1 First Law of hermodynamics 1 State Functions - A State Function is a thermodynamic quantity whose value deends only on the state at the moment, i. e., the temerature, ressure, volume, etc - he value of a state function is indeendent of the history of the system. - emerature is an examle of a state function. - he fact that temerature is a state function is extremely useful because it we can measure the temerature change in the system by knowing the initial temerature and the final temerature. - In other words, we don t need all of the nitty-gritty detail of a rocess to measure the change in the value of a state function. - In contrast, we do need all of the nitty-gritty details to measure the heat or the work of a system. Reversibility A reversible rocess is a rocess where the effects of following a thermodynamic ath can be undone be exactly reversing the ath. An easier definition is a rocess that is always at equilibrium even when undergoing a change. Phase changes and chemical equilibria are examles of reversible rocesses. Ideally the comosition throughout the system must be homogeneous. - his requirement imlies that the no gradients, currents or eddys can exist. - o eliminate all inhomogeneities, a reversible rocess must occur infinitely slow! - hus no truly reversible rocesses exist. However, many systems are aroximately reversible. And assuming reversible rocesses will greatly aid our calculations of various thermodynamic state functions. Reversibility during ressure changes ensures that = ex hat is, the ressure on the inside of the container is always equal to the ressure exerted on the outside of the container.
2 heorem of Maximum Work he maximum amount of work that can be extracted from an exansion rocess occurs under reversible conditions. hus the theorem imlies that during an irreversible exansion, some of the energy is lost as heat rather than work. he inhomogeneities (currents, gradients and eddys) in ressure that occur during an irreversible rocess are resonsible for the heat. Initial Definitions of Energy and Energy ransfer Internal Energy - Sum of the kinetic and otential energy in a samle of matter. Microscoic modes of Internal Energy (0 th century view) - degrees of freedom for energy storage translational rotational librational vibrations caused by intermolecular forces vibrational electronic nuclear, etc - more secifics in second semester - unnecessary for understanding of thermodynamics but sometimes helful. Macroscoic view of internal energy (19 th century view) - A reservoir of energy within the samle. - he details of the energy are unknown. Internal energy is a state function. Work Recall the definition of work from classical mechanics. w = F dl Microscoically, work involves the concerted motion of molecules (that is, molecules moving in one direction.) Work is a transfer of energy, not a quantity of energy. Work is not a state function. -calculation of work deends on thermodynamic ath (as seen from definition)
3 A more thermodynamically useful rearrangement (let F = A and d = dl A) is 3 w = d - the negative sign is a sign convention (discussed below) - ex is the external ressure (the ressure outside the container) - ex is used in definition because we really do not know what kind of ressures develo during a thermodynamic change. - i.e., ressure currents and eddies are likely for an arbitrary change. Other forms of work are ossible w = E dq Electrical work ( ) w = E d P Polarization work (e.g., iezoelectricity) w = σ da w = τ dθ Surface tension wisting work Sign conventions for work -w work done by system (exansion for work) +w work done on the system (comression for work) Heat - macroscoically heat is a thermal energy transfer - energy transfer usually characterized by temerature and the zeroth law of thermodynamics. - microscoically heat transfer comes from - inelastic collisions - energy transfer from one molecule to another in multile (every) direction Sign conventions for heat -q heat transferred away from body (heat lost) +q heat transferred into body (heat gained) ex
4 First Law of hermodynamics 4 he hysicists that studied energy changes recognized that the energy of an object could be changed via heat or work. James Joule demonstrated exerimentally the law of conservation of energy, that is, that energy is not gained or lost, but merely transferred from one object to another. he law of conservation of energy is also known as the first law of thermodynamics. Since energy changes can be exressed only as heat or work, the first law of thermodynamics has the mathematical exression U = q + w Subtle oints to make: 1) q and w are energy changes, writing q and w is imroer. - We can t refer to object having an amount of heat or work. he object has internal energy, enthaly, free energy, etc ) he form of the first law of thermodynamics deends on the sign convention chosen for heat and work. Older texts state that U = q w ; however, the sign convention for work is different, Differential Form of the First Law w = + d he differential form of the first law is written as ex du = ñq +ñw he differential forms of heat and work are written with the slash to emhasize that they are inexact differentials. Constant volume heat During a constant volume rocess, w = 0. herefore U = q v hat is, constant volume heat is equal to a change in the free energy.
5 Measureable thermodynamic quantities 5 hermal exansivity he thermal exansivity is the measure of how a material changes its volume as the temerature changes. 1 α = Isothermal comressibility 1 κ = he isothermal comressibility is the measure of how a material changes its volume as the ressure changes. Almost always, volume will decrease with increasing ressure, therefore, a negative sign is included in the definition to allow for the tabulation a ositive values. Both definitions modify an extensive change, or intensive change by dividing the change by the volume., into an Enthaly Background In order to fully exlain his ideas of free energy, Josiah Willard Gibbs needed to construct an energy state function that had the definition H = U + P Gibbs named the quantity the heat content because a change in the quantity corresonded to heat gained or lost by a system at constant ressure rovided no non work is being done, that is. H = q Kamerlingh Onnes eventually named the function, H the enthaly and the name stuck.
6 Heat caacity - Heat is roortional to mass. - Heat is roortional to temerature difference. - Proortionality constant for a secific substance is the heat caacity or secific heat. q = mc 6 Definition of Constant Pressure Heat Caacity C Since constant ressure heat is a change in enthaly, the constant ressure heat caacity can be rewritten as C H = his quantity is one of the easily measurable quantities that is very imortant in thermodynamics. Definition of Constant olume Heat Caacity C v Since constant volume heat is a change in internal energy, the constant volume heat caacity can be rewritten as C v U = v his quantity is another one of the easily measurable quantities that we will use a great deal in thermodynamics studies. Relationshi between C and C One way of examining the difference between internal energy and enthaly is by examining the difference between the constant ressure heat caacity and the constant volume heat caacity. C ( U P) ( P) H U + U U U C = = = + U U = + P
7 U At this oint, we want to find another exression for 7 since it is not an easily U measurable quantity. We can find an alternative exression for based the total differential of the internal energy. he internal energy is a function of temerature and volume, thus the total differential is U U du = d + d Dividing the total differential by d under constant ressure conditions yields U U U U U = + = + wo of the artial derivatives in this exression should be familiar: U = C and = α. Remember the thermal exansivity is 1 α = Let us substitute our result into the original equation for C C, U U U U U C C = + P = + + P U U U = + P = P P + = α + What is the relationshi between C C for an ideal gas? 1 1 nr 1 nr nr α = = = = U nr U nr U C C = α + = + = + nr U nr 1 U = + = nr + 1
8 U Remember that for an ideal gas = 0, therefore 1 U C C = nr + 1 = nr 8 he relationshi can be rewritten in terms of molar heat caacities. Cɶ Cɶ = R his relationshi is very imortant and worth memorizing. Why is C greater than C? - Heat caacity is substance s caacity to store energy. - A substance that can store energy via work has a greater heat caacity than a substance that has its volume ket constant. Heat Reexamined Isothermal Heat Isochoric Heat dq = Cd q = 0 dq C d q C d = = If C is indeendent of temerature, (fair assumtion for small temerature changes), then q = C d = C d = C = C Isobaric Heat ( ) 1 dq C d q C d = = Adiabatic Heat By definition, q = 0 All of the above results are free from assumtions (unless noted).
9 hermochemistry 9 Extent of Reaction o monitor the rogress, a variable known as the extant of reaction is defined as n ξ = n i i,0 ν i where n i is the number of moles of chemical secies i, n i,0 is the initial number of moles of chemical secies i and ν i is the stoichiometric coefficient of the chemical secies i for the secific reaction occurring. When the extent of reaction is defined this way, the value of the extant of reaction does not deend on the chemical secies used to examine the state. Examle: Consider the reaction P 4 (s) + 5 O (g) P O 5 (s). Suose we start with 3 moles of hoshorus and 15 moles of oxygen. a) What is the extant of reaction when 4 moles of P O 5 has been created? n n 4 mol 0 mol ν i i,0 ξ = = = i Note: How much P 4 has been consumed to create 4 moles of P O 5? A: moles. b) What is the extant of reaction when moles of hoshorus has reacted? n n 1mol 3mol ν 1 i i,0 ξ = = = i he definition includes the stoichiometric coefficient to ut the changes of the all the chemical secies on an equal footing. Soon we will be dealing reaction energies and other quantities secific to reactions. he reaction energies need to be intensive quantities and yet each chemical secies may have a different amount. hus reaction energies are defined with resect to the extent of reaction so that the secific chemical secies is not imortant, but the reaction as a whole is imortant.
10 Standard States 10 Standard Pressure - φ hermodynamic quantities are measured with resect to a ressure of 1 bar ( kpa) Standard emerature - φ hermodynamic quantities are measured with resect to a temerature of 5 C (98.15 K) Standard Concentration- c φ hermodynamic quantities are measured with resect to a concentration of 1m (that is 1 molal). - Molal is used as a concentration unit rather than molar because molal is indeendent of temerature. Biological Standard State he biological standard state is a ressure of 1 bar, a temerature of 37 C and H of 7, (that is [H + ] = ).
11 Hess Law 11 Since enthaly is a state function, we can choose more than one thermodynamic ath to calculate a state function. For chemical changes, Hess Law states that the enthaly of a reaction can be calculated from the enthalies of all of the chemical ste rocesses needed for the chemical reaction. In other words, if reactions can be added together to form a resultant reaction, then enthalies of the ste reactions can be added to find a resultant enthaly of reaction. Examle: Calculate the enthaly of reaction for the following reaction: Al(s) + 3 Cl (g) AlCl 3 (s), given the reactions below. Al(s) + 6 HCl(aq) AlCl 3 (aq) + 3 H (g) H = kj/mol HCl(g) HCl(aq) H = kj/mol H (g) + Cl (g) HCl(g) H = -185 kj/mol AlCl 3 (s) AlCl 3 (aq) H = -33 kj/mol Rearrange the equations such that their sum is the reaction of interest. Al(s) + 6 HCl(aq) AlCl 3 (aq) + 3 H (g) H = kj/mol 6[HCl(g) HCl(aq)] H = 6[-74.8 kj/mol] 3[H (g) + Cl (g) HCl(g)] H = 3[-185 kj/mol] [AlCl 3 (aq) AlCl 3 (s)] H = [+33 kj/mol] Al(s) + 6 HCl(aq) + 6 HCl(g) + 3 H (g) + 3 Cl (g) + AlCl 3 (aq) AlCl 3 (aq) + 3 H (g) + 6 HCl(aq) + 6 HCl(g) + AlCl 3 (s) H = kj/mol + 6[-74.8 kj/mol] + 3[-185 kj/mol] + [+33 kj/mol] = kj/mol Al(s) + 3 Cl (g) AlCl 3 (s) H = kj/mol
12 Enthaly of Formation 1 he enthaly of formation is the enthaly of a formation reaction for a articular substance. A formation reaction is that where a comound is formed from elements in the naturally occurring state. C(s) + 3 H (g) + ½ O (g) C H 5 OH (l) ½ P 4 (s) + 5/ O (g) P O 5 (s) 7 Fe(s) + 18 C(s) + 9 N (g) Fe 4 [Fe(CN) 6 ] 3 (s) Enthaly of formation of the elements By definition, the enthaly of formation of an element in its natural state is zero. f H (B(s)) = 0 kj/mol f H (C(grahite)) = 0 kj/mol f H (Br (l)) = 0 kj/mol f H (S 8 (s)) = 0 kj/mol f H (Ag(s)) = 0 kj/mol f H (Xe(g)) = 0 kj/mol Reaction enthalies can be calculated as a stoichiometric sum of enthalies of formation. his technique is an alication of Hess law. Examle: Calculate the enthaly of reaction for the following reaction: CH 3 COOH(l) + C 4 H 9 OH(l) C 4 H 9 OOCCH 3 (l) + H O(l), given the reactions below. C(s) + H (g) + O (g) CH 3 COOH(l) H = kj/mol 4 C(s) + 5 H (g) + ½ O (g) C 4 H 9 OH(l) H = - 38 kj/mol 6 C(s) + 6 H (g) + O (g) C 4 H 9 OOCCH 3 (l) H = kj/mol H (g) + ½ O (g) H O(l) H = kj/mol Note all of the above reactions are formation reactions. Rearrange the equations such that their sum is the reaction of interest. CH 3 COOH(l) C(s) + H (g) + O (g) H = kj/mol C 4 H 9 OH(l) 4 C(s) + 5 H (g) + ½ O (g) H = + 38 kj/mol 6 C(s) + 6 H (g) + O (g) C 4 H 9 OOCCH 3 (l) H = kj/mol H (g) + ½ O (g) H O(l) H = kj/mol CH 3 COOH(l) + C 4 H 9 OH(l) + 6 C(s) + 6 H (g) + O (g) + H (g) + ½ O (g) C(s) + H (g) + O (g) + 4 C(s) + 5 H (g) + ½ O (g) + C 4 H 9 OOCCH 3 (l) + H O(l) H = kj/mol + 38 kj/mol kj/mol kj/mol = - 84 kj/mol CH 3 COOH(l) + C 4 H 9 OH(l) C 4 H 9 OOCCH 3 (l) + H O(l) H = - 84 kj/mol
13 Calorimetry 13 Constant Pressure Calorimeter - coffee cu calorimeter - oen to atmoshere - aroriate for solution chemistry Constant olume Calorimeter - bomb calorimeter - sealed and isolated - aroriate for gas hase chemistry Adiabatic Calorimeter - heat measured by temerature needed to kee thermal energy constant. Process of measuring heat of combustion with bomb calorimeter 1. Mass wick used to start combustion.. Mass water used to absorb heat. 3. Mass standard (benzoic acid) to be combusted 4. Add water to combustion chamber to ensure that water from combustion will be in liquid hase. 5. Load standard and seal 6. Fill combustion chamber with oxygen ( 5 atm) 7. Begin combustion and record temerature change of water. 8. Calculate heat caacity of calorimeter from standard - need to account for heat of wick, and heat caacity of water. 9. Reeat rocess with samle. - with heat from wick, heat from the water and heat from the calorimeter, the heat of combustion of samle can be calculated. Hess law is often used to the enthaly of formation of a substance after its enthaly of combustion has been calculated. Examle: 0.53 g of the military exlosive, cyclotetramethylenetetranitramine (HMX), C 4 H 8 N 8 O 8 is combusted in a bomb (!) calorimeter and an internal energy change of 4.60 kj is measured. Calculate the enthaly of formation for HMX. First write a balanced equation for its comlete combustion. C 4 H 8 N 8 O 8 (s) + O (g) 4 CO (g) + 4 H O(l) + 4 N (g) Next calculate the molar internal energy of reaction from the exerimental data. 4.6 kj g = 616 kj mol 0.53g mol
14 o calculate the enthaly of the reaction, we need to return the definition of enthaly. H = U + ( P) = U + ( nr) = U + n ( R) Reexamining the balanced chemical equation, we see that 8 moles moles = 6 moles of gas has been roduced. C 4 H 8 N 8 O 8 (s) + O (g) 4 CO (g) + 4 H O(l) + 4 N (g) kJ H = U + n ( R) = 616 kj mol K mol K.478kJ = 616 kj mol + 6 = 616 kj mol kj mol = 601kJ mol mol 14 he molar enthaly of reaction can also be written in terms of the molar enthalies of formation. 4 f H (CO (g)) + 4 f H (H O(l)) - f H (C 4 H 8 N 8 O 8 (s)) = r H Rearranged, the equation becomes, f H (C 4 H 8 N 8 O 8 (s)) = 4 f H (CO (g)) + 4 f H (H O(l)) - r H From table, we find the enthalies of formation for CO (g) and H O(l) f H (C 4 H 8 N 8 O 8 (s)) = 4 ( kj/mol) + 4 (-85.8 kj/mol) - (-601 kj/mol) = -116 kj/mol emerature Deendence of Internal Energy and Enthaly Recall that C U = and C H = hese relationshis imly that we can find the internal energy or enthaly at a nonstandard temerature as long as we know the heat caacity. du = C d and dh = C d Integrating both sides of these equations yields U = C d and H = C d
15 For large temerature changes, we can t assume that the heat caacity is indeendent of temerature. hus to calculate the internal energy or enthaly at a nonzero temerature, we need the temerature deendence of the heat caacity. Examle: Calculate the change in enthaly of H (g) from 373 K to 1000 K, given that the constant ressure heat caacity has the form C = d + e + f where d = 7.8 J/K mol, e = J/K mol and f = J K/mol K 373 K ( ) H = H H = C d = d + e + f d = d d + e d + f d J K mol ( 7.8 J K mol)( 67 K) ( K ) e 1000 f 1000 e 1 1 = d = d ( ) + ( ) f = + 1 ( )( ) J K mol K = + + = = J mol 1403J mol J mol J mol 18.51kJ mol 15 Kirchoff s law 0 ( ) ( ) H = H 98 K + ν C d r r i,i 98 i hat is by taking a stoichiometric sum of the heat caacities and integrating over the temerature range, we can find the correction to the reaction enthaly at a non standard state temerature.
16 Examle: 471 kj/mol is the reaction enthaly under standard conditions for the following reaction: Fe O 3 (s) + 3 C(s) 4 Fe(s) + 3 CO (g). What is the reaction enthaly at 1000 K? According to the NIS Webbook Internet site, the constant ressure heat caacities of the chemical secies in the above smelting rocess can be fitted according to the 3 Shumate equation, C = A + B + C + D + E he values for A, B, C, D and E for each secies are given below. A B C D E Fe O C CO Fe Sum K 0 3 ( ) ( ) ( ) H 1000K = H 98K + ν A + B + C + D + E d r r i i i i i i 98K i B C D = H ( 98K) + ν A E i 1000 i i r i i i 98 i kJ ( ) ( ) ( )( ) ( r H 1000 K mol ) 5 8 ( ) ( ) ( ) ( ) = kJ 49.3kJ 1.6kJ 34.1kJ 11.kJ 6.5kJ 413.9kJ rh ( 1000 K) = + + = mol mol mol mol mol mol mol Kirchoff s Law can be stated with in a differential form as well. 16 U Include = C and H = C Bond Enthalies A bond enthaly is the energy needed to searate two atoms. abulated bond enthalies are average values calculated from the dissociation of many different comounds. hus tabulated bond enthalies are aroximate values. *However, they can be useful for aroximating enthalies of reaction because most of the chemical energy of a comound is held in its bonds.*
17 Work Reexamined 17 Isochoric Work work involves a volume change. w = d Since a constant volume rocess has no volume change, w = 0 if there is no other work (no non- work). No other assumtions have been made about the system. Isobaric Work Since the ressure is constant, it has no deendence on the volume. hus the ressure can ulled out of the integral. Subsequently, the integral of d is 1 = ex ( ) w = d = d = = ex ex ex 1 ex he only other assumtion made is that the system does not have any non- work. Isothermal Work In general we need to know the relationshi between ressure and volume to erform the calculation, i.e, we need an equation of state. Let us do the calculation for an ideal gas under reversible conditions. nr d w d d d nr nr ln = ex = = = = 1 Note restrictions on the alicability of the calculation. 1. no other non- work. ideal gas 3. reversible conditions Adiabatic Work No assumtions made! du = ñq + ñw du = ñw w = U
18 Refrigeration and the Joule-homson coefficient 18 Joule coefficient - - internal ressure µ J = U q = 0, w = 0 U = 0 isoenergetic rocess Joule-homson coefficient - µ J = - isoenthalic, adiabatic - µ = 0 for an ideal gas Classic Exerimental Aaratus - measure µ J directly. H 0 U = U U = d d = f f i i f i i f f i 0 U = U H = H f i i i f f i
19 Modern Exerimental Aaratus 19 H - measure µ J via - use isothermal conditions rather adiabatic conditions. isothermal conditions: 1 = H H H 1 1 = = = = H H H H C H H H µ J = C
20 Isoenthals on - lot. - inversion temerature - µ > 0 for substance to be used as a refrigerant - temerature must decrease as ressure decreases. 0
21 Real Gases Reexamined 1 Calculations on van der Waals gas - work w = d ex nr an = b nr an nr an w = ex d = d = d d d = + b b b 1 1 nr ln an = 1 b 1 - Difference in heat caacities C H U Cv = v U U 1 U U du = d + d du = d + d U U U U U = + = + U U U = ( ) H U U H = U + = + = + U U U C Cv = + U U = + = + U U = + = α + = α + π [ ]
22 Internal Energy Reexamined Why is U = q v under isochoric conditions? du = ñq + ñw du = ñq - d du = ñq U = q ex Assuming no non- work. Isobaric Internal Energy du = ñq + ñw du= C d d U = C d d = C d Note: no assumtions about the equation of state. v ex v ex v Isochoric Internal Energy du = ñq + ñw du = ñq d du = ñq U = C d = q Note: assumtion of no non- work ex v v Isothermal Internal Energy du = ñq +ñw = C d d = 0 du = 0 v
23 Enthaly Reexamined 3 Why is H = q under isobaric conditions? ( ) H = U + dh = du + d =ñq +ñw + d + v d dh =ñq + -d + d + v d dh = ñq H = q Assuming no non- work and reversible conditions Isobaric Enthaly ( ) H = U + dh = du + d =ñq +ñw + d + v d dh =ñq d + d dh = ñq H = C d = q Assuming no non- work and reversible conditions Isochoric Enthaly ( ) H = U + dh = du + d =ñq +ñw + d + v d dh =ñq d + d + v d dh = ñq + v d H = C d v d For ideal gas under reversible conditions nr H = Cd vd = Cd d Why not H = Cd nr ln? 1 - temerature has a deendence on the ressure. Isothermal Enthaly H = U + dh = du + d ( ) = Cd H = 0 Imortant: All of the above enthalies assume that the system has no non- work and reversible conditions (so that P = P ex )
24 Summary 4 Isothermal = 0 Isobaric = 0 Isochoric = 0 Adiabatic q = 0 Definition of work: w = d ex Definitions for heat: q = H q = U Sign conventions for heat and work. -w work done by system (exansion for work) +w work done on the system (comression for work) -q heat transferred away from body (heat lost) +q heat transferred into body (heat gained) Reversibility = ex Extent of Reaction n ξ = n i i,0 ν i Definition of Heat Caacities C q H = = C v q U = = v v First Law: du = ñq + ñw U = q + w U = C d ex d Partial Derivatives to be acquainted with: α, κ, C, C v, π, µ J, µ J 1 1 α = κ = C U and internal ressure π = H = C v U = (equals zero for ideal gas) v µ J = U µ J = H
25 Natural variables of U and H and total differentials 5 Internal energy can be exressed as a function of and U(,) Enthaly can be exressed as a function of and H(,) - both of these functional deendences are not unique - will consider other natural variables in the next chater. otal differential can be exressed as U U H H du = d + d dh = d + d Hess law emerature Deendence of Enthaly and Internal Energy U = C d and H = C d Kirchoff s law 0 ( ) ( ) H = H 98 K + ν C d r r i,i 98 i U = C and H = C Miscellaneous Alications of Heat Caacities Cɶ Cɶ = R C γ = C v
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AP Chem Lab 2 Quiz #1 Calorimetry Name Conceptual Understanding. Write complete sentences to show your understanding. Differentiate between kinetic energy and potential energy. Energy may be transferred
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