1

[1] 만약 O 3 의 dipole moment 가 0 이고그구조가선형이아니라면어떤 Lewis diagram 을그 릴수있는가? ( 총 5 점 ) O O O [] 아래문제에답하시오. ( 총 10점 ) (a) KF 가중성의원자들로쪼개지는데필요한분해에너지를계산하시오. KF 의결합길이 (R e ) 는.17 Å 이고다음의값들중필요하면사용하시오. (5점) K의 1차이온화에너지는 IE 1 (K)= 418.8 kj mol -1, K의 electron affinity는 EA(K) = 48.348 kj mol -1, IE 1 (F) = 1681.0 kj mol -1, EA(F) = 38.0 kj mol -1, ε 0 = 8.854 10-1 C J -1 m -1, Coulomb energy = [Q 1 Q ]/[4πε 0 R e ], N A = 6.0 10 3 mol -1, e = 1.60 10-19 C ΔE d = [{(Q 1 Q ) (4πε o R e )} (N A 10 3 )]-{IE 1 (K)-EA(F)} = [{(1.60 10-19 C) (6.0 10 3 mol -1 )} {(4)(3.14)(8.854 10-1 C J -1 m -1 )(.17 10-10 m)(10 3 J kj -1 )} {418.8-38.0} kj mol -1 = 640 kj mol -1 91 kj mol -1 = 549 kj mol -1 (b) 실제구해진결합에너지의실험값은 498 kj mol -1 유를설명하시오. (5점) 이다. 위에서구한값과차이가나는이 1) Repulsive energy is omitted ) It was assumed that KF is purely ionic 3) Also, it was assumed that K + - F - are point charges. - 1 -

[3] Van der Waals equation of state 는다음과같이표시된다. n + a V P ( V nb) = nrt 수증기의경우 a = 5.5 atm L mol -, b = 0.03 L mol -1 이라고한다. 500 K에서 1 L의부피안에있는 1 mol의수증기가있다고가정한다. R = 0.08 L atm mol -1 K -1 로계산한다. ( 총 15점 ) (a) 압력은몇기압인가? (3점) P= {(nrt) (V-nb)} a(n V ) =[{(1mol)( 0.08 L atm mol -1 K -1 )(500K)} {(1L)-(1mol 0.03 L mol -1 )}] [5.5 atm L mol - {(1mol) (1L) }] = 41. atm -5.5 atm =35.7 atm (b) 이상기체라면몇기압이예상되는가? ( 점 ) P= nrt/v ={(1mol)( 0.08 L atm mol -1 K -1 )(500K)}/1L = 40 atm (c) 실제기체는이상기체와비교하여기체분자간의 repulsive force와 attractive force에의하여압력은각각몇기압이증가또는감소하는가? (5점) Repulsive force에의해 41. atm- 40.0 atm = 1. atm 증가 Attractive force에의해 5.5 atm 감소 (d) 실제기체에서 attractive force에의해서일부의수증기가 분자씩 cluster를형성하고 3 분자이상의 cluster는형성되지않는다고가정하면 1 mol의수증기분자중상기의조건에서몇 % 가 분자 cluster를형성하는가? (5점) 압력은충돌하는숫자에비례한다. 실제충돌하는입자의개수 = 1mol x (35.7atm/41.atm) = 0.87 mol ( 여기서 35.7 atm은 repulsive +attractive 에의해가지는실제압력이고, 41.atm은 repulsive force 만있다면가지는압력 ) x만큼의 fraction이 cluster형성하면, (1-x)+(1/x)=0.87 x=0.6 6% [4] A gas mixture contains 4.5 mol Br and 33.1 mol F. ( 총10점 ) (a) Compute the mole fraction of Br in the mixture (5점) 4.5/(4.5+33.1)=0.1 (b) The mixture is heated above 150 o C and starts to react to give BrF 5. At a certain point in the reaction,. mol of BrF 5 is present. Determine the mole fraction of Br in the mixture at that point. (5점) Br (g)+ 5F (g) BrF 5 (g) 1.1mol 5.5 mol. mol 33.1mol -5.5 mol=7.6 mol F. 3.4mol Br +7.6 mol F +. mol BrF 5 = 33. mol이존재 The mole fraction of Br is 3.4/33. =0.10 - -

[5] 다음의질문에한줄혹은두줄로간략히핵심사항만기술하시오. ( 각 점 5 = 총 10 점 ) (a) Colligative properties 의특징을기술하라. ( 점 ) 용매의특성과용질의양에만의존한다. (b) Extensive properties 두개와 Intensive properties 두개를예로드시오.( 점 ) Extensive properties: V, m, E, H Intensive properties: T, P (c) State function 인것두개와아닌것두개를예로드시오. ( 점 ) State function 인것 : E, H, P, V, T State function 아닌것 : q, w (d) 내부에너지 (Internal energy) 를이루는것하나이상예를드시오. ( 점 ) 운동에너지 (kinetic energy), 회전, 진동, 화학결합에너지, 위치에너지 (e) 단열팽창 (adiabatic expansion) 에서온도가내려가는이유를설명하시오. (점) q=o 이므로외부와의열교환이없는상태에서팽창일을하게되면내부에너지가감소하여온도가감소한다. - 3 -

[6] 1.0 mol 의단원자 (monoatomic) 이상기체로이루어진시스템에대해다음과같은 state diagram 이주어졌다고하자. 그림위의모든점은평형상태를만족시키고따라서과정 a, b, c, d 는가역적과정이다. 단원자이상기체의 molar heat capacity 는 C v = (3/)R 이다. 1 L atm = 100 J의관계식을사용하라. R = 0.0806 L atm mol -1 K -1, C p -C v = R. ( 총 35점 ) 3.0 d A a B P (atm).0 1.0 D c b C 0.0 0.0 1.0.0 3.0 V (L) (a) A 에서 B 까지과정 a 를거쳐서도달할경우열과일을구하라. (5점) w AB = -PΔV = -6 L atm = -600 J q = nc P ΔT = 5/ nrδt =.5PΔV= 15 L atm = 1500 J (b) A 에서출발하여과정 a, b, c, d 를통하여다시 A 로돌아올경우시스템에가해지는열, 일, E 를구하라. (5점) A A는원래상태 Δ E = 0 w = -(ABCD의면적 ) = -3 L atm = -300 J q = - w = 300J (c) A 에서 C 까지비가역적과정 (irreversible process) 를통해도달할경우이비가역적과정을, 가능하다면위그림위에나타내시오. (5점) 비가역과정은위의그림위에나타낼수없다 - 4 -

(d) 위 (c) 의경우시스템에가해지는열, 일, 그리고열과일의합을가능한구하라. (5 점 ) q, w 는주어진경로에의존하므로, 비가역의경우구할수없다. 그러나 Δ E = (3/) nrδt=0 ( P A V A =P C V C ) (e) A 에서 C 로가는항온과정 (Isothermal process) 에서시스템에가해지는열, 일, 열과일의합을구하라. (5점) 단열팽창에서는 w = -nrt ln(v /V 1 ) = -P 1 V 1 ln(v /V 1 ) = -(3 atm)(1.0l) ln(3/1) = -3.3 L atm = -330 J q = -w = 330 J ΔE = 0 (f) A 에서가역적인단열팽창을통해 1기압으로압력이줄었다면이때부피와온도는어떻게변화하겠는가? (5점) 이때일을구하면? (5점) 단열팽창에서는 P 1 V γ 1 = P V γ, γ = (C P /C V ) = 5/3 A의경우, P 1 = 3 atm, V 1 = 1 L; P = 1 atm, 3 = 1 V γ, V = 3 3/5 = 1.93 L 이때, P 1 V 1 =RT 1 T 1 = (P 1 V 1 )/R P V =RT T = (P V )/R ΔT =-(1/R)( P 1 V 1 - P V ) = -(1/0.0806)(3-1.9) = -13.4 K (5점) ΔE=C V ΔT= 3/R ΔT= (1.5)(0.0806)(-13.4)= -1.65 L atm = -165J q= 0 w = ΔE= -165J - 5 -

[7] ( 총 10점 ) (a) Balance the following equation for the reaction that occurs when chlorine is dissolved in basic solution: Cl (g) ClO - 3 (aq) + Cl - (aq) (5점) Step1. We solve this by writing the Cl on the left sides of two half-equations: - Cl ClO 3 Cl Cl - - Step. Cl ClO 3 Cl Cl - Step3. The first half-equations becomes - Cl + 6H O ClO 3 Step4. Now the first half-equation becomes Cl + 1OH - ClO - 3 +6H O Step5. Cl + 1OH - ClO - 3 +6H O + 10e - Cl + e - Cl - Step6. Multiply the second equation by 5 and add: Cl + 1OH - ClO - 3 +6H O + 10e - 5Cl + 10e - 10 Cl - Total: 6Cl + 1OH - ClO - 3 + 10 Cl - + 6H O Dividing this equation by gives 3Cl + 6OH - ClO 3 - + 5 Cl - + 3H O (b) If 31.66 ml of 1.306 M NaOH was used, what will be the amount (in grams) of Cl (g) that was dissolved? Molar mass of Cl = 35.5 g mol -1 (5점) 31.55 ml x 1.306 M = 0.0413 mol NaOH 3Cl + 6OH - ClO - 3 + 5 Cl - + 3H O 0.007mol 0.0413 mol 0.007 mol Cl 0.007 mol 35.5 g/mol = 1.47 g - 6 -

[8] 다음그림은물과 HCl의혼합용액의끓는점 (T b ) 을 HCl의 mole fraction에대해 plot한것이다. 다음물음에대해답하라. 물의분자량은 18 g mol -1, HCl의분자량은 36.5 g mol -1 이다. ( 총 15점 ) T b ( o C) 108.58 100-111 0 0.111 1 X HCl (a) 위곡선과아래곡선은각각무엇을의미하는가? (5점) 위곡선 : vapor의조성아래곡선 : Liquid의조성 (b) 위와같이이상용액으로부터 deviation 을보이는이유는무엇인가? (5 점 ) 물 -HCl 의 Interaction 이물 - 물, HCl-HCl 의 interaction 보다더강하기때문. (c) 물 18 g과 HCl 36.5 g 이섞여있는용액을증류하면두가지온도의비등점에서물질이증류되어나온다. 먼저어떤성분몇 g이몇도에서끓어나오고다음어떤성분이몇 g이몇도에서끓어나오나? (5점) Azeotrope n HCl /(n HO+n HCl ) = 0.111 n HCl = 0.15, since n HO = 1 HCl : 36.5g 0.15 = 4.56g ( 분리안됨 ) 36.5g -4.56g = 31.94g ( 분리됨 ) 즉, 31.94g 의 HCl 이 -111 o C 에서분리, 108.56 o C 에서물 18g 과 HCl 4.56g 의 azeotrope 형성 - 7 -

[9] Barium chloride has a freezing point of 96 o C and a freezing point depression constant of 108 K kg mol -1. If 1g of an unknown substance dissolved in 56 g of barium chloride gives a solution with a freezing point of 937 o C, compute the molar mass of the unknown, assuming no dissociation takes place. ( 총 10점 ) ΔT = 937-96 = -5 o C = -5K m = -ΔT /K f = 5K / 108 K kg mol -1 = 0.31 mol kg -1 1g solute/0.56 kg solvent = 1.35 g kg -1 1.35/0.31= 9.4 g mol -1 [10] A metal reacts with aqueous hydrochloric acid to produce hydrogen. The hydrogen (H ) is collected over water at 5 o C under a total pressure of 0.9900 atm. The vapor pressure of water at this temperature is 0.0313 atm. Calculate the mass of hydrogen per liter of wet hydrogen above the water, assuming ideal gas behavior. ( 총10점 ) P H = P tot P HO = 0.9900 0.0313 = 0.9587 atm n H = (PV/RT) n H/V= (P/RT) n H/(1.000L) = (0.9587 atm)/{(0.0806 L atm mol -1 K -1 )(98.15K)} = 0.039185 mol L -1.0 g/mol x 0.039185 mol L -1 = 0.0783 g/l = 0.08 g/l - 8 -

[1] 아래문제에답하시오 ( 각 5 점 = 10 점 ) (a) 5.00 mol 의이상기체가일정 (constant) 온도 T = 98 K 에서 10.0 atm 에 1.00 atm 으로가역적으 로 (reversibly) 팽창할때, ΔS sys, ΔS surr, ΔS tot 를구하시오. 답 ) 일정온도 ΔS sys = q rev /T q rev = -w = nrt ln(v /V 1 ) = nrt ln(p 1 /P ) = (5.00 mol) (8.314 JK -1 mol -1 )(98K) ln(10.0) = +8.5 kj ΔS sys = q rev /T = 8500J/98K = +95.7 JK -1 (3 점 ) The surroundings give up the same amount of heat of the same temperature ΔS surr = - 8500J/98K = - 95.7 JK -1 (1 점 ) ΔS tot = ΔS surr + ΔS sys = 0 (1 점 ) (b) 만약위의과정이비가역적으로 (irreversibly) 일어날경우, 외부압력이 1.00 atm에고정되어있었다면, 이때 ΔS sys, ΔS surr, ΔS tot 를구하시오. (1 L atm = 100 J) 답 ) 마찬가지로 ΔS sys = q rev /T = 8500J/98K = +95.7 JK -1 왜냐하면 S 는 state function 이고처음과마지막상태가 a) 와같으므로 V 1 = (nrt)/p 1 = [(5.00 mol)(0.0806 L atm K -1 mol -1 )(98 K)]/ (10.0 atm) = 1.L V = (nrt)/p = [(5.00 mol)(0.0806 L atm K -1 mol -1 )(98 K)]/ (1.0 atm) = 1L For the system, q irrev = -w irrev = P ext ΔV =(1.00 atm)(1l-1.l) = 110 L atm = 11.0 kj (3 점 ) ( ΔE = 0) 따라서, surrounding 에는 q = -11.0 kj ΔS surr = q/t = (-11100 J) /(98K) = -37.0 JK -1 (1 점 ) ΔS tot = ΔS surr + ΔS sys = -37.0 JK -1 + 95.7 JK -1 = 58.7 JK -1 (1 점 )

[] 다음의반응에대해 5 o C 에서의 equilibrium constant 를구하시오. 3NO(g) N O(g) + NO (g) 필요하면아래값들을사용하시오. (5 점 ) ΔH f o (5 o C) kj mol -1 S o (5 o C) J K -1 mol -1 ΔG f (5 o C) C p (5 C) kj mol -1 J K -1 mol -1 NO(g) 90.5 10.65 86.55 9.84 N O(g) 8.05 19.74 104.18 38.45 NO (g) 33.18 39.95 51.9 37.0 답 ) ΔG o = ΔG f o (N O) + ΔG f o (NO ) - 3ΔG f o (NO) = (1.0 mol)(104.18 kj mol -1 ) + (1.0 mol)(51.9 kj mol -1 ) - 3(1.0 mol)(86.55 kj mol -1 ) = - 104.18 kj ( 점 ) lnk = - ΔG o / RT = [-(-104180 J)]/ [(8.314 JK -1 mol -1 )(98.15K)] = 4.03 ( 점 ) (To keep the units of the calculation correct, the ΔG o is rewritten as -104.18 kj mol -1, where per mole signifies per mole of the reaction as it is written.) K= e 4.03 = 1.8 10 18 (1 점 ) [3] 다음반응의 equilibrium constant가 600K에서 5.5 10 6 이다. CO(g) + 3H (g) CH 4 (g) + H O(g) CH 4, H O, CO 가스를 600K에서빈 Container에채운직후 partial pressure를측정하니, 각각 1.40 atm,.30 atm, 1.60 atm이었다. 반응이일어나서평형이이루어진후의 H 가스의 partial pressure 를구하시오. (10점) ( 아래와같이문제의의도와맞게문제를풀어야부분점수있습니다 ) 답 ) K = (P CH4P HO)/ (P CO P 3 H ) CO(g) + 3H (g) CH 4 (g) + H O(g) Initial 1.60 0 1.40.30 Change +y +3y -y -y (점) Final (1.60+y) +3y 1.40-y.30-y (점) [(1.40-y)(.30-y)]/[(1.60+y)(3y) 3 ] = 5.5 10 6 Since K is so big, y should be very small, (점) [(1.40)(.30)]/[(1.60)(3y) 3 ] = 5.5 10 6 (1점) y 3 = 1.34 10-8 y=.38 10-3 (점) P H = 3y = 7.1 10-3 atm (1점)

[4] 엔자임 (Enzyme) 의촉매작용에서의 kinetics 는 Michaelis-Menten 식으로주어지게된다. 다음메 커니즘을이용하고, Steady-State approximation 을사용하여반응속도 d[p]/dt 를엔자임의최초농 도 [E] O, [S], 그리고 K m = (k -1 +k )/k 1 등을이용하여유도하라 ( 과정을반드시쓸것 ). (10 점 ) E + S k 1 k -1 ES ES k E + P 답 ) d[es]/dt = 0 = k 1 [E][S] k -1 [ES] k [ES] [E] 0 = [E] + [ES] d[es]/dt = 0 = k 1 [E] 0 [S] k 1 [ES] [S] k -1 [ES] k [ES] [ES] = {k 1 [E] 0 [S] } {k 1 [S] + (k -1 + k )} = {[E] 0 [S] } {[S] + (k -1 + k )/ k 1 } 여기서 K m = (k -1 +k )/k 1 then, [ES] = {[E] 0 [S] } {[S] + K m } d[p]/dt = k [ES] = { k [E] 0 [S] }/{[S] + K m }

[5] 어떤화학반응의 activation energy 은 50 kj mol -1 이었다. R = 8.314 J mol -1 K -1 (a) 촉매를사용하여 activation energy 를 40 kj mol -1 로낮추면 300 K 에서반응은몇배로빨라지 는가? (5 점 ) ( 답 ) 촉매를사용하지않았을때의반응속도상수 k 1, activation energy 를 E a, 촉매를사용하였을때의반응속도상수를 k 라하면, () (1) Ea ln k1 = lna - (1) RT 4Ea ln k lna - 5 = RT () -1 k Ea 50 kj mol ln = = 1 k 5RT k k 1 1 = 55.1-1 - ( 5)( 8.314 J mol K )( 300 K) = 4.01 반응속도는 55.1 배빨라진다. (b) 촉매를사용하지않고온도만변화시켜반응속도를촉매를사용했을때의반응속도와같아지기위해서는몇도에서반응을시켜야하는가? (5점) 답 ) E a ln k1 = lna - (3) RT1 E a ln k = lna - (4) RT (4) (3) k ln k 1 Ea = R 1 T 1 = 4.01 T 1 1 T 1 T = 4.01 R 1 Ea 8.314 J mol 1-1 -1 1 R 1 3 = 4.01 = 4.01 =.67 10-1 T1 Ea 300 50 kj mol T K. K -1 T = 375 K 온도를 375 K 로올리면반응속도를 55.1 배빨라진다.

필요하면아래값들을사용하시오 N A = 6.0 10 3 h = 6.661 10-34 J s, c =.9979 10 8 m s -1, 1 ev = 1.60 10-19 J, R = 8.314 J mol -1 K -1 [6] Ne 원자의광전자스펙트럼을보면 9.890 10-11 m의 x-ray를사용하였을때광전자의운동에너지가각각 383.4 ev, 105. ev, 그리고 13.0 ev에서 peak가관찰되었다. 다음에답하시오. (5 4 = 0점 ) * 다음대로한것도만점, 9.89 10-10 m로고쳐서같은방식으로풀어더현실적인값을구한경우도맞게함. ( 빨간색이 Realistic values) (a) Ne의이온화에너지를 kj/mol의단위로구하시오. ε photon = hc/λ = [(6.661 10-34 J s)(.9979 10 8 m s -1 )] [(9.890 10-11 m)( 1.60 10-19 J ev -1 )] =1536 ev (153.6eV) IE = 1536 ev 13 ev = 11304 ev (1.6eV) In kj/mol: 11304 (1.60 10-19 J ev -1 ) (6.0 10 3 mol -1 ) = 1.09 10 6 kj/mol (084 kj/mol) (b) Ne을이온화하는데필요한최대한의빛의파장을 nm의단위로구하시오. λ = hc/ IE = [(6.661 10-34 J s)(.9979 10 8 m s -1 )] {11304 (1.60 10-19 J ev -1 )} = 1.09 10-10 m = 0.109 nm (57nm) (c) 위의정보로부터 Ne 전자들의 binding energy들을구하라. (ev) BE= ε photon (KE) electron 1536 ev 13 ev = 11304 ev 1536 ev 105. ev = 11331 ev 1536 ev 383.4 ev = 115.6 ev (ε photon = hc/λ = [(6.661 10-34 J s)(.9979 10 8 m s -1 )] [(9.890 10-10 m)( 1.60 10-19 J ev -1 )] =153.6 ev 383.4 ev 383.4-153.6 ev = -870. ev 105. ev -48.4 ev 13.0 ev -1.6 ev) (d) 광전자스펙트럼의세개의 peak는각각 Ne (1s s p 6 ) 의어떤오비탈에서나오는전자에 해당될까? KE BE 383.4 ev -870. ev 1s 105. ev -48.4 ev s 13.0 ev -1.6 ev p

[7] 다음에대해간단하게답하거나간략하게설명하라. ( 각 5 점 7 = 35 점 ) (a) Lanthanide contraction The general reduction in the radii of atoms of sixth-period elements that occurs during the filling of the 4f-orbitals (or 동일주기상원자번호가증가하면원자반지름의수축이일어나며특히 transition metal경우처럼 3d 혹은 4f 오비탈을채워야하는경우, 주기가커지면서오히려원자반지름의축소가일어날수있다.) (b) 4f orbital의 radial node 수는? n=4, l=3 n-1 = 3 total nodes. Of these, three are angular (l=3) and zero is radial. (c) O가 N에비해이온화에너지가낮은이유는? 7 N 1s s p 3 8 O 1s s p 4 산소의전자배치는 p 궤도에 3 개의전자가하나씩다채워진후에 1 개의전자가더채워져쌍을이루어야하고쌍을이룬전자들간의반발력때문에산소의이온화에너지가더작다. ( 즉산소의경우, 같은공간에전자두개가있어야하는요소가이온화가되면해소가되기때문에이온에서더욱안정한형태를유지한다.) (d) Pauli Exclusion Principle Two elements with the same spin may not occupy the same point in space at the same time. (or 어떤입자건모든양자수가동일한상태는있을수없다 )

(e) Zero-point energy The energy that remains in a quantum mechanical system even at the absolute zero of temperature E vib, v = hv(v+1/) Note that even in the ground state (v=0) the vibrational energy is not 0. (or 입자의움직임에제한이가해질때그에너지는정확히 0 이될수없다. 이는불확실성의원리때문이다 ) (f) Heisenberg Uncertainty Principle The product of the uncertainties in the position and the momentum of an object must exceed a lower limit. (ΔP)(Δx) (h/4π) 즉, 어떠한경우에도입자의운동량과위치를모두정확히측정하는것은불가능하다. 측정에의한불확실성이존재. (g) De Broglie wavelength formula The wavelength of a moving body equal to planck s constant divided by the momentum of the body λ = h/ m e v = h/p ( 가장중요 ) 모든운동하는입자는파동성을지니고있다.

[8] HO is a highly reactive chemical species that plays a role in atmospheric chemistry. The rate of the gasphase reaction HO (g) + HO (g) H O (g) + O (g) is second order in [HO ], with a rate constant at 5 o C of 1.4 10 9 L mol -1 s -1. Suppose some HO with an initial concentration of.0 10-8 M could be confined at 5 o C. Calculate the concentration that would remain after 1.0 s, assuming no other reactions take place. (10점) [9] The vapor pressure of butyl alcohol (C 4 H 9 OH) at 70.1 o C is 0.1316 atm; at 100.8 o C, it is 0.563 atm. (a) Calculate the molar enthalpy of vaporization (ΔH vap ) of butyl alcohol. (5 점 ) (b) Calculate the normal boiling point of butyl alcohol. (5 점 )

[10] Suppose 1.00 mol of water at 5 o C is flash-evaporated by allowing it to fall into an iron crucible maintained at 150 o C. Calculate S for the water, and S for the iron crucible, and S tot, if c p (H O(l)) = 75.4 J K -1 mol -1 and c p (H O(g)) = 36.0 J K -1 mol -1. Take H vap = 40.68 kj mol -1 for water at its boiling point of 100 o C. (10점) ΔS water (3 점 ) ΔS iron (3 점 ) ΔS total (4 점 )