2. Learning Goals Students will: understand the Rate Law Equation determine the Rate Law Equation given experimental data

3. Success Criteria Students will: Be able to write Rate Law expressions for chemical reactions

4. Purpose We understand: How the rate of a reaction changes during a reaction (the rate decreases as reactants are used up) How to calculate the rate of reaction based on data from a graph. (r = ∆c/∆t) which factors affect the rates of a reaction. What we want to do: Is develop a mathematical equation for rates of reaction (r) that incorporates some rate factors.

5. The Rate Law or Law of Mass Action The rate of a chemical reaction is proportional to the product of the concentrations of the reactants OR The rate, r, will always be proportional to the product of the initial concentrations of the reactants, where these concentrations are raised to some exponential values. For a typical chemical reaction: a A + b B  (products) This can be expressed as r α [A] m [B] n Where the exponents m and n are “orders of reaction” These exponents are NOT related to the co-efficients

6. Rate Law Equation The relationship r α [A] m [B] n can be converted to the Rate Law equation: r = k [A] m [B] n Where: [A] and [B] represent the concentrations of substance A and substance B k = rate law constant (which is not affected by the concentration) m and n are the “orders of reaction”

7. Determining Orders of Reaction The values m and n MUST be determined empirically (by conducting an experiment and mathematically analyzing the data) The exponents, m and n may be zero, fractions or integers. Some “experimental” data is provided below for the following equation: 2 X + 2 Y + 3 Z → products This means the rate law equation will be: r = k[X] m [Y] n [Z] p Let’s determine m, n and p.

8. 2 X + 2 Y + 3 Z → products Determine how the rate of reaction changes due to changes in concentration of X ([X]). To do this, we must keep the concentrations of Y and Z constant – they are controlled variables. Any changes in r must be entirely due to X. Experimental data showed that: As the concentration of X is doubled (x2), the rate of reaction also doubles (x2) As the concentration of X is tripled (x3), the rate of reaction also triples (x3)

9. Determining m We have discovered a linear relationship. r  [X] 1 (y = x 1 ) Linear relationships are known as first order relationships because the exponent is 1. Since the exponent is 1, then m=1

10. 2 X + 2 Y + 3 Z → products Determine how the rate of reaction changes due to changes in concentration of Y ([Y]). To do this, we must keep the concentrations of X and Z constant – they are controlled variables. Any changes in r must be entirely due to Y. Experimental data showed that: As the concentration of Y is doubled (x2), the rate of reaction multiplies by 4 (x4) (4 = 2 2 ) As the concentration of Y is tripled (x3), the rate of reaction multiplies by 9 (x9) (9 = 3 2 )

11. Determining n We have discovered an exponential relationship. r  [Y] 2 (y = x 2 ) Linear relationships are known as second order relationships because the exponent is 2. Since the exponent is 2, then n=2

12. 2 X + 2 Y + 3 Z → products Determine how the rate of reaction changes due to changes in concentration of Z ([Z]). To do this, we must keep the concentrations of X and Y constant – they are controlled variables. Any changes in r must be entirely due to Z. Experimental data showed that: As the concentration of Z is doubled (x2), the rate of reaction does not change (x1) (1 = 2 0 ) As the concentration of Z is tripled (x3), the rate of reaction does not change (x1) (1 = 3 0 )

13. Determining p We have discovered an line relationship. r  [Z] 0 (y = x o ) Line relationships are known as zeroth order relationships because the exponent is 0. Since the exponent is 0, then p=0

14. Rewrite the Rate Law r = k[X] 1 [Y] 2 [Z] 0 The overall order of reaction is determined by adding the individual orders of reaction. In this case 1 + 2 + 0 = 3

15. Let’s use some real data! Start by writing the Rate Law equation r = k[BrO 3 - (aq) ] m [HSO 3 - (aq) ] n Determine how the rate of reaction changes due to changes in concentration of BrO 3 - ([BrO 3 - (aq) ]). To do this, we must keep the concentration of HSO 3 - constant ([HSO 3 - (aq) ]) – it is a controlled variable. Any changes in r must be entirely due to BrO 3 - (aq). Let’s look at the data again – are there any 2 trials we can use for comparison in which [BrO 3 - (aq) ] changes and [HSO 3 - ] remains constant. Reaction: 2 BrO 3 - (aq) + 5 HSO 3 - (aq) = Br 2(g) + 5 SO 4 2- (aq) + H 2 O (l) + 3 H + (aq) trialInitial [BrO 3 - (aq) ](mmol/L)Initial [HSO 3 - (aq) ](mmol/L)Initial rate of Br 2(g) production (mmol/L∙s) 12.03.00.20 22.06.00.80 34.06.01.60

16. Determine m and n Let’s choose trials 2 and 3. [BrO 3 - (aq) ] changes and [HSO 3 - ] remains constant. As the concentration of [BrO 3 - (aq) ] is doubled (x2), the rate of reaction also doubles (x2) (2 = 2 1 ) This is a first-order reaction, m = 1

17. Determine m and n Determine how the rate of reaction changes due to changes in concentration of HSO 3 - ([HSO 3 - (aq) ]). To do this, we must keep the concentration of BrO 3 - constant ([BrO 3 - (aq) ]) – it is a controlled variable. Any changes in r must be entirely due to HSO 3 - (aq). Let’s look at the data again – are there any 2 trials we can use for comparison in which [HSO 3 - ] changes and [BrO 3 - (aq) ] remains constant.

18. Determine m and n Let’s choose trials 1 and 2. [HSO 3 - ] changes and [BrO 3 - (aq) ] remains constant. As the concentration of [HSO 3 - ] is doubled (x2), the rate of reaction multiplies by 4 (x4) (4 = 2 2 ) This is a second-order reaction, m = 2 Rewrite the Rate Law Equation; r = k[BrO 3 - (aq) ] 1 [HSO 3 - (aq) ] 2

19. Now let’s determine k (the Rate Law constant) Since r = k[BrO 3 - (aq) ] 1 [HSO 3 - (aq) ] 2 Then k = r [BrO 3 - (aq) ] 1 [HSO 3 - (aq) ] 2  Go back to the data and choose any trial. I will use trial 1. Input the r, [BrO 3 - (aq )], and [HS.O 3 - (aq) ] values. Therefore: k = 0.20 mmol/L·s [2.0 mmol/L] 1 [3.0 mmol/L] 2 k = 0.011 L 2 /mmol 2 ·s  Now: r = 0.011 L 2 /mmol 2 · s[BrO 3 - (aq) ] 1 [HSO 3 - (aq) ] 2  See page 376 for tips on the units for k

20. Applications Now that we have a completed rate Law Equation: r = 0.011 L 2 /mmol 2 · s[BrO 3 - (aq) ] 1 [HSO 3 - (aq) ] 2 We can input any concentrations of the reactants and determine a rate of reaction. Try This: What is the rate of reaction if: [BrO 3 - (aq) ] = 0.10 mmol/L and [HSO 3 - (aq) ]= 0.10 mmol/L?

21. Review of Order of Reaction

23. Determine the Rate Law Equation for these sets of data Reaction: 2 NO (g) + Br 2(g) → 2 NOBr (g) trialInitial [NO (g) ](mol/L)Initial [Br 2(g) ](mol/L)Initial rate of NOBr (g) production (mol/L∙s) 10.10 0.040 20.100.200.080 30.20 0.320 BrO 3 - (aq) + 5Br - (aq) + 6H + (aq) → 3 Br 2(g) + 3H 2 O (l) trial[BrO 3 - ] (mol/L)[Br - ] (mol/L)[H + ] (mol/L) Initial Rate (mol/L·s) 10.10 8.03 x 10 -4 20.200.10 1.62 x 10 -3 30.20 0.103.17 x 10 -3 40.10 0.203.22 x 10 -3

25. Experiment[BrO 3 - ] (M)[Br - ] (M)[H + ] (M) Initial Rate (M/s) 10.10 8.0 x 10 -4 20.200.10 1.6 x 10 -3 30.20 0.103.2 x 10 -3 40.10 0.203.2 x 10 -3