1. Calculate H for the reaction: 3 Fe 2 O 3 (s) 2 Fe 3 O 4 (s) + ½ O 2 (g) H =? (6 points)

Size: px

Start display at page:

Download "1. Calculate H for the reaction: 3 Fe 2 O 3 (s) 2 Fe 3 O 4 (s) + ½ O 2 (g) H =? (6 points)"

Lynn Richard
5 years ago
Views:

1 Chemitry 141 Name Exam 3 SHOW WORK for All Problem No electronic device allowed except a non-programmable cientific calculator. Data for problem below: 2 FeO() + ½ O 2 (g) Fe 2 O 3 () H = kj 3 FeO() + ½ O 2 (g) Fe 3 O 4 () H = kj FeO() Fe() + ½ O 2 (g) H = kj 1. Calculate H for the reaction: 3 Fe 2 O 3 () 2 Fe 3 O 4 () + ½ O 2 (g) H =? (6 point) 2. Write a thermochemical equation for H f [Fe 3 O 4 ()] AND calculate it numerical value. (8 point) 3. Ue H f [Fe(l)] = 90.5 kj/mol to calculate H for: Fe(l) + ½ O 2 (g) FeO() (6 point) 4. A 6.44 g iron nail left underground i completely ruted to green rut,feo(), When thi i expoed to the air it will convert to regular rut, Fe 2 O 3 (). Will heat be given off or taken in by thi proce ( green rut to brown rut)? How much heat? (8 point)

2 5. A g metal pot initially at 20.4 C ha pecific heat J/(g C ). (14 point) a. Determine it temperature after 32.7 kj of heat i added to the pot. b. Water with pecific heat J/(g C ) at an initial temperature of 17.8 C i added to the (hot) pot, Both water and pot come to the ame temperature 24.9 C. Calculate the ma of water added to the pot. 6. Standard Enthalpie of Formation in kj/mol: (12 point) NiO () H 2 O(l) NO(g) 90.4 Ni(NO 3 ) 2 6H 2 O() NiO(g) H 2 O(g) NO 2 (g) 33.8 Ni(NO 3 ) a. Write a thermochemical equation for the tandard enthalpy of formation for nickel(ii) nitrate hexahydrate, Ni(NO 3 ) 2 6H 2 O() Include phae and the H value. b. Determine the value of H for the reaction: 2 Ni(NO 3 ) 2 6H 2 O() 2 NiO() + 4 NO 2 (g) + O 2 (g) + 12 H 2 O(g)

3 7. For each of the following procee indicate if work and heat are poitive (+) negative( ) or zero (0). Aume ytem i 1 mole of reaction at tandard condition unle otherwie indicated. (10 point) a. H 2 O(l, 100 C) H 2 O(g, 100 C) b. Na(l) Na() c. NO(g) + ½ O 2 (g) NO 2 (g) d. O 2 (g, 1atm, 25 C) O 2 (g, 2 atm, 0 C) e. Cu(,100 C) Cu(, 350 K) Proce w = +,, or 0 q = +,, or 0 example He(g, 100 K, 1 liter) He(g, 100 K, 2 liter) 0 8. Calculate the frequency of a photon involved in a 8d 3p tranition in a hydrogen atom. I thi photon emitted or aborbed? (8 point) h = x J c = x 10 8 m/ 1 J = 1 kg m 2 / 2 ma of an e = m e = 9.109x10 28 g h λ c = E = h E = q + w w = P V H = q (at contant P) mu Hydrogen atom have energy level with energy E n = k/n 2 k = 2.179x10 18 J q = m T 1 L atm = J 1 cal = J

4 9. For the photoelectric effect hν = hν 0 + E K where E K = ½ m e u 2. (24 point) For lithium the threhold frequency i ν 0 = 5.952x10 16 Hz. with peed = u = 4.41x10 6 m/. A photon trike lithium o that a electron leave a. Determine the kinetic energy of the electron releaed in joule. b. Determine the wave length of the photon with jut enough energy to releae the electron with the indicated peed. c. Determine the de Broglie wavelength of the electron.

5 10. Match the following: (4 point) a. olved the ultraviolet catatrophe A. Eintein E. Heienberg b. explained the photoelectric effect B. Bohr F. de Broglie c. tated an Uncertainty Principle C. Hertz G. Rutherford d. explained hydrogen viible pectra D. Schrödinger H. Planck 11. In each et of three circle the bet anwer. (10 point) Highet energy: blue light IR Radio λ= 1 cm λ= 1 mm λ= 1 µm Lowet frequency: X-ray MW red light 12 MJ/mol 1200 kj/mol 2500 cal/mol Longet wavelength: gamma ray green light 1 GHz 12. In each line circle the bet anwer. (10 point) a. A 5d xy orbital ha angular momentum quantum number: b. The ubhell with exactly 5 orbital: 5 3p 4f 6d 7g c. Contain an orbital with magnetic quantum number = 3 : 7g 6d 8 9p 3f d. Shell with 25 orbital: n= 12 l = 12 m l = 25 n= 5 l = 4 e. With the ame energy a 4f: 5f 6p 4 7 5d f. Orbital ha exactly 2 planar node. 3p z 4 5f xyz 3d zx 4p x g. Orbital ha exactly 10 pherical node: 10 10p 12d 12p 15f h. Can aborb viible electromagnetic radiation: 2d 3p 1 2p 4f i. i a wavefunction for: 1 2 2p 3 3p j. i radial probability ditribution for: 2 2p 3p 3 4d

6 1. 3 Fe 2 O 3 () 6 FeO() + 3 ½ O 2 (g) H = kj 3 6 FeO() + 2 ½ O 2 (g) 2 Fe 3 O 4 () H = kj 2 3 Fe 2 O 3 () 2 Fe 3 O 4 () + ½ O 2 (g) ΔH = kj 2. 3 FeO() + ½ O 2 (g) Fe 3 O 4 () H = kj 3 Fe() + 3 ½ O 2 (g) 3 FeO() ΔH = kj 3 3 Fe() + 2 O 2 (g) Fe 3 O 4 () ΔH = kj = ΔH f [ Fe 3 O 4 () ] 3. Fe() + ½ O 2 (g) FeO() H = kj Fe(l) Fe() ΔH = 90.5 kj Fe(l) + ½ O 2 (g) FeO() ΔH = kj 1 mol Fe 1 mol FeO kj 4.?kJ = 6.44 g Fe = kj o kj given off g Fe 1 mol Fe 2 mol FeO 5.a. q J q = m ΔT o ΔT = m g J gc = C = T 2 T 1 = T C T 2 = C b. q P = g J gc (24.9 C 200.9) = = q w = m J gc (24.9 C 17.8 C) m = g water NiO () H 2 O(l) NO(g) 90.4 Ni(NO 3 ) 2 6H 2 O() NiO(g) H 2 O(g) NO 2 (g) 33.8 Ni(NO 3 ) a. Ni() + N 2 (g) + 6 O 2 (g) + 6 H 2 (g) Ni(NO 3 ) 2 6H 2 O() H = kj b. 2 Ni(NO 3 ) 2 6H 2 O() 2 NiO() + 4 NO 2 (g) + O 2 (g) + 12 H 2 O(g) { } + { } = kj 7. Proce w = +,, or 0 q = +,, or 0 a. H 2 O(l,100 C) H 2 O(g,100 C) + b. Na(l) Na() 0 c. NO(g) + ½ O 2 (g) NO 2 (g) + d. O 2 (g, 1atm, 25 C) O 2 (g, 2 atm, 0 C) + e. Cu(,100 C) Cu(, 350 K) 0

7 k 8. E = E 3 E 8 = k = k = 2.179x10 18 J ( ) = x10 19 J = hν Photon emitted ha energy x E x10 J J ν = = 3.140x h 6.626x10 J 9. a. E K = = ½ m e u 2 = x10 28 g (4.41x10 6 m 1 J 1 kg = 8.858x10 18 J 1 kg 1000 g )2 2 m 2 b. hν = hν 0 + E K = 6.626x10 34 J 5.952x x10 18 J = 3.945x10 17 J x10 18 J = 4.831x10 17 J hν = hc λ x10 J λ = hc 6.626x10 J x10 E 4.831x10 J 34 8 m 17 = 4.11x10 9 m c. λ = 1 kg 2 34 m h 6.626x10 J g 28 6 m mu 9.109x10 g 4.41x10 1 J 1 kg = 1.649x10 10 m 10. a. H b. A c. E d. B 11. Highet energy: blue light IR Radio λ= 1 cm λ= 1 mm λ= 1 µm Lowet frequency: X-ray MW red light 12 MHz 1200 khz Longet wavelength: gamma ray green light 1 GHz 12. In each line circle the bet anwer. (10 point) a. A 5d xy orbital ha angular momentum quantum number: b. The ubhell with exactly 5 orbital: 5 3p 4f 6d 7g c. Contain an orbital with magnetic quantum number = 3 : 7g 6d 8 9p 3f d. Shell with 25 orbital: n= 12 l = 12 m l = 25 n= 5 l = 4 e. With the ame energy a 4f: 5f 6p 4 7 5d f. Orbital ha exactly 2 planar node. 3p z 4 5f xyz 3d zx 4p x g. Orbital ha 10 pherical node: 10 10p 12d 12p 15f h. Can aborb viible electromagnetic radiation: 2d 3p 1 2p 4f i. i a wavefunction for: 1 2 2p 3 3p j. i radial probability ditribution for: 2 2p 3p 3 4d X4_F13a

Exam 2. Name: Class: Date:

Exam 2. Name: Class: Date: Class: Date: Exam 2 Multiple Choice Identify the choice that best completes the statement or answers the question. Record your name on the top of this exam and on the scantron form. Record the test ID