The Rate Expression. The rate, velocity, or speed of a reaction

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1 The Rate Expression The rate, velocity, or speed of a reaction Reaction rate is the change in the concentration of a reactant or a product with time. A B rate = - da rate = db da = decrease in concentration of A over time period db = increasee in concentration of B over time period Because [A] decreases with time, da is negative.

2 The Rate Expression d This expression gives the increase (+) or decrease (-) of concentration,, within a given time interval,. A + B + D R f da d B

3 The Rate Expression d This expression gives the increase (+) or decrease (-) of concentration,, within a given time interval,. A + B + D

4 The Rate Expression d This expression gives the increase (+) or decrease (-) of concentration,, within a given time interval,. A + 2B + D d ( A) 1 d ( B ) 2

5 The Rate Expression d This expression gives the increase (+) or decrease (-) of concentration,, within a given time interval,. A + 2B + D Reaction Rate and Stoichiometry

6 The rate law The rate of a chemical reaction is proportional to the product of the molar concentration of the reactants each raised to a power equal to the number of molecules of the substance undergoing reaction:

7 The rate law Effect of concentration on reaction rate A + B + D The rate equation can be written as: R f f da d B da d B k( ( A ))( ( BB) ) In which k is known as the reaction rate constant.

8 The order of reaction The overall order of a chemical reaction is the sum of the exponents of the concentration terms in a reaction kinetics equation that after integration gives a linear plot. a A + b B g G + h H The rate equation can be written as: 1 a d ( A) 1 b d ( B) 1 g d( G) 1 h d( H ) K( A) m ( B) n reaction is m th order in A reaction is n th order in B reaction is (m +n) th order overall

9 The order of reaction The overall order of a chemical reaction is the sum of the exponents of the concentration terms in a reaction kinetics equation that after integration gives a linear plot. a A + b B g G + h H The rate equation can be written as: 1 a d ( A) 1 b d ( B) 1 g d( G) 1 h d( H ) K( A) m ( B) n oncentration term exponents (m and n) are usually small whole numbers but may be fractional, negative or zero. They are unlikely to be the stoichiometric factors for the overall rate law.

10 Reaction Mechanisms Simple elementary step Elementary processes are reversible Sequence of elementary steps The reactions are unimolecular or bimolecular A B Exponents for concentration terms are the same as the stoichiometric factors for the elementary process

11 Reaction Mechanisms Simple elementary step Sequence of elementary steps One elementary step is usually slower than all the others and is known as the rate determining step Intermediates are produced in one elementary process and consumed in another. They do not appear in the overall chemical equation or the rate law.

12 Experimental determination of rate law F 2 (g) + 2lO 2 (g) 2FlO 2 (g) (method of initial rates) rate = k [F 2 ] x [lo 2 ] y Double [F 2 ] with [lo 2 ] constant Rate doubles x = 1 Quadruple [lo 2 ] with [F 2 ] constant Rate quadruples y = 1 F 2 (g) + 2lO 2 (g) 2FlO 2 (g) 1 rate = k [F 2 ][lo 2 ]

13 I. Zero Order Reaction: Are reactions, in which the rate of reaction is independent of the concentration of the reactants. The velocity of the reaction is constant.

14 I. Zero Order Reaction: Zero-order reactions are typically found when a material that is required for the reaction to proceed, such as a surface or a catalyst, is saturated by the reactants. 1) occurs in a closed system 2) there is no net build-up of intermediates 3) there are no other reactions occurring

15 I. Zero Order Reaction: Zero-order reactions are typically found when a material that is required for the reaction to proceed, such as a surface or a catalyst, is saturated by the reactants. Haber Process Decomposition of multi-sulfa drugs

16 I. Zero Order Reaction: rate = - d() d() = K rate = k () 0 = k The rate equation is a differential equation. It can be integrated to obtain an integrated rate equation that links concentrations of reactants or products with time.

17 I. Zero Order Reaction: rate = - d() rate = k () 0 = k d() = K - d = K On Integration t = o - Kt [ t ] is the concentration of at any time t [] 0 is the concentration of at time t=0 Intercept = o Slope = - K Time (t)

18 I. Zero Order Reaction: We can Also calculate the constant of reaction by rearranging the linear equation: t = o - Kt k 0 t t mole liter UNITS mloe?. liter Second / d ( ) K 0 onc. time mole Second / liter 1 1 mole. liter. S

19 I. Zero Order Reaction: The half-life (t ½ ) the time required for a quantity to fall to half its value as measured at the beginning of the time period. t ½ = t when t = 1/2 0

20 I. Zero Order Reaction: The half-life (t ½ ) t = o - k t Putting t = 1 2 o in the linear equation: t = o - t k t 1/2 = o o k t 1/2 = 1 2 o k

21 I. Zero Order Reaction: The shelf life (t 90% ) is the time required for 10% of the drug to decompose. In other word 90% of the drug remains. Putting t = 0.9 o in the linear equation: t 90% = o 0.9 o k t 90% = 0.1 o k

22 I. Zero Order Reaction: Time life (t life ) is the time required for 100% of the drug to decompose. Time for a drug to decompose completely. Putting t = zero in the linear equation: t life = o 0 k t life = o k

23 Example The decomposition of a multi-sulfa compound, was found to follow zero order kinetics. If its concentration was 0.47 mole/liter, when freshly prepared, and after 473 days its concentration reached mole/liter, calculate its degradation rate constant and t 1/2.

24 Example The decomposition of a multi-sulfa compound, was found to follow zero order kinetics. If its concentration was 0.47 mole/liter, when freshly prepared, and after 473 days its concentration reached mole/liter, calculate its degradation rate constant and t 1/2.

25 Solution k = o - t k = t 473 = mole/liter. days -1 t ½ = o = 0.47 = 452 days = 15 months 2K 2x

26 II. 1 st Order Reaction: The speed of the reaction depends on the concentration of only one reactant, raised to the first power. There can be other reactants present in the reaction, but their concentrations do not affect the rate. First-order equations are often seen in decomposition reactions. Absorption, metabolism, distribution and elimination (not linked with a carrier or a carrier with unsaturable state).

27 II. 1 st Order Reaction: The speed of the reaction depends on the concentration of only one reactant, raised to the first power. d ( ) K 1

28 II. 1 st Order Reaction: d ( ) K 1 Integrating the previous equation between 0 at time t = 0 and concentration at some later time t gives: ln ln 0 k t onverting to common logarithms yields: log log 0 k t 2.303

29 II. 1 st Order Reaction: ln ln 0 k t log log 0 k t Intercept = ln 0 Intercept = log 0 ln c Slope = - k log c Slope = - k t t

30 II. 1 st Order Reaction: ln ln 0 k t log log 0 The reaction rate constant k t k ln o t ln k t log o UNITS? 1 time d ( ) K 1 ( ) mole/ liter 1 S Sec( mole/ liter )

31 II. 1 st Order Reaction: The relation between concentration and time is exponential: t 0 e kt t 10 kt/ Initial concentration 1/2 0 1/4 0 1/8 0 oncentration at t t 1/2 t 1/2 t t 1/2

32 II. 1 st Order Reaction: The Half Life "t½ half life when = ½ o k t log o t k log o = log o = log 2 = k ½ o k T 1/2 k

33 II. 1 st Order Reaction: The Shelf Life (t 90% ) the shelf life when = 0.9 o k t log o t k log o t 90% = log o = (0.0457) = k 0.9 o k k

34 II. 1 st Order Reaction: Time life (t life ) is the time required for 100% of the drug to decompose. Putting t = zero in the linear equation:?

35 II. 1 st Order Reaction: A solution of a drug contained 500 units per ml when prepared. It was analyzed after a period of 40 days and was found to contain 300 units per ml. Assuming that the decomposition is first order, at what time will the drug have decomposed to one half of its original concentration? Solution: K = log o = log 500 = day -1 t t t ½ = = = 54 days K

36 II. 1 st Order Reaction: What is the half-life of N 2 O 5 if it decomposes with a rate constant of 5.7 x 10-4 s -1? t ½ = = = 1200 s = 20 minutes k 5.7 x 10-4 s -1