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Chapter 5

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  1. Chapter 5 Principles of Chemical Reactivity

  2. Basic Principles Thermodynamics:The science of heat and work Energy: the capacity to do work -chemical, mechanical, thermal, electrical, radiant, sound, nuclear -affects matter by raising its temperature, eventually causing a state change -All physical changes and chemical changes involve energy Potential Energy: energy that results from an object’s position -gravitational, chemical, electrostatic Kinetic Energy: energy of motion

  3. Basic Principles Law of Energy Conservation: Energy can neither be created nor destroyed -a.k.a. The first law of thermodynamics -The total energy of the universe is constant Temperature vs. Heat: • Temperature is the measure of an object’s heat energy • Heat ≠ temperature

  4. The Measurement of Heat Thermal Energy depends on temperature and the amount (mass or volume) of the object -More thermal energy a substances has the greater the motion its atoms/molecules have -Total thermal energy of an object is the sum of the individual energies of all atoms/molecules/ions that make up that object SI unit: Joule (J) 1 calorie = 4.184 J English unit = BTU

  5. Converting Calories to Joules Convert 60.1 cal to joules

  6. Basic Principles System: object or collection of objects being studied • In lab, the system is the chemicals inside the beaker Surroundings: everything outside of the system that can exchange energy with the system • The surroundings are outside the beaker Universe: system plus surroundings Exothermic:heat transferred from the system to the surroundings Endothermic: heat transferred from the surroundings to the system

  7. J g.K Specific Heat Capacity (C) amount of heat required to raise the temperature of 1gram of a substance by 1 kelvin SI Units: Specific heat capacity = J /g.K Specific heat of water = 4.184

  8. Heat Transfer Heat transfer equation used to calculate amounts of heat (q) in a substance K J J g.K g q1 + q2 + q3 … = 0 or qsystem + qsurroundings = 0

  9. Heat Transfer Calculate the amount of heat to raise the temperature of 200 g of water from 10.0 oC to 55.0 oC

  10. Heat Transfer Calculate the amount of heat energy (in joules) needed to raise the temperature of 7.40 g of water from 29.0°C to 46.0°C

  11. Heat Transfer A 1.6 g sample of metal that appears to be gold requires 5.8 J to raise the temperature from 23°C to 41°C. Is the metal pure gold? J g.K Specific heat of gold is 0.13 Therefore the metal cannot be pure gold.

  12. Changes of State occurs when enough energy is put into a substance to over come molecular interactions Solid-liquid: molecules in a solid when heated move about vigorously enough to break solid-solid molecular interactions to become a liquid Liquid-gas: molecules in a liquid when heated move about more vigorously enough to break liquid-liquid molecular interactions to become a gas Note: This happens in reverse by removing heat energy

  13. Energy and Changes of State Heat of fusion: heat needed to convert a substance from a solid to a liquid (at its melting/freezing point) 333 J/g for water Heat of vaporization: heat needed to convert a substance from a liquid to a gas (at its boiling/condensation point) 2256 J/g for water Example: Calculate the amount of heat involved to convert 100.0 g of ice at -50.0°C to steam at 200.0°C.

  14. The First Law of Thermodynamics This law can be stated as, “The combined amount of energy in the universe is constant” Also called-The Law of Conservation of Energy: • Energy is neither created nor destroyed in chemical reactions and physical changes.

  15. The First Law of Thermodynamics There are two basic ideas for thermodynamic systems: • Chemical systems tend toward a state of minimum potential energy Some examples of this include: -H2O flows downhill -Objects fall when dropped Epotential = mg(h)

  16. State Function The value of a state function is independent of pathway -An analogy to a state function is the energy required to climb a mountain taking two different paths: E1 = energy at the bottom of the mountain E1 = mgh1 E2 = energy at the top of the mountain E2 = mgh2 E = E2-E1 = mgh2 – mgh1 = mg(h) Examples of state functions: Temperature, Pressure, Volume, Energy, Entropy, and enthalpy Examples of non-state functions: Number of moles, heat, work

  17. The First Law of Thermodynamics • Chemical systems tend toward a state of maximum disorder Common examples of this are: - A mirror shatters when dropped and does not reform - It is easy to scramble an egg and difficult to unscramble it - Food dye when dropped into water disperses

  18. The First Law of Thermodynamics Thermodynamic questions: • Will these substances react when they are mixed under specified conditions? • If they do react, what energy changes and transfers are associated with their reaction? • If a reaction occurs, to what extent does it occur?

  19. Changes in Internal Energy (E or DU) all of the energy contained within a substance • all forms of energy such as kinetic, potential, gravitational, electromagnetic, etc. First Law of Thermodynamics states that the change in internal energy, E, is determined by the heat flow (q) and the work (w) E = q + w book: DU = q + w

  20. Changes in Internal Energy (E) DE = Eproducts – Ereactants DE = q + w at constant pressure: w = -P x DV q > 0 if heat is absorbed by the system q < 0 if heat is absorbed by the surroundings w > 0 if surroundings do work on the system w < 0 if system does work on the surroundings

  21. Changes in Internal Energy (E) If 1200 joules of heat are added to a system, and the system does 800 joules of work on the surroundings, what is the: • energy change for the system, Esys? • energy change of the surroundings, Esurr?

  22. Changes in Internal Energy (E) E is negative when energy is released by a system -Energy can be written as a product of the process

  23. Changes in Internal Energy (E) E is positivewhen energy is absorbed by a system undergoing a chemical or physical change • Energy can be written as a reactant of the process

  24. Exothermic reactions: release energy in the form of heat to the surroundings (DH < 0) -heat is transferred from a system to the surroundings Endothermic reactions: gain energy in the form of heat from the surroundings (DH > 0) -heat is transferred from the surroundings to the system For example, the combustion of propane: Combustion of butane: Enthalpy Changes for Chemical Reactions

  25. Enthalpy Change (DH) Heat content of a substance at constant pressure - Chemistry is commonly done in open beakers on a desk top at atmospheric pressure - Therefore enthalpy change (H) is the change in heat content: H = qp at constant pressure DE = qv at constant volume If DE and DH < 0: energy is transferred to the surroundings If DE and DH > 0: energy is transferred to the system - Enthalpy and energy differ by the amount of work DE = DH + w and w = -PDV

  26. Enthalpy Changes for Chemical Reactions Exothermic reactions generate specific amounts of heat • Because the potential energies of the products are lower than the potential energies of the reactants Endothermic reactions consume specific amounts of heat • Potential energies of the reactants are lower than the products DH for the reverse reaction is equal, but has the opposite sign to the forward reaction

  27. balanced chemical reaction with the H value for the reaction H < 0 designates an exothermic reaction: heat is a product, the container feels hot H > 0 designates an endothermic reaction: heat is a reactant, the container feels cold Thermochemical Equations

  28. Hess’s Law If a reaction is the sum of two or more other reactions, DH for the overall process is the sum of the DH for the component reactions • Hess’s Law is true because H is a state function If we know the following H values: ? Target: 1. 2.

  29. Hess’s Law We can calculate the H for the reaction by properly adding (or subtracting) the H for reactions 1 and 2 • Notice that the target reaction has FeO and O2 as reactants and Fe2O3 as a product • Arrange reactions 1 and 2 so that they also have FeO and O2 as reactants and Fe2O3 as a product • Each reaction can be doubled, tripled, or multiplied by a half, etc. • Hvalues are then doubled, tripled, etc. • If a reaction is reversed the Hvalueis changed to the opposite sign 4 FeO 4 Fe 2 O2 1. +1088 kJ + 2. 1

  30. Hess’s Law Given the following equations and H values calculate H for the reaction below:

  31. Hess’s Law -The + sign of the H value tells us that the reaction is endothermic. -The reverse reaction is exothermic, i.e.

  32. Standard Enthalpy of Formation Thermochemical standard state conditions The thermochemical standard T = 298.15 K The thermochemical standard P = 1.0000 atm - Be careful not to confuse these values with STP Thermochemical standard states of matter For pure substances in their liquid or solid phase the standard state is the pure liquid or solid - For aqueous solutions the standard state is 1.00 M concentration For gases the standard state is the gas at 1.00 atm of pressure - For gaseous mixtures the partial pressure must be 1.00 atm

  33. Standard Molar Enthalpies of Formation, Hfo The enthalpy change for the formation of one mole of a substance formed directly from its constituent elements in their standard states • The symbol for standard molar enthalpy of formation is Hfo (KJ/mol) The standard molar enthalpy of formation for MgCl2 is:

  34. Standard Molar Enthalpies of Formation (Hfo) Standard molar enthalpies of formation have been determined for many substances and are tabulated in Appendix L Standard molar enthalpies of elements in their most stable forms at 298.15 K and 1.000 atm are zero. Example: The standard molar enthalpy of formation for phosphoric acid is -1279 kJ/mol. Write the equation for the reaction for which Hof = -1279 kJ Note: P in standard state is P4 (s) Phosphoric acid in standard state is H3PO4(s)

  35. Standard Molar Enthalpies of Formation Calculate the enthalpy change for the reaction of one mole of H2(g) with one mole of F2(g) to form two moles of HF(g) at 25oC and one atmosphere.

  36. Standard Molar Enthalpies of Formation Calculate the enthalpy change for the reaction in which 15.0 g of aluminum reacts with oxygen to form Al2O3 at 25oC and one atmosphere. You do it!

  37. Hess’s Law Version II For chemical reaction at standard conditions: • the standard enthalpy change is the sum of the standard molar enthalpies of formation of the products minus the sum for the reactants - each enthalpy of formation is multiplied by its coefficient in the balanced chemical equation Final - Initial

  38. Hess’s Law  C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l) Calculate the DH°f for the following reaction from the data in appendix L:

  39. Hess’s Law Given the following information, calculate Hfo for H2S(g) You do it!

  40. Calorimetry An experimental technique that measures the heat transfer during a chemical or physical process Constant pressure calorimetry: A styrofoam coffee-cup calorimeter is used to measure the amount of heat produced (or absorbed) in a reaction • This is one method to measure qP (called DH) for reactions in solution qreaction + qsolution = 0 Note: Assuming no heat transfer to the surroundings

  41. If an exothermic reaction is performed in a calorimeter, the heat evolved by the reaction is determined from the temperature rise of the solution This requires a two part calculation When we add 25.00 mL of 0.500 M NaOH at 23.000oC to 25.00 mL of 0.600 M CH3COOH already in the calorimeter at the same temperature, the resulting temperature is observed to be 25.947oC. Determine heat of reaction and then calculate the change in enthalpy (as KJ/mol) for the production of NaCH3COO. CH3COOH(aq) + NaOH(aq) NaCH3COO(aq) + H2O(l) Calorimetry

  42. Calorimetry Constant volume calorimetry: Or Bomb calorimetry measures measure the amount of heat produced (or absorbed) in a chemical reaction -this method is used for measuring qv (DE) qreaction + qbomb + qwater = 0