Rates and Rate Laws

Factors that affect the Rate of a Reaction The Physical state of the reactants The Concentration of the Reactants The Temperature of the Reaction The Presence of a Catalyst

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

2NO2(g)  2NO(g) + O2(g) Reaction Rates: 1. Can measure disappearance of reactants 2. Can measure appearance of products 3. Are proportional stoichiometrically

2NO2(g)  2NO(g) + O2(g) [NO2] t Reaction Rates: 4. Are equal to the slope tangent to that point 5. Change as the reaction proceeds, if the rate is dependent upon concentration [NO2] t

Rate Law Equation 2A (g) + 3 B (aq)  4 C (g) As you write the rate equation remember that is based on changes in concentration so liquids or solids are not included in it. Rate Law = k [A]2[B]3 K is the rate constant and is dependent on Temperature The superscripts on the concentrations are the coefficients in the reaction. This is the best we can do without actual lab data to determine the effects of changing [A] and [B]

Rate Equivalent Equations 2A (g) + 3 B (aq)  4 C (g) Using the equation above we can relate the equivalent rate changes between compounds. Change in concentration over change in time -[A] / 2t = - [B] / 3t = [C] / 4t Notice the rates in the reactants are positive since they are disappearing and the products is positive since it is appearing.

Example 2A (g) + 3 B (aq)  4 C (g) So as an example; What would be the rate of C if you knew the rate of disappearance of A was 4.2 M/s -[A] / 2t = [C] / 4t - (-4.2)/2 = Rate of C/4 Rate of C = 8.4 M/s

Writing a (differential) Rate Law Problem - Write the rate law, determine the value of the rate constant, k, and the overall order for the following reaction: 2 NO(g) + Cl2(g)  2 NOCl(g) Experiment [NO] (mol/L) [Cl2] Rate Mol/L·s 1 0.250 1.43 x 10-6 2 0.500 5.72 x 10-6 3 2.86 x 10-6 4 11.4 x 10-6

Writing a Rate Law Part 1 – Determine the values for the exponents in the rate law: R = k[NO]x[Cl2]y Experiment [NO] (mol/L) [Cl2] Rate Mol/L·s 1 0.250 1.43 x 10-6 2 0.500 5.72 x 10-6 3 2.86 x 10-6 4 1.14 x 10-5 In experiment 1 and 2, [Cl2] is constant while [NO] doubles. The rate quadruples, so the reaction is second order with respect to [NO]  R = k[NO]2[Cl2]y

Writing a Rate Law Part 1 – Determine the values for the exponents in the rate law: R = k[NO]2[Cl2]y Experiment [NO] (mol/L) [Cl2] Rate Mol/L·s 1 0.250 1.43 x 10-6 2 0.500 5.72 x 10-6 3 2.86 x 10-6 4 1.14 x 10-5 In experiment 2 and 4, [NO] is constant while [Cl2] doubles. The rate doubles, so the reaction is first order with respect to [Cl2]  R = k[NO]2[Cl2]

Writing a Rate Law Part 2 – Determine the value for k, the rate constant, by using any set of experimental data: R = k[NO]2[Cl2] Experiment [NO] (mol/L) [Cl2] Rate Mol/L·s 1 0.250 1.43 x 10-6

Writing a Rate Law Part 3 – Determine the overall order for the reaction. R = k[NO]2[Cl2] 2 + 1 = 3  The reaction is 3rd order Overall order is the sum of the exponents, or orders, of the reactants

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

Solving an Integrated Rate Law Time (s) [H2O2] (mol/L) 1.00 120 0.91 300 0.78 600 0.59 1200 0.37 1800 0.22 2400 0.13 3000 0.082 3600 0.050 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)

Time vs. [H2O2] Time (s) [H2O2] 1.00 120 0.91 300 0.78 600 0.59 1200 1.00 120 0.91 300 0.78 600 0.59 1200 0.37 1800 0.22 2400 0.13 3000 0.082 3600 0.050 Regression results: y = ax + b a = -2.64 x 10-4 b = 0.841 r2 = 0.8891 r = -0.9429

Time vs. ln[H2O2] Time (s) ln[H2O2] 120 -0.0943 300 -0.2485 600 120 -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 r2 = 0.99978 r = -0.9999

Time vs. 1/[H2O2] Time (s) 1/[H2O2] 1.00 120 1.0989 300 1.2821 600 1.00 120 1.0989 300 1.2821 600 1.6949 1200 2.7027 1800 4.5455 2400 7.6923 3000 12.195 3600 20.000 Regression results: y = ax + b a = 0.00460 b = -0.847 r2 = 0.8723 r = 0.9340

And the winner is… Time vs. ln[H2O2] 1. As a result, the reaction is 1st order 2. The (differential) rate law is: 3. The integrated rate law is: 4. But…what is the rate constant, k ?

Finding the Rate Constant, k Method #1: Calculate the slope from the Time vs. ln[H2O2] table. Time (s) ln[H2O2] 120 -0.0943 300 -0.2485 600 -0.5276 1200 -0.9943 1800 -1.514 2400 -2.04 3000 -2.501 3600 -2.996 Now remember:  k = -slope k = 8.32 x 10-4s-1

Finding the Rate Constant, k Method #2: Obtain k from the linear regresssion analysis. Regression results: y = ax + b a = -8.35 x 10-4 b = -.005 r2 = 0.99978 r = -0.9999 Now remember:  k = -slope k = 8.35 x 10-4s-1

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

HW Problems [B] Rate of C in M/sec 1.00 4.00 2.00 8.00 3.00 9.00 1.50 A student has the following reaction; A(g) + 3 B (aq)  2 C(aq) Given that the rate of C was 12.2 M/min. Find the rate of B For the balanced equation above writer your predicted rate law. Given the lab data below, find the integrated rate law including value of k and identify order. [A] [B] Rate of C in M/sec 1.00 4.00 2.00 8.00 3.00 9.00 1.50 X 1.25 Y 25.00 4. Did the rate law that you obtained from the lab data match the one you predicted in #2? 5. Use the rate of C in Experiment 3 to find the rate of A.

HW Problems 1 cont 6. Given the lab Data below, write the rate law including value for k. Experiment # [A] [B] Rate of C m/min 1 0.10 2.00 x 10-3 2 0.20 4.00 x 10-3 3 0.30 4.86 x10-1

Homework #2 Given the following Lab Data, Identify order of the reaction and k value. Time (seconds) [A] 0.0 0.100 5.0 0.017 10.0 0.0090 15.0 0.0062 20.0 0.0047 2. In problem above find the Half Life of the reaction.

Homework #2 Continued Time ( Minutes) [CH4] 39 0.274 80. 0.238 140 3. Given the data below find the rate order of the reaction, the K value, Half Life and starting Concentration. Time ( Minutes) [CH4] 39 0.274 80. 0.238 140 0.190 210 0.146