Solution-07문제 풀이 Chapter 7.hwp

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1 <hapter 7> 3. Write the Lewis structures for the following molecules and polyatomic ions. In each case, the first atom is the central atom. (a) I (b) SiF 4 원자가전자수 : I = 7 원자가전자수 : Si = 4 = 6 F = 7 (c) BrI 3 (d) N 원자가전자수 : Br = 7 원자가전자수 : = 4 I = 7 N = Dinitrogen pentoxide, N 5, when bubbled into water can form nitric acid. Its skeleton structure has no NN or bonds. Write its Lewis structure. 9. Below are two different Lewis structures for nitrous acid (N ). Which is the better Lewis structure based only on formal charge? structure I structure II N N

2 * 형식전하 (formal charge) 구하는법 f = X (Y +Z/) X = 결합되지않은원자의원자가전자수 = 주기율표에서족번호의마지막자리수 Y = Lewis 구조에서원자에의해소유된비공유전자수 Z = Lewis 구조에서원자가공유한결합전자수 => structure I 1 N : X = 5, Y =, Z = 6. f = 5 (+6/) = 0 ( 왼쪽 ) : X = 6, Y = 4, Z = 4. f = 6 (4+4/) = 0 3 ( 오른쪽 ) : X = 6, Y = 4, Z = 4. f = 6 (4+4/) = 0 4 : X = 1, Y = 0, Z =. f = 1 (0+/) = 0 => structure II 1 N : X = 5, Y = 0, Z = 8. f = 5 (0+8/) = 1 ( 아래 ) : X = 6, Y = 6, Z =. f = 6 (6+/) = 1 3 ( 위 ) : X = 6, Y = 4, Z = 4. f = 6 (4+4/) = 0 4 : X = 1, Y = 0, Z =. f = 1 (0+/) = 0 => 형식전하측면에서 structure I 이더적합하다. 31. Predict the geometry of the following species. (a) S (b) IBr AX 의형태 AX E 3 의형태 (c) N 3 AX 3 의형태 (d) RnF 4 AX 4E 의형태

3 35. (a) octahedral (b) tetrahedral (c) seesaw (d) tetrahedral 47. There are the compounds with the formula l l l l l 1 l 3 l Which of these molecules are polar? l l l l 1 l 3 l => 1 과 3 의구조가 polar 하다. 49. Give the hybridization of the central atom in each species in Question 31. (a) S => 를중심으로 개의결합 : sp 혼성 (b) IBr => I 를중심으로 개의결합과 3개의비공유전자쌍 : sp 3 d 혼성 (c) N 3 => N 를중심으로 3개의결합 : sp 혼성 (d) RnF 4 => Rn 을중심으로 4개의결합과 개의비공유전자쌍 : sp 3 d 혼성 69. onsider vitamin. Its skeleton structure is

4 A 1 3 B (a) ow many sigma and pi bonds are there in vitamin? => 단일결합에는 1개의 sigma bond 존재하고이중결합에는 1개의 sigma bond 와 1개의 pi bond 존재하므로위의 Vitamin 는 0개의 sigma bonds 와 개의 pi bonds 가존재한다. (b) ow many unshared electron pairs are there? ( 위의 Lewis 구조로부터 ) 비공유전자쌍은 1 쌍이있다. (c) What are the approximate values of the angles marked (in blue) 1, and 3? (d) What is the hybridization of each atom marked (in red) A,B and? A : 를중심으로 개의결합과 개의비공유전자쌍 : sp 3 혼성 B : 를중심으로 1개의결합과 개의비공유전자쌍 : sp 혼성 : 를중심으로 3개의결합 : sp 혼성 75. Give the formula of an ion or molecules in which an atom of (a) N forms three bonds using sp 3 hybrid orbitals. + => N 을중심으로 4개의전자쌍이존재 : N 3, N 4 (b) N forms two pi bonds and one sigma bond.

5 => N 의삼중결합 : N, N (c) forms one sigma and one pi bond. => 의이중결합 : N (d) forms four bonds in three of which it uses sp hybrid orbitals. => 를중심으로 3개의전자쌍이존재 : 3 (e) Xe forms two bonds using sp 3 d hybrid orbitals. => Xe 를중심으로 5개의전자쌍이존재 : XeF 81. It is possible to write a simple Lewis structure for the S 4 ion, involving only single bonds, which follows the octect rule. owever, Linus Pauling and others have suggested an alternative structure, involving double bonds, in which the sulfur atom is surrounded by six electron pairs. (a) Draw the two Lewis structures. (b) What geometries are predicted for the two structures? => 둘모두 AX 4 의형태 : Tetrahedral (c) What is the hybridization of sulfur in each case? => 둘모두 S 를중심으로 4개의결합 : sp 3 (d) What are the formal charges of the atoms in the two structures? => * 첫번째구조 1 S : X = 6, Y = 0, Z = 8. f = 6 (0+8/) = + ( 네개모두동일 ) : X = 6, Y = 6, Z =. f = 6 (6+/) = 1 * 두번째구조 1 S : X = 6, Y = 0, Z = 1. f = 6 (0+1/) = 0 ( 왼쪽 ) : X = 6, Y = 6, Z =. f = 6 (6+/) = 1 3 ( 위 ) : X = 6, Y = 4, Z = 4. f = 6 (4+4/) = 0 4 ( 오른쪽 ) : X = 6, Y = 6, Z =. f = 6 (6+/) = 1 5 ( 아래 ) : X = 6, Y = 4, Z = 4. f = 6 (4+4/) = 0

6 hapter 8. Thermochemistry ( 열역학 ) 열의흐름은상태함수가아니므로어떤경로를따라진행되는가에의해결정 Endothermic (q>0) Exothermic (q<0) 열흐름의크기는 j 혹은 kj q = t = m c t = heat capacity ( 열용량 = 계의온도를 1 0 상승시키는데필요한열량, J/K ) c = specific heat ( 비열 = 순수물질 1g 을 1 0 상승시키는데필요한열량, J/g. 0 ) qre = qcal qre = cal t 일정한압력하에서는반응계의열을생성물과반응물사이의엔탈피차이로나타낼수있다. q = = 엔탈피는열함량이라불리우는일종의화학에너지 re product reac tant 발열반응 q = < 0 product < reac tant 흡열반응 q = > 0 product > reac tant < 열화학법칙 > 1. 의크기는반응물이나생성물의양에비례한다.. 한반응의 는역반응의 와부호는반대이고그크기는같다. 3. 반응에서의 값은반응이한단계로진행되거나, 여러단계를거쳐일어나거나상관없이같다. : ess s law 7. (a) what is q for the calorimeter? 열랑계내에서발생한열이얼마냐는질문이다. 열량계의열용량과온도차를곱한다. q = t = J 4 ( / ) ( ) 4 = ( / ) 4. J 4 = 8953J J 유효숫자, 4. 에맞춰 3 자 (b) what is q when 4.50g of fructose is burned? 4.50g의과당이연소될때발생하는열을구하는문제. 과당을반응으로보면열이 4 발생되었기에 q<0 이므로 J 이다. (c) what is q for the combustion of one mole fructose? 4 1mol의과당은 180.g 이므로, 4.50 g: ( J) = 180. g: x 로풀어내면, KJ 이다. 유효숫자는 4.50에맞춰 3자리이다.

7 15. (a) write the thermochemical equation for the reaction between one mole of nitrogen oxide and xygen. N + 1/ > N =57.0 kj (b) Is the reaction exothermic or endothermic? 57.0Kj of heat is evolved: 열이발생하였다 : 발열반응 : q<0: exothermic (c) Draw an energy diagram showing the path of this reaction. Reactant Product (d) What is when 5.00g of nitrogen oxide react? 5.00 N 5g은 mol 이므로 5.00 ( 57.0 KJ) = KJ (e) ow many grams of nitrogen oxide must react with an excess of oxygen to liberate 10KJ of heat? (30.01 g: 57.0 KJ = X :10.0 KJ) X= 5.6g 19. (a) write a balanced thermochemical equation for the reaction. 3 3 Sr(s)+(s)+ ( g) Sr 3( s) = kj (b) 반응된산소의몰수를구하면, 엔탈피를구할수있다. 압력과온도, 그리고부피가 나와있으므로이상기체방정식인 PV=NRT 를사용한다. 또한에너지가발하였으므로 엔탈피는마이너스값을갖는다. PV 10.00L 1atm N= = = 0.409mol RT ( K) (0.081L atm/mol K) 1.5 mol : 10kj = mol : x X= 33kj 9. 1mol의아세틸렌 ( ) 가분해할때의반응열을계산하라. 가분해하게되면, 만남게된다. 이러한반응식을얻어내기위해서는부산물인이산화탄소와물을제거하는반응식을구하자.

8 5 + + =199.5KJ + =+787KJ 1 + =+85.8KJ 이것을전부더하면, ( g) (s)+ ( g) =6.7 kj, 6.7KJ 이얻어지게된다. 37. 표준엔탈피변화, 는생성물들이갖는표준생성엔탈피의합에서반응물들이갖는표준생성엔탈피의합을빼준것과같다. 표 83에서생성엔탈피를찾고, 계수를고려하여차를구한다. (a) 58.9 kj (b) 108.5kj (c) 179.9kj 43. (a) 암모니아와반응하여질소, 염소, 플루오린화수소기체를발생한다하였으므로 0 lf 3( ) + N 3( ) N ( ) +l ( )+6F( ) = 1196 g g g g g kj (b) 6( 71.1) (lf 3의 f ) ( 46.1) = lf 3 는 lf ( g) = 169kJ 0 f (a) 프로페인연소시이산화탄소와물이생성되므로, ( g) + 5 ( g) 3 ( g) + 4 (l) 3 8 (b) = E+ ng RT, ng 는 ( 생성물기체의몰수반응물기체의몰수 ) 이다. 먼저엔탈피의변화를구하면, 3( 393.5) + 4( 85.8) ( 103.8) = 19.9kj ( 19.9 kj) + 3( )(5 + 73) kj = 1.5kj 분을걸었을때 50Kcal 가소모되므로, 10Kcal 를소모시키기위해서는, 60min : 50Kcal = x min :10Kcal, 8.8min 을걸어야한다. 69. (a) T (b) T.

9 (c) T. 결합엔탈피의정의 : 1mol의결합이기체상태에서끊어지는데필요한 (d) F (e) F 76. (a) ( ) ( 84.) = 851.5kj (b) q = t = m c t 를이용. 비열이문제에주어졌기에각질량을구하자. Al : g Fe:111.7g 이므로, KJ = [(101.96g * 0.77 j/gc) + (111.7g * 0.45j/gc)] t 가되어 상온에서반응이시작하였으므로 5를더하면 6637가된다. (c) 철의녹는점은 1535c이므로당연히녹는다. t =661이다.

10 hapter 9 Liquids and Solids 9. Mt. Mckinley in Alaska has an altitude of 0,30ft. Water( vap = 40.7kJ/mol) boils at 77 atop Mt. Mckinley. What is the normal atmospheric pressure at the summit? 답 ) ln(p /P 1 ) = vap /R[1/T 1/T 1 ] 이므로 ln(760mmg/x) = (40.7*10 3 J/mol)/(8.31J/mol K)[1/373K 1/350K] x = 30.6 = 31mmg 17. Given the following data about xenon, normal boiling point = 108 normal melting point = 11 triple point = 11 at 81mmg critical point = 16.6 at 58atm (a)onstruct an appropriate phase diagram for xenon. 답 ) (b)estimate the vapor pressure of xenon at 115 답 ) 약 500mmg 11도부터 108도까지 13도증가할때압력은 479mmg 증가했으므로비례식으로풀면 13:479=6:x x=1 1+81=50.xxx 약 500mmg

11 (c)is the density of solid Xe larger than that for liquid Xe? 답 ) yes 책 p.3 melting point 부분참고압력의증가는더욱높은밀도상의형성을유리하게한다. 이그림에서는고체액체평형선이오른쪽으로기울어져있어압력이증가할때고체는녹는점이상의온도에서도안정하게되므로고체가더욱높은밀도상을갖는다. 7.Explain in terms of forces between structural units why (a)i has a higher boiling point than Br 답 )dispersion force Br보다 I의분자크기가더큼 (b)ge 4 has a higher boiling point than Si 4 답 )dispersion force (c) has a higher melting point than 3 8 답 )ydrogen bonding (d)nal has a higher boiling point than 3 답 )Ionic Vs Molecular 33. What are ther strongest attractive forces that must be overcome to (a)boil silicon hydride Si 4 답 )dispersion forces (b)vaporize calcium chloride 답 )ionic bonds (c)dissolve l in carbon tetrachloride, l 4 답 )dispersion forces (d)melt iodine 답 )dispersion forces

12 37.f the four general types of solid, which one(s) (a)are generally lowboiling? 답 )molecular (b)are ductile and malleable? 답 )metallic (c)are generally soluble in nonpolar solvent? 답 )molecular table 9.5 참조 49. In the Lil structure shown in Figure 9.18, the chloride ions form a facecentered cubic unit cell 0.513nm on an edge. The ionic radius of l is 0.181nm? (a)along a cell edge, how much space is between the l ions? 답 )0.513nm0.181*nm= 0.151nm (b)would an Na + ion (r= 0.095nm) fit into this space? A K + ion (r= 0.133nm)? 답 )Na + r=0.190nm므로 0.151nm보다크므로 no fit K + r=0.66nm 역시 no fit 53. onsider the sodium chloride unit cell shown in Figure Looking only at the front face (five large l ions, four small Na + ions), (a)how many cubes share each of the Na + ions in this face? 답 )4개 (b)how many cubes share each of the l ions in this face? 답 ) 모서리의 l 는 8개의 cube와공유 face 가운데의 l 는 개의 cube와공유한다. 66. A flask with a volume of 10.0L contains 0.400g of hydrogen gas and 3.0g of oxygen gas. The mixture is ignited and the reaction (g) + (g)

13 goes to completion. The mixture is cooled to 7. Assuming 100% yield, (a)what physical state(s) of water is (are) present in the flask? 답 )PV= (mass/mm)rt 답 )liquid and vapor 총 3.60g의 water형성 mass = PV*MM/RT 이므로 # 물의 7 에서증기압은 6.74 mmg ( 부록1 참조 ) (6.74/760atm)(10.0L)(18.0g/mol)/(0.081L atm/mol K)(300K)= 0.57g이 liquid and vapor (b)what is the final pressure in the flask? 답 )6.74 = 6.7mmg (c)what is the pressure in the flask if 3.g of each gas is used? 답 )3.46atm 반응의질량비는 1:8:9 이므로각가스가 3.g씩사용되면산소가스는모두사용되고수소가스는.8g이남는다. 그리고 3.6g의물이생성된다. 이때의증기압이 6.74mmg이므로 6.74/760 = 0.035atm 여기에남은수소가스의압력을더해주면된다. P( ) = (.80g/.016g/mol)(0.081L atm/mol K)(300K)/10.0L = 3.4atm = 3.46atm 69. As shown in Figure 9.18, Li + ions fit into a closely packed array of l ions, but Na + ions do not. What is the value of the r cation /r anion ratio at which a cation just fits into a structure of this type? 답 )0.414 Lil 구조는각각의 l 이온이서로맞대고있으므로 1: = r cation + r anion : r anion r anion = r cation / ( ) r cation /r anion = 1 = 0.414

14 chap10. Solution 15. 1L의용액의무게는 1689g이다. 인산의무게 = g=1436g 물의무게 =1689g1436g=53g 1. (a) (b) 흡열반응이므로온도가늘어나면용해도도증가한다. 3. (a) (b) (c) g 벤젠당 0.435mol의나프탈렌존재. 나프탈렌의무게 = 순도 =

15 57.(a)BaS 4 (b) Ba + (aq)+s 4 (aq) BaS 4 (S) S 4 가더적으므로 mol의 BaS 4 를얻는다 BaS 4 의무게 = (c) 69. (a) 바닷물에녹아있는이온들이어는점을더낮게하므로어는점이낮다. (b) 설탕은차가워지면결정이된다. () 삼투압때문에혈관주사의영양액이혈액보다묽으면혈액세포안의액보다삼투압이낮아서혈액세포가파괴될때까지혈액세포안으로물이흘러들어가고너무진하면물은세포안에서흘러나와세포가죽게된다. (d) 가더많기때문에 (e) 뚜껑을열면압력이낮아져서녹아있던 가기체로나오기때문에. 77.(a) Molarity Na= (b)na 가한정되어있다면 :1.48mol Na>1.87mol Al 이한정되어있다면 :1.508mol Al>.95mol 1.87mol의 는반드시얻을수있다. (c)

16 hemistry Masterton & urley APTER (a) [A] Y 값 (b) instantaneous rate mol / L s (c) 0.10 mol / L average rate = = mol / L s 30 s (d) instantaneous rate < average rate 15. N N + (a) rate = k N [ ] (b) ( ) 1.17 mol / L min = k 0.50 mol / L k = 18.7 L/ mol min rate = k N = 18.7 L / mol min mol / L = 1.0 mol / L min (c) [ ] ( )

17 17. (a) [ ] [ ] rate = k N Br ( ) ( ) / min = 0.00 / / mol L k mol L mol L k = L mol / min (b) (c) [ ] [ ] [ ] rate = k N Br ( ) [ ] mol / L min = L / mol min mol / L Br Br = 0.15 mol / L [ ] [ ] [ ] [ ] [ ] N rate = k N Br = N [ ] mol / L min L / mol min N = N = 0.18 mol / L 4 [ N] 4 1. m n rate = k S 8 I m (a) Expt.,3 = m = Expt.3, 4 = n = n (b) rate = k S 8 I Expt.1 rate = k S 8 I 4 (c) mol / L min = k ( mol / L) ( mol / L) mol / L min k = = L/ mol min ( mol / L) ( mol / L) (d) rate = k S 8 I = L / mol min mol / L mol / L = mol / L min

18 7. m rate = k I (a) [ ] Expt.1, = m = Expt., 4 = n = rate = k I [ ] n m n (b) [ ] rate = k I Expt mol / L min = k mol / L mol / L k = 4.9 L/ mol min ( ) ( ) 5.0mL 0.100M = = 50.0mL I mL 1.00 g / ml 10% mol 34.0 = = 1.47M L (c) [ ] [ ] M rate = k = L mol mol L mol L = mol L I 4.9 / min / 1.47 / 0.36 / min 35. firstorder reactions [ A] 0 ln = kt [ A ] mg 43.mg (a) ln = 1.4 = k( hr) k = 1.65hr 1 t = ln hr k = 1.65hr = (b) 1/ 1 (c) 1 ln 1.65 hr 0.05 t = 1.8hr 1 = t

19 45. secondorder reactions 1 t1/ = k[ A] 0 (a) 1 k = =.65 L/ mol min 1.51min 0.50 mol / L ( )( ) (b) 1 1 = kt [ A] [ A] =.65t t = =.11min.65 rate = k.65 L / mol min mol / L = 0.97 mol / L min (c) [ ] ( ) k E 1 1 a ln = k1 R T1 T.8 Ea 1 1 (a) ln = 5.84 = = = 5 Ea J kj (b) ln = = 3.79 k = 44.3 k 3 k = L/ mol min (c) ln = = T = = T T = 979 K

20 8. + I I (a) f 6.44 / f 6.48 / o = 6.48kJ 6.44kJ = 9.48kJ E = 165kJ kJ 174kJ a, reverse = kj mol = kj mol (b) E / RT a, reverse kreverse Ae ( Ea Ea, reverse )/ RT 9/ ( ) 1.1 = = e = e = e = 0.33 Ea / RT k Ae k = L/ mol s = 46 L/ mol s reverse rate = k I = 46 L / mol s 0.00 mol / L = 1.8 mol / L s (c) [ ] ( ) da [ ] = ka [ ] a dt da [ ] = akdt [ A] [ A] 0 ln = akt [ A ]

7 7.1 Chapter 7 Covalent Bonding

7 7.1 Chapter 7 Covalent Bonding 7 7.2? Lewis : Lewis 7 7.3 (valance electron) # of valance e- = # of group (core electron) (Bond) (ionic bond) : metal-nonmetal (covalent bond) : nonmetal-nonmetal (metallic