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PRINCIPLES OF CHEMISTRY I CHEM 1211 CHAPTER 5

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PRINCIPLES OF CHEMISTRY I CHEM 1211 CHAPTER 5. DR. AUGUSTINE OFORI AGYEMAN Assistant professor of chemistry Department of natural sciences Clayton state university. CHAPTER 5 THERMOCHEMISTRY. THERMOCHEMISTRY. - The study of the relationship between heat and chemical reactions. ENERGY.

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  1. PRINCIPLES OF CHEMISTRY I CHEM 1211CHAPTER 5 DR. AUGUSTINE OFORI AGYEMAN Assistant professor of chemistry Department of natural sciences Clayton state university

  2. CHAPTER 5 THERMOCHEMISTRY

  3. THERMOCHEMISTRY - The study of the relationship between heat and chemical reactions

  4. ENERGY - The ability to do work or to transfer heat - Energy is necessary for life: humans, plants, animals, cars - Forms of energy are interconvertible Two major categories - Kinetic energy - Potential energy

  5. ENERGY Kinetic Energy (Ek) - Energy of motion - Atoms and molecules possess kinetic energy since they have mass and are in motion m = mass of the object (kg) v = velocity (speed) of the object (m/s) - Units: The joule (J), 1 J = 1 kg-m2/s2, 4.184 J = 1 cal

  6. ENERGY Calculate the kinetic energy of an object of mass 38 g, which is moving with a constant velocity of 54 m/s. (a) in joules and (b) in calories

  7. ENERGY Potential Energy (Ep) - Energy by virtue of its position relative to other objects - Arises when there is a force operating on an object Ep = mgh m = mass of the object (kg) g = gravitational constant (9.8 m/s2) h = height of the object relative to a reference point (m) - Units: The joule (J), 1 J = 1 kg-m2/s2, 4.184 J = 1 cal

  8. ENERGY A 25-g marble is thrown upward and travels through a vertical distance of 10.0 m from the ground. Calculate the potential energy of the marble at a height of 5.0 m from the ground.

  9. ENERGY Electrostatic Potential Energy (Eel): - Due to interactions between charged particles κ = constant of proportionality (8.99 x 109 J-m/C2) C = coulomb, a unit of electrical charge Q1 and Q2 = electrical charges on two interacting objects d = distance between the two objects

  10. ENERGY Electrostatic Potential Energy (Eel): - Due to interactions between charged particles - For molecular-level objects, Q1 and Q2 are on the order of magnitude of electron charge (1.60 x 10-19 C) - Same sign Q1 and Q2 (both positive or both negative) causes repulsion and Eel is positive - Opposite signs Q1 and Q2 (one positive and one negative) cause attraction and Eel is negative - The lower the energy of a system, the more stable the system - Opposite charges interact more strongly and the system is more stable

  11. ENERGY Work (w) - The energy transferred when a force moves an object - The product of force (F) and distance (d) through which the object moves w = F x d Force - Any kind of push or pull exerted on an object

  12. ENERGY Heat (q) - Energy used to cause the temperature of an object to change - A form of energy necessary to change the temperature of a substance Chemical Energy - Potential energy resulting from forces that hold atoms together

  13. SYSTEM AND SURROUNDINGS System - The limited and well-defined portion of the universe under study Surroundings - Everything else in the universe Studying energy changes in a chemical reaction - The reactants and products make up the system - The reaction container and everything else make up the surroundings

  14. SYSTEM AND SURROUNDINGS Open System - Matter and energy can be exchanged with the surroundings (water boiling on a stove without a lid) Closed System - Energy but not matter can be exchanged with the surroundings (two reactants in a closed cylinder reacting to produce energy) Isolated System - Neither matter nor energy can be exchanged with the surroundings (insulated flask containing hot tea)

  15. INTERNAL ENERGY (E) - Sum of all potential and kinetic energies of all components - Change in internal energy = final energy minus initial energy E = Efinal - Einitial - Energy can neither be created nor destroyed - Energy is conserved

  16. INTERNAL ENERGY (E) E = Efinal - Einitial If Efinal> Einitial E is positive and system has gained energy from its surroundings If Efinal< Einitial E is negative and system has lost energy to its surroundings

  17. INTERNAL ENERGY (E) Energy Diagram Efinal Einitial E < 0 Internal energy, E Internal energy, E E > 0 Energy lost to surroundings Energy gained from surroundings Einitial Efinal E of system decreases E of system increases

  18. INTERNAL ENERGY (E) E = q + w q = heat added to or liberated from a system w = work done on or by a system Internal energy of a system increases when - Heat is added to the system from surroundings (positive q) - Work is done on the system by surroundings (positive w) +w +q system

  19. INTERNAL ENERGY (E) E = q + w q = heat added to or liberated from a system w = work done on or by a system Internal energy of a system decreases when - Heat is lost by the system to the surroundings (negative q) - Work is done by the system on the surroundings (negative w) -w -q system

  20. INTERNAL ENERGY (E) Endothermic Process - Process in which system absorbs heat (endo- means ‘into’) - Heat flows into system from its surroundings (melting of ice - the reason why it feels cold) - Heat is a reactant and E is positive N2(g) + O2(g) + heat → 2NO(g)

  21. INTERNAL ENERGY (E) Exothermic Process - Process in which system loses heat - Heat flows out of the system (exo- means ‘out of’) (combustion of gasoline) - Heat is a product and E is negative CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) + heat

  22. INTERNAL ENERGY (E) State Function - Property that depends on initial and final states of the system - Does not depend on path or how a change occurs - Internal Energy depends on initial and final states - Internal Energy is a state function - q and w, on the other hand, are not state functions

  23. INTERNAL ENERGY (E) Internal energy is influenced by - Temperature - Pressure - Total quantity of matter - Internal Energy is an extensive property

  24. INTERNAL ENERGY (E) Calculate E for a system absorbing 58 kJ of heat from its surroundings while doing 19 kJ of work on the surroundings. State whether it is an endothermic or an exothermic process q = +58 kJ (heat is added to the system from surroundings) w = -19 kJ (work is done by the system on the surroundings) E = q + w E = (+ 58 – 19) kJ = +39 kJ Endothermic

  25. ENTHALPY (H) - Heat flow in processes occurring at constant pressure - Only work-pressure (P-V work) are performed At constant pressure w = - PV w = work p = pressure V = change in volume = Vfinal - Vinitial

  26. ENTHALPY (H) Expansion of Volume - V is a positive quantity and w is a negative quantity - Energy leaves the system as work - Work is done by the system on the surroundings Compression of Volume - V is a negative quantity and w is a positive quantity - Energy enters the system as work - Work is done on the system by the surroundings

  27. ENTHALPY (H) Calculate the work associated with the expansion of a gas from 32 L to 58 L at a constant pressure of 12 atm w = - PV w = - (12 atm)(58 L - 32 L) = - 310 L.atm Gas expands hence work is done by system on surroundings

  28. ENTHALPY (H) Calculate the work associated with the compression of a gas from 58 L to 32 L at a constant pressure of 12 atm w = - PV w = - (12 atm)(32 L - 58 L) = 310 L.atm Gas compresses hence work is done on system by surroundings

  29. ENTHALPY (H) H = E + PV H, E, P, and V are all state functions Change in Enthalpy H = (E + PV)

  30. ENTHALPY CHANGE (H) Change in Enthalpyat Constant Pressure H = E + PV H = (qp + w) - w = qp qp = heat at constant pressure E = q + w PV = - w

  31. ENTHALPY CHANGE (H) H = qp Change in enthalpy = heat gained or lost at constant pressure Positive H - System gains heat from the surroundings - Endothermic process Negative H - System releases heat to the surroundings - Exothermic process

  32. ENTHALPY CHANGE (H) H = Hfinal - Hinitial - Enthalpy change is a state function - Enthalpy is an extensive property

  33. ENTHALPY CHANGE (H) Standard Enthalpy Change (Ho) - When all reactants and products are in their standard states Standard State - Pure form of a substance at standard temperature and pressure (STP) Conditions of STP - Standard temperature: 273 K or 0 oC - Standard pressure: 1.00 atm (101.325 kPa or 100 kPa)

  34. ENTHALPY CHANGE (H) • Enthalpy of Reaction (Hrxn) • H accompanying a chemical reaction • Enthalpy of Formation (Hf) • - H for forming a substance from its component elements • Enthalpy of Vaporization • - H for converting liquids to gases • Enthalpy of Fusion • - H for melting solids • Enthalpy of Combustion • - H for combusting a substance in oxygen

  35. ENTHALPY OF REACTION(Hrxn) - Heat of reaction - Enthalpy change accompanying a chemical reaction Hrxn = Hproducts – Hreactants Horxn = standard enthalpy of reaction Horxn = Hoproducts – Horeactants Thermochemical Equation - A chemical equation for which H is given

  36. ENTHALPY OF REACTION(Hrxn) Consider combustion of methane CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) H = -890 kJ H is an extensive property (depends on amount) From balanced equation 1 mol CH4 reacts with 2 mol O2 to release 890 kJ of heat Similarly 3 mol CH4 reacts with 6 mol O2 to release (3 x 890 kJ) of heat

  37. ENTHALPY OF REACTION(Hrxn) For the reverse reaction H is equal in magnitude but opposite in sign CO2(g) + 2H2O(l) → CH4(g) + 2O2(g) H = +890 kJ H depends on state of the reactants and products For instance, enthalpy of H2O(g) > enthalpy of H2O(l) Heat is absorbed when converting from liquid to gas

  38. ENTHALPY OF REACTION(Hrxn) Consider the following reaction CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) H = -890 kJ/mol Calculate the enthalpy change if 2.0 g methane is burned in excess oxygen When 1 mol of methane is burned, 890 kJ of heat is released

  39. CALORIMETRY - Measurement of heat flow - Device used to measure heat flow is the calorimeter

  40. CALORIMETRY Heat Capacity - Amount of heat required to raise the temperature of a substance by 1 K (or 1 oC) Units: cal/K (cal/oC) or J/K (J/oC) 1 cal = 4.184 J cal = calorie and J = Joule - Greater heat capacity requires greater amount of heat to produce a given temperature increase

  41. CALORIMETRY Specific Heat Capacity (Cs) - The quantity of heat energy necessary to raise the temperature of 1 gram of a substance by 1 K (or 1 oC) Units: cal/g.K (cal/g.oC) or J/g.K (J/g.oC) Calorie - The amount of heat energy needed to raise the temperature of 1 gram of water by 1 Kelvin (or 1 degree Celsius) Heat = [mass(g) of solution] x [specific heat of solution] x [T] q = mCs T

  42. CALORIMETRY Constant-Pressure - Heat lost by reaction (qrxn) = heat gained by solution (qsoln) - qrxn= qsoln qsoln = [mass(g) of solution] x [specific heat of solution] x [T] - Specific heat of dilute aqueous solutions are approximately equal to that of water (4.18 J/g.K)

  43. CALORIMETRY Constant-Volume (Bomb Calorimetry) - Mostly used to study combustion reactions - Heat capacity of the calorimeter (Ccal) is first determined using a substance that releases a known quantity of heat qrxn = - Ccal x T

  44. CALORIMETRY Calculate the amount of heat needed to increase the temperature of 140 g of a solution from 28 oC to 72 oC. Take the specific heat capacity of the solution as that of water, 4.18 J/g.K qsoln = mCs T T = (72 - 28) oC = 44 oC = 44 K q = (4.18 J/g.K)(140 g)(44 K) = 26,000 J

  45. CALORIMETRY When 50.0 mL of 0.100 M AgNO3 and 50.0 mL of 0.100 M HCl are mixed in a constant-pressure calorimeter, the temperature of the mixture increases from 22.20 oC to 23.11 oC as per the following reaction AgNO3(aq) + HCl(aq) → AgCl(s) + HNO3(aq) Calculate H (kJ/mol) for the reaction if the combined solution has a mass of 100.0 g and a specific heat of 4.18 J/g.oC

  46. CALORIMETRY T = (23.11 - 22.20) oC = 0.91 oC - qrxn = qsoln Solution temperature increases so qrxn is negative (exothermic) qrxn = - (4.18 J/g.oC)(100.0 g)(0.91 oC) = - 380 J = - 0.38 kJ mol AgNO3 = (0.100 M)(0.0500 L) = 0.00500 mol H = (-0.38 kJ)/(0.00500 mol) = -76 kJ/mol

  47. CALORIMETRY When a 0.5865-g sample of lactic acid (HC3H5O3) is burned in a bomb calorimeter whose heat capacity is 4.812 kJ/oC, the temperature increases from 23.10 oC to 24.95 oC. Calculate the heat of combustion of lactic acid (a) per gram and (b) per mole T = (24.95 - 23.10) oC = 1.85 oC qrxn = - Ccal x T = - (4.812 kJ/oC)(1.85 oC) = - 8.90 kJ (a) H = (8.90 kJ)/(0.5865 g) = - 15.2 kJ/g (b) H = (15.2 kJ/g)(90.09 g/mol) = - 1370 kJ/mol

  48. HESS’S LAW If a reaction is carried out in a series of steps - Enthalpy change for the overall reaction will equal the sum of the enthalpy changes for the individual steps - Overall H is independent of the number of steps - Overall H is independent of the reaction path - Useful for calculating energy changes that are very difficult to measure

  49. HESS’S LAW Useful Hints - Work backward from the required reaction, manipulating reactants and products of any given reactions - Reverse any reactions as needed - Multiply reactions by appropriate factors as needed - Use trial and error but allow the final reaction to guide you

  50. HESS’S LAW To determine H for A → 2C, given A → 2B HA C → B HC Reverse second reaction and multiply by 2 A → 2B HA 2B → 2C -2HC A → 2C H = HA + -2HC