CHEM 101 Thermchemistry Lect-07 Heat (q) is a frm f energy (E), and is expressed in Jule r kil Jule units. Temperature is a measure f the directin f heat flw; heat flws spntaneusly frm an bject having higher temperature t anther f lwer temperature. Heat effects are quantitative, and in that respect they can be used in stichimetric calculatins. 1. Heat Capacity The amunt f heat [ q ] necessary t raise the temperature f sme >whle thing= per degree kelvin is the Heat Capacity [ C ] f that 'whle thing'. C = q(jules)/ T s C has units f Jules/deg K rearrange t q = C * T where T = ( T final - T initial ) Heat Capacity is a prperty unique t every item, even fr things made frm the same material. The heat capacity f a 3 inch irn blt will be much smaller in magnitude than the heat capacity f a 20 ft sectin f irn rail, even thugh bth are made frm the same material, because they differ s much in mass. 2. Specific heat is a rati relating heat capacity t mass., i.e., specific heat is the heat capacity fr an item divided by its mass. S, sp.ht. = Heat Capacity (in Jules/deg K) / mass (in grams) sp.ht. has units f Jules/(gram deg K ) The specific heat f the irn blt and rail are the same, because bth heat effects are n a per gram basis. When the prduct f mass and specific heat is substituted fr Heat Capacity, the first expressin becmes mre useful: q = ( specific heat ) ( mass) ( T ) Specific heat is an identifying and characteristic prperty fr each type f material. Liquid water has a specific heat f 4.18 J/gN (please remember this value and its units). Metals have much lwer specific heats than water. Cnsider three examples invlving specific heat: ex. 1 a. Benzene has a specific heat f 1.72 J/gN. Hw many Jules f heat are required t change the temperature f 10.0 g f benzene frm 18.6 t 21.3NC? What is the heat capacity f 10.0 g f benzene? b. Carbn dixide gas has a specific heat f 0.843 J/gN. Hw many g f CO 2 can be heated frm 43.6 t 51.4NC, by supplying 100 Jules f heat energy? c. Chlrine gas (Cl 2 ) has a specific heat f 0.478 J/gN. When 0.38 mle f chlrine at 18NC are cled s that 25.0 Jules are remved, what is the final temperature? Imprtant Science Fact: Each individual example f matter has its wn unique value f specific heat. (This is because each individual example f matter has a unique set f bnded atms, and ultimately energetics are assciated with making/breaking f chemical bnds.) Imprtant Science Fact: The specific heat f a given individual example f matter depends upn which state r phase it is in. The specific heat f water as a liquid is 4.18 J/g-deg, slid water (ice) has a specific heat f 2.03 J/gN, and steam (water in the gas r vapr state) has a specific heat f 2.02 J/gN. (This is because changing t/frm ne state t anther invlves additin/remval f heat energy.) Please remember these values fr water. 3. Thermal Effects fr Phase Changes Every item, at any cnditin, has a certain heat cntent, t which heat can be added r remved by varius prcesses. A Change in heat cntent fr a prcess ccurring under cnditins f cnstant pressure ( q p ) is nt necessarily the same as when cnditins are at cnstant vlume ( q v ). Mst labratry prcesses are pen t the atmsphere and take place at a cnstant pressure f abut 1 atm. Under usual labratry cnditins f cnstant pressure ( q p ), the heat assciated PER MOLE f item is called the ENTHALPY CHANGE ( H ), q p = ÄH. Sme familiar changes in heat cntent that ccur at cnstant pressure include phase changes, i.e., physical changes between slid, liquid and gas states. (s) = (R), melting, H (fusin) ; (R) = (s), freezing, - H (fusin) ; fr water, H (fusin) = 6000 J/ml,
(R) = (g), biling, H (vaprizatin) ; (g) = (R), cndensatin, - H (vaprizatin). H (vaprizatin)= 40.7 kj/ml. Cnsider the fllwing table shwing nrmal melting and biling temperatures (i.e., at ne atmsphere pressure) and enthalpies f phase change prcesses fr varius substances... name frmula nrmal m.p. H ( fusin) nrmal b.p. H ( vaprizatin) benzene C 6 H 6 5 9.84 kj / mle 80 30.8 kj / mle brmine Br 2-7 10.8 59 29.6 mercury Hg -39 2.33 357 59.4 naphthalene C 10 H 8 80 19.3 218 43.3 water H 2 O 0 6.00 100 40.7 ex. 2 Write thermchemical a. biling f brmine, b. cndensatin f naphthalene, equatins fr: c. freezing f benzene, d. biling f Hg. ex. 3 Hw many Jules wuld be required t cnvert 50.00 g f liquid mercury t mercury vapr, at its biling temperature? ex. 4 When 40.0 g f steam at 125EC are mixed with 125 g f ice at -15EC, what is the final state and final temperature f the mixture? (apple prblem) Imprtant Science Fact: Phase changes ccur at a cnstant temperature. Energies f phase changes are expressed in Jules r kil Jules per mle and require interpretatin f ( ± ) sign. HISTORY f adding HEAT energy at a cnstant rate t a quantity f WATER 100 temperature C Thermchemistry f Water (@ 1.0 atm. pressure) +/- 40.7 kj/mle (c) phase change Liquid Phase sp.ht.=4.18 J/g-deg sp.ht.=2.02 J/g-deg Gas Phase (liq) (d) (gas) (e) 0 (s) (b) (liq) phase change +/- 6000 J/mle (a) Slid Phase sp.ht.=2.03 J/g-deg time
(a) (b) (c) (d) (e) ice (slid phase), nly ONE phase present, the relatin q = (sp.ht. slid)(mass)(tf-ti) can be applied. PHASE CHANGE: slid and liquid phases are bth present and in equilibrium at a cnstant melting pint temperature. TWO phases are present and temperature is cnstant q = (sp.ht.)(mass)(tf-ti) cannt be used. Must use H ( fusin) (n a mle basis) and furnish sign. liquid (liquid phase), nly ONE phase present, the relatin q = (sp.ht. liquid)(mass)(tf-ti) can be applied. PHASE CHANGE: liquid and gas phases are bth present and in equilibrium at a cnstant biling pint temperature. TWO phases are present and temperature is cnstant q = (sp.ht.)(mass)(tf-ti) cannt be used. Must use H ( vapr.) (n a mle basis) and furnish sign. steam (gas phase), nly ONE phase present, the relatin q = (sp.ht. gas)(mass)(tf-ti) can be applied. 4. Differentiating between SYSTEM and SURROUNDINGS. Cnsider the fllwing questins: When heat is added t smething, where des it cme frm? When is lst by smething, where des it g? A fundamental law f thermdynamics is that the energy f the universe is cnstant. Cnsequently, the "smething" f majr interest t the experimenter it is labeled as the "system", and everything else in the universe (except the system) is labeled as the "surrundings". A scheme fr defining energy terms and algebraic signs is based n the delineatin f a system frm its surrundings. S, when directin f heat flw is FROM system TO surrundings then sign f heat ( q ) fr the system is negative (-) change in temperature f system is T final is LOWER than T initial term used fr the system prcess is exthermic crrespnding sign f heat fr surrundings is psitive (+) change in temperature f surrundings is T final is HIGHER than T initial term used fr prcess w/r t surrundings is endthermic when directin f heat flw is FROM surrundings TO system - then all is the reverse f the abve 5. Calrimetry A cmmn methd used t measure heat gain/lss in thermchemistry is t immerse the system in a knwn quantity (mass) f liquid water, and measure (indirectly) the heat gain/lss fr the system, via the crrespnding temperature change f the water surrunding it. The mass f surrunding water is easily determined, and its temperature change readily measured, s the calculatin f heat transferred between system and surrunding water is given by: q (surrunding water) = (4.18 J/gN) x (mass H 2 O (surr) ) x ( T H 2 O (surr)) = q (calrimeter) Recgnize if heat is transferred frm the system t the surrunding water, then the prcess w/r t the system is exthermic, the system will have lst heat (q sys is negative), and surrunding water will have gained heat and thereby experience an increase in its temperature. Thus, in calrimetry, heat effects f system and surrunding always have same magnitude, but ppsite sign, s q = q ( system) ( surrundings)
ex. 5 When 12.78 g f a metal at 87.6EC is placed in 51 ml f liquid water at 17.4EC, the highest temperature reached by the water and metal is 18.1EC. What is the specific heat f this metal? A labratry reactr that des NOT allw vlume t change is called a bmb calrimeter. Reactins in bmb calrimeters are cnstant vlume prcesses. (SHOW a bmb calrimeter) Energies invlved in cnstant vlume prcesses are given symbls f q v (per gram) and E (TOTAL ENERGY per mle). E is always assciated with cnstant vlume prcesses, as fr example measured in bmb calrimeters. ex. 6 A 1.374 g sample f an rganic cmpund was sealed in a "bmb calrimeter" and immersed in a liquid water calrimeter at a temperature f 18.32 C. The sample was ignited and burned, and the final temperature f the liquid water in the calrimeter was 21.53 C. (In a separate experiment the Heat Capacity f the calrimeter, i.e., the surrundings, was measured as 10.12 kj/n.) If the mlecular weight f the rganic cmpund is 109 g/ml, what is the TOTAL ENERGY (symbl E) fr this cmbustin? Clearly, chemical reactins cause changes in heat cntent. Buildings are Aheated@ by energy released when rganic cmpunds are cmbusted. What quantity f heat is invlved in a reactin? The quantity invlved has t be the difference between heat cntent f prducts, and heat cntent f reactants. S, E( reactin) = [ E( prducts) ] [ E( reactants) ] (at cnstant vlume) Fr prcesses ccurring under cnditins f cnstant pressure and temperature, (i.e., vlume changes, but nt P and T) the quantity f energy invlved (per mle) is defined as ENTHALPY (symbl H). In a manner similar t the abve expressin fr E, an expressin fr H is H = [ H ] [ H ] (at cnstant P and T) ( reactin) ( prducts) ( reactants) Fr example, given that the reactin P (s) + 3/2 Cl 2(g) = PCl 3(g) EXOTHERMIC prcess can be represented as: has an enthalpy ( H ) f -287.0 kj/ml, this H ( reactin) = [ heat cntent f prduct(s) ] - [ heat cntent f reactants ] = - 287.0 kj/ml, r H( reactin) = [ H( PCl )] [ H( P) + 3/ 2 H( Cl )] = -287 kj / mle 3 2 This Rx will be used t define a standard enthalpy f frmatin, which is represented by H f 6. Enthalpies f Frmatin Enthalpy values fr chemical substances cannt be determined withut sme reference pint, just as lcatins n Earth cannt be specified withut sme reference pint - fr example the ZERO meridian line established at Greenwich, England. If ZERO meridian were lcated in Paris, then EACH pint n Earth wuld have different crdinates, BUT differences between the crdinates f any TWO pints n Earth wuld still be the same. In an analgus manner nly differences in enthalpies, H, between tw substances are infrmative in chemical prcesses. Between what tw substances? The difference is always between PRODUCTS and REACTANTS. A special cnsideratin applies when the prduct is a single cmpund, and the reactants are all elements that frm the cmpund. H is then called the enthalpy f frmatin and is given the symbl H f. The superscript ZERO is used t specify standard state cnditins (1 atm and 25 deg.c). The reference pint in chemical thermdynamics cmes frm defining the enthalpy fr elements in their standard state as ZERO. Cnsider the cnsequences f this definitin: Phsphrus trichlride is frmed frm the elements phsphrus and chrine accrding t the reactin, P (s) + 3/2 Cl 2 (gas) = P Cl 3 (gas) and the heat effects invlved with frmatin f ne mle f the cmpund under a cnstant pressure f ne atm - and at a temperature f 25 deg.c - can be readily measured. As nted abve, 287 kj f heat energy are liberated fr every mle f PCl 3 frmed. H( reactin ) = - 287.0 kj/ml
In terms f enthalpy differences between prducts and reactants, this expressin may be recast as H( reactin) = [ H( PCl )] [ H( P) + 3/ 2 H( Cl )] = - 287.0 kj/mle 3 2 Applicatin f the reference pint regarding enthalpy f frmatin f elements gives the expressin H( reactin) = [ H( PCl )] [ 0+ 3/ 2 x 0 ] = - 287.0 kj / mle which gives the value fr the enthalpy f frmatin f phsphrus trichlride H( reactin) = Hf = - 287.0 kj/ml 3 A table f enthalpy f frmatin values is listed in all general chemistry texts. The usefulness f values f H f t the wrking chemist is remarkable, and they are mst ften applied in the fllwing relatin: H( reactin) = [ Hf ( prducts) ] [ H( f ( reactants) ] ex. 7 ex. 8 ex. 9 Write thermchemical equatins fr the enthalpy f frmatin, H f, f the fllwing cmpunds carbn dixide, magnesium carbnate, ethyl alchl, ammnium perchlrate When 25.0 g f barium carbnate is heated t frm barium xide and carbn dixide (a) write a balanced thermchemical equatin fr the prcess. (b) hw many jules f heat will be invlved, per mle? fr 25.0 g barium carbnate? Determine H( reactin ) fr cmbustin f ethane, C 2H 6(g), t carbn dixide and liquid water. Anther practical applicatin fr the abve expressin, is t find H f values fr a specific substance, given measurement f enthalpy fr a thermchemical prcess (i.e., H( reactin ) ), and when H f values are knwn fr all remaining substances present. ex. 10 Cmbustin f 1.243 g f dimethyl ether at cnstant pressure, C 2 H 6 O (g), liberates 39.3 kj f Determine H f f dimethyl ether. (given qp = 39.3 kj fr 1.243 g cmpd.) 7. The algebra f thermchemical equatins: Hess' Law Thermchemical equatins apply t bth physical and chemical changes, and they bey all rules f algebra. Cnsider the next three thermchemical equatins which are "algebraic" variatins f the same chemical reactin heat. Mg (s) + 2 O 2(g) = MgO (s) H Rx = - 601.7 kj, MgO(s) = Mg (s) + 2 O 2(g) H Rx = + 601.7 kj 2 Mg (s) + O 2(g) = 2 MgO (s) H Rx = - 1203.4 kj ex. 11 Given P (s) + 3/2 Cl 2(g) = PCl 3(g) H Rx = -287 kj P (s) + 5/2 Cl 2(g) = PCl 5(g) H Rx = -347.9 kj Determine ÄH fr the prcess, PCl 3(s) + Cl 2(g) = PCl 5(g). ex. 12 Given 2 N 2(g) + 2 O 2(g) = NO (g) H Rx = +90.2 kj NH 4 NO 3(s) = 2 N 2(g) + NO (g) + 2 H 2 O (R) H Rx = -115.7 kj H 2(g) + 2 O 2(g) = H 2 O (R) H Rx = -285.8 kj Calculate the enthalpy f frmatin fr ammnium nitrate (s).
Answers t Prblems Wrked Out in Class ex. 1 a. q = (sp.ht.)(mass)( Ä T )? q = ( 1.72 J / g-deg)(10.0 g)(21.3-18.6 deg) = 46.4 Jule b. +100 Jule = (0.843 J / g-deg)(? gram)(51.4-43.6 deg)? gram = 15.2 c. 71 g first cnvert frm mle t gram:? g = 0.38 mle 1 mle = 27 gram -25.0 Jule = (0.478 J / g-deg)( 27 gram)(? T f - 18) T final = 18-1.94 = 16.06 deg.c ex.2 a. Br 2 (liquid) = Br 2 (gas) H = + 29.6 kj/ml b. C 10 H 8 (gas) = C 10 H 8 (liquid) H = - 43.3 kj/ml c. C 6 H 6 (liquid) = C 6 H 6 (slid) H = - 9.84 kj/ml d. Hg (liquid) = Hg (gas) H = + 59.4 kj/ml ex. 3 Nte: temperature DOES NOT CHANGE during phase transitins, s can=t use q =(sp.ht)(mass)( T ). Wrk directly ff the fact that enthalpy always has units f J/mle r kj/mle, and scale kj fr given masses that are nt equal t the mass f ne mle.. Cnvert frm mass t mle:? mle = 50.00 g Hg 1 mle = 0.249 mle 201g Cnvert frm listed value f enthalpy f vaprizatin, as 59.4 kj / ne mle, t mle value f prblem? kj = 0.249 mle 594. kj 1 mle? kj = 14.8 kj ex. 5 a. system: metal T(initial) = 87.6 deg.c T(final) = 18.1 deg.c mass = 12.78 gram sp.ht. =? metal LOST heat, prcess is exthermic can=t determine sp.ht. metal b/c heat lss ( -q ) is nt knwn b. surrundings: water T(initial) = 17.4 deg.c T(final) = 18.1 deg.c mass = 51 gram sp.ht. = 4.18 J/g-deg water GAINED heat, prcess is endthermic. find q(water)? q(surrundings) = ( J / g ) 417. deg (51 g)( 181. 17. 4deg) = + 149.2 Jule c. The magnitude f heat gained by surrundings (water) must be the same as heat lst by system (metal) but has ppsite sign in rder t cnserve energy. S endthermic fr surrundings must mean exthermic fr system, r q(system) = - q(surrundings) q(metal) = - (+ 149.2 Jule) = - 149.2 Jule d. nw find sp.ht. metal: q(metal) = (sp.ht. metal) (mass metal) [ T(final) - T(initial) metal ] -149.2 J = (? sp.ht. ) ( 12.78 g ) ( 18.1-87.6 deg) sp.ht.metal = 0.168 J/g-deg (nte: sp.ht. cannt have a negative value. Why?)
ex. 6 a. system: a chemical reactin T (system) really nt applicable here, mass cmpd. = 1.372 b. surrundings: calrimeter T(initial) = 18.32 T(final) = 21.53 heat capacity : 10.12 kj/deg. This is a cnstant vlume prcess ( q v ) temperature f calrimeter increased, therefre it gained heat energy. Calrimeter prcess is endthermic. Hw much heat was gained by the calrimeter? q v (calrimeter) = (calrimeter heat capacity) [T(final) - T(initial) calrimeter] q v (surr) = (10.12 kj / deg ) ( 21.53-18.32 deg ) = + 32.5 kj q v (reactin) = - q v (calrimeter); q v (reactin) = - 32.5 kj c. q v (reactin) = - 32.5 kj (fr 1.374 g cmpd) scale up fr ne mle cmpd (109 g) 325. kj? q v (reactin per mle) = 109 g / mle = - 2580 kj/mle 1374. g Use upper case thermchemical symbls fr per mle quantities. B./c this prcess tk place in a bmb calrimeter it is a cnstant vlume prcess, s q v (n a per mle base) gives the TOTAL ENERGY (E) E = - 2580 kj/mle ex. 7 a. C (s) + O 2 (g) = CO 2 (g) H f = - 393.5 kj/mle b. Mg (s) + C (s) + 3/2 O 2 (g) = MgCO 3 (s) H f = - 1113 kj/mle c. 2 C (s) + 3 H 2 (g) + 1/2 O 2 (g) = C 2 H 5 OH (liq) H f = - 278 kj/mle d. N 2 (g) + 2 H 2 (g) + 1/2 Cl 2 (g) + 2 O 2 (g) = NH 4 ClO 4 (s) H f = - 295 kj/mle ex. 8 a. balanced chemical equatin: BaCO 3 (s) = BaO (s) + CO 2 (g) H( reactin ) =? b. H Rx = [ H f BaO H f CO H ( ) + ( ) ] [ f ( BaCO ) ] 2 3 H( reactin ) = [ ( - 582 ) + ( - 394 ) ] - [ - 1219 ] = + 243 kj/mle fr 25.0 g barium carbnate? kj = 25.0 g + 243 kj 1 mle 1 mle 197 g = + 30.8 kj ex. 9 a. balanced chemical equatin: C 2 H 6 (g) + 7/2 O 2 (g) = 2 CO 2 (g) + 3 H 2 O (liq) H( reactin ) ) =? b. H = [ 2 H + 3 H ] [ H + ( 7 / 2) H ] ( reactin) f ( CO2 ) f ( H2O) f ( C2H6) f ( O2 ) = [ 2 ( - 394) + 3 ( - 286) ] - [ ( - 85) + ( ZERO) ] = - 1561 kj/mle
ex. 10 a. balanced chemical equatin: C 2 H 6 O (g) + 3 O 2 (g) = 2 CO 2 (g) + 3 H 2 O (liq) H( reactin ) =? b. cmbustin f 1.243 g (at cnstant pressure) liberates 39.3 kj, s q p = -39.3 kj cnvert t H( reactin ) )? H =??? = - 1454 kj/mle c. H = [ 2 H + 3 H ] [ H + 3 H ] ( reactin) f ( CO2 ) f ( H2O ) f ( C2 H6O) f ( O2 ) ( 2 6 ) - 1454 kj/mle = [ 2 ( - 394) + 3 ( - 286) ] - [? H f C H O + 3 ( ZERO) ] H f ( C 2 H 6 O ) = - { ( - 1454 ) - [ - 788-858 ] } = - 192 kj/mle ex. 11 given: (eq 1.) P (s) + 3/2 Cl 2 (g) = PCl 3 (g) H f = - 287 kj/mle (eq 2.) P (s) + 5/2 Cl 2 (g) = PCl 5 (g) H f = - 348 kj/mle find H (reactin) fr PCl 3 (s) + Cl 2 (g) = PCl 5 (g) alter given chemical equatins and assciated enthalpies t yield desired reactin. reverse (eq 1.) PCl 3 (g) = P (s) + 3/2 Cl 2 (g) H f = + 287 kj/mle add this t (eq 2.) P (s) + 5/2 Cl 2 (g) = PCl 5 (g) H f = - 348 kj/mle net equatin is: PCl 3 (s) + Cl 2 (g) = PCl 5 (g) H( reactin ) = - 61 kj/mle ex. 12 given: (eq 1.) 1/2 N 2 (g) + 1/2 O 2 (g) = NO (g) H f = + 90 kj/mle (eq 2.) NH 4 NO 3 (s) = 1/2 N 2 (g) + NO (g) + 2 H 2 O (liq) H f = - 116 kj/mle (eq 3.) H 2 (g) + 1/2 O 2 (g) = H 2 O (liq) H f = - 286 kj/mle find: N 2 (g) + 2 H 2 (g) + 3 / 2 O 2 (g) = NH 4 NO 3 (s) H( reactin ) =??? reverse (eq 2.) 1/2 N 2 (g) + NO (g) + 2 H 2 O (liq) = NH 4 NO 3 (s) H f = + 116 kj/mle add (eq 1.) 1/2 N 2 (g) + 1/2 O 2 (g) = NO (g) H f = + 90 kj/mle add TWO (eq 3.) 2 H 2 (g) + O 2 (g) = 2 H 2 O (liq) H f = - 572 kj/mle net equatin is: N 2 (g) + 2 H 2 (g) + 3 / 2 O 2 (g) = NH 4 NO 3 (s) H f = - 366 kj/ml