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CH 6: Thermochemistry

Vanessa Prasad-Permaul Valencia College CHM 1045. CH 6: Thermochemistry. THERMOCHEMISTRY ENERGY. INTRODUCTION TO THERMODYNAMICS KEY DEFINITIONS FIRST LAW OF THERMODYNAMICS HEAT, WORK & INTERNAL ENERGY ENTHALPY THERMODYNAMIC SYSTEM HESS’ LAW STANDARD ENTHALPIES OF FORMATION

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CH 6: Thermochemistry

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  1. Vanessa Prasad-Permaul Valencia College CHM 1045 CH 6: Thermochemistry

  2. THERMOCHEMISTRYENERGY • INTRODUCTION TO THERMODYNAMICS • KEY DEFINITIONS • FIRST LAW OF THERMODYNAMICS • HEAT, WORK & INTERNAL ENERGY • ENTHALPY • THERMODYNAMIC SYSTEM • HESS’ LAW • STANDARD ENTHALPIES OF FORMATION • ENERGY AND THE ENVIRONMENT

  3. THERMOCHEMISTRYENERGY • Energy is the capacity to do work, or supply heat. Energy = Work + Heat • Kinetic Energy is the energy of motion. Ek = 1/2mv2Ek = kinetic energy m = mass (kg) v = velocity (m/s) (1 Joule = 1 kgm2/s2) (1 calorie = 4.184 J) • Potential Energyis stored energy.

  4. THERMOCHEMISTRYENERGY EXAMPLE 6.1 A regulation baseball weighing 143 grams travels 75 miles per hour. What is the kinetic energy of this baseball in Joules? Convert to calories. Ek = ½ mv2 m = 143g x 1kg = 0.143kg 1000g v = 75miles x 1hr x 1min x 1609.3m = 33.527 m/s 1 hour 60min 60sec 1 mile Ek = ½ x 0.143kg x (33.527m/s)2 = 80.3kg.m2/s2 = 80.3J 80.3J x 1 cal = 19.2cal 4.184J

  5. THERMOCHEMISTRYENERGY EXERCISE 6.1 An electron whose mass is 9.11 x 10-31 kg is accelerated by a positive charge to a speed of 5.0 x 106 m/s. What is the kinetic energy of the electron in Joules? In calories? Ek = ½ mv2

  6. THERMOCHEMISTRYHEAT OF REACTION THE FIRST LAW OF THERMODYNAMICS: • The law of the conservation of energy:Energy cannot be created or destroyed. • The energy of an isolated system must be constant. • The energy change in a system equals the work done on the system + the heat added. DE = Efinal – Einitial = E2 – E1 = q + w q = heat, w = work

  7. THERMOCHEMISTRYHEAT OF REACTION • Thermal Energyis the kinetic energy of molecular motion • Thermal energy is proportional to the absolute temperature. Ethermal T(K) • Heat is the amount of thermal energy transferred between two objects at different temperatures.

  8. THERMOCHEMISTRYENTHALPY • Enthalpies of Physical Change: Enthalpy is a state function, the enthalpy change from solid to vapor does not depend on the path taken between the two states. Hsubl = Hfusion + Hvap

  9. THERMOCHEMISTRYHEAT OF REACTION ILLUSTRATION OF A THERMODYNAMIC SYSTEM

  10. THERMOCHEMISTRYHEAT OF REACTION • In an experiment:Reactants and products are the system; everything else is the surroundings. • Energy flow from the system to the surroundings has a negative sign (loss of energy). (-E or - H) • Energy flow from the surroundings to the system has a positive sign (gain of energy). (+E or +H)

  11. THERMOCHEMISTRYHEAT OF REACTION • Enthalpies of Chemical Change:Often called heats of reaction (DHreaction). Endothermic:Heat flows into the system from the surroundings, heat is absorbed. • DH has a positive sign. • Energy added • q is (+) Exothermic:Heat flows out of the system into the surroundings, heat is evolved. • DH has a negative sign. • Energy subtracted • q is (-)

  12. THERMOCHEMISTRYENERGY SIGN CONVENTIONS FOR HEAT, WORK AND INTERNAL ENERGY

  13. THERMOCHEMISTRYHEAT OF REACTION EXAMPLE 6.2 Barium hydroxide octahydrate reacts with ammonium nitrate: Ba(OH)2.8H2O(s) + 2NH4NO3(s) 2NH3(g) + 10H2O(l) + Ba(NO3)2(aq) When 1mol of barium hydroxide octahydrate reacts with 2 mol of ammonium nitrate, the reaction mixture absorbs 170.8kJ of heat. Is this reaction exothermic or endothermic? What is the heat of reaction (q)? Energy is absorbed so this reaction is endothermic. q is +170.8kJ 13

  14. THERMOCHEMISTRYHEAT OF REACTION EXERCISE 6.2 Ammonia burns in the presence of a platinum catalyst to give nitric oxide 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(l) 4 mol of ammonia is burned and 1170kJ of heat is evolved. Is the reaction endothermic or exothermic? What is the value of q? Pt

  15. THERMOCHEMISTRYENTHALPY • The amount of heat exchanged between the system and the surroundings is given the symbolq. q = DE + PDV At constant volume (DV = 0): qv = DE At constant pressure: qp = DE + PDV = DH Enthalpy of Reaction: DH = Hproducts – Hreactants

  16. THERMOCHEMISTRYENTHALPY Work = -atmospheric pressure * area of piston * distance piston movesw = -PDV PRESSURE-VOLUME WORK

  17. THERMOCHEMISTRYHEAT OF REACTION EXAMPLE 6.2A If a balloon is inflated from 0.100L to 1.85L against an external pressure of 1.00atm, how much work is done (J)? w = -PDV = -1.00atm (1.85L – 0.100L) = -1.75 L.atm Conversion 1L.atm = 101.3J -1.75 L.atm x 101.3J 1L.atm -177J 17

  18. THERMOCHEMISTRYENTHALPY EXERCISE 6.2A A cylinder equipped with a piston expands against an external pressure of 1.58 atm. If the initial volume is 0.485L and the final volume is 1.245L, how much work is done (in J)?

  19. THERMOCHEMISTRYTHERMOCHEMICAL EQUATIONS EXAMPLE 6.3 Aqueous sodium hydrogen carbonate reacts with hydrochloric acid to produce aqueous sodium chloride, water and carbon dioxide gas. The reaction absorbs 12.7kJ of heat at constant pressure for each mole of NaHCO3. Write the thermochemical equation. NaHCO3(aq) + HCl NaCl(aq) + H2O(l) + CO2(g) DH = +12.7kJ

  20. THERMOCHEMISTRYTHERMOCHEMICAL EQUATIONS EXERCISE 6.3 A propellant for rockets is obtained by mixing the liquids hydrazine (N2H4) and dinitrogen tetroxide. These compounds react to give gaseous nitrogen and water vapor and 1049kJ of heat is evolved (at constant pressure when 1 mol reacts). Write the thermochemical equation for this reaction 20

  21. THERMOCHEMISTRYTHERMOCHEMICAL EQUATIONS • Reversing a reaction changes the sign of DH for a reaction. C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(l) DH = –2219 kJ 3 CO2(g) + 4 H2O(l)  C3H8(g) + 5 O2(g) DH = +2219 kJ • Multiplying a reaction increases DH by the same factor. 3 [C3H8(g) + 15 O2(g)  9 CO2(g) + 12 H2O(l)]: DH= 3(-2219) kJ DH = -6657 kJ

  22. THERMOCHEMISTRYTHERMOCHEMICAL EQUATIONS EXAMPLE 6.4 When 2 mol H2(g) and 1mol O2 react to give liquid water, 572kJ of heat evolves: 2H2(g) + O2(g)2H2O(l) ; DH = -572kJ Write this equation for 1 mol of liquid water. Give the reverse equation, in which 1 mol of liquid water dissociates into hydrogen and oxygen. 2H2(g) + O2(g)2H2O(l) ; DH = -572kJ 2 1H2(g) + ½ O2(g)1H2O(l) ; DH = -286kJ Reversing the equation: 1H2O(l) 1H2(g)+ ½ O2(g) ; DH = +286kJ

  23. THERMOCHEMISTRYTHERMOCHEMICAL EQUATIONS EXERCISE 6.4 Write the thermochemical equation for the reaction described in Exercise 6.3 for the case involving 1 mol N2H4. Write the reverse reaction.

  24. THERMOCHEMISTRYHEATS OF REACTION Applying Stoichiometry to Heats of Reaction Grams of A (reactant or product) Conversion Factor: grams of A to mols of A (using molar mass) Conversion Factor: mols of A to kJ (DH) x x kiloJoules of Heat =

  25. THERMOCHEMISTRYHEATS OF REACTION EXAMPLE 6.5 How much heat is involved when 9.007x105g of ammonia is produced according to the following reaction (assuming the reaction is at constant pressure)? N2(g) + 3H2(g) 2NH3(g); DH = -91.8kJ 9.07 x 105g x 1 mol NH3 x -91.8kJ = -2.45 x 106 kJ 17.0g NH3 2 mol NH3

  26. THERMOCHEMISTRYHEATS OF REACTION EXERCISE 6.5 How much heat evolves when 10.0g of hydrazine reacts according to the reaction described in EXERCISE6.3?

  27. THERMOCHEMISTRYHEAT CAPACITY AND SPECIFIC HEAT • Heat capacity (C)is the amount of heat required to raise the temperature of an object or substance a given amount. qcal = Ccal x DT • Specific Heat:The amount of heat required to raise the temperature of 1.00 g of substance by 1.00°C at constant pressure. q = s x m x t q = heat required (energy) s = specific heat m = mass in grams t = Tf - Ti

  28. THERMOCHEMISTRYHEATS OF REACTION SPECIFIC HEATS AND MOLAR CAPACITIES FOR SOME COMMON SUBSTANCES @ 25oC

  29. THERMOCHEMISTRYCALORIMETRY & HEAT CAPACITY • Molar Heat:The amount of heat required to raise the temperature of 1.00 mole of substance by 1.00°C. q = MH x n x t q = heat required (energy) MH = molar heat n = moles t = Tf - Ti

  30. THERMOCHEMISTRYHEAT CAPACITY AND SPECIFIC HEAT EXAMPLE 6.6 Calculate the heat absorbed by 15.0g of water to raise the temperature from 20.0oC to 50.0oC (at constant pressure). The specific heat of water is 4.18 J/g.oC Dt = 50.0oC – 20.0oC = 30.0oC q = s x m x Dt q = 4.18J x 15.0g x 30.0oC = 1.88 x 103 J g.oC

  31. THERMOCHEMISTRYHEAT CAPACITY AND SPECIFIC HEAT EXERCISE 6.6 Iron metal has a specific heat of 0.449 J/(g.oC). How much heat is transferred to a 5.00 g piece of iron, initially at 20.0oC when placed in a pot of boiling water? (Assume that the temperature of the water is 100.0oC and the water remains at this temperature, which is the final temperature of the iron).

  32. THERMOCHEMISTRYCALORIMETRY • Calorimetry is the science of measuring heat changes (q) for chemical reactions. There are two types of calorimeters: • Bomb Calorimetry: A bomb calorimeter measures the heat change at constant volume such that q = DE. • Constant Pressure Calorimetry: A constant pressure calorimeter measures the heat change at constant pressure such that q = DH.

  33. THERMOCHEMISTRYCALORIMETRY Constant Pressure Bomb

  34. THERMOCHEMISTRYCALORIMETRY EXAMPLE 6.7 Suppose 0.562g of graphite is placed in a calorimeter with an excess of oxygen at 25.00oC and 1atm pressure. Excess O2 ensures that all carbon burns to form CO2. The graphite is ignited, and it burns according to the following equation C(graphite) + O2(g) CO2(g) On reaction, the calorimeter temperature rises from 25.00oC to 25.89oC. The heat capacity of the calorimeter and it’s contents was determined in a separate experiment to be 20.7 kJ/oC. What is the heat of reaction at 25.00oC and 1 atm pressure? Express in a thermochemical equation.

  35. THERMOCHEMISTRYCALORIMETRY EXAMPLE 6.7 continued… qrxn = -CcalDt = -20.7kJ/oC x (25.89oC – 25.00oC) = -20.7kJ/oC x 0.89oC = -18.4kJ 0.562g x 1 mol = 0.0468 mol 12 g C -18.4kJ = -3.9 x 102 kJ/mol 0.0468mol C(graphite) + O2(g) CO2(g) ; DH = -3.9 x 102 kJ

  36. THERMOCHEMISTRYCALORIMETRY EXERCISE 6.7 Suppose 33mL of 0.120M HCl is added to 42mL of a solution containing excess sodium hydroxide in a coffee cup calorimeter. The solution temperature, originally at 25oC, rises to 31.8oC. Give the enthalpy change DH for the reaction: HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) Assume that the heat capacity and the density of the final solution in the cup are those of water. s = 4.184kJ/g.oC d = 1.000g/mL Express the answer as a thermochemical equation.

  37. THERMOCHEMISTRYHESS’S LAW • 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.(not a physical change, chemical change) 3 H2(g) + N2(g)  2 NH3(g) DH° = –92.2 kJ

  38. THERMOCHEMISTRYHESS’S LAW • Reactants and products in individual steps can be added and subtracted to determine the overall equation. (1) 2 H2(g) + N2(g)  N2H4(g) DH°1 = ? (2) N2H4(g) + H2(g)  2 NH3(g) DH°2 = –187.6 kJ (3) 3 H2(g) + N2(g)  2 NH3(g) DH°3 = –92.2 kJ DH°1 + DH°2 = DH°reaction Then DH°1 = DH°reaction - DH°2 DH°1 = DH°3 – DH°2 = (–92.2 kJ) – (–187.6 kJ) = +95.4 kJ

  39. THERMOCHEMISTRYHESS’S LAW EXAMPLE 6.8 What is the enthalpy of reaction, DH, for the formation of tungsten carbide (WC) from the elements? W(s) + C(graphite) WC(s) 2W(s) + 3O2(g) 2WO3(s) ; DH = -1685.8 kJ C(graphite) + O2(g) CO2(g) ; DH = -393.5 kJ 2WC(s) + 5O2(g) 2WO3(s) + 2CO2(g) ; DH = -2391.8kJ LAST EQUATION NEEDS TO BE REVERSED 2CO2(g) + 2WO3(s) 2WC(s) + 5O2(g) ; DH = 2391.8kJ

  40. THERMOCHEMISTRYHESS’S LAW EXAMPLE 6.8 cont… W(s) + C(graphite) WC(s) 2W(s) + 3O2(g) 2WO3(s) ; DH = -1685.8 kJ 2 2 C(graphite) + O2(g) CO2(g) ; DH = -393.5 kJ 2CO2(g) + 2WO3(s) 2WC(s) + 5O2(g) ; DH = 2391.8kJ 2 2

  41. THERMOCHEMISTRYHESS’S LAW EXAMPLE 6.8 cont… W(s) + C(graphite) WC(s) W(s) + 3/2 O2(g) WO3(s) ; DH = -842.9 kJ C(graphite) + O2(g) CO2(g) ; DH = -393.5 kJ CO2(g) + WO3(s) WC(s) + 5/2 O2(g) ; DH = 1195.9kJ W(s) + C(graphite) WC(s) DH = -40.5kJ

  42. THERMOCHEMISTRYHESS’S LAW EXERCISE 6.8 Manganese metal can be obtained by the reaction of manganese dioxide and aluminum 4Al(s) + 3MnO2(s) 2Al2O3(s) + 3Mn(s) What is the DH for this reaction? Use the following data: 2Al(s) + 3/2O2(g) Al2O3(s) ; H = −1676 kJ Mn(s) + O2(g) MnO2(s) ; H = −520 kJ

  43. THERMOCHEMISTRYSTANDARD ENTHALPIES OF FORMATION • Standard Heats of Formation (DH°f):The enthalpy change for the formation of 1 mole of substance in its standard state from its constituent elements in their standard states. • The standard heat of formation for any element in its standard state is defined as being ZERO. DH°f = 0 for an element in its standard state

  44. CO(g) -111 C2H2(g) 227 Ag+(aq) 106 CO2(g) -394 C2H4(g) 52 Na+(aq) -240 H2O(l) -286 C2H6(g) -85 NO3-(aq) -207 NH3(g) -46 CH3OH(g) -201 Cl-(aq) -167 N2H4(g) 95.4 C2H5OH(g) -235 AgCl(s) -127 HCl(g) -92 C6H6(l) 49 Na2CO3(s) -1131 THERMOCHEMISTRYSTANDARD ENTHALPIES OF FORMATION Some Heats of Formation, Hf° (kJ/mol)

  45. THERMOCHEMISTRYSTANDARD ENTHALPIES OF FORMATION • Thermodynamic Standard State:Most stable form of a substance at 1 atm pressure and 25°C; 1 M concentration for all substances in solution. • These are indicated by a superscript ° to the symbol of the quantity reported. • Standard enthalpy changeis indicated by the symbol DH°.

  46. THERMOCHEMISTRYSTANDARD ENTHALPIES OF FORMATION • Calculating DH° for a reaction: DH° = DH°f (Products) – DH°f (Reactants) • For a balanced equation, each heat of formation must be multiplied by the stoichiometric coefficient. aA + bB  cC + dD DH° = [cDH°f(C) + dDH°f(D)] – [aDH°f(A) + bDH°f(B)]

  47. THERMOCHEMISTRYSTANDARD ENTHALPIES OF FORMATION EXAMPLE 6.9 Use the values of DHof to calculate the heat of vaporization DHovap of carbon disulfide at 25oC. The vaporization process is: CS2(l) CS2(g) DHof= 89.7kJ/mol DHof= 116.9kJ/mol DHovap = SnDHof(products) – SmDHof(reactants) DHof[CS2(g)] - DHof[CS2(l)] 116.9kJ - 89.7kJ = 27.2kJ

  48. THERMOCHEMISTRYSTANDARD ENTHALPIES OF FORMATION EXERCISE 6.9 Calculate the heat of vaporization, DHovap of water using standard enthalpies of formation

  49. THERMOCHEMISTRYSTANDARD ENTHALPIES OF FORMATION EXAMPLE 6.10 Large quantities of ammonia are used to prepared nitric acid. The first consists of the catalytic oxidation of ammonia to nitric oxide 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g) 45.9kJ 0kJ 90.3 6kJ -241.8kJ What is the standard enthalpy change for this reaction? DHovap = SnDHof(products) – SmDHof(reactants) = [4(90.3) + 6(-241.8)]kJ - [4(-45.9) + 5(0)]kJ = -906kJ

  50. THERMOCHEMISTRYSTANDARD ENTHALPIES OF FORMATION EXERCISE 6.10 Calculate the enthalpy change for the following reaction: 3NO2(g) + H2O(l) 2HNO3(aq) + NO(g)

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