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Thermochemistry. Specific Heat Formula c p c p = Specific Heat Q Q = Energy (heat) lost or gained  T  T = Temperature change m m = Mass.

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Presentation on theme: "Thermochemistry. Specific Heat Formula c p c p = Specific Heat Q Q = Energy (heat) lost or gained  T  T = Temperature change m m = Mass."— Presentation transcript:

1 Thermochemistry

2 Specific Heat Formula c p c p = Specific Heat Q Q = Energy (heat) lost or gained  T  T = Temperature change m m = Mass

3 Two Types of Thermal Reactions Exothermic: Releases Thermal Energy (heat) sign is – Endothermic: Absorbs Thermal Energy (heat) sign is +

4 Exothermic Processes Reactants  Products + energy Processes in which energy is released as it proceeds, and surroundings become warmer

5 Endothermic Processes Reactants + energy  Products Processes in which energy is absorbed as it proceeds, and surroundings become colder

6 Enthalpy =  H In a chemical reaction, Enthalpy (  H) is equal to the energy that flows as heat. (at a constant pressure) Example: When 1 mol of methane gas is burned it releases 890 kJ of energy CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(g) + heat  H = -890kj = exothermic reaction

7 Calculate  H for a process in which 5.8 g of methane are burned.  H = -890kJ per mol CH 4 Rxn: CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(g) +heat Solution: Molar mass of CH 4 = g / mol 5.8 g CH 4 1 mol CH 4 = 0.36 mol CH g CH mol CH kJ = -320 kJ 1 mol CH 4

8 Calculate  H when 12.8 g of sulphur dioxide reacts with excess oxygen to form sulphur trioxide.  H = kJ per mol SO 2 Solution: Molar mass of SO 2 = g / mol 12.8 g SO 2 1 mol SO 2 = mol SO g SO mol SO kj = kJ 1 mol SO 2

9 Hydrogen peroxide decomposes according to the following thermochemical reaction: H 2 O 2 (l) → H 2 O(l) + 1/2 O 2 (g) ΔH = kJ Calculate the change in enthalpy (ΔH) when 4.00 grams of hydrogen peroxide decomposes. Calculate the change in enthalpy (ΔH) when grams of water are created…

10 The reaction that occurs in hand warmers is: 4Fe(s) + 3O 2 (g)  2Fe 2 O 3 (s) + heat  H of reaction = -1652kJ How much heat is released when 2.5 grams of Fe(s) is reacted with excess oxygen?

11 Miller + 2 Stitts  LadyGaga - $600 2 Winters + Miller  Justin Bieber + $300 Stitt  2 Elvis + Winter - $500 Lady Gaga  Justin Bieber + 4 Elvis $_____ ??? Law of Concert Tickets!

12 You can add KNOWN equations to solve UNKOWN equations Law of Concert Tickets = Hess’s Law FLIP (reverse) the equation – you must FLIP the sign (- +) MULTIPY / DIVIDE equation – must Multiply / Divide the ΔH! Two Rules:

13 2C(s) + H 2 (g) ---> C 2 H 2 (g)ΔH° = ??? kJ C 2 H 2 (g) + (5/2)O 2 (g) ---> 2CO 2 (g) + H 2 O(l)ΔH° = kJ C(s) + O 2 (g) ---> CO 2 (g)ΔH° = kJ H 2 (g) + (1/2)O 2 (g) ---> H 2 O(l)ΔH° = kJ Example #1: Calculate the enthalpy for this reaction: Given the following thermo chemical equations : a) first eq: flip it so as to put C 2 H 2 on the product side b) second eq: multiply it by two to get 2C c) third eq: do nothing kJ + (-787 kJ) + ( kJ) = kJ

14 C 2 H 2 (g) + (5/2)O 2 (g) ---> 2CO 2 (g) + H 2 O(l)ΔH° = kJ C(s) + O 2 (g) ---> CO 2 (g)ΔH° = kJ H 2 (g) + (1/2)O 2 (g) ---> H 2 O(l)ΔH° = kJ 2C(s) + H 2 (g) ---> C 2 H 2 (g)ΔH° = ??? kJ

15 SrO(s) + CO 2 (g) ---> SrCO 3 (s)ΔH = -234 kJ 2SrO(s) ---> 2Sr(s) + O 2 (g)ΔH = kJ 2SrCO 3 (s) ---> 2Sr(s) + 2C(s) + 3O 2 (g)ΔH = kJ Example #2: Given the following data: Find the ΔH of the following reaction: C(s) + O 2 (g) ---> CO 2 (g) a) first equation - flip it b) second equation - divide by two c) third equation - flip it, divide by two (+592) + (-1220) = -394

16 SrO(s) + CO 2 (g) ---> SrCO 3 (s)ΔH = -234 kJ 2SrO(s) ---> 2Sr(s) + O 2 (g)ΔH = kJ 2SrCO 3 (s) ---> 2Sr(s) + 2C(s) + 3O 2 (g)ΔH = kJ Example #2: Given the following data: Find the ΔH of the following reaction: C(s) + O 2 (g) ---> CO 2 (g) a) first equation - flip it b) second equation - divide by two c) third equation - flip it, divide by two (+592) + (-1220) = -394

17 Hess's Law Δ In a reaction, the change in enthalpy (ΔH) is the same - regardless if the reaction occurs in a single step or in several steps. Δ If a series of reactions are added together, the net change in ΔH is the sum of the enthalpy changes for each step.

18 Rules for using Hess's Law Δ If the reaction is multiplied (or divided) by some factor, Δ H must also be multiplied (or divided) by that same factor. Δ If the reaction is reversed (flipped), the sign of Δ H must also be reversed.

19 Water phase changes & Energy Temperature remains __________ during a phase change. constant

20 Phase Change Diagram Processes occur by addition of energy   Processes occur by removal of energy

21 Phase Diagram  Represents phases as a function of temperature and pressure.  Critical temperature: temperature above which the vapor can not be liquefied.  Critical pressure: pressure required to liquefy AT the critical temperature.  Critical point: critical temperature and pressure (for water, T c = 374°C and 218 atm).

22 Phase Changes

23 Effect of Pressure on Boiling Point

24 Phase Diagram  Represents phases as a function of temperature and pressure.  Critical temperature: temperature above which the vapor can not be liquefied.  Critical pressure: pressure required to liquefy AT the critical temperature.  Critical point: critical temperature and pressure (for water, T c = 374°C and 218 atm).

25 Phase changes by Name

26 Water

27 Carbon dioxide Phase Diagram for Carbon dioxide

28 Carbon

29 Phase Diagram for Sulfur

30 Reaction Pathway Shows the change in energy during a chemical reaction

31 Exothermic Reaction reaction that releases energy products have lower PE than reactants 2H 2 (l) + O 2 (l)  2H 2 O(g) + energy energy released

32 Endothermic Reaction reaction that absorbs energy reactants have lower PE than products 2Al 2 O 3 + energy  4Al + 3O 2 energy absorbed

33 Latent Heat of Phase Change Molar Heat of Fusion The energy that must be absorbed in order to convert one mole of solid to liquid at its melting point. The energy that must be removed in order to convert one mole of liquid to solid at its freezing point. Molar Heat of Solidification

34 Latent Heat of Phase Change #2 Molar Heat of Vaporization The energy that must be absorbed in order to convert one mole of liquid to gas at its boiling point. The energy that must be removed in order to convert one mole of gas to liquid at its condensation point. Molar Heat of Condensation

35 Latent Heat – Sample Problem Problem: The molar heat of fusion of water is kJ/mol. How much energy is needed to convert 60 grams of ice at 0  C to liquid water at 0  C? Mass of ice Molar Mass of water Heat of fusion

36 Heat of Solution Heat of Solution The Heat of Solution is the amount of heat energy absorbed (endothermic) or released (exothermic) when a specific amount of solute dissolves in a solvent.

37 Energy is the capacity to do work Thermal energy is the energy associated with the random motion of atoms and molecules Chemical energy is the energy stored within the bonds of chemical substances Nuclear energy is the energy stored within the collection of neutrons and protons in the atom Electrical energy is the energy associated with the flow of electrons Potential energy is the energy available by virtue of an object’s position 6.1

38 Thermochemistry is the study of heat change in chemical reactions. The system is the specific part of the universe that is of interest in the study. open mass & energyExchange: closed energy isolated nothing SYSTEM SURROUNDINGS 6.2

39 Exothermic process is any process that gives off heat – transfers thermal energy from the system to the surroundings. Endothermic process is any process in which heat has to be supplied to the system from the surroundings. 2H 2 (g) + O 2 (g) 2H 2 O (l) + energy H 2 O (g) H 2 O (l) + energy energy + 2HgO (s) 2Hg (l) + O 2 (g) 6.2 energy + H 2 O (s) H 2 O (l)

40 Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure.  H = H (products) – H (reactants)  H = heat given off or absorbed during a reaction at constant pressure H products < H reactants  H < 0 H products > H reactants  H > 0 6.3

41 Thermochemical Equations H 2 O (s) H 2 O (l)  H = 6.01 kJ Is  H negative or positive? System absorbs heat Endothermic  H > kJ are absorbed for every 1 mole of ice that melts at 0 0 C and 1 atm. 6.3

42 Thermochemical Equations CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (l)  H = kJ Is  H negative or positive? System gives off heat Exothermic  H < kJ are released for every 1 mole of methane that is combusted at 25 0 C and 1 atm. 6.3

43 H 2 O (s) H 2 O (l)  H = 6.01 kJ The stoichiometric coefficients always refer to the number of moles of a substance Thermochemical Equations If you reverse a reaction, the sign of  H changes H 2 O (l) H 2 O (s)  H = kJ If you multiply both sides of the equation by a factor n, then  H must change by the same factor n. 2H 2 O (s) 2H 2 O (l)  H = 2 x 6.01 = 12.0 kJ 6.3

44 H 2 O (s) H 2 O (l)  H = 6.01 kJ The physical states of all reactants and products must be specified in thermochemical equations. Thermochemical Equations 6.3 H 2 O (l) H 2 O (g)  H = 44.0 kJ How much heat is evolved when 266 g of white phosphorus (P 4 ) burn in air? P 4 (s) + 5O 2 (g) P 4 O 10 (s)  H = kJ 266 g P 4 1 mol P g P 4 x 3013 kJ 1 mol P 4 x = 6470 kJ

45 The specific heat (s) of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius. The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius. C = ms Heat (q) absorbed or released: q = ms  t q = C  t  t = t final - t initial 6.4

46 How much heat is given off when an 869 g iron bar cools from 94 0 C to 5 0 C? s of Fe = J/g 0 C  t = t final – t initial = 5 0 C – 94 0 C = C q = ms  t = 869 g x J/g 0 C x –89 0 C= -34,000 J 6.4

47 Constant-Volume Calorimetry No heat enters or leaves! q sys = q water + q bomb + q rxn q sys = 0 q rxn = - (q water + q bomb ) q water = ms  t q bomb = C bomb  t 6.4 Reaction at Constant V  H ~ q rxn  H = q rxn

48 Constant-Pressure Calorimetry No heat enters or leaves! q sys = q water + q cal + q rxn q sys = 0 q rxn = - (q water + q cal ) q water = ms  t q cal = C cal  t 6.4 Reaction at Constant P  H = q rxn

49 Prentice-Hall © 2002General Chemistry: Chapter 7Slide 49 of 50 Terminology Energy, U – The capacity to do work. Work – Force acting through a distance. Kinetic Energy – The energy of motion.

50 Prentice-Hall © 2002General Chemistry: Chapter 7Slide 50 of 50 Energy Kinetic Energy ek =ek = 1 2 mv 2 [e k ] = kg m 2 s2s2 = J w = Fd [w ] = kg m s2s2 = J m Work

51 Prentice-Hall © 2002General Chemistry: Chapter 7Slide 51 of 50 Energy Potential Energy – Energy due to condition, position, or composition. – Associated with forces of attraction or repulsion between objects. Energy can change from potential to kinetic.

52 Prentice-Hall © 2002General Chemistry: Chapter 7Slide 52 of 50 Energy and Temperature Thermal Energy – Kinetic energy associated with random molecular motion. – In general proportional to temperature. – An intensive property. Heat and Work – q and w. – Energy changes.

53 Prentice-Hall © 2002General Chemistry: Chapter 7Slide 53 of 50 Heat Energy transferred between a system and its surroundings as a result of a temperature difference. Heat flows from hotter to colder. – Temperature may change. – Phase may change (an isothermal process).

54 Prentice-Hall © 2002General Chemistry: Chapter 7Slide 54 of 50 Units of Heat Calorie (cal) – The quantity of heat required to change the temperature of one gram of water by one degree Celsius. Joule (J) – SI unit for heat 1 cal = J

55 Prentice-Hall © 2002General Chemistry: Chapter 7Slide 55 of 50 Heat Capacity The quantity of heat required to change the temperature of a system by one degree. – Molar heat capacity. System is one mole of substance. – Specific heat capacity, c. System is one gram of substance – Heat capacity Mass  specific heat. q =mc  T q =CTCT


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