Lecture 9 (2) PDF - Chemical Kinetics

 The reaction rate is Directly proportional (Dependent) to: o Second power of concentration of single reactant [(A)2]. o First power of concentration of two reactants [(A)1(B)1].  Rate equation: 𝒅𝒅𝑨𝑨 𝒅𝒅𝒅𝒅 𝑹𝑹𝑹𝑹𝑹𝑹𝑹𝑹 = 𝒌𝒌 (𝑨𝑨)𝟐𝟐...

 The reaction rate is Directly proportional (Dependent) to: o Second power of concentration of single reactant [(A)2]. o First power of concentration of two reactants [(A)1(B)1].  Rate equation: 𝒅𝒅𝑨𝑨 𝒅𝒅𝒅𝒅 𝑹𝑹𝑹𝑹𝑹𝑹𝑹𝑹 = 𝒌𝒌 (𝑨𝑨)𝟐𝟐 OR 𝑹𝑹𝑹𝑹𝑹𝑹𝑹𝑹 = − =− = 𝒌𝒌 (𝑨𝑨)(𝑩𝑩) 𝒅𝒅𝒅𝒅 𝒅𝒅𝒅𝒅  IF: o a → Initial conc. of reactant A. o b → Initial conc. of reactant B. o x → Conc. reacting of A & B at time t. 𝒅𝒅𝒅𝒅 ∴ = 𝒌𝒌 (𝑨𝑨)(𝑩𝑩) = 𝒌𝒌 (𝒂𝒂 − 𝒙𝒙)(𝒃𝒃 − 𝒙𝒙) 𝒅𝒅𝒅𝒅 o (a-x) → (A) → Conc. remaining of A at time t. o (b-x) → (B) → Conc. remaining of B at time t.  A + B → Product 𝒅𝒅𝒅𝒅  Rate equation: ∴ = 𝒌𝒌 (𝑨𝑨)(𝑩𝑩) = 𝒌𝒌 (𝒂𝒂 − 𝒙𝒙)(𝒃𝒃 − 𝒙𝒙) 𝒅𝒅𝒅𝒅 Suppose the initial conc. of (A) and (B) are equal ∴a=b 𝒅𝒅𝒅𝒅 ∴ = 𝒌𝒌 (𝒂𝒂 − 𝒙𝒙)(𝒃𝒃 − 𝒙𝒙) = 𝒌𝒌 (𝒂𝒂 − 𝒙𝒙)𝟐𝟐 𝒅𝒅𝒅𝒅  Integrated Rate Equation Graphical Plot 𝟏𝟏 𝟏𝟏  Plot “1/(a-x)” on y-axis and “t” on = + 𝒌𝒌𝒌𝒌 x-axis → straight line (linear). (𝒂𝒂 − 𝒙𝒙) 𝒂𝒂 𝟏𝟏 𝟏𝟏 = + 𝒌𝒌𝒌𝒌 𝑪𝑪𝒕𝒕 𝑪𝑪𝒐𝒐 K 𝟏𝟏  Linear equation. (𝒂𝒂 − 𝒙𝒙)  Where: o (a-x) → Ct → Conc. 𝟏𝟏 remaining at time t. 𝒂𝒂 o a → Co → Initial conc. t o x → Conc. reacting at time t. o Intercept → 1/a. o Slope → k. o k → Second order rate constant.  Equation for “k” Unit of “k” calculation  Conc.-1 Time-1. 𝟏𝟏 𝟏𝟏  Ex: − o Liter. mole-1. sec-1. 𝑪𝑪𝒕𝒕 𝑪𝑪𝒐𝒐 𝒌𝒌 = o ml. mg-1. min-1. 𝒕𝒕 𝟏𝟏 𝟏𝟏 𝟏𝟏 𝒌𝒌 = − 𝒙𝒙 𝑪𝑪𝒕𝒕 𝑪𝑪𝒐𝒐 𝒕𝒕  Half life (t1/2 or t50%) Shelf life or Expiry date (t90%)  ∴ (a-x) = ½ a OR Ct = ½ Co.  ∴ (a-x) = 0.9 a OR Ct = 0.9 Co. 𝟏𝟏 𝟏𝟏 ∴ 𝒕𝒕𝟏𝟏 = ∴ 𝒕𝒕𝟗𝟗𝟗𝟗% = 𝟗𝟗𝒂𝒂𝒂𝒂 𝟐𝟐 𝒂𝒂𝒌𝒌 o Both half life & shelf life are Inversely proportional to initial conc.  Ex: Ampicillin, amoxicillin and cycloserine.  Ex: Ethyl acetate in alkaline medium. 1) The saponification of ethyl acetate at 25 °C was investigated. The initial concentration of both ethyl acetate and NaOH in the mixture was 0.01 M. the change in concentration of alkali during 20 minutes was 0.00566 mole/liter. Compute the rate constant and the half-life of the reaction. Solution: - a = 0.01M and x = 0.00566 M. - ∴ (a-x) = 0.01 - 0.00566 = 0.00434 M. 𝟏𝟏 𝟏𝟏 = + 𝒌𝒌𝒌𝒌 (𝒂𝒂 − 𝒙𝒙) 𝒂𝒂 𝟏𝟏 𝟏𝟏 𝟏𝟏 𝟏𝟏 ∴ = + 𝒌𝒌𝒙𝒙𝒙𝒙𝒙𝒙 ∴ = + 𝒌𝒌𝒙𝒙𝒙𝒙𝒙𝒙 (𝟎𝟎. 𝟎𝟎𝟎𝟎 − 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎) 𝟎𝟎. 𝟎𝟎𝟎𝟎 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎 𝟎𝟎. 𝟎𝟎𝟎𝟎 ∴ 𝒌𝒌 = 𝟔𝟔. 𝟓𝟓𝟓𝟓 𝑳𝑳𝑳𝑳𝑳𝑳𝑳𝑳𝑳𝑳. 𝒎𝒎𝒎𝒎𝒍𝒍𝒍𝒍−𝟏𝟏. 𝒎𝒎𝒎𝒎𝒎𝒎−𝟏𝟏 - T1/2: 𝟏𝟏 𝟏𝟏 𝒕𝒕𝟏𝟏 = = = 𝟏𝟏𝟏𝟏. 𝟑𝟑 𝒎𝒎𝒎𝒎𝒎𝒎. 𝟐𝟐 𝒂𝒂𝒌𝒌 𝟎𝟎. 𝟎𝟎𝟎𝟎 𝒙𝒙 𝟔𝟔. 𝟓𝟓𝟓𝟓  In second order reactions (A+B → product) → IF the conc. of one reactant is in large excess → its contribution to rate equation is regarded constant (its conc. apparently not change) → reaction rate follow Pseudo 1st order.  The conc. of the solvent is very large → regarded constant. ∴ Drug decomposition follow Pseudo 1st order.  Ex: Hydrolysis of drugs by water in dilute solutions.  The conc. of [H+] or [OH-] is maintained constant by buffer. ∴ Drug decomposition follow Pseudo 1st order.  In some cases the shelf life is regarded as time required for 1% of drug to decompose (99% remaining).  Examples: 1) Urea IV solution: o Because urea hydrolyze to: - Ammonia → Highly toxic. - Carbon dioxide → Cause ampoule explosion. 2) Epinephrine solution: o Darken due to oxidation → polymerization of degradation product gives black melanin pigment. 3) IV pro-drug of phenytoin: o Hydrolysis of pro-drug to the insoluble parent drug lead to precipitation.  a A + b B → Product  Acc. to law of mass action: Rate α (A)a(B)b - Reaction rate is directly proportional to conc. of (A)a(B)b at any time. - ∴ At the beginning of the reaction → the INITIAL rate is directly proportional to the INITIAL conc. of (A)a(B)b.  To determine the reaction order of reactant (A) → “a”: o Do some experiments with different initial conc. of (A). o BUT initial conc. of (B) must be held CONSTANT.  To determine the reaction order of reactant (B) → “b”: o Do some experiments with different initial conc. of (B). o BUT initial conc. of (A) must be held CONSTANT.  Overall order = Order of (A) + Order of (B). Given below in the initial rate data of the following reaction, Determine the rate law and rate constant of the reaction. 2 MnO4- + 5 ClO3- + 6 H+ → 2 Mn+2 + 5 ClO4- + 3 H2O Initial rate Exp. no. [MnO4-] [ClO3-] [H+] (M/S) 1 0.1 M 0.1 M 0.1 M 5.2 x 10-3 2 0.25 M 0.1 M 0.1 M 3.3 x 10-2 3 0.1 M 0.3 M 0.1 M 1.6 x 10-2 4 0.1 M 0.1 M 0.2 M 7.4 x 10-3  Rate = k [MnO4-]a [ClO3-]b [H+]c To determine order of MnO4-: - Use data in exp. no. (1) & (2) → Why?? o Because initial conc. of [ClO3-] & [H+] are constant. - Take the ratio between rate law in exp. (1) & (2): 𝑹𝑹𝑹𝑹𝑹𝑹𝑹𝑹𝟐𝟐 𝒌𝒌 [𝑴𝑴𝑴𝑴𝑶𝑶𝟒𝟒−]𝒂𝒂 [𝑪𝑪𝑪𝑪𝑶𝑶𝟑𝟑−]𝒃𝒃 [𝑯𝑯+]𝒄𝒄 → 𝑬𝑬𝑬𝑬𝑬𝑬. 𝟐𝟐 = 𝑹𝑹𝑹𝑹𝑹𝑹𝑹𝑹𝟏𝟏 𝒌𝒌 [𝑴𝑴𝑴𝑴𝑶𝑶𝟒𝟒−]𝒂𝒂 [𝑪𝑪𝑪𝑪𝑶𝑶− 𝒃𝒃 𝟑𝟑 ] [𝑯𝑯 ] + 𝒄𝒄 → 𝑬𝑬𝑬𝑬𝑬𝑬. 𝟏𝟏 𝟑𝟑. 𝟑𝟑 𝒙𝒙 𝟏𝟏𝟏𝟏−𝟐𝟐 𝒌𝒌 [𝟎𝟎. 𝟐𝟐𝟐𝟐]𝒂𝒂 [𝟎𝟎. 𝟏𝟏]𝒃𝒃 [𝟎𝟎. 𝟏𝟏]𝒄𝒄 → 𝑬𝑬𝑬𝑬𝑬𝑬. 𝟐𝟐 = 𝟓𝟓. 𝟐𝟐 𝒙𝒙 𝟏𝟏𝟏𝟏−𝟑𝟑 𝒌𝒌 [𝟎𝟎. 𝟏𝟏]𝒂𝒂 [𝟎𝟎. 𝟏𝟏]𝒃𝒃 [𝟎𝟎. 𝟏𝟏]𝒄𝒄 → 𝑬𝑬𝑬𝑬𝑬𝑬. 𝟏𝟏 [𝟎𝟎. 𝟐𝟐𝟐𝟐]𝒂𝒂 𝟎𝟎. 𝟐𝟐𝟐𝟐 𝒂𝒂 𝟔𝟔. 𝟐𝟐𝟐𝟐 = ∴ 𝟔𝟔. 𝟐𝟐𝟐𝟐 = [ ] = 𝟐𝟐. 𝟓𝟓𝒂𝒂 [𝟎𝟎. 𝟏𝟏]𝒂𝒂 𝟎𝟎. 𝟏𝟏 ∴ a = 2 → ∴ order of MnO4- is second order.  Rate = k [MnO4-]a [ClO3-]b [H+]c To determine order of ClO3-: - Use data in exp. no. (1) & (3) → Why?? o Because initial conc. of [MnO4-] & [H+] are constant. - Take the ratio between rate law in exp. (1) & (3): 𝑹𝑹𝑹𝑹𝑹𝑹𝑹𝑹𝟑𝟑 𝒌𝒌 [𝑴𝑴𝑴𝑴𝑶𝑶𝟒𝟒−]𝒂𝒂 [𝑪𝑪𝑪𝑪𝑶𝑶− 𝒃𝒃 + 𝒄𝒄 𝟑𝟑 ] [𝑯𝑯 ] → 𝑬𝑬𝑬𝑬𝑬𝑬. 𝟑𝟑 = 𝑹𝑹𝑹𝑹𝑹𝑹𝑹𝑹𝟏𝟏 𝒌𝒌 [𝑴𝑴𝑴𝑴𝑶𝑶𝟒𝟒−]𝒂𝒂 [𝑪𝑪𝑪𝑪𝑶𝑶− 𝒃𝒃 + 𝒄𝒄 𝟑𝟑 ] [𝑯𝑯 ] → 𝑬𝑬𝑬𝑬𝑬𝑬. 𝟏𝟏 𝟏𝟏. 𝟔𝟔 𝒙𝒙 𝟏𝟏𝟏𝟏−𝟐𝟐 𝒌𝒌 [𝟎𝟎. 𝟏𝟏]𝒂𝒂 [𝟎𝟎. 𝟑𝟑]𝒃𝒃 [𝟎𝟎. 𝟏𝟏]𝒄𝒄 → 𝑬𝑬𝑬𝑬𝑬𝑬. 𝟑𝟑 = 𝟓𝟓. 𝟐𝟐 𝒙𝒙 𝟏𝟏𝟏𝟏−𝟑𝟑 𝒌𝒌 [𝟎𝟎. 𝟏𝟏]𝒂𝒂 [𝟎𝟎. 𝟏𝟏]𝒃𝒃 [𝟎𝟎. 𝟏𝟏]𝒄𝒄 → 𝑬𝑬𝑬𝑬𝑬𝑬. 𝟏𝟏 [𝟎𝟎. 𝟑𝟑]𝒃𝒃 𝟎𝟎. 𝟑𝟑 𝒃𝒃 𝟑𝟑. 𝟏𝟏 = ∴ 𝟑𝟑. 𝟏𝟏 = [ ] = 𝟑𝟑𝒃𝒃 [𝟎𝟎. 𝟏𝟏]𝒃𝒃 𝟎𝟎. 𝟏𝟏 ∴ b = 1 → ∴ order of ClO3- is first order.  Rate = k [MnO4-]a [ClO3-]b [H+]c To determine order of H+: - Use data in exp. no. (1) & (4) → Why?? o Because initial conc. of [MnO4-] & [ClO3-] are constant. - Take the ratio between rate law in exp. (1) & (4): 𝑹𝑹𝑹𝑹𝑹𝑹𝑹𝑹𝟒𝟒 𝒌𝒌 [𝑴𝑴𝑴𝑴𝑶𝑶𝟒𝟒−]𝒂𝒂 [𝑪𝑪𝑪𝑪𝑶𝑶− 𝒃𝒃 + 𝒄𝒄 𝟑𝟑 ] [𝑯𝑯 ] → 𝑬𝑬𝑬𝑬𝑬𝑬. 𝟒𝟒 = 𝑹𝑹𝑹𝑹𝑹𝑹𝑹𝑹𝟏𝟏 𝒌𝒌 [𝑴𝑴𝑴𝑴𝑶𝑶𝟒𝟒−]𝒂𝒂 [𝑪𝑪𝑪𝑪𝑶𝑶− 𝒃𝒃 + 𝒄𝒄 𝟑𝟑 ] [𝑯𝑯 ] → 𝑬𝑬𝑬𝑬𝑬𝑬. 𝟏𝟏 𝟕𝟕. 𝟒𝟒 𝒙𝒙 𝟏𝟏𝟏𝟏−𝟑𝟑 𝒌𝒌 [𝟎𝟎. 𝟏𝟏]𝒂𝒂 [𝟎𝟎. 𝟏𝟏]𝒃𝒃 [𝟎𝟎. 𝟐𝟐]𝒄𝒄 → 𝑬𝑬𝑬𝑬𝑬𝑬. 𝟒𝟒 = 𝟓𝟓. 𝟐𝟐 𝒙𝒙 𝟏𝟏𝟏𝟏−𝟑𝟑 𝒌𝒌 [𝟎𝟎. 𝟏𝟏]𝒂𝒂 [𝟎𝟎. 𝟏𝟏]𝒃𝒃 [𝟎𝟎. 𝟏𝟏]𝒄𝒄 → 𝑬𝑬𝑬𝑬𝑬𝑬. 𝟏𝟏 [𝟎𝟎. 𝟐𝟐]𝒄𝒄 𝟎𝟎. 𝟐𝟐 𝒄𝒄 𝟏𝟏. 𝟒𝟒𝟒𝟒 = ∴ 𝟏𝟏. 𝟒𝟒𝟒𝟒 = [ ] = 𝟐𝟐𝒄𝒄 [𝟎𝟎. 𝟏𝟏]𝒄𝒄 𝟎𝟎. 𝟏𝟏 ∴ c = 1/2 → ∴ order of H+ is 1/2. ∴ Order of MnO4- → 2 & order of ClO3- → 1 & order of H+ → 0.5. ∴ Overall Order of reaction = 2 + 1 + 0.5 = 3.5. ∴ Rate Law: Rate = k [MnO4-]2 [ClO3-]1 [H+]0.5 Determine rate constant “k”: - Use data in any exp. with the calculated orders. - Take exp. (1): 5.2 x 10-3 = k [0.1]2 [0.1] [0.1]0.5 𝟓𝟓. 𝟐𝟐 𝒙𝒙 𝟏𝟏𝟏𝟏−𝟑𝟑 𝒌𝒌 = = 𝟏𝟏𝟏𝟏. 𝟒𝟒 [𝟎𝟎. 𝟏𝟏]𝟐𝟐 [𝟎𝟎. 𝟏𝟏] [𝟎𝟎. 𝟏𝟏]𝟎𝟎.𝟓𝟓  Data (C & t) are substituted in integrated equation of different orders to calculate “K”.  The equation that keep calculated “K” CONSTANT at all time intervals → Is equation of right order. Reaction Order Rate Equation 𝑪𝑪𝟎𝟎 − 𝑪𝑪𝒕𝒕 Zero order 𝑲𝑲 = 𝒕𝒕 𝟐𝟐. 𝟑𝟑𝟑𝟑𝟑𝟑 𝑪𝑪𝟎𝟎 First order 𝑲𝑲 = 𝑳𝑳𝑳𝑳𝑳𝑳 𝒕𝒕 𝑪𝑪𝒕𝒕 𝟏𝟏 𝑿𝑿 Second order 𝒌𝒌 = 𝒂𝒂𝒂𝒂 (𝒂𝒂 − 𝒙𝒙)  Plot of (a-x) vs t → straight line  ∴ Zero order.  Plot of log (a-x) vs t → straight line  ∴ First order.  Plot of 1/(a-x) vs t → straight line  ∴ Second order. Zero order First order Second order C0 Log C0 K - K/2.303 𝟏𝟏 -K (𝒂𝒂 − 𝒙𝒙) Ct log Ct 𝟏𝟏 𝒂𝒂 t t t  Order Half life (t1/2) t1/2 relation to “a”  T1/2 directly proportional to 𝑪𝑪𝟎𝟎 initial conc. Zero order 𝒕𝒕𝟏𝟏/𝟐𝟐 = 𝟐𝟐𝟐𝟐 ∴ t1/2 α a1  T1/2 Independent on initial 𝟎𝟎.𝟔𝟔𝟔𝟔𝟔𝟔 conc. First order 𝒕𝒕𝟏𝟏/𝟐𝟐 = 𝒌𝒌 ∴ t1/2 α a0  T1/2 inversly proportional to 𝟏𝟏 initial conc. Second order 𝒕𝒕𝟏𝟏/𝟐𝟐 = 𝒂𝒂𝒂𝒂 ∴ t1/2 α a -1 𝟏𝟏 ∴ 𝒕𝒕𝟏𝟏/𝟐𝟐 ∝ 𝒂𝒂𝒏𝒏−𝟏𝟏  n → order of the reaction. If 2 reactions are initiated at 2 different initial conc. a1 & a2. 𝟏𝟏 𝟏𝟏 𝒕𝒕𝟏𝟏⁄𝟐𝟐 (𝟏𝟏) ∝ 𝒕𝒕𝟏𝟏⁄𝟐𝟐 (𝟐𝟐) ∝ 𝒂𝒂𝒏𝒏−𝟏𝟏 𝟏𝟏 𝒂𝒂𝒏𝒏−𝟏𝟏 𝟐𝟐 Divide both equations 𝒕𝒕𝟏𝟏 (𝟏𝟏) 𝟐𝟐 𝒂𝒂𝒏𝒏−𝟏𝟏 𝟐𝟐 𝒂𝒂𝟐𝟐 = 𝒏𝒏−𝟏𝟏 = ( )𝒏𝒏−𝟏𝟏 𝒕𝒕𝟏𝟏 (𝟐𝟐) 𝒂𝒂𝟏𝟏 𝒂𝒂𝟏𝟏 𝟐𝟐 Logarithmic form 𝒕𝒕𝟏𝟏 (𝟏𝟏) 𝟐𝟐 𝒂𝒂𝟐𝟐 𝒍𝒍𝒍𝒍𝒍𝒍 = (𝒏𝒏 − 𝟏𝟏)𝒍𝒍𝒍𝒍𝒍𝒍 ( ) 𝒕𝒕𝟏𝟏 (𝟐𝟐) 𝒂𝒂𝟏𝟏 𝟐𝟐 Finally calculate “n” (order) 𝒍𝒍𝒍𝒍𝒍𝒍 [𝒕𝒕𝟏𝟏 (𝟏𝟏)/ 𝒕𝒕𝟏𝟏 (𝟐𝟐)] 𝟐𝟐 𝟐𝟐 𝒏𝒏 = + 𝟏𝟏 𝒍𝒍𝒍𝒍𝒍𝒍 [𝒂𝒂𝟐𝟐 /𝒂𝒂𝟏𝟏 ]  The 2 half lives are calculated from plotting (a-x) vs t for two different initial conc. a1 & a2.  Two conc. during a single run may be taken as a1 & a2.  Determine time at ½ a1 & ½ a2.  Substitute in the equation & obtain the order “n”.  Example: From the following plot, determine the order of sulphadiazine degradation.  𝒍𝒍𝒍𝒍𝒍𝒍 [𝒕𝒕𝟏𝟏 (𝟏𝟏)/ 𝒕𝒕𝟏𝟏 (𝟐𝟐)] 𝒍𝒍𝒍𝒍𝒍𝒍 [𝟏𝟏𝟏𝟏. 𝟖𝟖/ 𝟏𝟏𝟏𝟏. 𝟓𝟓] 𝟐𝟐 𝟐𝟐 ∴ First order 𝒏𝒏 = + 𝟏𝟏 𝒏𝒏 = + 𝟏𝟏 = 𝟎𝟎. 𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗 ≅ 𝟏𝟏 𝒍𝒍𝒍𝒍𝒍𝒍 [𝒂𝒂𝟐𝟐/𝒂𝒂𝟏𝟏] 𝒍𝒍𝒍𝒍𝒍𝒍 [𝟖𝟖𝟖𝟖/𝟏𝟏𝟏𝟏𝟏𝟏] reaction.  They are reactions involving more than 1 step elementary reactions.  Can’t be expressed by simple zero, first or second order equations.  A reversible reaction may be expressed as: Kf A+B C+ D Kr  Where: Kf → rate constant of forward reaction. Kr → rate constant of reverse reaction.  Simplest reversible reaction → forward & reverse reactions are first order. Kf A B Kr  Kf A B Kr  The rate of decrease of reactant “A” equals: Rate of “A” decrease in forward step MNUS Rate of “A” increase in reverse step  The rate equation is: 𝒅𝒅𝒅𝒅 − = 𝒌𝒌𝒇𝒇 𝑨𝑨 − 𝒌𝒌𝒓𝒓 𝑩𝑩 𝒅𝒅𝒅𝒅  Example of first order reversible decomposition reaction → Epimerisation of tetracycline.  A reaction in which the decomposition of a drug involve two or more pathways.  The overall rate equation → the SUM of constants of each pathway.  Example: Decomposition of prednisolone → Parallel pseudo 1st order reaction with appearance of acidic & neutral steroidal products. K1 A (Acidic product) (Prednisolone) X K2 B (Neutral product)  K1 A (Acidic product) (Prednisolone) X K2 B (Neutral product)  The rate equation is: 𝒅𝒅𝑿𝑿 − = 𝒌𝒌𝟏𝟏 𝑿𝑿 + 𝒌𝒌𝟐𝟐 𝑿𝑿 = 𝒌𝒌𝒆𝒆𝒆𝒆𝒆𝒆 𝑿𝑿 𝒅𝒅𝒅𝒅  Where: k1 → rate constant for formation of “A”. k2 → rate constant for formation of “B”. kexp → experimental rate constant. kexp = k1 + k2.  A reaction in which two first order reactions occurs simultaneously → one after the other.  Simple consecutive reaction: K1 K2 A B C Rate of Rate of “A” Rate of change in formation of decomposition conc. of “B” Rate “C” equation 𝒅𝒅𝒅𝒅 𝒅𝒅𝒅𝒅 𝒅𝒅𝒅𝒅 − = 𝒌𝒌𝟏𝟏 𝑨𝑨 = 𝒌𝒌𝟏𝟏 𝑨𝑨 − 𝒌𝒌𝟐𝟐 𝑩𝑩 = 𝒌𝒌𝟐𝟐 𝑩𝑩 𝒅𝒅𝒅𝒅 𝒅𝒅𝒅𝒅 𝒅𝒅𝒅𝒅  Example → hydrolysis of chlordiazepoxide. Chlordiazepoxide K1 Lactum K2Benzophenone  A Complex reaction involves all the different types of complex reactions (reversible, parallel and consecutive). K2 Polysaccharides K-2 K3 Glucose Coloured materials 5 HMF K1 K4 Formic & Levulinic acids