GENERAL PRINCIPLES OF CHEMISTRY - CHEM110/CHEM195 TEST 3

Transcription

1 School of Chemistry, University of KwaZulu-Natal, Westville Campus GENERAL PRINCIPLES O CHEMISTRY - CHEM110/CHEM195 TEST 3 Date: WEDNESDAY, 4 May 2011 Total marks: 25 Time: 17h45 18h30 Examiners: Ms H Govender Dr B Moodley IMPORTANT: Complete this part immediately. Surname: Initial Student No: Tutorial Day: Tutorial Group: Tutor s Name: INSTRUCTIONS: 1. Answer ALL questions. 2. or Section A which contains the multiple choice questions, write your answers on the multiple choice answer sheet and follow the instructions given in the question. 3. Calculators may be used but all working must be shown. 4. The pages of this test must not be unpinned. 5. Your answers for Section B must be written on the question paper in the spaces provided. The left-hand pages may be used for extra space or for rough work. 6. Marks will be deducted for the incorrect use of significant figures and the omission of units. 7. You must write legibly in black or blue ink. Pencils and Tipp-Ex are not allowed. 8. This test consists of 7 pages. Please check that you have them all. 9. A data sheet and a periodic table are also provided. Question No Total Mark

2 School of Chemistry, University of KwaZulu-Natal, Westville Campus, Durban SECTION A - Multiple Choice Questions or each of the following questions, select the correct answer from the list provided. There is only one correct answer for each question. Indicate your answer on the multiple choice answer sheet provided. Make a dark heavy mark with HB pencil that fills the block of the appropriate letter completely. 1. The ground-state electron configuration of which element is [Kr]5s 1 4d 5. A Nb B C D Mo Cr Mn 2. The condensed ground state electron configuration for K + is and the ion is A K + = [Ar]4s 1, diamagnetic. B C D K + = [Ar]4s 1, paramagnetic. K + = [Ar], diamagnetic. K + = [Ar], paramagnetic. 3. Which of the following is a valid set of four quantum numbers (n, l, ml, ms)? A 2, 2, -1, -½ B 1, 0, 1, +½ C 1, 1, 0, -½ D 2, 1, 0, +½ 2

3 School of Chemistry, University of KwaZulu-Natal, Westville Campus, Durban 4. Which TWO bonds are most similar in polarity? A B C D O and Cl. B and Cl Al Cl and I Br I Br and Si Cl 5. In the Lewis structure of BeCl 2 why are we unable to draw double bonds between the Be atom and the Cl atoms in BeCl 2? A That would give positive formal charges to the chlorine atoms and a negative formal charge to the beryllium atom. B C D There aren't enough electrons. That would result in more than eight electrons around beryllium. That would result in more than eight electrons around each chlorine atom. [5] End of Section A 3

4 QUESTION 1 School of Chemistry, University of KwaZulu-Natal, Westville Campus, Durban SECTION B (Answer in ink on the question paper) Calculate and show which of the following photons has the highest energy? (a) A photon with a frequency of s 1 (b) A photon with a wavenumber of cm 1 (3) a) ν = s 1 E = hν = J s s 1 = J b) ν = cm 1 E = hc λ since ν = λ 1 so E = hcν E = J s cm s cm 1 E = J 4

5 QUESTION 2 School of Chemistry, University of KwaZulu-Natal, Westville Campus, Durban Draw THREE possible Lewis structures for the fulminate ion, CNO, where nitrogen is the central atom and use formal charges to show which structure is the most plausible. (5) C = 4 N = 5 O = 6 Total = = 16 C-N-O 16 4 = 12 Distribution of remaining electrons can be done in 2 possible ways C N O OR C N O To achieve the octet, the following 3 structures can be obtained: C N O C N O C N O C (C) = 4-(2+3) = -1 = 4-(4+2) = -2 = 4-(6+1) = -3 C (N) = 5-(0+4) = +1 = 5-(0+4) = +1 = 5-(0+4) = +1 C (O) = 6-(6+1) = -1 = 6-(4+2) = 0 = 6-(2+3) = C N O C N O C N O Structure (a) is the most suitable structure because the negative charge is on the most electronegative atom, O, and the most electropositive atom has the least non-zero charge. 5

6 School of Chemistry, University of KwaZulu-Natal, Westville Campus, Durban QUESTION 3 Draw the Lewis structure for Se 4 and give its molecular geometry and bond angles. (2½) Se = 6 = 7 Total = 6+(4x7) = = 26 Se Se Molecular geometry is seesaw and bond angles are 90 and 120. QUESTION 4 (a) Arrange the following substances in increasing order of melting point. MgO Ne C 3 H 8 CH 3 CH 2 OH CH 2 OHCHOHCH 2 OH Ne < C 3 H 8 < CH 3 CH 2 OH < CH 2 OHCHOHCH 2 OH < MgO lowest highest (b) Give an explanation for your answer in (a). (3) MgO has ionic bonding which is an intramolecular force which is the strongest bond and would have the highest melting point. Both CH 3 CH 2 OH and CH 2 OHCHOHCH 2 OH have hydrogen bonding but CH 2 OHCHOHCH 2 OH has more OH substituents and would therefore have more hydrogen bonds and a higher melting point than CH 3 CH 2 OH. C 3 H 8 has London dispersion forces and has a higher molar mass than Ne and would therefore have a higher melting point than Ne. 6

7 QUESTION 5 School of Chemistry, University of KwaZulu-Natal, Westville Campus, Durban A piece of titanium metal with a mass of 20.8 g is heated in boiling water to 99.5 C and then dropped into a coffee-cup calorimeter containing 75.0 g of water at 21.7 C. When thermal equilibrium is reached, the final temperature is 24.3 C. Calculate the specific heat capacity of titanium. (Specific heat of water is J g -1 K -1 ) (3) q calor = - q metal m x s x T = - (m x s x T) (75.0 g)(4.184 J g -1 K -1 )(24.3 C 21.7 C) = - [(20.8 g)(s metal )(24.3 C 99.5 C) (75.0 g)(4.184 J g -1 K -1 )(2.6 K) = - [(20.8 g)(s metal )(-75.2 K)] J = - ( x s metal ) s metal = J/ g K = 0.52 J g -1 K -1 QUESTION 6 The first step in the production of nitric acid from ammonia involves the oxidation of NH 3. 4NH 3 (g) + 5O 2 (g) 4NO(g) + 6H 2 O(g) Calculate the standard enthalpy change for this reaction if the standard enthalpies of formation are as follows: H f [NH 3 (g)] = kj mol -1 H f [NO(g)] = kj mol -1 H f [H 2 O(g)] = kj mol -1 (2½) H f = [4(90.29) + 6( )] [4( ) + 0] = [ ] [-183.6] = kj mol -1 7

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