0) T system = (T final T initial ) < 0

13 Directionality of Energy Transfer When energy enters the system and from the surroundings, the process is said to be ENDOTHERMIC. In the case of thermal energy, the temperature of the system increases. (q system > 0) T system > 0

14 Heat & Changes in Internal Energy Heat flows between the system and surroundings. U is defined as the internal energy of the system. q = the heat absorbed or lost U = U final U initial = ± q

15 Heat & Changes in Internal Energy If heat enters the system: If heat leaves the system: U = U final U initial = ± q U > 0 therefore q is positive (+) U < 0 therefore q is negative ( ) The sign of q is a convention, it designates the direction of heat flow between the system and surroundings.

16 Heat & Changes in Internal Energy U = U final U initial = ± q Surroundings Usystem heat in q system > 0 (+) U system > 0 System heat out q system < 0 ( ) U system < 0 U system increases decreases q = the heat absorbed or lost by the system.

17 Thermodynamics is the scientific study of the interconversion of heat and other kinds of energy. State functions are properties that are determined by the state of the system, regardless of how that condition was achieved. energy, pressure, volume, temperature E = E final - E initial P = P final - P initial V = V final - V initial Potential energy of hiker 1 and hiker 2 is the same even though they took different paths. T = T final - T initial 17

18 First Law of Thermodynamics The total internal energy of an isolated system is constant. The system is that which we are interested in. The surroundings are everything in contact with the system. Together: System + Surroundings = Universe

19 First Law of Thermodynamics Energy Energy Any energy lost by the system is transferred to the surroundings and vice versa. Any change in energy is related to the final and initial states of the system. U system = U final U initial The same holds for the surrounding!

20 Relating U to Heat and Work Energy cannot be created or destroyed. Energy of the universes (system + surroundings) is constant. Any energy transferred from a system must be transferred to the surroundings (and vice versa). From the first law of thermodynamics: When a system undergoes a physical or chemical change, the change in internal energy is given by the heat added to or absorbed by the system plus the work done on or by the system: U q w change in system energy = heat lost or gained by the system + work done by or on the system

21 energy transfer in (endothermic), +q energy transfer out (exothermic), -q SYSTEM U = q + w w transfer in (+w) w transfer out (-w)

22 energy U Final Energy in U f > U i work and heat can balance! Energy out U f < U i U initial work in Work work out q in Heat q out U initial U Final U system > 0 (+) U system = 0 U system < 0 ( )

23 Energy is the capacity to do work. Work equals a force applied through a distance. When you do work, you expend energy. When you push down on a bike pump, you do work. P = force force = area 2 x change in volume V = x 3 F x 3 P V = x = F x 2 P V = work Therefore work is equal to a change of volume at constant pressure.

24 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 V (a) V = 5.4 L 1.6 L = 3.8 L P = 0 atm W = -0 atm x 3.8 L = 0 L atm = 0 joules (b) V = 5.4 L 1.6 L = 3.8 L P = 3.7 atm w = -3.7 atm x 3.8 L = L atm w = L atm x J 1L atm = J Try one more! A gas does 135 J of work while expanding, and at the same time it absorbs 156 J of heat. What is the change in internal energy? Answer: + 21J 24

25 Heat Capacity When heat is absorbed or lost by a body, the temperature must change as long as the phase (s, g or l) remains constant. The amount of heat (q) transfer related to an object and temperature is given by: q = C DT J C or K q = heat lost or gained (J) C = Heat Capacity of an object T = T final T initial is the temperature change ( C or K)

26 Heat & Specific Heat Capacity When heat is absorbed or lost by a body, the temperature must change as long as the phase (s, g or l) remains constant. The amount of heat (q) transfer per unit mass of a substance is related to the mass and temperature by: q = m S T q = heat lost or gained (J) m = mass of substance (g) J S= the Specific Heat Capacity of a compound g C or K T = T final T initial is the temperature change ( C or K)

27 Does it matter if we calculate a temperature change in Kelvin or degrees C? Recall that T = T f T in let T in = 25.0 C and T f = 50.0 C T f = 50.0 C = 50.0 C = K T in = 25.0 C = (25.0 C ) = K T = 25.0 C T = 25.0 K The are the same!

28 Determine the final temperature of a 25.0 g block of metal that absorbs 255 cal of energy. The initial temperature of the block was 17.0 C J 2.72 The specific heat capacity of the metal is: g C Strategy Map: Find T final of the metal after it absorbs the energy Data Information: Mass, initial temp, heat capacity of metal, heat (q) absorbed. Solve q = ms T for T final, plug in data. Final temperature of the metal is determined.

29 Determine the final temperature of a 25.0 g block of metal that absorbs 255 cal of energy. The initial temperature of the block was 17.0 C J 2.72 The specific heat capacity of the metal is: g C q = m S T q = m S (T Final T initial ) rearranging: q T final= +T m C initial T f = +255 cal 25.0 g J 1 cal C = 32.7 C J 2.72 g C T f > T in as expected

30 Example Problem: 55.0 g of iron at 99.8 C is placed into into 225 g water at initially at 21.0 C. At thermal equilibrium, the water and iron are both at 23.1 C What is the specific heat capacity of the metal?

31 Example Problem: 55.0 g of iron at 99.8 C is placed into into 225 g water at initially at 21.0 C. At thermal equilibrium, the water and iron are both at 23.1 C What is the specific heat capacity of the metal? Solution: Heat is transferred from the hot metal to the colder water. Energy is conserved so: q + q = 0 Fe q Fe water = -q water

32 Example Problem: 55.0 g of iron at 99.8 C is placed into into 225 g water at initially at 21.0 C. At thermal equilibrium, the water and iron are both at 23.1 C What is the specific heat capacity of the metal? Solution: Heat lost by metal = heat gained by water -q Fe = q w m Fe S Fe T Fe = -m w S w T w

33 Example Problem: 55.0 g of iron at 99.8 C is placed into into 225 g water at initially at 21.0 C. At thermal equilibrium, the water and iron are both at 23.1 C What is the specific heat capacity of the metal? Solution: S Fe = (-m w S w T w )/ [m Fe T Fe] ]

34 Example Problem: 55.0 g of iron at 99.8 C is placed into into 225 g water at initially at 21.0 C. At thermal equilibrium, the water and iron are both at 23.1 C What is the specific heat capacity of the metal? Solution: S Fe

35 Energy & Changes of State When matter absorbs heat, its temperature will rise until it undergoes a Phase Change. The matter will continue to absorb energy, however during the phase change its temperature remains constant: Phase changes are Isothermal processes.

36 Energy Transfer & Changes of State Some changes of state (phase changes) are endothermic: When you perspire, water on your skin evaporates. This requires energy. Heat from your body is absorbed by the water as it goes from the liquid state to the vapor state, as a result you cool down. + energy H2O(l) + Heat H2O(g)

37 Energy Transfer & Changes of State Some changes of state (phase changes) are exothermic: When it is muggy outside, water condenses on your skin. This releases energy. Heat from condensation is absorbed by your skin as water in the vapor state coverts to the liquid state. As a result you feel hot. + energy H2O(g) H2O(l) + Heat

38 Heating/Cooling Curve for Water Note the isotherms at each phase transition.

39 Heat & Changes of State The energy associated with a change of state is given by the Enthalpy or Heat of the phase change. Since there is no temperature change associated with the process, the units are most often in J/g or J/mol. Sublimation: subh > 0 (endothermic) Vaporization: vaph > 0 (endothermic) Melting or Fusion: fush > 0 (endothermic) Deposition: deph < 0 (exothermic) Condensation: conh < 0 (exothermic) Freezing: freh < 0 (exothermic) Where H refers to the Heat of a phase change

40 Energy Change Calculations Heating & Cooling: q (heating or cooling) = m S T Heat absorbed or lost in a Phase change: q (phase change) = (phase change) H n (n = moles or grams)

41 Problem: What quantity of heat is required to melt 500. g of ice at 0 C then heat the resulting water to steam at 100 C? +333 J/g J/g melt the ice form liquid water at 0 C heat the water to 100 C boil water fush S water Vap H

42 Problem: What quantity of heat is required to melt 500. g of ice at 0 C then heat the resulting water to steam at 100 C? Constants Needed: J Heat of fusion of ice = 333 g Specific heat of water = 4.18 J g K J Heat of vaporization = 2260 g Equations: q phase change = m H q heat = m S T

43 Problem: What quantity of heat is required to melt 500. g of ice at 0 C then heat the resulting water to steam at 100 C? Ice H 2 O(s) (0 C) fush water H 2 O(l) (0 C) C water water H 2 O(l) (100 C) vaph steam H 2 O(g) (100 C)

44 Problem: What quantity of heat is required to melt 500. g of ice at 0 C then heat the resulting water to steam at 100 C? Ice H 2 O(s) (0 C) fush water H 2 O(l) (0 C) C water water H 2 O(l) (100 C) vaph steam H 2 O(g) (100 C) melt ice heat water boil water q total = q1 = nice DfusH + q2 = mwater Cwater DT + q3 = nwater DvapH

45 Problem: What quantity of heat is required to melt 500. g of ice at 0 C then heat the resulting water to steam at 100 C? Ice H 2 O(s) (0 C) fush water H 2 O(l) (0 C) C water water H 2 O(l) (100 C) vaph steam H 2 O(g) (100 C) J qtotal 500. g 333 g J 500. g 4.18 (100.0 C 0.0 C) g C J g g J

46 Enthalpy Enthalpy, H is the heat transferred between the system and surroundings under conditions of constant pressure. if no PV work is done by the system, V = 0 H = q p H = U + PV ( ) D H = D U + PV = D U + PDV H = U p 0 the subscript p indicates constant pressure the change in enthalpy is the change in internal energy at constant pressure

47 Enthalpy is a State Function No matter which path is taken (A B) the results are the same: Final Initial B Since individual Enthalpies A cannot be directly measured, we only deal with enthalpy changes ( H = H H ) f i For a system the overall change in Enthalpy is path independent.

48 Enthalpy Conditions Since Enthalpies are state functions, one must specify the conditions at which they are measured. H(T,P): Enthalpy is a function of temperature and pressure. H indicates that the Enthalpy is taken at Standard State conditions. Standard State Conditions are defined as: 1 atm = 760 mm Hg or 760 torr & K or 25 C

49 Enthalpy Diagrams Values of H are measured experimentally. Negative values indicate exothermic reactions. Positive values indicate endothermic reactions. A decrease in enthalpy during the reaction; H is negative. An increase in enthalpy during the reaction; H is positive.

50 Enthalpies & Chemical Reactions: Energy H r Reactants Products H = H final H initial Enthalpy of reaction = H r = H products H reactants H (Products) H (Reactants) H r > 0 H r < 0 H (Reactants) Endothermic H (Products) Exothermic

51 Thermochemical Equations Just like a regular chemical equation, with an energy term. CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g) rh o = 802 kj CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g) kj Energy is a product just like CO 2 or H 2 O! What does this imply? CONVERSION FACTORS!!! From the equation: energy out Exothermic kj of Energy Released +802 kj of Energy Released 1 mol CH 4(g) consumed 2 mol H2O (g) produced

52 Problem: How many kj of energy are released when g of methane, CH 4 (g) is combusted? CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g) rh o = 802 kj g mols J molar mass Reaction enthalpy 1 mol CH g CH g 1 mol rxn 1 mol CH kj = 1mol rxn kj

53 Thermochemical Equations When the reaction is reversed, the sign of H reverses: CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g) H r o = 802 kj Exothermic CO 2 (g) + 2H 2 O(g) CH 4 (g) + 2O 2 (g) H r o = +802 kj Endothermic

54 Thermochemical Equations H scales with the reaction: 1 2 [ ] [ ] CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g) H r o = 802 kj CH 4(g) + O 2(g) 1 CO 2(g) + H2O(g) 2 2 H r o = 401 kj Yes you can write the reaction with fractions, so long as you are writing it on a mole basis

55 Thermochemical Equations Change in enthalpy depends on state: CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g) H r = 802 kj CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(l) H r = 890 kj From your text: H 2 O(g) H 2 O(l) [ H = 44 kj/mol ] 2 = 88kJ = 88kJ This means that water s liquid state lies 44 kj/mol lower than the gas state The difference in the rxn H is due to the change in state!!

56 Enthalpy (H) Comparing the reaction with water as a gas or liquid: CH 4 (g) + 2O 2 (g) reactants H r o = 802 kj either pathway gives the same results! products CO 2 (g) + 2H 2 O(g) The difference is the 88 kj released when 2 mols of water go from gas to liquid kj H r o = 890 kj products CO 2 (g) + 2H 2 O(l)

57 Constant Pressure Calorimetry, Measuring H A constant pressure calorimeter can be used to measure the amount of energy transferred as heat under constant pressure conditions, that is, the enthalpy change for a chemical reaction.

58 Constant Pressure Calorimetry, Measuring H The constant pressure calorimeter used in general chemistry laboratories is often a coffee-cup calorimeter. This inexpensive device consists of two nested Styrofoam coffee cups with a loose-fitting lid and a temperature-measuring device such as a thermometer or thermocouple.

59 Constant Pressure Calorimetry, Measuring H Because the coffee-cup calorimeter. is an isolated system, What happens in the coffee cup, stays in the coffee cup! No mass loss to the surroundings. No heat loss to the surroundings.

60 Problem: 5.44 g of ammonium nitrate was added to 150.0mL of water in a coffeecup calorimeter. This resulted in a decrease in temperature from 18.6 C to 16.2 C. Calculate the enthalpy change for dissolving NH 4 NO 3 (s) in water in kj/mol. Assume the solution has a specific heat capacity of 4.18 J/g? K.

61 Problem: 5.44 g of ammonium nitrate was added to 150.0mL of water in a coffeecup calorimeter. This resulted in a decrease in temperature from 18.6 C to 16.2 C. Calculate the enthalpy change for dissolving NH 4 NO 3 (s) in water in kj/mol. Assume the solution has a specific heat capacity of 4.18 J/g? K. Data Information: Mass of reactant, C, water & temperature change. Step 1: Cals. Moles of NH 4 NO 3 Step 3: q solution q solution + q rxn = 0 Eq. gives mole ratios (stoichiometry) q rxn = q(nh 4 NO 3 ) Step 2: Determine q solution = mc T Step 4: rh =q rxn /mol NH 4 NO 3 Enthalpy per mole of reactant

62 Problem: 5.44 g of ammonium nitrate was added to 150.0mL of water in a coffeecup calorimeter. This resulted in a decrease in temperature from 18.6 C to 16.2 C. Calculate the enthalpy change for dissolving NH 4 NO 3 (s) in water in kj/mol. Assume the solution has a specific heat capacity of 4.18 J/g? K. 1 mol NH NO 5.44 g NH NO moles NH NO g solution q m C T solution J qsolution g 4.18 (16.2 C 18.6 C) g C 3 qsolution J

63 Problem: 5.44 g of ammonium nitrate was added to 150.0mL of water in a coffeecup calorimeter. This resulted in a decrease in temperature from 18.6 C to 16.2 C. Calculate the enthalpy change for dissolving NH 4 NO 3 (s) in water in kj/mol. Assume the solution has a specific heat capacity of 4.18 J/g? K. 3 q = J solution q + q = 0 rxn solution 3 q = - q = J rxn solution 3 qrxn J 1kJ kj D rh = = = moles of reaction moles NH NO 10 J mol 4 3 The sign is positive indicating an endothermic process.

64 Constant Volume Calorimetry, Measuring U Under conditions of constant volume, any heat transferred is equal to a change of internal energy U r. q V = U r

65 Constant Volume Calorimetry, Measuring U Heats of combustion ( r U o combustion ) are measured using a device called a Bomb Calorimeter. A combustible sample is reacted with excess O 2 The heat capacity of the bomb is constant. The heat of reaction is found by: q = - C DT rxn cal C = C + C cal water bomb

66 Calculating Heat in an Exothermic Reaction Octane, the primary component of gasoline combusts by the reaction: C 8 H 18 (l) + 25/2 O 2 (g) 8 CO 2 (g) + 9H 2 O(l) A 1.00 g sample of octane is burned in a bomb calorimeter that contains 1.20 kg of water surrounding the bomb. The temperature of the water rises to C from C when the octane is reacted. If the heat capacity of the bomb is 837 J/ C, calculate the heat of reaction per mole of octane. Since the temperature of the water rose, the reaction must have been exothermic: Therefore one can write: q rxn = q water + q bomb

67 Calculating Heat in an Exothermic Reaction - q RXN = q rxn = m wate r S water T water + q bomb T water 4.184J g C g C q water J 837 C C q bomb q RXN = 4803 J or 48.0 kj Heat transferred per mole q V : kj 3 kj = mol mol 1.00g 114.2g

68 Hess s Law The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for the individual steps in the reaction. Intermediate Reaction so we can write rh 1 + rh 2 = rh Reactants rh =? unknown! Products What if the enthalpy changes through another path are know? The sum of the H s in one direction must equal the sum in the other direction. Why? Because enthalpy is a state function Path independent!

69 Hess s law & Energy Level Diagrams Forming CO 2 can occur in a single step or in a two steps. r H total is the same no matter which path is followed.

70 Hess s Law Problem: Example: Determine the rh for the reaction: 3H 2 (g) + N 2 (g) 2NH 3 (g) H ro =??? Given: (1) 2 H 2 (g) + N 2 (g) N 2 H 4 (g) H 1 = kj (2) N 2 H 4 (g) + H 2 (g) 2NH 3 (g) H 2 = kj Notice that the path from reactants to products in the desired reaction goes through an intermediate compound in the given reactions. This means that the path for hydrogen and nitrogen to produce ammonia goes through hydrazine (N 2 H 4 ). Therefore, the path to the enthalpy of the reaction must be a sum of the two given reactions!

71 Hess s Law Problem: Example: Determine the rh for the reaction: 3H 2 (g) + N 2 (g) 2NH 3 (g) H ro =??? Given: (1) 2 H 2 (g) + N 2 (g) N 2 H 4 (g) H 1 = kj (2) N 2 H 4 (g) + H 2 (g) 2NH 3 (g) H 2 = kj adding equations (1) & (2) yields: 2 H 2 (g) + N 2 (g) + N 2 H 4 (g) + H 2 (g) N 2 H 4 (g) + 2NH 3 (g)

72 Hess s Law Problem: Example: Determine the rh for the reaction: 3H 2 (g) + N 2 (g) 2NH 3 (g) H ro =??? Given: (1) 2 H 2 (g) + N 2 (g) N 2 H 4 (g) H 1 = kj (2) N 2 H 4 (g) + H 2 (g) 2NH 3 (g) H 2 = kj adding equations (1) & (2) yields: Look what happens / / 2 H 2 (g) + N 2 (g) + N 2 H 4 (g) + H 2 (g) N 2 H 4 (g) + 2NH 3 (g) 3H 2 (g) + N 2 (g) 2NH 3 (g)

73 Hess s Law Problem: Example: Determine the rh for the reaction: 3H 2 (g) + N 2 (g) 2NH 3 (g) H ro =??? Given: (1) 2 H 2 (g) + N 2 (g) N 2 H 4 (g) H 1 = kj (2) N 2 H 4 (g) + H 2 (g) 2NH 3 (g) H 2 = kj adding equations (1) & (2) yields: Look what happens therefore / / 2 H 2 (g) + N 2 (g) + N 2 H 4 (g) + H 2 (g) N 2 H 4 (g) + 2NH 3 (g) 3H 2 (g) + N 2 (g) 2NH 3 (g) H r = H 1 + H 2 = kj + ( kj) = 92.2 kj

74 intermediates H 2 (g) + N 2 H 4 (g) reactants 3H 2 (g) + N 2 (g) (Step 1) H o = 95.4kJ (Step 2) H o = 187.6kJ (Overall) H o = 92.2kJ products 2N 3 (g) Reaction (1) is endothermic by 95.4 kj Reaction (2) is exothermic by kj Since the exothermicity has a greater magnitude than the endothermicity, the overall process is exothermic ( r H < 0).

75 Standard Enthalpies of Formation When 1 mole of compound is formed from its elements, the enthalpy change for the reaction is called the enthalpy of formation, fh o (kj/mol). These enthalpies are always reported at Standard conditions: 1 atm and 25 C (298 K). The standard enthalpies of formation of the most stable form of any element is zero: H f (element) = 0 H f O 2 (g) = 0 H f O(g) 0 elemental form NOT the elemental form

76 Enthalpy Values Since Enthalpy is a state function, enthalpy values depend on the reaction conditions in terms of the phases of reactants and products. H 2 (g) + ½ O 2 (g) H 2 O(g) H = -242 kj H 2 (g) + ½ O 2 (g) H 2 O(liq) H = -286 kj Same reaction, different phases, different enthalpies.

77 Formation Reactions A chemical reaction that describes the formation of one mole of a compound from its elements at standard state conditions is known as a formation reaction. The formation of water is given by the reaction: H 2 (g) + ½ O 2 (g) H 2 O(l) Each element and the compound are represented by the physical state they take on at 25.0 C and 1 atm pressure. (Standard State conditions)

78 Question: What is the formation reaction for potassium permanganate? elements K (s) + Mn (s) + 2 O 2 (g) compound KMnO 4 (s) salts are solids at standard state conditions. metals are solids at standard state conditions. oxygen is a gas at standard state conditions. balance for one mole of the product standard state conditions = 25 o C and 1 atm

79 Enthalpy Changes for a Reaction: Using Standard Enthalpy Values All components of a reaction can be related back to their original elements. Each compound has an enthalpy of formation associated with it. Reactants require energy to return to component elements. Products release energy when formed form component elements. Since enthalpy is a state function, the sum of the above must relate somehow to the overall enthalpy of a reaction.

80 Consider the Combustion of Propane 3C(s) + 8H 2 (g) + 5O 2 (g) C 3 H 8 (g) + 5O 2 (g) 3CO 2 (g) + 4 H 2 O(l) In order to make CO 2 (g) and H 2 O(l) one must break the propane up into its elements. This takes energy. The elements carbon, hydrogen and oxygen then combine to make the new compounds, CO 2 and H 2 O. This process releases energy.

81 Enthalpy Changes for a Reaction: Using Standard Enthalpy Values The sum of the fh for the products multiplied by the respective coefficients, subtracted by the sum of the fh for the reactants multiplied by the respective coefficients, yields the H f for the reaction. In other words: Energy gained Energy spent = Net Energy ( ) ( ) D = å D -å D H n H products m H reactants r f f n and m are the stoichiometric balancing coefficients.

82 Problem: Calculate the H r for CaCO 3 (s) formed from CaO and CO 2 given the following H f CaO(s) + CO 2 (g) H f [CaO(s)] = kj/mol H f [CO 2 (g)] = kj/mol H f [CaCO 3 (s)] = 1207 kj/mol CaCO 3 (s)

83 Problem: Calculate the rh for CaCO 3 (s) given the following H f CaO(s) + CO 2 (g) CaCO 3 (s) H f [CaO(s)] = kj/mol D rh = åndfh ( products) H f [CO 2 (g)] = kj/mol åmdfh reactants H f [CaCO 3 (s)] = 1207 kj/mol ( )

84 Problem: Calculate the rh for CaCO 3 (s) given the following fh CaO(s) + CO 2 (g) CaCO 3 (s) fh [CaO(s)] = kj/mol D rh = åndfh ( products) fh [CO 2 (g)] = kj/mol åmdfh reactants fh [CaCO 3 (s)] = 1207 kj/mol ( ) rh = 1mol CaCO 3 1mol CaO 1mol CO kj/mol ( kj/mol kj/mol) rh = 178 kj/mol

85 Problem: The combustionh for naphthalene, C 10 H 8 (l) is 5156 kj/mol. Calculate the H f for naphthalene given the following enthalpies of formation: H f [CO 2 (g)] = kj/mol H f [H 2 O(l)] = kj/mol

86 Problem: The combustionh for naphthalene, C 10 H 8 (l) is 5156 kj/mol. Calculate the fh for naphthalene given the following enthalpies of formation: fh [CO 2 (g)] = kj/mol fh [H 2 O(l)] = kj/mol Step 1: Write the balanced equation for the reaction C 10 H 8 (l) + 12 O 2 (g) 10CO 2 (g) + 4H 2 O(l)

87 Next recall that: ( ) ( ) D = å D -å D H n H products m H reactants r f f C 10 H 8 (l) + 12 O 2 (g) 10CO 2 (g) + 4H 2 O(l) combustionh = 10 fh [CO 2 (g)] + 4 fh [H 2 O(l)] { f H [C 8 H 10 (l)] + 12 fh [O 2 (g)]} From the problem, all quantities are know but fh C10H8

88 Problem: The combustionh for naphthalene, C 10 H 8 (l) is 5156 kj/mol. Calculate the fh for naphthalene given the following enthalpies of formation: fh [CO 2 (g)] = kj/mol fh [H 2 O(l)] = kj/mol fh [C 10 H 8 (l)] = 10 fh [CO 2 (g)] + 4 fh [H 2 O(l)] { comb H + 12 fh [O 2 (g)]} elements = 0

89 Problem: The combustionh for naphthalene, C 10 H 8 (l) is 5156 kj/mol. Calculate the fh for naphthalene given the following enthalpies of formation: fh [CO 2 (g)] = kj/mol fh [H 2 O(l)] = kj/mol fh [C 10 H 8 (l)] = 10 fh [CO 2 (g)] + 4 fh [H 2 O(l)] H f [C 10 H 8 (l)] = { comb H + 12 fh [O 2 (g)]} elements = 0 10 ( kj/mol + 4 ( kj/mol) ( 5156 kj/mol) = + 79 kj/mol