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Thermochemistry1 The energy aspect of chemistry. Thermochemistry2 Thermochemical Data and Services Energy is a quantity that can be stored in a chemical.

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Presentation on theme: "Thermochemistry1 The energy aspect of chemistry. Thermochemistry2 Thermochemical Data and Services Energy is a quantity that can be stored in a chemical."— Presentation transcript:

1 Thermochemistry1 The energy aspect of chemistry

2 Thermochemistry2 Thermochemical Data and Services Energy is a quantity that can be stored in a chemical form, and chemical reaction involve energy. There is an extensive thermochemical data for various scientific and engineering applications. NIST site has thermochemical data for over 6000 organic and small inorganic compounds. It also has reaction thermochemistry data for over 9000 reactions and ion energetics data for over 16,000 compounds. Thermochemical services you may need or provide Calculation services Consulting Experimental services Technical literature search service

3 Thermochemistry3 Temperature Energy & Heat Temperature is a measure of hot and cold – an intensive property. Thermometers in various scales are used to compare temperature. Recall: u rms = ( 3 k T / m ) 1/2 = (3 RT / M ) 1/2 Energy is the driving force for changes. A change often is associated with a certain amount of energy and amounts of energy can be measured according to the quantities changed. Like other quantities, energy is an extensive property, unlike temperature. Heat is energy in transfer or energy transferred and heat is often used as another term for energy. In this aspect, it is the quantity of energy transferred or involved. Describe hotness and heat

4 Thermochemistry4 A system for Thermochemistry For the study of energy, there must be a reference for its amounts transferred into or out. Such a reference is called a system. The boundary isolates the system from its surrounding, so that the amount of energy or heat transferred into or out can be identified. Picture a system with boundary, surrounding, and heat transfer in your mind. Surrounding System Environment

5 Thermochemistry5 Measuring Heat (Energy) Heat = Energy = Work Heat is energy – the ability to do work. Units for work, heat or energy is J 1 J = 1 N m ( in honor of James Joule 1818 – 1889 ) 1 cal = J (defined. The amount of heat to raise 1 g water for 1 K ) Gaining heat increase T, as the kinetic energy of molecules increase. Remember these from discussion of gases? Kinetic energy = ½ m u 2 3 R T = 1 / 3 N A m u 2 Heat capacity: the amount of heat required to raise T by 1 K Specific heat: heat capacity (no unit) compare to water (1.000 cal K -1 g -1 ) Meaning of T ?

6 Thermochemistry6 Conservation of Energy Energy can be converted from one form to another at a fixed rate, and it cannot be destroyed or created. In other words, energy is conserved. Quantity of heat q follows q system + q surrounding = 0 What can be created and destroyed? 1 cal = J (heat-work conversion) 1 J = 1 kPa L Has this principle been proved? What is a perpetual motion? How can this principle be disproved? What’s the principle of conservation of matter? no creation & no destruction gain of q is +ve loss of q is -ve

7 Thermochemistry7 Calorimeter Experiment Heat gained Heat capacity Temperature difference Heat capacity of water = 1 cal deg -1 g -1 molar heat capacity of water = 18 * J deg -1 mol -1 = 75.3 J deg -1 mol -1 Specific Heat (no unit) water 1 copper aluminum What to use for a miracle thaw? q = C *  T

8 Thermochemistry8 Bomb and Coffee-cup Calorimeter q = C *  T Use a known amount of q and measure  T to determine C. By measuring  T, calculate C = q /  T (corection) Once C is known, and a known amount of reactants, heat of reaction,  H reaction, can be evaluated

9 Thermochemistry9 How do you solve this? Two mL solutions of NaOH and HCl are both 1.00 mol L -1. When they are mixed in a coffee cup, the temperature raised from 20.0 to 26.7 o C. What is the heat released. Solution : What’s the reaction? What assumptions to make? Steps?

10 Thermochemistry10 Determine Heat of Reaction Two mL solutions of NaOH and HCl are both 1.00 mol L -1. When they are mixed in a coffee cup, the temperature raised from 20.0 to 26.7 o C. What is the heat released. Solution : Assumptions: no heat loss, consider solution as water, C = g * 4.18 J g –1 K –1 = 836 J K –1. Heat of reaction = 836 J K –1.* (26.7 – 20.0) K = 5601 J = 5.6 kJ. Mole of H + and OH –1 reacted = 1.0 mol L –1 * L = 0.10 mol Molar heat of neutralization,  H reaction = 5.6 kJ / 0.1 mol = 65 kJ mol –1 Reaction: H + + OH –1 = H 2 O  H reaction = 65 kJ mol –1

11 Thermochemistry11 Pressure-Volume Work Pressure = P (N m -2 ) V = A Area * h distance (m 3 = 1000 L) F force = P * A W Work = F * h distance = P * A * h distance = P V 1 J= 1 N m -2 m 3 () = Pa L = 1 kPa L P AhAh 1000 L 1 m 3 Show P V is work and 1 J = 1 kPa L

12 Thermochemistry12 The Gas Constant R The ABC laws of gases can be combined into one and the result is an ideal gas equation. P V = n R T 1 atm * L R = = L atm mol –1 K –1 1 mol K kPa * L R = = L kPa mol –1 K –1 = J mol –1 K –1 1 mol K 1 kPa L = __?__ J1 L = __?__ m 3 1 L atm = __?__ J1 m 3 = __?__ L

13 Thermochemistry13 The Gas Constant R You blow a balloon for a volume of 1.00 L per breath. If the pressure you are blowing against is 1.10 atm, how much work, in J, have you done? Solution : P * V = 1.10 atm * 1.00 L = 1.10 L atm 1.10 L atm = J J mol –1 K – L atm mol –1 K –1 = R is used for unit conversion

14 Thermochemistry14 First Law of Thermodynamics To accommodate the principle of conservation of energy, we define an Internal Energy U, as the total kinetic and potential energy. The U cannot be measured or determined, thus we usually are only interested in  U, the difference in internal energy before and after a change.  U = q + w The w : work done onto the system The q : heat absorbed by the system The heat absorbed by the system and the work done onto the system are all used to increase the internal energy of the system.

15 Thermochemistry15 Evaluating Internal Energy A balloon expanded against the atmosphere pressure performed 25 J of work, and absorbed 30 J of heat, what is the change in internal energy? Solution :  U = q + w = (30 – 25) J = 5 J Discussion : A change in internal energy results in change in temperature T. How does the average kinetic energy of the molecule change?

16 Thermochemistry16 State and State Function The state of a system is the description of T, P, amounts and kinds of substances in the system, etc. The state defines all these quantities, and these quantities are specific for the state. A property that has a specific value for a state is called a state function (or function of state). The internal energy, U, is a state function, and the difference  U is evaluated from the final and initial state. It is independent of the path.  U = U f – U i I will further explain the meaning of path independence using examples. Heat q and work w depend on path, for example. Explain state functions!

17 Thermochemistry17 Heats of Reaction,  U &  H For a reactionReactants  Products The change in internal energy is  U = U Reactants – U products = q rxn + w The change in internal energy  U is the heat of reaction under constant volume, q v,  U = q v, because w = 0, no P-V work. Enthalpy  H is the heat of reaction under constant pressure q p,  H = q p Under this condition, work done on or by the system, is w = – P  V. Thus,  U = q p - P  V =  H – P  V or  H =  U + P  V Explain  U and  H.  H  U + P  V reactants products

18 Thermochemistry18 An  U,  H Example For the reaction, 2CO (g) + O 2 (g)  2 CO 2 (g), at 298 K q p = kJ, calculate q v and w. Solution : The work is due to change of volume, w = P  V = ( n f – n i ) R T = ( 2 – 3) * J mol –1 K –1 * 298 K = – 2500 J = – 2.5 kJ mol –1  H = – kJ mol –1 (of equation)  U = – – (– 2.5) kJ = – kJ Note: 3 moles of gas is reduced to 2 moles, work must be done to the system to compress it back to the same P, w is positive. HH UU w 2CO (g) + O 2 (g) 2CO 2 (g)

19 Thermochemistry19 Application of Enthalpy Given, C 12 H 22 O 11 (s) + 12 O 2 (g)  12 CO 2 (g) + 11 H 2 O (l),  H = –5650 kJ How much heat is released when 10.0 g of sucrose, (molar mass 342.3), after you have ingested, digested, and completely oxidized.? Solution : The condition given here applies to the human utilization of sucrose (S), 10.0 g S= – kJ 1 mol S g S – 5650 kJ 1 mol S – ve, energy released Evaluate  U and w Given per rxn eqn

20 Thermochemistry20 Enthalpy Diagram for States Explain these changes of state: H 2 O (s)  H 2 O (l)  H = 6.01 kJ at 273 K H 2 O (l)  H 2 O (g)  H = 44.0 kJ at 298 K How much energy is required to convert 72 kg of ice at 273 K to vapor at 298 K? Solution : 72e3 g ice = __x_ ? ___ 72e3 g water * J (298 – 273) K = __y_ ? ___ 4e3 mol * 44.0e3 J / mol = __z__ ? ___ Total = _x_ + _y_ + _z_ = ________________ 18 g Ice at 273 K Water at 298 K vapor 44.0 kJ 6.01 kJ Water at 273 K 1 mol 18 g 6.01 kJ 1 mol

21 Thermochemistry21 Endothermic and Exothermic Reactions Use an energy level diagram to help visualize endothermic and exothermic reactions. Explain the sign changes in the following: H 2 O (s)  H 2 O (l)  H = 6.01 kJ at 298 K H 2 O (l)  H 2 O (s)  H = – 6.01 kJ at 298 K H 2 O (l)  H 2 O (g)  H = 44.0 kJ at 273 K H 2 O (g)  H 2 O (l)  H = – 44.0 kJ at 273 K 2 NO (g)  N 2 (g) + O 2 (g)  H = 181 kJ N 2 (g) + O 2 (g)  2 NO (g)  H = – 181 kJ Explain why the sign changes when the process is reversed

22 Thermochemistry22 Hess’s Law Like internal energy, enthalpy is also a state function. Hess’s Law is based on the principle of conservation of energy. It implies that if a process occurs in stages (even if only hypothetical), the enthalpy change for the overall process is the sum of the enthalpy changes for the individual steps. Another statement: the heat evolved or absorbed in a chemical process is the same whether the process takes place in one or in several steps. Germain Hess ( ) Explain Hess’s law

23 Thermochemistry23 Hess’s Law Illustrated A + B  AB  H 1 AB + B = AB 2  H 2 __add___________________ A + 2 B = AB 2  H 1 +  H 2 =  H 1 2  H 1 2  H 2 H1H1 A + 2 B AB 2 AB + B

24 Thermochemistry24 Chemical Energy and Hess’s Law C(graphite) O 2  CO  H ° = – 110 kJ/mol. 2 C(graphite) + O 2  2 CO  H ° = – 220 kJ/mol ( multiplied by 2 ) 6 C(graphite) + 3 O 2  6 CO  H ° = – 660 kJ/mol ( multiplied by 6 ) 2 CO  C(graphite) + O 2  H ° = 220 kJ/mol ( +ve ) CO (g) O 2 (g)  CO 2 (g)  H ° = – 283 kJ/mol. 1 mol C and O 2 1 mol CO 2 1 mol CO & 0.5 mol O 2 – 393 kJ – 283 kJ A quantity difficult to measure: – 110 kJ

25 Thermochemistry25 Standard State Standard temperature and pressure (STP) for gases are K and 1 atm. Standard state refers to states at pressure of 1.0 bar, and temperature must be specified for data at standard state. Standard enthalpy of reaction, standard enthalpy of formation, standard enthalpy of combustion, and in general standard energy of changes are referring to standard states. Reactants at standard states Reaction at other states Products at standard states

26 Thermochemistry26 Standard Enthalpy of Formation The standard enthalpy of formation,  H f °, is the energy of reaction for the formation of the substance at its standard state from its elements at their standard states. Enthalpies of elements are 0 at their standard state, these are reference points.  H f ° kJ mol –1 of compounds H 2 O (l)– H 2 O (g) – CO (g) – CO 2 (g) – CH 4 (g) – 74.8 C 2 H 2 (g) C 2 H 4 (g) C 2 H 6 (g) – 84.7 CH 3 OH (l) – C 2 H 5 OH (l) – Discuss the significance and value in everyday life of the data in the Table on the right!

27 Thermochemistry27 Meaning of Some Special Enthalpies of Formation Cpd  H f. kJReaction (Explain) C diamond 1.9 C (graphite)  C (diamond) (phase transition) Br 2 (g)30.9Br 2 (l)  Br 2 (g) (evaporation at 298 K) P 4 (red) -17.6P 4 (s, white)  P 4 (s, red) (White phosphorus is considered the standard state) H (g)217.8½ H 2 (g)  H (g) (½ H–H bond energy)

28 Thermochemistry28 Find Standard Enthalpy of Formation The enthalpy of combustion for H 2, C(graphite) and CH 4 are , , and kJ/mol respectively. Calculate the standard enthalpy of formation dH f for CH 4. (correction made) Solution :C (graphite) + 2 H 2  CH 4  H f ° = _____ Explain these from the data 2 *(-285.8) 2 H 2 (g) + O 2 (g)  2 H 2 O(l)  H = kJ C(graphite) + O 2 (g)  CO 2 (g)  H = kJ CO 2 (g) + 2H 2 O(l)  CH 4 (g) + 2O 2 (g)  H= kJ (+ve) (1) + (2) + (3) C (graphite) + 2 H 2  CH 4  H f ° = kJ Make an energy level diagram from these to see the relationship.

29 Thermochemistry29 Energy Level Diagram C(graphite) + 2 H 2 (g) + 2 O 2 (g) CO H 2 O CH 4 (g) + 2 O 2 (g)  H = kJ  H f ° = kJ

30 Thermochemistry30 Application of Energy Level Diagram and Enthalpies of Formation C(graphite) + 2 H 2 (g) + 2 O 2 (g) CO H 2 O CH 4 (g) + 2 O 2 (g)  H = kJ  H f ° = kJ If you know any two of the three thermodynamic data, you can evaluate the third.

31 Thermochemistry31 Applications of  H f °  H f ° can be very useful for the calculation of enthalpy of reaction, by applying Hess’s Law. Calculate the  H for the reaction: C 2 H 6 (g) + 7 / 2 O 2 (g)  2 CO 2 (g) + 3 H 2 O (g)  H = ____? Solution : look up H f °s for C 2 H 6 etc  H rxn =  products  H f ° –  reacts  H f °  H = 2*(– 393.5) + 3*( – 285.8) – {1 * (– 84.7) + ( 7 / 2 )*(0)} = _______ kJ 2 C (s) + 3 H 2 (g) + 7 / 2 O 2 (g) C 2 H / 2 O 2 2 CO H / 2 O 2  H f C 2 H 6 = *  H f CO 2 2*(-393.5) = *  H f H 2 O 3*(-285.8) = Explain the generalized formula 2 CO 2 (g) + 3 H 2 O (g)  H = ______

32 Thermochemistry32 Enthalpy of Reaction and Fuel 6 CO H 2 O Chlorophyll Sun light h v C 6 H 12 O O 2,  H = 2.8e3 kJ Animal digestion Oxidation Fuels:coal, petroleum, natural gas, methanol, ethanol, hydrogen, gasified coal, waste, organic matter, etc. Please read about fuel and environmental impacts.

33 Thermochemistry33 Chemistry: relationship of material and energy Chemistry is the study of behavior of materials under the influence of energy, which drives all changes. Without energy, there is no behavior to study, and energy not only causes all changes, it is also the source of life. Thus, life must constantly utilizes energy, almost exclusively chemical energy. Furthermore, energy is also required to keep and maintain a comfortable and pleasant living. Chemical energy, usually known as food and fuel, is one of the most important energy sources. Thus, the skill to use thermodynamic data is important.

34 Thermochemistry34 Review questions – 0 For an endothermic reaction, how does the internal energy,  U, change, positive, zero, or negative? What is the units for heat capacity? Look for the data and name  H for the processes H 2 O (l)  H 2 O(g), H 2 O (g)  H 2 O(l), H 2 O (s)  H 2 O(g), H 2 O (s)  H 2 O(l) at STP? For which of the above has the smallest difference of  U -  H ? The reaction in a bomb calorimeter is carried out at what condition? constant P, constant V, constant T, at STP or at 1 atm and 25 K When 2.00 g of CH 4 are burned in a bomb calorimeter (Cv = 2.677e3 J / K), the temperature rises from to o C. What is the q absorbed by the calorimeter? What is the molar heat of combustion of CH 4 under this condition?

35 Thermochemistry35 Review questions - 1 When mol HCl reacts with mol NaOH contained in 25.0 mL of solution, the temperature of the water increases by 13.7 o C. What is the heat of the reaction? (limiting reagent, heat capacity) What is the molar heat of neutralization reaction? (stoichiometry and thermal energy) Which of the following are state functions? U, H, q v, P, T, w (work), V The standard enthalpy of formation for C and CO are –393 and 287 kJ respectively. What is the standard enthalpy of formation for CO? (Hess’s law, standard enthalpy of formation, energy level diagram slide 24, given any two of these three, you should be able to find the third one.) How much energy is released when g CO is formed from reaction of carbon and oxygen? (stoichiometry)

36 Thermochemistry36 Review questions - 2 The standard enthalpy of formation of water (liquid H 2 O) is –286 kJ mol –1. What is standard enthalpy of combustion (or oxidation) of H 2 ? What is the energy for the reaction, H 2 O (l)  H 2 (g) + ½O 2 (g)? How much energy is release when 0.15 g H 2 at 273K and 1 bar is ignited in excess O 2 in a bomb calorimeter in which the water vapor so formed is condensed into liquid at at the same temperature? (calculation gives –21.5 kJ, and the negative sign indicated energy released) What are  U,  W and  H for the reaction 0.1 H 2 O (l) = 0.1 H 2 O (g, 300 K and 1 bar) ? (Note, the internal energy increases,  W = P  V =  n R T;  n = 0.1 mol; if hydrogen is burned in air, this amount of energy and this energy is not released immediately from the burning of H 2 gas.)

37 Thermochemistry37 Review questions - 3 When g of MgCl 2 is formed by burning Mg(s) in Cl 2 (g), 32.1 kJ of energy is released. What is the enthalpy of formation of MgCl 2 ? The standard enthalpy of formation is zero for which ones of the following, O 2 (g), O 3 (g), O(g), C(graphite), C(diamond), C(g), P 4 (white), P 4 (red), Br 2 (s), Br 2 (l), Br 2 (g), I 2 (s), I 2 (l), I 2 (g), Hg(s), Hg(l), S(s), S(g)? (standard states, which ones have positive enthalpies of formation?) Given these data at STP, S(s) + O 2 (g)  SO 2 (g)  H = –3 95 kJ S(g) + O 2 (g)  SO 2 (g)  H = –618 kJ What is the standard enthalpy of formation of S(g)? (+223 kJ mol -1 )

38 Thermochemistry38 Review questions - 4 For applications of standard enthalpy of formation and enthalpy of reaction, try these: The standard enthalpy of formation for CH 3 CHO(g), H 2 O(l) and CO 2 (g) are –166, –286, and –394 kJ mol -1 respectively. Calculate the heat of combustion of CH 3 CHO(g). (-1194 kJ mol -1, Hess’s law) The standard enthalpy of formation of C 2 H 2 (g) is 227 kJ mol -1. What is the enthalpy of combustion of C 2 H 2 (g)? (use data from above, Hess’s law) Calculate the energy evolved in the complete oxidation of 5.55 g of Al at 298 K and 1 atm pressure. (  Ho f of Al 2 O 3 is –1767 kJ mol -1 ) What is the energy required to reduce 5.55 g of Al 2 O 3 ? (use data from the above question)

39 Thermochemistry39 Review questions - 5 When concentrated sulfuric acid is diluted with water, the solution becomes hot. Is the reaction of H 2 SO 4 with H 2 O exothermic or endothermic? A synthetic gas contains 55.0% CO, 33.0% H 2, and 12% CO 2. If all the energy from the combustion of 1.0 L of this gas is absorbed by 1.0 L of H 2 O initially at 300 K, what is the final temperature? (find the required data and carried out the calculation) A and B are two different compounds with the same molecular formula, C5H10. Both compounds burn in oxygen to give CO 2 (g) and H 2 O(g). The enthalpies of combustion for A and B are –3050 and 3025 kJ mol- 1, respectively, at 298 K. What is the  Ho for the reaction A  B?

40 Thermochemistry40 Review – 6 (see slide 27 and 28) (see slide 7) (review  Ho and  uo difference, see slides ) (see slide 25-29) The standard enthalpy of formation of diamond is 1.9 kJ mol -1, and the standard enthalpy of combustion at 298 K is kJ mol -1 for graphite. What is the theoretical enthalpy of combustion of diamond at 298 K? The heat capacity of platinum and liquid water are J g-1, K-1 and respectively. What mass of liquid water at 283 K, is needed to cool 50.0 g of platinum from 323 K to 300 K when the platinum is placed in the water? The thermochemical equation of the reaction at 298 K is, Al(s) + 3 / 2 O 2 (g)  Al 2 O 3 (s)  H o = –1767 kJ What is the standard internal energy change,  U o, at 298 K for this reaction? With the following information, what is the standard enthalpy of ormation of C 2 H 4 (g) C 2 H 4 (g) + 3 O 2  2CO H 2 O(l),  H o = – x kJ Data:  H o f = – y kJ mol for CO 2  H o f = – z kJ mol for H 2 O Solution:  H o f = _______ kJ mol for C 2 H 4

41 Thermochemistry41 Weakness shown in Test 2 problems 3. A sample of gas absorbs exactly 200 J from its surroundings. The gas expands from 2.75 L to 13.5 L against a constant atm pressure. What is the internal energy change for the gas? 5. Radiation with a wavelength of 217 nm causes electrons to be ejected from the surface of lithium metal. If the maximum kinetic energy of ejected electrons is 4.420e-19 J, what is the longest wavelength radiation that can be used to disloge electrons from the surface of lithium? 8. Number of humps p, d orbitals in 4  r2  2 versus r2 plot. (4  r2 R 2 vs r2 ) 11. What is the electronic configuration of the transition ions, Fe 3+ ?

42 Thermochemistry42 Final Exam info Chem120 Dec. 11, 14 – 17 pm PAC all areas and MC4020, 4021, 4045, 4058, 4059, 4060, 4061


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