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Rates and Rate Laws. Reaction Rate The change in concentration of a reactant or product per unit of time.

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Presentation on theme: "Rates and Rate Laws. Reaction Rate The change in concentration of a reactant or product per unit of time."— Presentation transcript:

1 Rates and Rate Laws

2 Reaction Rate The change in concentration of a reactant or product per unit of time

3 2NO 2 (g)  2NO(g) + O 2 (g) Reaction Rates: 2. Can measure appearance of products 1. Can measure disappearance of reactants 3. Are proportional stoichiometrically

4 2NO 2 (g)  2NO(g) + O 2 (g) Reaction Rates: 4. Are equal to the slope tangent to that point  [NO 2 ] tt 5. Change as the reaction proceeds, if the rate is dependent upon concentration

5 Rate Laws Differential rate laws express (reveal) the relationship between the concentration of reactants and the rate of the reaction. Integrated rate laws express (reveal) the relationship between concentration of reactants and time The differential rate law is usually just called “the rate law.”

6 Writing a (differential) Rate Law 2 NO(g) + Cl 2 (g)  2 NOCl(g) Problem - Write the rate law, determine the value of the rate constant, k, and the overall order for the following reaction: Experiment[NO](mol/L) [Cl 2 ] (mol/L)RateMol/L·s 10.2500.250 1.43 x 10 -6 20.5000.250 5.72 x 10 -6 30.2500.500 2.86 x 10 -6 40.5000.500 11.4 x 10 -6

7 Writing a Rate Law Part 1 – Determine the values for the exponents in the rate law: Experiment[NO](mol/L) [Cl 2 ] (mol/L)RateMol/L·s 10.2500.250 1.43 x 10 -6 20.5000.250 5.72 x 10 -6 30.2500.500 2.86 x 10 -6 40.5000.500 1.14 x 10 -5 In experiment 1 and 2, [Cl 2 ] is constant while [NO] doubles. R = k[NO] x [Cl 2 ] y The rate quadruples, so the reaction is second order with respect to [NO]  R = k[NO] 2 [Cl 2 ] y

8 Writing a Rate Law Part 1 – Determine the values for the exponents in the rate law: Experiment[NO](mol/L) [Cl 2 ] (mol/L)RateMol/L·s 10.2500.250 1.43 x 10 -6 20.5000.250 5.72 x 10 -6 30.2500.500 2.86 x 10 -6 40.5000.500 1.14 x 10 -5 R = k[NO] 2 [Cl 2 ] y In experiment 2 and 4, [NO] is constant while [Cl 2 ] doubles. The rate doubles, so the reaction is first order with respect to [Cl 2 ]  R = k[NO] 2 [Cl 2 ]

9 Writing a Rate Law Part 2 – Determine the value for k, the rate constant, by using any set of experimental data: Experiment[NO](mol/L) [Cl 2 ] (mol/L)RateMol/L·s 10.2500.250 1.43 x 10 -6 R = k[NO] 2 [Cl 2 ]

10 Writing a Rate Law Part 3 – Determine the overall order for the reaction. R = k[NO] 2 [Cl 2 ] Overall order is the sum of the exponents, or orders, of the reactants 2+1 = 3  The reaction is 3 rd order

11 Determining Order with Concentration vs. Time data (the Integrated Rate Law) Zero Order: First Order: Second Order:

12 Solving an Integrated Rate Law Problem: Find the integrated rate law and the value for the rate constant, k A graphing calculator with linear regression analysis greatly simplifies this process!! (Click here to download my Rate Laws program for theTi-83 and Ti-84) (Click here to download my Rate Laws program for theTi-83 and Ti-84)

13 Time vs. [H 2 O 2 ] Time (s) [H 2 O 2 ] 01.00 1200.91 3000.78 6000.59 12000.37 18000.22 24000.13 30000.082 36000.050 y = ax + b a = -2.64 x 10 -4 b = 0.841 r 2 = 0.8891 r = -0.9429 Regression results:

14 Time vs. ln[H 2 O 2 ] Time (s) ln[H 2 O 2 ] 00 120-0.0943 300-0.2485 600-0.5276 1200-0.9943 1800-1.514 2400-2.04 3000-2.501 3600-2.996 Regression results: y = ax + b a = -8.35 x 10 -4 b = -.005 r 2 = 0.99978 r = -0.9999

15 Time vs. 1/[H 2 O 2 ] Time (s) 1/[H 2 O 2 ] 01.00 1201.0989 3001.2821 6001.6949 12002.7027 18004.5455 24007.6923 300012.195 360020.000 y = ax + b a = 0.00460 b = -0.847 r 2 = 0.8723 r = 0.9340 Regression results:

16 And the winner is… Time vs. ln[H 2 O 2 ] 1. As a result, the reaction is 1 st order 2. The (differential) rate law is: 3. The integrated rate law is: 4. But…what is the rate constant, k ?

17 Finding the Rate Constant, k Method #1: Method #1: Calculate the slope from the Time vs. ln[H 2 O 2 ] table. Now remember:  k = -slope k = 8.32 x 10 -4 s -1

18 Finding the Rate Constant, k Method #2: Method #2: Obtain k from the linear regresssion analysis. Now remember:  k = -slope k = 8.35 x 10 -4 s -1 Regression results: y = ax + b a = -8.35 x 10 -4 b = -.005 r 2 = 0.99978 r = -0.9999

19 Rate Laws Summary Zero Order First Order Second Order Rate Law Rate = kRate = k[A]Rate = k[A] 2 Integrated Rate Law [A] = -kt + [A] 0 ln[A] = -kt + ln[A] 0 Plot the produces a straight line [A] versus tln[A] versus t Relationship of rate constant to slope of straight line Slope = -k Slope = k Half-Life


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