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Transcription:

1

2

3 2009年度 環境地球化学 大河内 温度上昇による炭酸水の発泡 気泡 温度が高くなると 溶けきれなくなった 二酸化炭素が気泡として出てくる

4 2009年度 環境地球化学 圧力上昇による炭酸水の発泡 栓を開けると 瓶の中の圧力が急激に 小さくなるので 発泡する 大河内

5 CO 2 K H CO 2 H 2 O K H + 1 HCO 3- K 2 H + CO 3 2- (M) [CO 2 H 2 O] = K H p CO2 K H (M/atm) p CO2 CO 2 (atm) P CO2 =360 ppm K H =0.04 M/atm K 1 =4.0 x 10-7 M ph = 5.6

6 z P (atm) PV z = RT z=1 = 0 = 0

7 N 2 0.78 78 % O 2 0.21 21 % Ar 0.0093 0.93 % CO 2 380 10-6 380 ppm Ne 18 10-6 18 ppm He 5.2 10-6 5.2 ppm CH 4 1.5 10-6 1.5 ppm

8

9 X (mixing ratio) C X C X = (moles of X) / (mole of air) = (volume of X) / (volume of air) A B A B 9 A/B =0.5 A/B =0.5

10 (%) = (ml) X 102 (ml) (ml) (ppm v ) = X 106 (ml) ppm = part per million = 10-6 (ml) (ppb v ) = X 109 (ml) ppb = part per billion = 10-9

11 X (number density) n X n X = X (molecules) (cm 3 ) X + Y P + Q d dt [X] = k[x][y] = ( ) (1 )

12 X n X C X n X = C x n a n a PV = NRT n a = A vn R(8.314 J mol -1 K -1 ) A v (6.022 x 10 23 molecules mol -1 ) V

13 n = A v P a RT n = A v P X RT C X P T X

14 (mass concentration) X M X (kg mol -1 ) ρ X ρ X = n X M X A v

15 partial pressure P X P X X C X P X = C X P n X P X = n X A v RT

16

17 P B =K HB X B P A =K HA X A P P * A P B A P B =P B* X B X B P B* B P A P A =P A* X A X B 0 A B X B 1 B A

18 A ( ) A (g) + H 2 O A H 2 O A (g) A(aq) K = [A(aq)] [A(g)] [A(aq)] = K H p A K H (M atm -1 )

19 O 2 O 3 CO 2 SO 2 NO NO 2 HNO 2 HNO 3 NH 3 K H (M atm -1 ) at 298 K 1.3 10-3 1.1 10-2 3.4 10-2 1.2 10 1 1.9 10-3 1.0 10-2 4.9 10 1 2.1 10 5 6.2 10 1

20 HNO 3 CH 3 COOH HCl OH SO 2 H 2 S O 3 NO O 2 10 6 10 5 10 4 10 3 10 2 10 1 10 0 10-1 10-2 10-3 H 2 O 2 HCOOH NH 3 DMS CO 2 NO 2 N 2 K H (M atm -1 )

21 ex. etc p HA HA (g) K H HA (aq) H + + A - K a HA (g) HA (aq) [HA(aq)] = K p H HA HA (aq) H + + A - [H + ][A ] K = a [HA]

22 [HA(aq)] tot [HA(aq)] tot = = [HA(aq)] + [A (aq)] [HA(aq)] 1+ K a [H + ] [HA(aq)] tot 1+ K a [H + ] = K H p HA K eff

23 CO 2 + H 2 O ( ) CO 2 H 2 O (aq) [CO 2 H 2 O] = K H (CO 2 ) p CO2 CO 2 H 2 O (aq) HCO - 3 + H + (aq) K = [HCO ][H + ] 3 a1 [CO H O] 2 2 HCO 3 - CO 3 2- + H + (aq) K = [CO 2 ][H + ] 3 a2 [HCO ] 3

24 [CO 2 (aq)] tot = [CO 2 H 2 O(aq)]+ [HCO 3 (aq)]+ [CO 2 3 (aq)] = [CO 2 H 2 O(aq)] 1+ K a1 [H + ] + K a1k a 2 [H + ] 2 K H (CO 2 ) 1+ K a1 [H + ] + K a1k a 2 = [H + ] 2 p CO2 K eff (CO 2 )

25 K eff (CO 2 ) = K H (CO 2 ) 1+ K a1 [H + ] + K a1k a 2 [H + ] 2 K eff (CO 2 ) (M atm -1 ) [CO 2 ] tot (µm) ph 5 K eff (CO 2 ) ph 5 K eff (CO 2 ) K H (CO 2 ) K H (CO 2 )

26

27 aa + bb cc + dd v = 1 f a = 1 c d[a] = 1 d[b] dt b dt d[c] = 1 d[d] dt d dt = k [A] m [B] n f [A], [B], [C], [D] A, B, C, D k f m+n

28 1 a d[a] dt = k d[a] = k a = k dt t = 0[A] 0 [A] = kt + [A] 0

29 1 a d[a] dt d[a] = k 1 a[a] = k 1 [A] dt = k 1 [A] t = 0[A] 0 t = 0 t [A ] [A ] 0 d[a] [A] t 0 = k 1 dt

30 ln [A] [A] 0 = k 1 t [A] = [A] 0 e k 1t ln[a] = kt + ln[a] 0

31 [A] 0 t 1/2 [A] 0 [A] 0 e k 1t 1/2 2 = t 1/ 2 = ln(2) k = τ ln(2) = 0.693τ t 1/2

32 A B t = 0 a 0 t = t a - x x A a t = 0 t dx dt = k 1 (a x) t = 0 t x t dx = k (a x) 1 dt 0 0

33 (a-x) = X - dx = dx ln t dx = k X 1 dt a x a (a x) a = k 1 t ln(a x) = k 1 t + lna ln (a x) t k 1 0 log (a-x) log a t k 1

34 v f = k 2 [A] 2 A B t = 0 a 0 t = t a - x x A a t = 0 t dx dt = k 2 (a x) 2 x t dx = k (a x) 2 2 dt 0 0

35 (a-x) = X - dx = dx t dx = k 2 dt a x a X 2 1 X a x a 1 a x 0 = k 2 t = k 2t + 1 a 1/(a-x) t k 2 1/ (a-x) 1/a t k 2

36 2 v f = k 2 [A][B] A + B product t = 0 a b 0 t = t a - x b - x x A B a b t = 0 t dx dt = k 2 (a x)(b x)

37, 1 (b a) dx (a x)(b x) = k 2 dt 1 a x 1 b x dx = k 2 dt t = 0 t 1 (b a) x 0 1 a x dx x 0 1 b x dx = t 0 k 2 dt

38 (a-x) = X (b-x) =Y - dx = dx, -dx = dy 1 (b a) a x a 1 X dx b x b 1 Y dy = t 0 k 2 dt 1 (b a) (a x) ln a ln (b x) b = k 2 t 1 (b a) ln a(b x) b(a x) = k 2 t

39 A + B A + B C C + A CA + B k product k 1 C k 1 A + B k 2 CA k 3 CB + A C, CA, CB d[c] = k 1 [A][B] k 1 [C] k 2 [C][A] dt d[ca] = k 2 [C][A] k 3 [CA][B] dt

40 d[cb] dt d[pr od] dt = k 3 [CA][B] k 4 [CB] = k 4 [CB] d[c] dt = d[ca] dt = d[cb] dt = 0

41 k 1 [A][B] k 1 [C] k 2 [C][A] k 2 [C][A] k 3 [CA][B] k 3 [CA][B] k 4 [CB] ʼ [C] = ʼ [CA] = k 1 [A][B] k 1 + k 2 [A] k 2 [C][A] k 3 [B] = k 1 k 2 [A] 2 k 3 (k 1 + k 2 [A]) = 0 ʼ = 0 ʼ = 0 ʼ

42 ʼ [CB] = k 3 [CA][B] k 4 = k 1 k 2 [A]2 [B] k 4 (k 1 + k 2 [A]) d[pr od] dt = k 4 [CB] = k 1 k 2 [A] 2 [B] k 1 + k 2 [A] d[pr od] k 2 [A] k -1 = k dt 1 [A][B] d[pr od] k k 2 [A] k -1 = 1 k 2 [A] 2 [B] dt k 1

43 = ( ) ( ) =

44 Svante Arrhenius (1889 ) k = Aexp E A RT lnk = E A R 1 T + ln A A E A R ln A ln k (1/T) ln k - E A /R 1/T

45

46 Van der Waals( ) P + n2 a V 2 V - n b = nrt a b ( )

47 V n

48 n V n V V n P real = P ideal - a n V 2

49 4 π(2r)2 3 1 v 8 r r r 1 v 1 mol b = 4N A v = (16/3)N A πr 3 N A

50 PV (atm ) 22.414 P (atm)

51 1 mol PV = R ( ) or T n mol PV = RT PV = nrt P (atm or Pa) n (mol) V ( or m 3 ) T (K) R 0.08206 atm mol -1 K -1 8.314 J mol -1 K -1

52 0, 1 atm 22.414 mol -1 PV = 1 atm x 22.414 mol -1 T 273.15 K = 0.082058

53 A (n A mol) B (n B mol) C (n C mol) P T V= n T RT P T ( ) n T = n A + n B + n C

54 P T = n TRT (n A + n B + n C )RT = V V = n ART n B RT n C RT + + V V V = P A + P B +P C Dalton

55 P A ( )= P T n A RT/V n T RT/V = n A n T =C A ( ) P A = C A P T, P B = C B P T, P C = C C P T ΣC i = 1

56 V T = n TRT (n A + n B + n C )RT = P P = n A RT n B RT n C RT + + P P P = V A + V B +V C P V A V B V C