일반화학 I 기말고사 Useful P

일반화학 I 기말고사 2004. 12. 14. ------------------------------------------------------------------------------------------------------------------- Useful Physical Constants and Equations Universal gas constant R = 0.08206 L atm / K mol = 8.314 J / K mol) Plank s constant h = 6.626 x 10-34 Js Mass of electron m e = 9.109 x 10-31 kg Avogadro s number N A = 6.022 x 10 23 Speed of light c = 2.9979 x 10 8 m s -1 Permittivity of the vacuum ε 0 = 8.854 x 10-12 C 2 J -1 m -1 Electron charge q e = 1.062 x 10-19 C Plank s formula E = hν de Broglie s wavelength λ = h / mv = h / p Bohr s formula E n = - (2.178 x 10-18 J)(Z 2 /n 2 ) G o = H o - T S o Arrhenius Equation k = A e -Ea/RT 1. (10 점 ) 다음물음에간단히답하시오. (a) (2 점 ) 20 Ca 의 ground-state electronic configuration 을 1s orbital 부터쓰시오. (b) (2점) many electron atom에서 3s, 3p, 3d orbital은 nondegenerate 상태이다. 그이유를설명하시오. (c) (2 점 ) 4p z orbital 의 radial node 와 angular node 의수는각각몇개인가? (d) (4점) O가 N에비해이온화에너지가낮은이유를설명하시오. 7 N 1s 2 2s 2 2p 3 8 O 1s 2 2s 2 2p 4 1

2. (10 점 ) 수업시간에는주로 ideal gas (PV = nrt) 의경우에대하여 molar Gibbs energy 에관한표현을배웠다. 그렇다면다음식으로표현되는 non-ideal gas 에관해서 생각해보자. PV = nrt + nbp (constant b 0, b = 0 일때는 ideal gas) 이때 molar Gibbs energy 와압력 (P) 의상관관계는다음과같다. ΔG o vap(g) = - RT ln P - b(p-1) 이상기체의경우와달리위식에나타난 non-ideal gas의 nbp term 이 vapor pressure와온도의상관관계 에어떤영향을미칠것인지를식을이용하여설명하시오. 3. (5 점 ) 물 (H 2 O) 의 entropy of fusion 과 vaporization 은아래와같다. ΔS o fus = 20.5 J K -1 mol -1 ΔS o vap = 109.1 J K -1 mol -1 ΔS o vap 값이 ΔS o fus 보다훨씬큰값을갖는이유를설명하시오. 4. (8점) Constant pressure 1 atm 하에서다음각반응이자발적으로일어나는온도범위를구하시오. 필요하면아래값들을사용하시오. o ΔH f (25 o C) kj mol -1 o S (25 o C) J K -1 mol -1 o ΔG f (25 o C) C p (25 C) kj mol -1 J K -1 mol -1 Fe (s) 0 27.28 0 25.10 O 2 (g) 0 205.03 0 29.36 Fe 2 O 3 (s) - 824.2 87.40-742.2 103.85 NH 4 NO 3 (s) - 365.56 151.08-184.02 139.3 N 2 O (g) 82.05 219.74 104.18 38.45 H 2 O (g) - 241.82 188.72-228.59 35.58 (a) (4 점 ) 4 Fe (s) + 3 O 2 (g) 2 Fe 2 O 3 (s) (b) (4 점 ) NH 4 NO 3 (s) N 2 O (g) + 2 H 2 O (g) 2

5. (12점) Suppose H 2 (g) and I 2 (g) are sealed in a flask at T = 400K with partial pressures P(H 2 )= 1.320 atm and P(I 2 ) = 1.140 atm. At this temperature H 2 and I 2 do not react rapidly to form HI(g), although after a long enough time they would produce HI(g) at its equilibrium partial pressure. Suppose, instead, that the gases are heated in the sealed flask at 600K, a temperature at which they quickly reach equilibrium: H 2 (g) + I 2 (g) 2HI (g) The equilibrium constant for the reaction is 92.6 at 600K: P 2 (HI) / P(H 2 )P(I 2 ) = 92.6 (a) (4 점 ) What are the equilibrium values of P(H 2 ), P(I 2 ), and P(HI) at 600K? (b) (4 점 ) What percentage of I 2 originally present has reacted when equilibrium is reached? (c) (4점) Enough H 2 is added to increase its partial pressure to 2.000 atm at 600K before any reaction takes place. The mixture then once again reaches equilibrium at 600K. What are the final partial pressures of the three gases? 6. (10점) Dinitrogen tetraoxide (N 2 O 4 ) 는실온에서다음과같은 first-order 반응을따라분해된다. N 2 O 4 (g) 2 NO 2 (g), - d[n 2 O 4 ] / dt = k [N 2 O 4 ] 30 o C 에서의 k = 5.1 x 10 6 s -1 이고, 이반응의 activation energy 는 54.0 kj mol -1 이다. (a) (5점) 30 o C 에서, N 2 O 4 (g) 의 partial pressure가 0.10 atm에서 0.010 atm으로줄어드는데필요한시간 (in seconds) 을계산하시오. (b) (5 점 ) 300 o C 일때 (a) 의조건을만족시키는데걸리는시간을계산하시오. 3

7. (15 점 ) Ce (IV) 에의한 Fe (II) 의산화반응은아래와같고, 각각의반응물을초기농 도를변화시키면서반응속도를측정한값은아래표와같다. Ce 4+ (aq) + Fe 2+ (aq) Ce 3+ (aq) + Fe 3+ (aq) [Ce 4+ ] (mol L -1 ) [Fe 2+ ] (mol L -1 ) Rate (mol L -1 s -1 ) 1.1 X 10-5 1.8 X 10-5 2.0 X 10-7 1.1 X 10-5 2.8 X 10-5 3.1 X 10-7 3.4 X 10-5 2.8 X 10-5 9.5 X 10-7 (a) (5 점 ) 이반응의 rate 는어떻게표현되는가? (b) (5 점 ) rate constant k 를구하고, 단위를표시하시오. (c) (5점) [Ce 4+ ] = 2.6 x 10-5 M, 구하시오. [Fe 2+ ] = 1.3 x 10-5 M 에서의용액의초기반응속도를 8. (10점) Silver ion (Ag 2+ ) 에의한다음산화-환원촉매반응의 mechanism은다음과같다. Reactive intermediate Ag 2+ 의농도에대하여 steady-state approximation ( 정류상태근사법 ) 을써서이반응 mechanism에대한 rate law를구하시오. Tl + (aq) + 2 Ce 4+ (aq) Tl 3+ (aq) + 2 Ce 3+ (aq) (Mechanism) k 1 Ag + (aq) + Ce 4+ (aq) Ag 2+ (aq) + Ce 3+ (aq) (fast) k -1 k 2 Tl + (aq) + Ag 2+ (aq) Tl 2+ (aq) + Ag + (aq) (slow) k 3 Tl 2+ (aq) + Ce 4+ (aq) Tl 3+ (aq) + Ce 3+ (aq) (fast) 4

9. (10 점 ) Hydrogen atom 의 3s 와 3p wave function 은다음과같다. σ = Zr / a 0 이고, a 0 = 0.529 X 10-10 m 일때, (a) (5 점 ) 3s wave function 에대해서 node 가존재하는 r 값을구하시오. (b) (5 점 ) 3p wave function 에대해서 node 가존재하는 r 값을구하시오. 10. (10점) 탄소-탄소이중결합 (C=C) 의길이는약 1.34 Å 이다. 이이중결합에서의 electron 의 motion은 particle in a one-dimensional box 의경우로간주될수있다. d 2 Ψ dx 2 + 2m e E Ψ (x) h 2 = 0 h = h 2π 0 x a (a = 1.34 Å) (a) (5점) Calculate the energy of an electron in each of its three lowest allowed states (E 1, E 2, and E 3 ) if it is confined to move in a one-dimensional box of length 1.34 Å. (b) (5점) Calculate the wavelength (λ) of light necessary to excite the electron from its ground state to the first excited state. ( 한학기동안수고하셨습니다.) 5

일반화학 I 기말고사 2004. 12. 14. ------------------------------------------------------------------------------------------------------------------- Useful Physical Constants and Equations Universal gas constant R = 0.08206 L atm / K mol = 8.314 J / K mol) Plank s constant h = 6.626 x 10-34 Js Mass of electron m e = 9.109 x 10-31 kg Avogadro s number N A = 6.022 x 10 23 Speed of light c = 2.9979 x 10 8 m s -1 Permittivity of the vacuum ε 0 = 8.854 x 10-12 C 2 J -1 m -1 Electron charge q e = 1.062 x 10-19 C Plank s formula E = hν de Broglie s wavelength λ = h / mv = h / p Bohr s formula E n = - (2.178 x 10-18 J)(Z 2 /n 2 ) G o = H o - T S o Arrhenius Equation k = A e -Ea/RT 1. (10 점 ) 다음물음에간단히답하시오. (a) (2 점 ) 20 Ca 의 ground-state electronic configuration 을 1s orbital 부터쓰시오. 답 ) 20 Ca = 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 (b) (2점) many electron atom에서 3s, 3p, 3d orbital은 nondegenerate 상태이다. 그이유를설명하시오. 답 ) They are shielded by the core electron to different degree. (c) (2 점 ) 4p z orbital 의 radial node 와 angular node 의수는각각몇개인가? 답 ) 2 radial nodes & 1 angular node. 1

(d) (4 점 ) O 가 N 에비해이온화에너지가낮은이유를설명하시오. 7 N 1s 2 2s 2 2p 3 8 O 1s 2 2s 2 2p 4 답 ) 산소의전자배치는 p궤도에 3개의전자가하나씩다채워진후에 1개의전자가더채워져쌍을이루어야하고쌍을이룬전자들간의반발력때문에산소의이온화에너지가더작다. ( 즉산소의경우, 같은공간에전자두개가있어야하는요소가이온화가되면해소가되기때문에이온에서더욱안정한형태를유지한다.) 2. (10 점 ) 수업시간에는주로 ideal gas (PV = nrt) 의경우에대하여 molar Gibbs energy 에관한표현을배웠다. 그렇다면다음식으로표현되는 non-ideal gas 에관해서 생각해보자. PV = nrt + nbp (constant b 0, b = 0 일때는 ideal gas) 이때 molar Gibbs energy 와압력 (P) 의상관관계는다음과같다. ΔG o vap(g) = - RT ln P - b(p-1) 이상기체의경우와달리위식에나타난 non-ideal gas의 nbp term 이 vapor pressure와온도의상관관계 에어떤영향을미칠것인지를식을이용하여설명하시오. 답 ) ΔG o vap = ΔH o vap - TΔS o vap = - RT ln P - b(p-1) (ΔH o vap, ΔS o vap f (T)) RT 1 ln P 1 b(p 1 1) = ΔH o vap T 1 ΔS o vap 으로부터 R ln P 1 b(p 1 1) / T 1 = ΔH o vap /T 1 ΔS o vap ------ (a) R ln P 2 b(p 2 1) / T 2 = ΔH o vap /T 2 ΔS o vap ------ (b) (a) (b) : R ln (P 1 / P 2 ) b[{(p 1 1) / T 1 } {(P 2 1) / T 2 }] = ΔH o vap [(1/T 1 ) (1/T 2 )] ln (P 1 / P 2 ) = (ΔH o vap / R) [(1/T 1 ) (1/T 2 )] (b/r)[{(p 1 1) / T 1 } {(P 2 1) / T 2 }] (van t Hoff Equation for ideal gas) + (extra term) The extra term clearly changes P vap (T). 2

3. (5 점 ) 물 (H 2 O) 의 entropy of fusion 과 vaporization 은아래와같다. ΔS o fus = 20.5 J K -1 mol -1 ΔS o vap = 109.1 J K -1 mol -1 ΔS o vap 값이 ΔS o fus 보다훨씬큰값을갖는이유를설명하시오. 답 ) The S liq for water is smaller than in other liquids because it has smaller degree of freedom due to the hydrogen-bonding. 4. (8점) Constant pressure 1 atm 하에서다음각반응이자발적으로일어나는온도범위를구하시오. 필요하면아래값들을사용하시오. o ΔH f (25 o C) kj mol -1 o S (25 o C) J K -1 mol -1 o ΔG f (25 o C) C p (25 C) kj mol -1 J K -1 mol -1 Fe (s) 0 27.28 0 25.10 O 2 (g) 0 205.03 0 29.36 Fe 2 O 3 (s) - 824.2 87.40-742.2 103.85 NH 4 NO 3 (s) - 365.56 151.08-184.02 139.3 N 2 O (g) 82.05 219.74 104.18 38.45 H 2 O (g) - 241.82 188.72-228.59 35.58 (a) (4 점 ) 4 Fe (s) + 3 O 2 (g) 2 Fe 2 O 3 (s) 답 ) ΔH o = 2 x (-824.2 kj mol -1 ) 4 x 0 3 x 0 = - 1648.4 kj mol -1 ΔS o = 2 x (87.40 J K -1 mol -1 ) 4 x (27.28) 3 x (205.03) = - 549.41 J K -1 mol -1 Boyle Temperature T* = ΔH o / ΔS o = (- 1648.4 x 10 3 J mol -1 ) / (- 549.41 J K -1 mol -1 ) 3000 K G o = H o - T S o 에서 ΔH o < 0, ΔS o < 0 이므로, 자발적반응이일어날조건 ( G o < 0) 은 T < T* 이반응은 (0 < ) T < 3000K 범위에서자발적이다. (b) (4 점 ) NH 4 NO 3 (s) N 2 O (g) + 2 H 2 O (g) 답 ) ΔH o = (82.05 kj mol -1 ) + 2 x (-241.82) (- 365.56) = - 36.03 kj mol -1 ΔS o = (219.74 J K -1 mol -1 ) + 2 x (188.72) 151.08 = 446.1 J K -1 mol -1 G o = H o - T S o 에서 ΔH o < 0, ΔS o > 0 이면항상 G o < 0 이다. 이반응은모든 temperature range ( 온도범위 ) 에서자발적이다. 3

5. (12점) Suppose H 2 (g) and I 2 (g) are sealed in a flask at T = 400K with partial pressures P(H 2 )= 1.320 atm and P(I 2 ) = 1.140 atm. At this temperature H 2 and I 2 do not react rapidly to form HI(g), although after a long enough time they would produce HI(g) at its equilibrium partial pressure. Suppose, instead, that the gases are heated in the sealed flask at 600K, a temperature at which they quickly reach equilibrium: H 2 (g) + I 2 (g) 2HI (g) The equilibrium constant for the reaction is 92.6 at 600K: P 2 (HI) / P(H 2 )P(I 2 ) = 92.6 (a) (4점) What are the equilibrium values of P(H 2 ), P(I 2 ), and P(HI) at 600K? 답 ) refer to Chapter 9 example 9.10 (p289) (b) (4점) What percentage of I 2 originally present has reacted when equilibrium is reached? 답 ) refer to Chapter 9 example 9.10 (p290) (c) (4점) Enough H 2 is added to increase its partial pressure to 2.000 atm at 600K before any reaction takes place. The mixture then once again reaches equilibrium at 600K. What are the final partial pressures of the three gases? 답 ) refer to Chapter 9 example 9.15 (p295) 6. (10점) Dinitrogen tetraoxide (N 2 O 4 ) 는실온에서다음과같은 first-order 반응을따라분해된다. N 2 O 4 (g) 2 NO 2 (g), - d[n 2 O 4 ] / dt = k [N 2 O 4 ] 30 o C 에서의 k = 5.1 x 10 6 s -1 이고, 이반응의 activation energy 는 54.0 kj mol -1 이다. (a) (5점) 30 o C 에서, N 2 O 4 (g) 의 partial pressure가 0.10 atm에서 0.010 atm으로줄어드는데필요한시간 (in seconds) 을계산하시오. 답 ) -dc/dt = kt 로부터 ln (c/c 0 ) = - kt ln (0.010 atm / 0.10 atm) = - (5.1 x 10 6 s -1 ) t t = 4.5 x 10-7 s 4

(b) (5점) 300 o C 일때 (a) 의조건을만족시키는데걸리는시간을계산하시오. 답 ) ln (k 2 / k 1 ) = (E a / R) [(1/T 2 ) (1/T 1 )] ln [ k 300 / (5.1 x 10 6 )] = [(54 x 10 3 J mol -1 ) / (8.314 J K -1 mol -1 )] / [(1 / 573.15 K ) (1 / 303.15 K )] k 300 = 1.2 x 10 11 s -1 ln (0.010 atm / 0.10 atm) = - (1.2 x 10 11 s -1 ) t t = 1.9 x 10-11 s 7. (15점) Ce (IV) 에의한 Fe (II) 의산화반응은아래와같고, 각각의반응물을초기농도를변화시키면서반응속도를측정한값은아래표와같다. Ce 4+ (aq) + Fe 2+ (aq) Ce 3+ (aq) + Fe 3+ (aq) [Ce 4+ ] (mol L -1 ) [Fe 2+ ] (mol L -1 ) Rate (mol L -1 s -1 ) 1.1 X 10-5 1.8 X 10-5 2.0 X 10-7 1.1 X 10-5 2.8 X 10-5 3.1 X 10-7 3.4 X 10-5 2.8 X 10-5 9.5 X 10-7 (a) (5점) 이반응의 rate는어떻게표현되는가? 답 ) 1) [Ce 4+ ] 가 constant 일때, [Fe 2+ ] 가 1.6배증가함에따라반응속도도 1.6배증가 [Fe 2+ ] 에대하여 first-order. 2) [Fe 2+ ] 가 constant 일때, [Ce 4+ ] 가 3.1배증가함에따라반응속도도 3.1배증가 [Ce 4+ ] 에대하여 first-order. rate = k [Fe 2+ ][Ce 4+ ] (b) (5점) rate constant k 를구하고, 단위를표시하시오. 답 ) 2.0 x 10-7 mol L -1 s -1 = k (1.8 x 10-5 mol L -1 )(1.1 x 10-5 mol L -1 ) k = 1.0 x 10 3 L mol -1 s -1 (c) (5점) [Ce 4+ ] = 2.6 x 10-5 M, [Fe 2+ ] = 1.3 x 10-5 M 에서의용액의초기반응속도를구하시오. 답 ) rate = k [Fe 2+ ][Ce 4+ ] = (1.0 x 10 3 L mol -1 s -1 ) (1.3 x 10-5 M )(2.6 x 10-5 M) = 3.4 x 10-7 mol L -1 s -1 5

8. (10 점 ) Silver ion (Ag 2+ ) 에의한다음산화 - 환원촉매반응의 mechanism 은다음과같 다. Reactive intermediate Ag 2+ 의농도에대하여 steady-state approximation ( 정류상태 근사법 ) 을써서이반응 mechanism 에대한 rate law 를구하시오. Tl + (aq) + 2 Ce 4+ (aq) Tl 3+ (aq) + 2 Ce 3+ (aq) (Mechanism) k 1 Ag + (aq) + Ce 4+ (aq) Ag 2+ (aq) + Ce 3+ (aq) (fast) k -1 k 2 Tl + (aq) + Ag 2+ (aq) Tl 2+ (aq) + Ag + (aq) (slow) k 3 Tl 2+ (aq) + Ce 4+ (aq) Tl 3+ (aq) + Ce 3+ (aq) (fast) 답 ) Steady-state approximation 에따라, d [Ag 2+ ] / dt = k 1 [Ag + ][Ce 4+ ] k -1 [Ag 2+ ][Ce 3+ ] k 2 [Tl + ][Ag 2+ ] = 0 로부터 [Ag 2+ ] = k 1 [Ag + ][Ce 4+ ] / (k -1 [Ce 3+ ] + k 2 [Tl + ]) rate = k 2 [Tl + ][ Ag 2+ ] = k 1 k 2 [Tl + ][Ag + ][Ce 4+ ] / (k -1 [Ce 3+ ] + k 2 [Tl + ]) 9. (10 점 ) Hydrogen atom 의 3s 와 3p wave function 은다음과같다. σ = Zr / a 0 이고, a 0 = 0.529 x 10-10 m 일때,, (a) (5점) 3s wave function에대해서 node가존재하는 r값을구하시오. 답 ) A node occurs when R(3p) = 0. The function equals zero when 6σ σ 2 = 0, which means it equals zero when σ = 6. Therefore, the node is at r = 6 a 0 = 6 (0.529 x 10-10 m) = 3.17 x 10-10 m = 3.17 Å (b) (5점) 3p wave function에대해서 node가존재하는 r값을구하시오. 답 ) Node occurs at the roots of 27-18σ + 2σ 2 = 0 r / a 0 = σ = {18 ± (324 8 x 27)} / 4 = 1.902 & 7.098 r 1 = 1.902 (0.529 x 10-10 m) = 1.01 Å r 2 = 7.098 (0.529 x 10-10 m) = 3.75 Å 6

10. (10 점 ) 탄소 - 탄소이중결합 (C=C) 의길이는약 1.34 Å 이다. 이이중결합에서의 electron 의 motion 은 particle in a one-dimensional box 의경우로간주될수있다. d 2 Ψ dx 2 + 2m e E Ψ (x) h 2 = 0 0 x a (a = 1.34 Å) (a) (5 점 ) Calculate the energy of an electron in each of its three lowest allowed states (E 1, E 2, and E 3 ) if it is confined to move in a one-dimensional box of length 1.34 Å. h = h 2π (b) (5 점 ) Calculate the wavelength (λ) of light necessary to excite the electron from its ground state to the first excited state. ( 한학기동안수고하셨습니다.) 7