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Chapter 5 Principles of Chemical Reactivity. Basic Principles Thermodynamics: The science of heat and work Energy: the capacity to do work -chemical,

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Presentation on theme: "Chapter 5 Principles of Chemical Reactivity. Basic Principles Thermodynamics: The science of heat and work Energy: the capacity to do work -chemical,"— Presentation transcript:

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 objects 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 objects 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 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 Basic Principles

7 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 Jg.KJg.K

8 Heat Transfer Heat transfer equation used to calculate amounts of heat (q) in a substance J g Jg.KJg.K K q 1 + q 2 + q 3 … = 0 or q system + q surroundings = 0

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

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 Specific heat of gold is 0.13 Therefore the metal cannot be pure gold. 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? Jg.KJg.K

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: 1.Chemical systems tend toward a state of minimum potential energy Some examples of this include: -H 2 O flows downhill -Objects fall when dropped E potential = 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: E 1 = energy at the bottom of the mountain E 1 = mgh 1 E 2 = energy at the top of the mountain E 2 = mgh 2 E = E 2 -E 1 = mgh 2 – mgh 1 = 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 2.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: 1.Will these substances react when they are mixed under specified conditions? 2.If they do react, what energy changes and transfers are associated with their reaction? 3.If a reaction occurs, to what extent does it occur?

19 Changes in Internal Energy ( E or U) 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: U = q + w

20 Changes in Internal Energy ( E) E = E products – E reactants E = q + w at constant pressure: w = -P x V 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: 1.energy change for the system, E sys ? 2.energy change of the surroundings, E surr ?

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 positive when energy is absorbed by a system undergoing a chemical or physical change –Energy can be written as a reactant of the process

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

25 Enthalpy Change ( H) 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 = q p at constant pressure E = q v at constant volume If E and H < 0: energy is transferred to the surroundings If E and H > 0: energy is transferred to the system - Enthalpy and energy differ by the amount of work E = H + w and w = -P V

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 H for the reverse reaction is equal, but has the opposite sign to the forward reaction

27 Thermochemical Equations 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

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

29 Hesss 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 O 2 as reactants and Fe 2 O 3 as a product –Arrange reactions 1 and 2 so that they also have FeO and O 2 as reactants and Fe 2 O 3 as a product Each reaction can be doubled, tripled, or multiplied by a half, etc. H values are then doubled, tripled, etc. If a reaction is reversed the H value is changed to the opposite sign 1. 2. + 4 FeO 4 Fe 2 O 2 +1088 kJ 1

30 Hesss Law Given the following equations and H values calculate H for the reaction below:

31 Hesss 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, H f o 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 H f o (KJ/mol) The standard molar enthalpy of formation for MgCl 2 is:

34 Standard Molar Enthalpies of Formation ( H f o ) 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 H o f = -1279 kJ Note: P in standard state is P 4 (s) Phosphoric acid in standard state is H 3 PO 4 (s)

35 Standard Molar Enthalpies of Formation Calculate the enthalpy change for the reaction of one mole of H 2(g) with one mole of F 2(g) to form two moles of HF (g) at 25 o C 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 Al 2 O 3 at 25 o C and one atmosphere. You do it!

37 Hesss 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 Hesss Law C 3 H 8 (g) + 5O 2 (g) 3CO 2 (g) + 4H 2 O(l) Calculate the H° f for the following reaction from the data in appendix L:

39 Hesss Law Given the following information, calculate H f o for H 2 S (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 q P (called H) for reactions in solution q reaction + q solution = 0 Note: Assuming no heat transfer to the surroundings

41 Calorimetry 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 23.000 o C to 25.00 mL of 0.600 M CH 3 COOH already in the calorimeter at the same temperature, the resulting temperature is observed to be 25.947 o C. Determine heat of reaction and then calculate the change in enthalpy (as KJ/mol) for the production of NaCH 3 COO. CH 3 COOH (aq) + NaOH (aq) NaCH 3 COO (aq) + H 2 O (l)

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 q v ( E) q reaction + q bomb + q water = 0


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