Chem 115 POGIL Wrksheet - Week 8 - Answers Thermchemistry (Cntinued), Electrmagnetic Radiatin, and Line Spectra Key Questins & Exercises 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) 2 2 2 ÄH = 2599.2 kj ÄH = 1410.9 kj ÄH = 285.8 kj The (a) equatin has C2H 2(g) n the left, where we need it in the target equatin, but its cefficient is 2. If we take ne half f equatin (a) we will have the crrect cefficient. The ÄH cntributin will be (½)( 2599.2 kj) = 1299.6 kj. Equatin (b) has C2H 2(g) with the crrect cefficient fr ur target equatin, but we need it n the right, nt the left as given. Therefre, we will reverse it, making its ÄH cntributin be +1410.9 kj. Finally, if we take equatin (c) as given, it will intrduce the H 2(g) with the crrect cefficient, and it will result in cancellatins f the CO 2(g), H2O(l), and O 2(g) intrduced by ur manipulatins f the first tw equatins, which are absent in the target equatin. The fllwing sum results: (½a) C2H 2(g) + 5/2 O 2(g) 2 CO 2(g) + H2O(l) ÄH = 1299.6 kj ( b) 2 CO 2(g) + 2 H2O(l) C2H 4(g) + 3 O 2(g) ÄH = +1410.9 kj (c) H 2(g) + ½ O 2(g) H2O(l) ÄH = 285.8 kj C H (g) + H (g) C H (g) ÄH = 174.5 kj 2 2 2 2 4 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) 106.7 C(s) + 2 Cl 2(g) CCl 4(l) Fe2O 3(s) 822.16 2 Fe(s) + 3/2 O 2(g) Fe2O 3(s) HNO 3(g) 134.3 1/2 H 2(g) + 1/2 N 2(g) + 3/2 O 2(g) HNO 3(g) NaHCO 3(s) 947.7 Na(s) + 1/2 H 2(g) + C(s) + 3/2 O 2(g) NaHCO 3(s)
3. i. Given: (a) N2O 4(g) + 1/2 O 2(g) N2O 5(g) ÄH = +1.67 kj (b) HNO (g) 1/2 N O (g) + 1/2 H O(l) ÄH = 2.96 kj 3 2 5 2 Calculate ÄH fr the reactin N2O 4(g) + H2O(l) + 1/2 O 2(g) 2 HNO 3(g) ÄH =? The tw given equatins, manipulated as shwn belw, will add t give the target reactin and its enthalpy: (a) N2O 4(g) + 1/2 O 2(g) N2O 5(g) ÄH = +1.67 kj ( 2b) N2O 5(g) + H2O(l) 2 HNO 3(g) ÄH = +5.92 kj N O (g) + H O(l) + 1/2 O (g) 2 HNO (g) ÄH = +7.59 kj 2 4 2 2 3 ii. Given fllwing standard enthalpy f frmatin data: Cmpund ÄH (kj/ml) f N2O 4(g) +9.66 HNO 3(g) 134.31 H2O(l) 285.83 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. The enthalpy fr the reactin is the sum f the enthalpies f frmatin fr prducts minus thse fr reactants, each multiplied by its stichimetric cefficient in the balanced equatin: ÄH = 2ÄH f (HNO 3) {ÄH f (N2O 4) + ÄH f (H2O) + ½ÄH f (O 2)} = (2)( 134.31 kj) {+9.66 kj + ( 285.83 kj) + 0 kj} = 7.55 kj Nte that the value f ÄH f(o 2) is taken as 0, because O 2(g) is the standard state fr elemental xygen.
As expected by the Law f Hess, the answers are virtually the same. In many cases there is a slight difference, due t minr differences in the values frm the experimental data. 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: CH 4(g) 74.85 kj CO 2(g) 393.5 kj H O(l) 285.8 kj 2 This can be calculated as fllws: ÄH rxn = {ÄH f(co 2) + 2ÄH f(h2o)} {ÄH f(ch 4) + 2ÄH f(o 2)} = {( 393.5 kj) + (2)( 285.8 kj)} {( 74.85 kj) + 0} = 890.2 kj = 890.2 kj 5 Nte that the value f ÄH f(o 2) is taken as 0, because O 2(g) is the standard state fr elemental xygen. 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 393.5 kj and 285.8 kj, respectively, and that the heat f cmbustin f C H (l) is 3267.7 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, C6H 6(l). 6 C(graphite) + 3 H 2(g) C6H 6(l) ÄH f(c6h 6) =? ii. Write the balanced thermchemical equatin fr the heat f cmbustin f benzene, C6H 6(l). C6H 6(l) + 15/2 O 2(g) 6 CO 2(g) + 3 H2O(l) ÄH cmb = 3267.7 kj 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. We can express the heat f cmbustin as ÄH cmb = 6ÄH f(co 2) + 3ÄH f(h2o) ÄH f(c6h 6)
We have values fr everything in this equatin except ÄH f(c6h 6). Therefre, we can rearrange and slve fr it: ÄH f(c6h 6) = 6ÄH f(co 2) + 3ÄH f(h2o) ÄH cmb = (6)( 393.5 kj) + (3)( 285.8 kj) ( 3267.7 kj) = +49.3 kj 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 (í = c/ë) 19 19 light with energy f 3.0 x 10 J r light with energy f 4.5 x 10 J (E = hí) 8. An argn laser emits green light with a wavelength f 514.5 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. (a) Nte that t cnvert frm nanmeters t Ångstrms we need nly mve the decimal pint t the right ne place. (b) Use í = c/ë. We need wavelength in meters, because the units n the speed f light are -1 m s. We culd cnvert nm t m first, but with a calculatr it is simpler t enter 9 nanmeters as 10 m; i.e., 514.5E-9". (c) Given the preceding results, we can calculate energy by either E = hí r E = hc/ë. Using the first f these, 34 14 1 19 E = hí = (6.626 x 10 J s)(5.827 x 10 s ) = 3.861 x 10 J 9. Is energy emitted r absrbed when the fllwing transitins ccur in hydrgen: (a) frm n = 1 t n = 3 The transitin is t a higher energy state, s energy crrespnding t the difference between the states must be absrbed. (b) frm n = 5 t n = 2 The transitin is t a lwer energy state, s energy crrespnding t the difference between the states must be emitted.
+ (c) an H in acquires an electrn int the n = 2 state. + An H in is essentially a hydrgen atm with an electrn an infinite distance away, the state fr which n =. By adding an electrn t the n = 2 state, it has lwer energy (the electrn is clser t the nucleus t which it is attracted), s energy will be emitted 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 All transitins in the Balmer series are t n f = 2 frm n i > 2. The energy f light increases ging frm red t vilet, s the red line is the smallest state-t-state separatin, and the vilet line is the largest. red (656 nm) 3 2 blue-green (486 nm) 4 2 blue (434 nm) 5 2 vilet (410 nm) 6 2 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? 18 E phtn = 1.635 10 J A wavelength f 121 nm falls well within the ultravilet regin.