1. 1 7-3-2011

2. 2 Reaction Rates and Stoichiometry In general, for the reaction aA + bB → cC + dD Rate = - (1/a)Δ[A]/Δt = - (1/b)Δ[B]/Δt = (1/c) Δ[C] /Δt = (1/d) Δ[D] /Δt

3. 3 Write the rate expression for the following reaction: CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (g) rate = -  [CH 4 ] tt = -  [O 2 ] tt 1 2 =  [CO 2 ] tt =  [H 2 O] tt 1 2

4. 4 Depending on the time interval between measurements, the rates are called : 1. average rate: rate measured between long time interval 2. instantaneous rate: rate measured between very short interval (at specific time and conc) 3. initial rate: instantaneous rate at the beginning of an experiment

5. 5 slope of tangent slope of tangent slope of tangent

6. 6 Reaction Rates Example: How is the rate of disappearance of N 2 O 5 related to the rate of appearance of NO 2 in the following reaction? 2 N 2 O 5 (g) 4 NO 2 (g) + O 2 (g)

7. 7 Reaction Rates Example: If the rate of decomposition of N 2 O 5 in the previous example at a particular instant is 4.2 x 10 -7 M /s, what is the rate of appearance of NO 2 ? 2 N 2 O 5 (g) 4 NO 2 (g) + O 2 (g)

8. 8 Rate = - 1  [N 2 O 5 ] = 1  [NO 2 ] 2  t 4  t  [NO 2 ] = - 4  [N 2 O 5 ]  t 2  t = 2 x 4.2 x 10 -7 M /s = 8.4 x 10 -7 M/s

9. 9 How do we measure Rate of a Reaction experimentally ? 1.For reactions in solution: Changes in concentration can be measured spectroscopically 2.For reactions involving gases: Changes in pressure can be measured 3.For reactions in solution with ions present: Change in concentrations can be measured through electrical conductance

10. 10 Rate of a Reaction So if we have an aqueous solution of molecular bromine and formic acid, how do we determine the reaction rate? Br 2 (aq) +HCOOH (aq) → 2Br – (aq) +2H + (aq) +CO 2 (g) time

11. 11 Rate Calculations How do we calculate the rate of a reaction? –We first need this information: Time (s) [reactant]

12. 12 Rate Calculations Br 2(aq) + HCOOH (aq) → 2Br – (aq) + 2H + (aq) + CO 2(g)

13. 13 Using this information, calculate the average rate of the bromine reaction over the first 50s of the reaction. Average Rate = -Δ[Br 2 ]/Δt = -[Br 2 ] final – [Br 2 ] initial /[t] final – [t] initial Instantaneous Rate = rate for specific instance in time [Br 2 ] / t

14. 14 Rate Calculations Average Rate = - [Br 2 ] final – [Br 2 ] initial / [t] final – [t] initial Average Rate = - (0.0101- 0.0120)M / (50s – 0s) Average Rate = 0.002M / 50s Average Rate = 3.80 x 10 -5 M/s

15. 15 Rate Laws In the reaction: A  B We may write: Rate  [A] x where x is called the order of the reaction. When x = 1 the reaction is called a first order reaction; when x = 2 the reaction is called a second order reaction, and so on. X can be an integer, a fraction or zero (a zero order reaction has a rate independent on concentration).

16. 16 For more complex reactions, we may have: A + B  Products Rate  [A] x [B] y Rate = k[A] x [B] y Where k is the rate constant. The reaction is said to be of the x th order with respect to A and y th order with respect to B. the overall reaction order is equal to x+y.

17. 17 It should be clear that the order of the reaction with respect to some reactant is not the number of moles in a stoichiometric reaction (it is not the molecularity). For instance, in the reaction: NO 2(g) + CO (g)  CO 2(g) + NO (g) The rate below 225 o C was found to be independent on [CO] and the rate law is given by the relation: Rate = k[NO2] 2 Which means that the reaction is second order with respect to NO 2 and zero order with respect to CO.

18. 18 Another example can be presented by considering the following reaction: 2 ICl (g) + H 2(g)  I 2(g) + 2 HCl (g) The rate law for this reaction was experimentally determined at 230 o C to be: Rate = 0.163 mol L -1 s -1 [ICl][H 2 ] This means that the reaction is first order with respect to both ICl and H 2 and the overall order is two (second order).

19. 19 Rate laws are always determined experimentally Reaction order is always defined in terms of reactants Reactant order is not related to the stoichiomteric coefficient in the overall reaction. F 2 (g) + 2ClO 2 (g) 2FClO 2 (g) rate = k [F 2 ][ClO 2 ]

20. 20 Determining the Rate Law Rate laws (and reaction orders) must be determined experimentally The key is to calculate the initial rate of reaction for different concentrations of reactants 1. If a reaction is zero order for a particular reactant, then changing its concentration will have no effect upon the reaction rate

21. 21 2.If a reaction is first order for a particular reactant, then changing its concentration will cause a direct, proportional change in the reaction rate. In other words, doubling the concentration will double the reaction rate, etc.

22. 22 3.If a reaction is second order for a particular reactant, then changing its concentration will cause an exponential change in the reaction rate. In other words, doubling the concentration will result in a four-fold increase (2 2 ) in reaction rate; tripling the concentration will result in a nine- fold increase (3 2 ) in reaction rate.

23. 23 Example Find the rate law and the rate constant for the reaction: NH 4 + (aq) + NO 2 - (aq)  N 2(g) + 2H 2 O (l) The following data were collected, keeping [NO 2 - ] constant

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26. 26 From the first table we can observe that the initial rate is directly proportional to [NH 4 + ] which suggests that the rate is first order with respect NH 4 +. The same conclusion can be reached from the second table where as the concentration of NO 2 - was doubled, the rate was also doubled which means that the reaction is first order with respect to NO 2 - as well.

27. 27 The overall order of the reaction is thus 2 (a second order reaction) and the rate law is: Rate = k [NH 4 + ][NO 2 - ] To fully determine the rate law we need to determine k. This is done by substitution using concentrations of [NH 4 + ] and [NO 2 - ] with the corresponding reaction rate: 10.8 x 10 -7 = k*(0.20*0.20) k = 2.7x10 -4 mol L -1 s -1 Therefore, the rate law will be: Rate = 2.7x10 -4 mol L -1 s -1 [NH 4 + ][NO 2 - ]

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30. 30 It is clear from the first two experiments that when the concentration of O 3 was doubled, the rate was doubled as well. Therefore, the reaction is first order with respect to O 3 From experiments 2 and 3 keeping the concentration of O 3 constant at 2.0x10 -5 M, decreasing the concentration of NO 2 by one-half results in a decrease of the initial rate by the same value. The reaction is therefore first order with respect to NO 2 The rate law can be written as: Rate = k [NO 2 ][O 3 ]

31. 31 The rate constant can be calculated by substitution for the concentrations of reactants and corresponding rate value, taking the first experiment, for instance, will give: 0.022 mol L -1 s -1 = k * 5.0x10 -5 (mol L -1 )* 1.0x10 -5 (mol L -1 ) k = 4.4x10 7 L mol -1 s -1