1. “K” Chemistry (part 1 of 3) Chapter 13: Reaction Rates and Kinetics

2. Reaction Rates The rate of a chemical reaction is a measure of how fast the reaction occurs A slow reaction will have a small fraction of molecules reacting and forming products A fast reaction will have a large fraction of molecules reaction and forming products

3. As the reactants react and form the products, the concentrations of each change as a function of time

4. Rate Calculations For a balanced chemical reaction: aA + bB  cC + dD

5. Practice Consider the following balanced chemical equation: H 2 O 2(aq) + 3 I - (aq) + 2 H + (aq)  I 3 - (aq) + 2 H 2 O (l) In the first 10.0 seconds of the reaction, the concentration of I - dropped from 1.000 M to 0.868 M. (a)Calculate the average rate of this reaction in this time interval (b)Predict the rate of change in the concentration of H + (Δ[H + ] / Δ t) during this time interval

6. Practice Using the above reaction, predict the rate of change in concentration of hydrogen peroxide and the I 3 -1 during the same time interval

7. Rate Laws The rate of a reaction depends on the concentration of one or more of the reactants For example (a simple decomposition reaction): A  Products As long as the reverse reaction is negligibly slow (or non existent), we can write a relationship (a rate law) Rate = k [A] n

8. Rate Laws Rate = k [A] n k is the proportionality constant called the rate constant n is a number called the reaction order – IT IS NOT NUMBER OF MOLES OR THE STOICHIOMETRY COEFFICIENT!!! – It MUST be determined experimentally!!

9. Reaction Order, n If n = 0, the reaction is zero order and the rate is independent of the concentration of A – By mathematical definition, [A] 0 = 1 so the rate law (Rate = k [A] 0 ) is equal to k regardless of the concentration of A If n = 1, the reaction is first order and the rate is directly proportional to the concentration of A If n = 2, the reaction is second order and the rate is proportional to the square of the concentration of A

10. Zero-Order Reaction Rate = k [A] 0 = k For a zero order reaction, the concentration of the reactant decreases linearly with time The rate is constant because it does not slow down as the [A] decreases

11. First-Order Reaction Rate = k [A] 1 For a first order, the rate is directly proportional to the concentration of the reactant Consequently, the rate slows as the concentration of the reactant decreases

12. Second-Order Reaction Rate = k [A] 2 For a second order reaction, the rate of the reaction is proportional to the square of the concentration of the reactant Consequently, the rate is even more sensitive to the reaction concentration

13. Quick Tip The “order of the reactant” refers to only the reactant of focus A  Products Rate = k [A] n This reaction is n th order with respect to “reactant A” A “reaction order” is the sum of all the exponents

14. Determining the Order of the Reaction The order of the reaction can only be determined by experiment Uses a method called the method of initial rates – The rate for a short period of time at the beginning of the reaction is measured using different initial reactant concentrations to determine the effect of [A] has on the rate

15. Method of Initial Rates Notice for this data: – When the concentration of A doubles, the rate is directly proportional… the rxn is therefore first order with respect to “reactant A” [A] (M)Initial Rate (M/s) 0.100.015 0.200.030 0.400.060

16. The Value of the Rate Constant, k The rate constant, k, can be calculated using this experimentally determined data after the determination of the orders (the exponents) Rate = k [A]1 k = rate/[A] = (0.015 M*s -1 ) / (0.10M) = 0.15s -1 NOTICE THAT: the rate constant for a first-order reaction is s -1

17. Data for Zero-Order Notice for this data: – The initial rate is independent for the reactant concentration – the rate is the same at all measured initial concentrations [A] (M)Initial Rate (M/s) 0.100.015 0.200.015 0.400.015

18. Data for Second-Order Notice for this data: – The initial rate quadruples for a doubling of a reactant concentration. The relationship between concentration and rate is quadratic. [A] (M)Initial Rate (M/s) 0.100.015 0.200.060 0.400.240

19. KEY THINGS!! The rate constants for zero- and second-order reactions have different units than first-order NOTICE THAT: – the rate constant for a zero-order reaction is M*s -1 – the rate constant for a first-order reaction is s -1 – the rate constant for a second-order reaction is M -1 *s -1

20. Reaction Order for Multiple Reactants (not a simple decomp.) aA + bB  cC + dD Again, as long as the reverse is negligibly slow (or non-existent), the rate law can be defined as: Rate = k [A] m [B] n m is the reaction order with respect to A n is the reaction order WRT B The overall order is the sum of the exponents (m+n)

21. Example The reaction between hydrogen gas and iodine has been experimentally determined to be first order WRT both reactants… and thus second order OVERALL H 2(g) + I 2(g)  2HI (g) Rate =k[H 2 ] 1 [I 2 ] 1 The exponents ARE NOT THE COEFFICIENTS!!

22. Example The reaction between hydrogen and nitrogen monoxide has been experimentally determined to be first order WRT hydrogen and second order WRT to nitrogen monoxide 2H 2(g) + 2NO (g)  N 2(g) + 2H 2 O (g) Rate = k[H 2 ] 1 [NO] 2 **Notice the exponents are the ‘experimentally determined’ orders… NOT THE COEFFICIENTS!!

23. Practice Calculating Orders Consider the following reaction between NO 2 and CO: NO 2(g) + CO (g)  NO (g) + CO 2(g) The initial rate of the reaction was measured at several different concentrations if the reactants with the following results: Calculate the rate law for the reaction AND the rate constant (k) [NO 2 ] (M)[CO] (M)Initial Rate (M/s) 0.10 0.0021 0.200.100.0082 0.20 0.0083 0.400.100.033

24. Look for anything held constant!! – If you compare the rates while one reactant is held constant than you can assume the reaction rate is CAUSED by the reactant that is changing!! The [NO 2 ] doubled while the rate quadrupled – Implies this rxn is second order WRT to NO 2 The [CO] (while the [NO2] is constant) doubles while the rate is unchanged… implying zero order [NO 2 ] (M)[CO] (M)Initial Rate (M/s) 0.10 0.0021 0.200.100.0082 0.20 0.0083 0.400.100.033

25. Rate = k[NO 2 ] 2 [CO] 0 = k[NO 2 ] 2 Now, calculate the k – plug in any two corresponding data points = 0.21 M -1 *s -1 **Notice the UNITS!! [NO 2 ] (M)[CO] (M)Initial Rate (M/s) 0.10 0.0021 0.200.100.0082 0.20 0.0083 0.400.100.033

26. Practice #2 Consider the following reaction: CHCl 3(g) + Cl 2(g)  CCl 4(g) + HCl (g) Calculate the rate law for the reaction AND the rate constant (k) [CHCl 3 ] (M)[Cl 2 ] (M)Initial Rate (M/s) 0.010 0.0035 0.0200.0100.0069 0.020 0.0127 0.040 0.027

27. Half Life (t 1/2 ) Read through pages 584-587 and work out the half life problems – In a nutshell, the half life is the time it takes for half of a sample to disappear (break down / decompose)

28. RATE LAW SUMMARY

29. Effects of Temperature – Effects of Pressure (for gases) – Activation Energy – Catalysis – – Enzymes – Collision Theory – Rate Determining Step –

30. End of Chapter 13 WORK THROUGH THE PRACTICE PROBLEMS!!!