Examples: 1. How much heat is given off by a 50.0 g sample of copper when it cools from 80.0 to 50.0 C?

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1 NOTES: Thermchemistry Part 1 - Heat HEAT- TEMPERATURE - Thermchemistry: the study f energy (in the frm f heat) changes that accmpany physical & chemical changes heat flws frm high t lw (ht cl) endthermic reactins: absrb energy in the frm f heat; shw a psitive value fr quantity f heat (q > 0) exthermic reactins: release energy in the frm f heat; shw a negative value fr quantity f heat (q < 0) Magnitude f Heat Flw: Units f heat energy: Fr a pure substance f mass m, the expressin f q can be written as: where, q = m = c = t = Specific heat = the amunt heat that must be added t raise the temp. f 1 g f a substance by 1 C, with n change in state. Specific heat values (in J / g C): Examples: 1. Hw much heat is given ff by a 50.0 g sample f cpper when it cls frm 80.0 t 50.0 C?

2 2. Irn has a specific heat f J/g C. When a 7.55 g piece f irn absrbs J f heat, what is the change in temperature? If it was riginally at rm temp. (22.0 C), what is the final temperature? 3. The specific heat f cpper is J/g C. Hw much heat is absrbed by a cpper plate with a mass f g t raise its temperature frm 25.0 C t ven temperature (420 F)? Thermchemistry Part 2 - Calrimetry Q: If yu leave yur keys and yur chemistry bk sitting in the sun n a ht summer day, which ne is htter? Q: Why is there a difference in temperature between the tw bjects? Heat required t melt ice (a.k.a. latent heat f fusin) cannt be measured directly, but calrimetry prvides an experimental methd allwing this heat transfer t be measured indirectly. Calrimetry: measurement f the amunt f heat evlved r absrbed in a chemical reactin, change f state r frmatin f a slutin. The enthalpy change assciated with a chemical reactin r prcess can be determined experimentally. Measure the during a reactin at CONSTANT pressure. A is a device used t measure the heat absrbed r released during a chemical r physical prcess Simple/Cffee Cup Calrimeter The cup is filled with water, which absrbs the heat evlved (r given ff) by the reactin. Picture f cffee cup calrimeter: q rxn = -q cal

3 What happens in a calrimeter? One bject will, and the ther will the heat System lses heat t surrundings = System absrbs heat frm surrundings = When a ht chunk f metal is drpped in a cl glass f water, the metal cls ff. Where did the heat frm the metal g? Did the metal lse mre heat than the water gained? Magnitude f = (ALWAYS!) T d calrimetry prblems First, make a chart - Example 1 : A small pebble is heated and placed in a fam cup calrimeter cntaining 25.0 g f water at 25.0 C. The water reaches a maximum temperature f 26.4 C. Hw many jules f heat were released by the pebble? The specific heat f water is J/g C. Measurement Water (cal) Pebble (rxn) Heat (q) Mass (m) Specific Heat (c) Final Temp (T f) Initial Temp (T i) Example 2: When 1.00 g f ammnium nitrate, NH 4NO 3, is added t 50.0 g f water in a cffee cup calrimeter, it disslves, NH 4NO 3 (s) NH 4+ (aq) + NO 3- (aq), and the temperature f the water drps frm C t C. Calculate q fr the reactin system. Example 3: Suppse that g f water at 22.4 C is placed in a calrimeter. A g sample f Al is remved frm biling water at a temperature f 99.3 C and quickly placed in a calrimeter. The substances reach a final temperature f 32.9 C. Determine the SPECIFIC HEAT f the metal.the specific heat f water is J/g C.

4 Bmb Calrimeter NOTE: In a bmb calrimeter, heat is transferred frm the sample t the xygen-enriched chamber, t the metal that makes up the chamber, t the water thus we cannt just use the specific heat f water; instead heat capacity f the calrimeter, Ccal, can be used r calculated. It is pssible t calculate the amunt f heat absrbed r evlved by the reactin if yu knw the heat capacity, Ccal, and the temp. change, Δt, f the calrimeter. Everything else is the same (remember, the heat lst frm the reactin ges int the calrimeter) Picture f bmb calrimeter: EXAMPLE 4: The reactin between hydrgen and chlrine, H 2 + Cl 2 2HCl, can be studied in a bmb calrimeter. It is fund that when a 1.00 g sample f H 2 reacts cmpletely, the temp. rises frm C t C. Taking the heat capacity f the calrimeter t be 9.33 kj/ C, calculate the amunt f heat evlved in the reactin. EXAMPLE 5: When 1.00 ml f caffeine (C 8H 10N 4O 2) is burned in air, 4.96 x 10 3 kj f heat is evlved. Five grams f caffeine is burned in a bmb calrimeter. The temperature is bserved t increase by C. What is the heat capacity f the calrimeter in J/ C? EXAMPLE 6: When twenty milliliters f ethyl ether, C 4H 10O. (d=0.714 g/ml) is burned in a bmb calrimeter, the temperature rises frm 24.7 C t 88.9 C. The calrimeter heat capacity is kj/ C. (a) What is q fr the calrimeter? (b) What is q when 20.0 ml f ether is burned? (c) What is q fr the cmbustin f ne mle f ethyl ether?

5 Thermchemistry Part 3 Enthalpy and Thermchemical Equatins Enthalpy Enthalpy = a type f chemical energy (thermdynamic ptential), smetimes referred t as heat cntent Enthalpies f Reactin ( H rxn): the enthalpy change (reprted in kj/ml) that accmpanies a chemical reactin is called the enthalpy f reactin. Als called heat f reactin If H rxn = negative Exthermic Heat is evlved, r given ff Under cnditins f cnstant pressure, q = ΔH < 0 (negative sign) If H rxn = psitive Endthermic Heat is absrbed Under cnditins f cnstant pressure, q = ΔH > 0 (psitive sign) Thermchemical Equatins Thermchemical equatins are balanced chemical equatins that shw the assciated enthalpy change ( H) balanced equatin enthalpy change (ΔH rxn) Rules f Themchemistry: Rule #1) The magnitude f H is directly prprtinal t the amunt f reactant cnsumed and prduct prduced. Example 1: H 2 + Cl 2 2Hcl H = kj Calculate H when 1.00 g f Cl 2 reacts. Example 2: When an ice cube weighing 24.6 g f ice melts, it absrbs 8.19 kj f heat. Calculate H when 1.00 ml f slid water melts. Example 3: Methanl burns t prduce carbn dixide and water: 2CH 3OH + 3O 2 2CO 2 + 4H 2O kj What mass f methanl is needed t prduce 1820 kj? Example 4: Hw much heat is prduced when 58.0 liters f hydrgen (at STP) are als prduced? Zn + 2HCl ZnCl 2 + H kj

6 Rule #2) H fr a reactin is equal in the magnitude but ppsite in sign t H fr the reverse reactin. (If 6.00 kj f heat absrbed when a mle f ice melts, then 6.00 kj f heat is given ff when 1.00 ml f liquid water freezes) Example: Given: H 2 + ½ O 2 H 2O H = kj Calculate H fr the equatin: 2H 2O 2H 2 + O 2 Rule #3) The value f H fr a reactin is the same whether it ccurs in ne step r in a series f steps. fr the verall equatin is the sum f the H s fr the individual equatins: Hess s Law: H = H 1 + H 2 + Example 1: Calculate H fr the reactin: C + ½ O 2 CO Given: C + O 2 CO 2 H = kj 2CO + O 2 2CO 2 H = kj Example 2: Find the heat f reactin (enthalpy) fr the fllwing reactin: NO + ½ O 2 NO 2 H =? Given the fllwing equatins. ½ N 2 + ½ O 2 NO H = kj ½ N 2 + O 2 NO 2 H = Thermchemistry Part 4 Phase Changes and Heats f Frmatin specific heat = the amt f heat that must be added t a stated mass f a substance t raise its temp by 1 C, with n change in state. Ex: Hw much heat is released by g f H 2O as it cls frm 85.0 C t 40.0 C? (specific heat f water = 4.18 J/g C)

7 Heat changes invlving phase changes LATENT HEAT OF FUSION, Hfus : the enthalpy change (energy absrbed) when a cmpund is cnverted frm a slid t a liquid withut a change in temperature. Latent means hidden; the heat absrbed/released during a phase change des nt cause the temperature t change. Nte: Hfus fr water is LATENT HEAT OF VAPORIZATION, liquid t a gas withut a change in temperature. Nte: Hvap - the enthalpy change (energy absrbed) when a cmpund is cnverted frm a Hvap fr water is When substances change state, they ften have different specific heats: c ice= 2.09 J/g C c water= 4.18 J/g C c steam= 2.03 J/g C Example 1: Hw much heat is released by g f H 2O as it cls frm C t C Five steps 1. Cl the steam m c steam T = 2. Cndense m(- H vap) = 3. Cl the liquid water m c water T = 4. Freeze m(- H fus) = 5. Cl the slid ice m c ice T = Example 2: Hw much heat energy is required t bring g f water at 55.0 C t its biling pint (100 C) and then vaprize it? Example 3: Hw much heat energy is required t cnvert 15.0 g f ice at 12.5 C t steam at C?

8 Enthalpies f Frmatin H f = enthalpy f frmatin usually exthermic see table fr H f values enthalpy f frmatin f an element in its stable state = these can be used t calculate H fr a reactin standard enthalpy change, H, fr a given thermchemical equatin is = t the sum f the standard enthalpies f frmatin f the prduct the standard enthalpies f frmatin f the reactants. elements in their standard states can be mitted: 2 Al (s) + Fe 2O 3(s) 2 Fe (s) + Al 2O 3(s) the cefficient f the prducts and reactants in the thermchemical equatin must be taken int accunt: 2 Al (s) + 3 Cu 2+ (aq) 2 Al 3+ (aq) + 3 Cu (s) Example: Calculate H fr the cmbustin f ne mle f prpane: C 3H 8 (g) + 5O 2 (g) 3CO 2 (g) + 4H 2O (l) Example: The thermchemical equatin fr the cmbustin f benzene, C 6H 6, is: C 6H 6 (l) + 15/2 O 2 (g) 6CO 2 (g) + 3H 2O (l) H = kj Calculate the standard heat f frmatin f benzene. Example: When hydrchlric acid is added t a slutin f sdium carbnate, carbn dixide gas is frmed. The equatin fr the reactin is: 2H + (aq) + CO 3 2- (aq) CO 2(g) + H 2O(l) Calculate H fr this thermchemical equatin.

9 Thermchemistry Part 5 Spntaneity THERMODYNAMICS = the study f energy changes that accmpany physical and chemical changes. Enthalpy (H): ): the ttal energy stred within a substance Enthalpy change ( H): a cmparisn f the ttal enthalpies f the prduct & reactants. Exthermic v. Endthermic: Exthermic reactins/changes: release energy in the frm f heat; have negative H values H 2O(g) H 2O(l) ΔH = kj Endthermic reactins/changes: absrb energy in the frm f ehat; have psitive H values. H 2O(l) H 2O(g) ΔH = kj Changes that invlve a decrease in enthalpy are favred! Reactin pathways: Entrpy (S): the measure f the degree f disrder in a system; in nature, things tend t increase in entrpy, r disrder. All physical & chemical changes invlve a change in entrpy, r S. (remember that high entrpy is favrable) Enthalpy and entrpy are DRIVING FORCES fr spntaneus reactins (rxns that happen at nrmal cnditins) It is the interplay f these 2 driving frces that determines whether r nt a physical r chemical change will actually happen. Free Energy (G): relates enthalpy and entrpy in a way that indicates which predminates; the quantity f energy that is available r stred t d wrk r cause change. where: G = H = T = S =

10 G: ps value means change is NOT spn. G: neg value means change IS spn. Relating Enthalpy and Entrpy t Spntaneity Example f reactin H S Spntaneity H 2O (g) H 2O (l) H 2O (s) H 2O (l) Examples: 1) Fr the decmpsitin f O 3 (g) t O 2(g): 2O 3(g) 3O 2(g) H = kj/ml and S = J/ml K at 25 C. a) Calculate G fr the reactin. b) Is the reactin spntaneus? c) Is H r S (r bth) favrable fr the reactin? 2) What is the minimum temperature (in C) necessary fr the fllwing reactin t ccur spntaneusly? Fe 2O 3 (s) + 3CO(g) 2Fe(s) + 3CO 2 (g) H = kj/ml; S = J/K ml (Hint: assume G = -0.1 kj/ml)

11 Standard Heats f Frmatin, Gibbs Free Energy, and Entrpy Substance (cmpunds at 1 atm, aqueus ins at 1M) H f ΔG f S at 25 C (J/ml K) at 25 C (kj/ml) at 25 C (kj/ml) Al 3+ (aq) Al2O3 (s) Br2 (g) Br2 (l) C (s, diamnd) C (s, graphite) CH4 (g) C3H8(g) CO (g) CO2 (g) CO3 2- (aq) CaCO3 (s) CaO (s) Cl2 (g) Cu (s) Cu + (aq) Cu 2+ (aq) F2 (g) Fe (s) Fe2O3 (s) H + (aq) H2 (g) H2O (g) H2O (l) H2O2 (l) HCl (g) H2S (g) I2 (g) I2(s) Mg(OH)2(s) N2 (g) NH3 (g) NO (g) NO2 (g) Na2CO3 (s) NaCl (s) O2 (g) O3 (g) P (s, white) P (s, Red) S (s, rhmbic) S (s, mnclinic) SO2 (g) SO3 (g)