Chem 115 POGIL Worksheet - Week 8 Thermochemistry (Continued), Electromagnetic Radiation, and Line Spectra

Transcription

1 Chem 115 POGIL Wrksheet - Week 8 Thermchemistry (Cntinued), Electrmagnetic Radiatin, and Line Spectra Why? As we saw last week, enthalpy and internal energy are state functins, which means that the sum f the heats f any set f steps that adds t give an verall reactin will have the same heat as ding the reactin directly. This is Hess s Law. We will revisit this tday and g n t see that if we use a special kind f thermchemical reactin, called the standard enthalpy f frmatin, we can calculate enthalpies f reactins withut having t manipulate a series f individual thermchemical equatins fr each step. T understand the mdern mdel f atmic structure, we need t lk at the nature f light and ther frms f electrmagnetic radiatin. One f the mst imprtant ideas t emerge at the start f the twentieth century was that energy f subatmic particles and electrmagnetic radiatin is nt cntinuus, but rather is quantized in discrete allwable values. An understanding f the basic relatinships amng the characteristic prperties f electrmagnetic radiatin and an appreciatin f the nature f state-t-state transitins in quantized systems will lay a fundatin fr understanding the mdern quantum mechanical mdel f the atm. Learning Objective Understand Hess s Law Understand the definitin and use f standard enthalpy f frmatin Understand the fundamental relatinships f electrmagnetic radiatin and the electrmagnetic spectrum Understand the relatinship between line spectra and the quantum cncept Success Criteria Be able t calculate enthalpy f a target reactin frm a series f given reactins Be able t apply standard enthalpies f frmatin t calculate the enthalpy f a reactin Be able t d cnversins between energy, wavelength, and frequency f electrmagnetic radiatin Be able t crrelate state-t-state transitins with regin f the electrmagnetic spectrum and calculate the energy f a state-t-state transitin Prerequisite Have read sectins 5.6 and 5.7 Have read sectins 6.1 thrugh 6.3

2 Infrmatin Revisited (Standard Cnditins, State Functins and Hess's Law) Remember that the measured value f ÄH depends n the states f all reactants and prducts (s, l, g, aq) and the temperature and pressure under which the reactin ccurs. In rder t make meaningful direct cmparisns, it is useful t define a set f standard cnditins. By internatinal agreement, standard cnditins are defined as T = 25 C; P = 1 atm; all substances in their usual states fr these cnditins (the standard state). The standard state f an element is its mst stable state; e.g., H 2(g), C(s) graphite, S 8(s), P 4(s). Fr cmpunds, the standard state is the mst prevalent state under standard cnditins; e.g., H O(l), CO (g), C H (g), C H (l) Enthalpy is a state functin, which nly depends upn current cnditins (the state f the system) fr its value, nt n hw the current state was reached. As applied t ÄH = H f H i, the value f the enthalpy change fr any prcess depends nly n the difference between the final and initial states, nt n the path chsen. This means that any set f steps, whether real r imagined, that take the system frm the initial state t the final state f interest will have a sum f ÄH values fr all the steps that is equal t the value f ÄH fr the verall prcess dne directly. This principle, called Hess's Law f Cnstant Heat Summatin, was first established by G. H. Hess in 1840: The enthalpy change fr a reactin is independent f path. In applying Hess's Law, a set f given thermchemical equatins is manipulated such that they add t give a balanced thermchemical equatin fr the prcess f interest (the target equatin). In ding this, whenever a given thermchemical equatin is multiplied (usually by an integer r ratinal fractin), its ÄH is likewise multiplied. Whenever the directin f a given thermchemical equatin is reversed, its ÄH value changes sign. Review Exercise 1. Calculate ÄH fr the reactin, Given: C2H 2(g) + H 2(g) C2H 4(g) (a) 2 C2H 2(g) + 5 O 2(g) 4 CO 2(g) + 2 H2O(l) (b) C2H 4(g) + 3 O 2(g) 2 CO 2(g) + 2 H2O(l) (c) H (g) + ½ O (g) H O(l) ÄH = kj ÄH = kj ÄH = kj Infrmatin (Standard Enthalpies f Frmatin, ÄH f) Hess s Law calculatins are dne s frequently that it is cnvenient t have tabulated data fr a large number f reactins. The mst generally useful data fr these kinds f calculatins are standard enthalpies f frmatin.

3 The standard enthalpy f frmatin, ÄH f, f a cmpund is the enthalpy change fr the reactin in which ne mle f the cmpund in its standard state is made frm the stichimetric amunts f its elements in their standard states. Fr all elements in their standard states, ÄH 0, by definitin. f Fr example, the fllwing thermchemical equatins define standard enthalpies f frmatin fr C H (g) and C H (g), respectively (a) 2 C(s) + H (g) C H (g) (b) 2 C(s) + 2 H (g) C H (g) f f ÄH = kj ÄH = kj Ntice that each f these equatins gives the heat prduced when exactly ne mle f the prduct cmpund is made. There are n thermchemical equatins fr the frmatin f substances such as C(s) as graphite r H 2(g) as the gaseus element, because these are their nrmal elemental states. Their standard enthalpies f frmatin are set as zer by definitin. Let s use the tw enthalpy f frmatin equatins given abve t calculate the enthalpy f the fllwing reactin: C2H 2(g) + H 2(g) C2H 4(g) ÄH =? Yu calculated the enthalpy f this reactin in Review Exercise 1, but nw we are ging t d it with the thermchemical equatins that define ÄH f fr C2H 2(g) and C2H 4(g). T d this, we simply need t take the reverse f equatin (a) with the negative f its given enthaply, and add it t equatin (b), using its given enthalpy. The resulting sum is: ( a) C2H 2(g) 2 C(s) + H 2(g) ÄH f = kj (b) 2 C(s) + 2 H 2(g) C2H 4(g) ÄH f = kj C H (g) + H (g) C H (g) ÄH = kj This is the same answer yu btained in exercise 1 (if yu did it crrectly), as expected n the basis f Hess s Law. But ntice that the answer we btain here is the fllwing sum, where ÄH f (H ) = 0, because H (g) is an element in its standard state: 2 2 ÄH rxn = ÄH f(c2h 4) { H f(c2h 2) + ÄH f(h 2)} = {+52.3 kj} { kj + 0 kj} This is the sum f the enthalpies f frmatin f the prducts, multiplied by their stichimetric cefficients (here, 1), minus the sum f the enthalpies f frmatin f the reactants, multiplied by their stichimetric cefficients (here again, bth 1). This is a general result, which we can summarize as ÄH rxn = ÓnpÄH p ÓnrÄH r where n p and n r are the stichimetric cefficients f each f the prducts and each f the reactants, respectively. [The Ó (Greek sigma) means take the sum f.] Nte that this relatinship can nly be used if all the data are enthalpies f frmatin. Fr a general reactin

4 we wuld have aa + bb cc + dd, ÄH rxn = [cäh f(c) + däh f(d)] [aäh f(a) + bäh f(b)] Key Questins 2. Write the balanced thermchemical equatins that pertain t the standard enthalpies f frmatin f the given cmpunds. Cmpund ÄH f (kj/ml) Thermchemical Equatin CCl 4(g) Fe2O 3(s) HNO 3(g) NaHCO 3(s) Exercises 3. i. Given: (a) N2O 4(g) + 1/2 O 2(g) N2O 5(g) ÄH = kj (b) HNO (g) 1/2 N O (g) + 1/2 H O(l) ÄH = 2.96 kj Calculate ÄH fr the reactin N2O 4(g) + H2O(l) + 1/2 O 2(g) 2 HNO 3(g) ÄH =? ii. Given fllwing standard enthalpy f frmatin data: Cmpund ÄH (kj/ml) f N2O 4(g) HNO 3(g) H2O(l) Calculate ÄH fr the reactin N2O 4(g) + H2O(l) + 1/2 O 2(g) 2 HNO 3(g) ÄH =? Cmpare yur answer t yur answer in part i.

5 4. Calculate the heat f cmbustin f methane, CH 4(g), defined by the fllwing thermchemical equatin: CH 4(g) + 2 O 2(g) CO 2(g) + 2 H2O(l) ÄH rxn =? Given the fllwing standard enthalpies f frmatin: Cmpund ÄH (kj/ml) f CH 4(g) CO 2(g) H2O(l) kj 5. Calculate the enthalpy f frmatin, ÄH f, fr benzene, C6H 6(l), given that the heats f frmatin f CO 2(g) and H2O(l) are kj and kj, respectively, and that the heat f cmbustin f C H (l) is kj. T d this, carry ut the fllwing steps. 6 6 i. Write the balanced thermchemical equatin that defines the enthalpy f frmatin f benzene, C H (l). 6 6 ii. Write the balanced thermchemical equatin fr the heat f cmbustin f benzene, C H (l). 6 6 iii. Based n yur answer t questin ii, write an expressin fr the heat f cmbustin f benzene, ÄH cmb, in terms f the enthalpies f frmatin f the reactants and prducts. Using the data given in the prblem, slve this fr the unknwn value f ÄH f(c6h 6), the enthalpy f frmatin f benzene. Infrmatin (Electrmagnetic Radiatin) Light and ther frms f radiant energy can be thught f a being prpagated as a wave f scillating electric and magnetic cmpnents. Hence, we call radiant energy electrmagnetic radiatin. Electrmagnetic radiatin is characterized by its energy (E), wavelength (ë Greek lambda), and frequency (í Greek nu). The speed f prpagatin in a vacuum, cmmnly called 8 1 the speed f light (c), is x 10 m s. The relatinships between E, ë, and í are í = c/ë E = hí = hc/ë 34 where h is Plank s cnstant, equal t x 10 J s. These equatins apply t all kinds f electrmagnetic radiatin. The units f wavelength are usually nanmeters (nm), where 1 nm = 9 10 m, althugh ther length units may be used, depending n the radiatin. An lder unit called 10 the Ångstrm (Å) is smetimes used with light and with x-rays, where 1 Å = 10 m = 0.1 nm.

6 1 Frequency usually has units f sec, called the hertz (Hz). This means the number f waves that pass a reference pint per secnd, but the wrd wave is nt part f the unit and des nt factr int the dimensinal analysis. The fllwing chart shws the wavelength ranges f the varius kinds f electrmagnetic radiatin. Key Questins 6. Fr each f the fllwing, indicate which kind f radiatin has higher energy. red light r blue light infrared radiatin r radi waves x-rays r visible light 7. Fr each f the fllwing, indicate which has higher frequency light with ë = 490 nm r light with ë = 520 nm light with energy f 3.0 x 10 J r light with energy f 4.5 x 10 J Exercise 8. An argn laser emits green light with a wavelength f nm. Calculate the fllwing fr -1 this light: (a) the wavelength in Å; (b) the frequency in Hz (s ); (c) the energy in jules, J.

7 Infrmatin (Line Spectra and the Bhr Mdel) Electrns bmbarding hydrgen gas (H 2) in a discharge tube cause the frmatin f mnatmic H atms, sme f which are in high energy states (excited states). The atms can lse a prtin f their excess energy by emitting electrmagnetic radiatin. Rather than a cntinuus spectrum cntaining all wavelengths within a range, the emitted radiatin f hydrgen and ther mnatmic gasses cnsists f nly certain characteristic wavelengths, which are typically bserved as discrete lines in the spectrum, falling in the range frm the ultravilet regin thrugh the infrared regin. Hence, these spectra are called line spectra. Fr hydrgen, fur f these wavelengths fall in the visible regin (where they can be seen by a human eye) and give lines that are vilet (410 nm), blue (434 nm), blue-green (486 nm), and red (656 nm). In 1885, Balmer determined that these lines fit a relatively simple equatin that relates the reciprcal f their wavelength (1/ë) t pairs f integers (n 1 and n 2). When additinal lines in the invisible infrared and ultravilet regins were discvered, they and Balmer s visible lines were all fund t fit the fllwing general equatin: 7-1 where, the Rydberg cnstant, is x 10 m, and the n values are integers with n 2 > n 1. This equatin was nly valid fr the emissins frm hydrgen, a ne-electrn atm. In the case f the visible lines f the Balmer series, the value f n 1 is 2. When n 1 = 1, the assciated lines fall in the ultravilet regin (the Lyman series). Values f n 1 3 crrespnd t successive series that fall in the infrared regin (the Paschen, Brackett, and Pfund series), with increasing wavelength and decreasing frequency and energy as n increases. Neils Bhr, inspired by the Balmer equatin and Max Plank s ideas f quantized energy, devised a mdel fr the hydrgen atm. In Bhr s mdel the single electrn is thught t be cnfined t certain rbits arund the nucleus, where each rbit crrespnds t a particular, quantized energy, given by the equatin 1 Here n, the principal quantum number, can take n integer values frm 1 t. This equatin can + 2+ be extended t ther ne-electrn atms (e.g., He, Li ) in the frm 2 2 E = BZ /n n = 1, 2, 3,... where B is a cnstant similar t the Rydberg cnstant and Z is the atmic number (number f prtns). In either frm, the Bhr equatin predicts that the energy f the ne-electrn atm is inversely prprtinal t the square f the quantum number f its state. This gives rise t the energy level diagram shwn n the last page f this wrk sheet, in which the spacing between energy states gets smaller as n increases.

8 In Bhr s mdel, as n increases, the electrn is in an rbit that is further frm the nucleus and therefre has a higher (less negative) energy. Nte that the zer f energy crrespnds t n =, which has the electrn an infinite distance away frm the nucleus. Any lwer value f n places the negative electrn clser t the psitive nucleus t which it is attracted, giving rise t a mre favrable (negative) energy. The atm absrbs r emits energy in the frm f electrmagnetic radiatin by changing frm an initial state (E i) with a value n i t a final state (E f) with a value n f. These changes in state are called transitins. With absrptin, the atm acquires energy, resulting in a higher energy state. Thus, fr an absrptin transitin, n f > n i. With emissin, the atm lses energy, resulting in a lwer energy state. Thus, fr emissin, n f < n i. It is transitins frm higher energy states t lwer energy states that give rise t the bserved line spectra. Fr any transitin, the energy f the electrmagnetic radiatin (a phtn) absrbed r emitted must exactly match the energy difference between the tw states; i.e., ÄE = E f E i = E phtn = hí Using the Bhr equatin fr the allwed energies in states n f and n i, we can write an equatin t calculate the phtn s energy as This equatin has the same frm as the experimentally determined Balmer equatin, if we assume n f and n i f the Bhr equatin crrespnd t n 1 and n 2 f the Balmer equatin. Regardless f whether absrptin r emissin is invlved, the energy f a phtn is always taken as a psitive number, s any sense f sign t the energy calculated frm this equatin is rutinely drpped. As the diagram n the last page shws, the Balmer series, which gives rise t the fur lines in the visible spectrum, crrespnds t transitins frm higher states dwn t n f = 2 (i.e., 3 2, 4 2, 5 2, 6 2). The energy gaps between n f = 2 and n i > 2 are such that the wavelengths f the emitted phtns fall in the visible regin. In the Lyman series (see diagram), the transitins are frm upper states dwn t the lwest state (n f = 1), called the grund state f the atm. The gaps between n = 1 and n > 1 states are bigger, and s the emitted phtns have higher energies, falling in the ultravilet regin. By cntrast, in the Paschen series (see diagram), the transitins are frm higher states dwn t n f = 3. Because the energy separatins between states with higher n values are smaller, the transitins have lwer energies, falling in the infrared regin. The energy differences are even smaller when the transitins are frm upper states t either n f = 4 r n f = 5, making the emitted radiatin fall even further int the infrared regin fr these series (cf. Brackett and Pfund series n the diagram).

9 Key Questins 9. Is energy emitted r absrbed when the fllwing transitins ccur in hydrgen: (a) frm n = + 1 t n = 3, (b) frm n = 5 t n = 2, (c) an H in acquires an electrn int the n = 2 state. Exercise 10. The fur visible lines f the Balmer series in the emissin spectrum f hydrgen are vilet (410 nm), blue (434 nm), blue-green (486 nm), and red (656 nm). Assign these as state-t- state transitins f the type n n, giving the n values invlved fr each line. i f 11. Calculate the energy f the first line in the Lyman series fr the hydrgen atm, which arises frm a transitin frm n i = 2 t n f = 1. What is the wavelength f the radiatin emitted? In what regin f the electrmagnetic spectrum des it fall?

10 Energy Level Mdel fr a One-Electrn Atm (Named series are fr the hydrgen atm.)