Kinetics Part IV: Activation Energy Jespersen Chapter 14 Sec 5 & 6 Dr. C. Yau Spring 2014

The Collision Theory The rate of a reaction is proportional to the number of effective collisions per second among the reactant molecules. We already know concentration plays an important part in rxn rate: Conc Freq Collision Rxn Rate Only EFFECTIVE collisions lead to products. Only a small fraction of collisions lead to products, based on two other factors: 1) Activation Energy 2) Molecular Orientation

Kinetic Energy Distribution Ea = Activation energy = minimum energy needed for collision to be effective REMEMBER: This is the graph for Kinetic Energy. Don’t confuse with graph for Potential Energy. Fig 14.11 p.666 Activation Energy is not affected by increase in temperature.

Collision Theory Of Reactions For a reaction to occur, three conditions must be met: Reactant particles must collide. Collision energy must be enough to break bonds/initiate. Particles must be oriented so that the new bonds can form. e.g. NO2Cl + Cl NO2 + Cl2 Figure 13.9 The importance of molecular orientation during a collision in a reaction. The key step in the decomposition of NO2Cl to NO2 and Cl2 is the collision of a Cl atom with a NO2Cl molecule. (a) A poorly oriented collision. (b) An effectively oriented collision.

Eqn Summarizing 3 Factors in Collision Theory Particulate Level: Rxn Rate (molecules L-1 s-1) = N x forientation x fKE N = # collisions per second per liter of mixture forientation = fraction of collisions with effective orientation fKE = fraction of collisions with sufficient kinetic energy for effective collision (area under the curve with KE  Ea

has been found to be related to Ea and T in this equation: Mathematically, fKE has been found to be related to Ea and T in this equation: Still remember what fKE stands for?

Eqn Summarizing 3 Factors in Collision Theory Macroscopic Level: Equation has to be in terms of moles instead of molecules. Conversion factor is So we divide the previous equation by Avogadro’s number to get reaction rate in units of mol L-1 s-1

Temperature Effects Changes in temperature affect the rate constant, k, according to the Arrhenius equation: p is the steric factor Z is the frequency of collisions. Ea is the activation energy R is the Ideal Gas Constant (8.314 J/(mol K) T is the temperature (K) We substitute A (the frequency factor) for (pZ)) Activation Energy (Ea) : The energy needed for the reaction to proceed. Includes the energy needed to break the bonds in the reactant molecules. Frequency Factor (A) : The relative probability that the reactant molecules will collide in the proper orientation to initiate product formation. This is an important equation to remember!

Graphical Determination of Ea You are expected to be able to derive this yourself. How exactly do we determine Ea? What do we plot on the x-axis? on the y-axis? How do we find Ea on the graph?

Example 14.12 p.670: Determine Ea in kJ/mol What do we do with this data?

Ln k vs. 1/T Ln k Then what?... How do we find Ea?

Determination of Ea from k at 2 temperatures Ratio form: Can be used when A isn’t known. You should be able to derive this equation for yourself. We did a similar derivation earlier this semester for Hvap, VP and T. ChemFAQ: Calculate k at a selected temperature, given Ea and k at another temperature.

Example Given that k at 25°C is 4.61×10-1 M/s and that at 50°C it is 4.64×10-1 M/s, what is the activation energy for the reaction? ChemFAQ: Calculate k at a selected temperature, given Ea and k at another temperature. Ea= 188 J/mol = 2 x102 J/mol Can you think of a reason why the graphical method would give a more accurate value for Ea?

Working With The Arrhenius Equation k (M/s) T °C 0.000886 25 0.000894 50 0.000908 100 0.000918 150 Given the following data, predict k at 75°C using the graphical approach. Trendline gives Ea/R and lnA. Substitute in 1/T in kelvin and solve for k. Trendline: y = -36.025x – 6.908 k = ? at T = 75oC ANS k=9.01×10-4M/s

T deg C k 1/T (in K-1) ln k 25 0.000886 0.003355705 -7.028793607 50 0.000894 0.003095975 -7.019804783 150 0.000918 0.002364066 -6.993313167 100 0.000908 0.002680965 -7.004266179

In the reaction 2N2O5(g) 4 NO2(g) + O2(g) the following temperature and rate constant information is obtained. What is the activation energy of the reaction? 102 kJ mol-1 -102 kJ mol-1 1004 kJ mol-1 -1004 kJ mol-1 none of these T (K) k (s-1) 338 328 318 4.87(10-3) 1.50(10-3) 4.98(10-4) Slope of plot = -12246.62383; slope = -Ea/R ; R=8.314e-3 kJ/molK. Incorrect answers from other version of R If we are to determine Ea graphically, what do we graph? Slope = -1.224x104 K and y-intercept = 30.9, what are the units? What is Ea? Practice with Example 14.12 p.672, Exer.26,27,28

Potential Energy Diagrams                                                                                                                                                                                                                                                                                                  The product is said to be “thermodynamically favored” over the reactant. LEARN THIS TERMINOLOGY. Figure 13.11 Potential-energy diagram for an exothermic reaction.

Potential Energy Diagrams demonstrate the energy needs and products as a reaction proceeds tell us whether a reaction is exothermic or endothermic tell us if a reaction occurs in one step or several steps show us which step is the slowest Do not confuse PE diagram with KE diagram! Learn the terminology! So, remember which is the KE diagram?

Potential Energy Diagram What would the potential energy diagram look like for an endothermic reaction? Make a sketch of a PE diagram for an endothermic reaction. Where do we look to find the activation energy? Where do we look to find the heat absorbed during the reaction? What is thermodynamically favored?

Catalysts speed a reaction, but are not consumed by the reaction PE Graph speed a reaction, but are not consumed by the reaction may appear in the rate law lower the Ea for the reaction may be heterogeneous or homogeneous PE Reaction Coordinate Ea of uncatalyzed rxn Ea of catalyzed rxn Figure 13.16 Effect of a catalyst on a reaction. (a) The catalyst provides an alternative, low-energy path from the reactants to the products.

KE Graph: Effect of Catalyst Fraction of Molecules Kinetic Energy Ea without catalyst Ea with catalyst 000 000

Catalytic Actions may serve to weaken bonds through induction may serve to change polarity through amphipathic/surfactant effects may reduce geometric orientation effects Heterogeneous catalyst: reactant and product exist in different states. Homogeneous catalyst: reactants and catalyst exist in the same physical state

Example of a heterogeneous catalyst Well-known “The Haber Process.” Fe 3H2 (g) + N2 (g) 2NH3 (g) Fe Figure 13.17 The Haber process. Catalytic formation of ammonia molecules from hydrogen and nitrogen on the surface of a catalyst. Note: Fe is never consumed. Catalysts do not have to be in large amounts.