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# Kinetics_Math_ver1.0

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### Kinetics_Math_ver1.0

1. 1. 速度論の数学的基礎 1. 変数分離形積分法 (同時方程式) 2. 定数係数１階線形微分方程式の公式 3. 零次反応 4. 一次反応 5.二次反応 6. 二成分反応 7. 可逆一次反応Ⅰ 8. 可逆一次反応Ⅱ 9. 半減期 Presented by S. Kume 140326 ver 1.0
2. 2. 1. 変数分離形積分法 (同時方程式) ネイピア数 e ≅ 2.718 ex = exp(x) ln(ex) = x ! dy dx = f (x) " g(y) # dy g(y) = f (x)dx dy g(y) \$ = f (x)dx\$ + C ! dy g(y) " = ln y 例題 ! dy dt = "ay # dy y = "adx dy y \$ = "a dx\$ + C 変数分離形積分法により ! dy y " = ln y,#a dx" + C = #ax + C ! dy y " = ln y ,#a dx" + C = #ax + C ln y = "ax + C # y = e"ax+C
3. 3. 2. 定数係数１階線形微分方程式の公式 ! dy dx + ay = Q(x) "y = e#ax eax Q(x)dx\$ + C{ } 導入方法 ! dy dx + ay = Q(x) " eax dy dx + aeax y = eax Q(x) " eax dy dx + d dx eax # \$ % & ' (y = eax Q(x) eax の微分はaeaxとなる性質を利用して ! d dx f (x)g(x) = df (x) dx g(x) + f (x) dg(x) dx 積の微分公式より ! eax dy dx + d dx eax " # \$ % & 'y = d dx eax y( ) ! d dx eax y( ) = eax Q(x) つまり 両辺を x で積分すると eax y = eax Q(x)dx" + C # y = e\$ax eax Q(x)dx" + C{ }
4. 4. ! A k " #" P 零次反応: 実験的な反応速度が濃度に比例しない ! v = d[P] dt = " d[A] dt = k ! " d[A] dt = k # d[A] = "kdt d[A] =\$ " k dt + C #\$ [A] = "kt + C t = 0ならば、[A] = [A]0 ! [A] = "kt + C # [A]0 = "k \$ 0 + C # C = [A]0 すなわち [A] = [A]0 " kt 3. 零次反応
5. 5. ! A k " #" P 一次反応: 実験的な反応速度が濃度の１乗に比例する ! v = d[P] dt = " d[A] dt = k[A] ! " d[A] dt = k[A] # d[A] = "k[A]dt # d[A] [A] = "kdt ! ln[A] = "kt + C # [A] = e"kt+C 変数分離形積分法により ! d[A] [A] " = #k dt" + C \$ ln[A] = #kt + C ! log M = P " M = eP ex+y = ex # ey t = 0ならば、[A] = [A]0 ! [A]0 = e"k#0+C \$ [A]0 = eC [A] = e"kt +C # [A] = e"kt eC # [A] = [A]0e"kt # [A] = [A]0 exp "kt( ) すなわち 4. 一次反応
6. 6. 二次反応: 実験的な反応速度が濃度の２乗に比例する ! A k " #" P ! v = d[P] dt = " d[A] dt = k[A]2 " d[A] dt = k[A]2 # " d[A] [A]2 = kdt 変数分離形積分法により ! " d[A] [A]2# = k dt# + C \$ " " 1 [A] % & ' ( ) * = kt + C t = 0ならば、[A] = [A]0 すなわち ! " " 1 [A]0 # \$ % & ' ( = k ) 0 + C * 1 [A]0 = C " " 1 [A] # \$ % & ' ( = kt + C ) 1 [A] = kt + 1 [A]0 ) [A] = 1 kt + 1 [A]0 5. 二次反応
7. 7. 実験的な反応速度が濃度[A]および [B]に比例する ! A + B k " #" P ! v = d[P] dt = " d[A] dt = " d[B] dt = k[A][B] ! dx dt = k[A][B] " dx [A][B] = kdt " dx [A]0 # x( ) [B]0 # x( ) = kdt [A] = [A] 0 – x, [B] = [B]0 – x とおくと ! [A]0 " [B]0 6. 二成分反応 のとき € 1 [A]0 − x( ) [B]0 − x( ) = − 1 [A]0 −[B]0( ) 1 [A]0 − x( ) − 1 [B]0 − x( ) ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = − 1 [A]0 −[B]0( ) [B]0 − x( )− [A]0 − x( ) [A]0 − x( ) [B]0 − x( ) ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ € = − 1 [A]0 −[B]0( ) [B]0 −[A]0 [A]0 − x( ) [B]0 − x( ) ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = − 1 [A]0 −[B]0( ) −([A]0 −[B]0) [A]0 − x( ) [B]0 − x( ) ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = 1 [A]0 ! x( ) [B]0 ! x( )
8. 8. dx [A]0 ! x( ) [B]0 ! x( ) = ! 1 [A]0 ![B]0( ) 1 [A]0 ! x( ) ! 1 [B]0 ! x( ) ! " # # \$ % & & dx dx [A]0 ! x( ) [B]0 ! x( ) = kdt 1 [A]0 ! x( ) dx の積分を考える。 ここで 1 [A]0 ! x( ) dx 0 x " # ! 1 z dz [A]0 [A]0!x " [A]0 ! x = z と置くと ! 1 z dz [A]0 [A]0!x " = ! [A]0 [A]0!x ln z#\$ %& = ! ln([A]0 ! x)! ln[A]0[ ] = !ln [A]0 ! x [A]0 " # \$ % & ' = !ln [A] [A]0 " # \$ % & ' したがって の左辺は
9. 9. 1 [A]0 ! x( ) [B]0 ! x( ) dx" = ! 1 [A]0 ![B]0( ) 1 [A]0 ! x( ) ! 1 [B]0 ! x( ) # \$ % % & ' ( ( dx" = ! 1 [A]0 ![B]0( ) !ln [A] [A]0 " # \$ % & '! !ln [B] [B]0 " # \$ % & ' ( ) * + , - . / 0 10 2 3 0 40 = 1 [A]0 ![B]0( ) ln [A] [A]0 " # \$ % & '! ln [B] [B]0 " # \$ % & ' . / 0 10 2 3 0 40 = 1 [A]0 ![B]0( ) ln [A][B]0 [A]0[B] " # \$ % & ' [B]も同様に行い ! [A]0 = [B]0 1 [A] " 1 [A]0 = 1 [B] " 1 [B]0 = kt のとき
10. 10. 実験的な反応速度において逆反応が無視できない場合 ! A ! B ! k1 ! k"1 7. 可逆一次反応Ⅰ ! v = d[B] dt = " d[A] dt = k1[A] " k"1[B] # d[A] dt = k"1[B] " k1[A] # d[A] dt = k"1 [A]0 "[A]( )" k1[A] # d[A] dt = " k1 + k"1( )[A]+ k"1[A]0 # d[A] dt = " k1 + k"1( ) [A] " k"1 k1 + k"1 [A]0 \$ % & ' ( ) ! K = [B]eq [A]eq ! [A]0 = [A]eq +[B]eq であるので ! K = [A]0 "[A]eq [A]eq = k1 k"1 ! [A]0 "[A]eq = k1 k"1 [A]eq # [A]0 = k1 k"1 [A]eq +[A]eq # [A]0 = k1 + k"1 k"1 [A]eq # k"1 k1 + k"1 [A]0 = [A]eq 時間 t 後における濃度を [A] とすると " d[A] dt = # k1 + k#1( ) [A] #[A]eq( )" d[A] [A] #[A]eq = # k1 + k#1( )dt
11. 11. ! d[A] [A] "[A]eq # = " k1 + k"1( ) dt# + C \$ ln [A] "[A]eq( )= " k1 + k"1( )t + C 変数分離形積分法により t = 0ならば、[A] = [A]0となるので ! C = ln [A]0 "[A]eq( ) つなわち ! ln [A] "[A]eq( )= " k1 + k"1( )t + ln [A]0 "[A]eq( ) # ln [A] "[A]eq( )" ln [A]0 "[A]eq( )= " k1 + k"1( )t # ln [A] "[A]eq [A]0 "[A]eq = " k1 + k"1( )t # [A] "[A]eq [A]0 "[A]eq = e" k1 +k"1( )t # [A] "[A]eq = [A]0 "[A]eq( )e" k1 +k"1( )t # [A] = [A]eq + [A]0 "[A]eq( )e" k1 +k"1( )t kobs = k1 + k"1 とおく および A = [A]0 "[A]eq [A] = [A]eq + Ae"kobs t
12. 12. 実験的な反応速度において逆反応が無視できない場合 ! A ! B ! k1 ! k"1 ! v = d[B] dt = " d[A] dt = k1[A] " k"1[B] 時間 t 後に、[A]が x mol/dm3に変化したとすると ! dx dt = k1 [A] " x( ) " k"1 [B]+ x( ) # dx dt = k1[A] " k"1[B] " x k1 + k"1( ) # dx dt = k1 + k"1( ) k1[A] " k"1[B] k1 + k"1 " x \$ % & ' ( ) ! m = k1[A] " k"1[B] k1 + k"1 とおく ! dx dt = k1 + k"1( ) m " x( ) # dx m " x = k1 + k"1( )dt 変数分離形積分法により dx m " x # = k1 + k"1( ) dt# + C \$ "ln m " x( ) = k1 + k"1( )t + C 8. 可逆一次反応Ⅱ
13. 13. ! "ln m " x( ) = k1 + k"1( )t + C # "ln m " 0( ) = k1 + k"1( )\$ 0 + C # C = "ln m( ) 時間 t = 0ならば、 x = 0 となる ! ln m( ) " ln( ) = k1 + k"1( )t # ln m m " x \$ % & ' ( ) = k1 + k"1( )t # m m " x = e k1 +k"1( )t # m " x m = e" k1 +k"1( )t # m " x = me" k1 +k"1( )t # x = m + me" k1 +k"1( )t C = -ln (m)を代入すると ! kobs = k1 + k"1 とおく ! "x = m + me#kobs t \$ x = m + m % exp #kobst( )
14. 14. 9. 反応の半減期 [A]=[A0]/2になる時間をt=t1/2とすると € ln [A] [A0] = –kt ⇔ [A] = [A0]exp −kt( ) € [A] = [A0] 2 を代入すると € ln [A0] 2 [A0] = –kt1/ 2 ⇔ ln 1 2 = –kt1/ 2 ⇔ t1/ 2 = − 1 k ln 1 2 = ln2 k ≈ 0.693 k 一次反応における半減期は、初期濃度に依存しない。 一次反応
15. 15. [A]=[A0]/2になる時間をt=t1/2とすると € 1 [A] = kt + 1 [A0] € [A] = [A0] 2 を代入すると 二次反応 1 [A0] 2 = kt1/ 2 + 1 [A0] ⇔ t1/ 2 = 1 [A0]k 一次反応以外の他の反応半減期はすべて濃度に依存する。 零次反応 € [A0] −[A] = kt [A]=[A0]/2になる時間をt=t1/2とすると € [A] = [A0] 2 を代入すると € [A0] − [A0] 2 = kt1/ 2 ⇔ [A0] 2 = kt1/ 2 ⇔ t1/ 2 = [A0] 2k