Kinetics Past Paper PDF

Sucrose, C12H22O11, reacts with water in acid solution to give glucose and fructose, which have the same chemical formula. (8 marks) C12H22O11 (aq) + H2O (l)  2 C6H12O6 (aq) Kinetic data were obtained at room temperature for sucrose and the following plots were constructed:...

Sucrose, C12H22O11, reacts with water in acid solution to give glucose and fructose, which have the same chemical formula. (8 marks) C12H22O11 (aq) + H2O (l)  2 C6H12O6 (aq) Kinetic data were obtained at room temperature for sucrose and the following plots were constructed: I & & - &Eperfect - (a) What is the overall order of the reaction? (b) What is the rate constant for the reaction? (c) Starting with [A]0 = 0.0082M, how long will it take for [A]t = 7.5 x 10-5 M? (d) Calculate the half-life for the reaction described in part c). fit Chemical Kinetics (a) What is the overall order of the reaction? (b) What is the rate constant for the reaction? a) 1st order => h[] us to is a straight line h[3 = let + h[Jo slope -le = ~ ~ -0 0037 -h = b ymx. + min Chemical ↓ = 0 0037. Kinetics c) Starting with [A]0 = 0.0082M, how long will it take for [A]t = 7.5 x 10-5 M? - - > d) Calculate the half-life for the reaction described in part c). Ch [ T = = kt + h[T h = 0 0037. min & (7. 5x10 - 5) = - 0. 0037t + h(0. 0082) t min for l = = 0 0037. mil (t = 1 74. min for l = 2 7. min") Chemical Kinetics c) Starting with [A]0 = 0.0082M, how long will it take for [A]t = 7.5 x 10-5 M? d) Calculate the half-life for the reaction described in part c). d) +z = ha - b +z h2 177 min = - = 1 0 0037 min -. Chemical Kinetics Temperature and Rate Generally, as temperature increases, so does the reaction rate. Molecular View: higher T = more Energy -- = more reactive collisions Where does this show up in the rate law? e.g. Rate = k[A] Chemical Kinetics  The Collision Model In a chemical reaction, bonds are broken and new bonds are formed. Molecules can only react if they collide - with each other. - Not collision results in every a reaction !! Chemical Kinetics  The Collision Model Furthermore, molecules must collide with the correct orientation and with enough energy - - to cause bond breakage and formation. N2O(g) + NO(g) N2(g) + N2 2O(g) boad needs to form K Chemical Kinetics  Activation Energy In other words, there is a minimum amount of energy - required for reaction: the activation energy, Ea. - - Just as a ball cannot get over a hill if it does not roll up the hill with enough energy, a reaction cannot occur unless the molecules possess sufficient energy to get over the activation energy barrier. ↑ 2 Chemical Kinetics  Reaction Coordinate Diagrams highest · It is helpful to visualize energy changes throughout a process on a reaction coordinate diagram like this one for the my reaction progress rearrangement of methyl isonitrile. break C-N boud Chemical form C-2 bond Kinetics  Reaction Coordinate Diagrams It shows the energy of the reactants and & products (and, therefore, ∆E). The high point on the diagram is the transition state. Ea ETs ER = - The species present at the transition state is called the activated complex. The energy gap between the reactants and the activated complex is the activation energy barrier. Chemical Kinetics  Distributions of Molecular Kinetic Energies Temperature is defined as a measure of the average kinetic - energy of the molecules in a sample. At any temperature there is a wide distribution of kinetic energies. (some molecules are moving fast while others move slow) The curves above are so-called “Boltzmann - Chemical Kinetics distributions” - Maxwell–Boltzmann Distributions As the temperature # increases, the curve flattens and - broadens. - Thus, at higher temperatures, a larger population of molecules has higher energy. Chemical Kinetics Maxwell–Boltzmann Distributions If the dotted line represents the activation energy, as the temperature increases, so does the fraction of molecules that can overcome the activation energy barrier. & high t As a result, the more molecules reaction rate - have KEY Ea increases. Chemical Kinetics -  Maxwell–Boltzmann Distributions This fraction of molecules can be found through the expression: - activation E - ↑ Temperature where R is the gas constant and T is the temperature. - Chemical Kinetics  Arrhenius Equation Svante Arrhenius developed a mathematical relationship between k and Ea: and T ! How many molecules How many molecules have are oriented properly enough energy to overcome when they collide energy barrier where A is the frequency factor, a number that - represents the likelihood that collisions would occur with the proper orientation for reaction. Chemical Kinetics  Arrhenius Equation Arrhenias Plot Taking the natural logarithm of both sides, the equation becomes 1 RT y = mx + b When k is determined experimentally at several temperatures, Ea can be calculated from the slope of a plot of ln k vs. 1/T. y=ln(k); x=1/T; m=-Ea/R; b=ln(A) - Chemical Kinetics Ea=-slope*R -  1 ln k = -Ea ( ) + ln A E.g. Arrhenius Plot - RT N2O5(CCl4) → N2O4(CCl4) + ½ O2(g) y = mx +b Determine Ea for the rxn: => h() = - Ec + + hA R. W 2x104k slope = -1. Ea 1 2x1044 - - = R. Ea ( 2x104k) (R) =. R = 8 314. K Ea ( 2x10"K)(8 31k) 9976] Fro 99 7 l =.. : :. Ea can also be obtained (less accurately) using just two data points (i.e. measure k @ two temperatures):  − Ea  1  − Ea  1 ln k 2 =   + ln A ln k1 =   + ln A  R  T2  R  T1 Subtract one equation from the other  − Ea  1   − Ea  1  ln k 2 − ln k1 =   + ln A −   + ln A - ↑  R  T2   R  T1   − Ea  1  Ea  1 ln k 2 − ln k1 =   + ln A +   − ln A  R  T2  R  T1 k 2  Ea  1  Ea  1  Ea  1 1  ln =   −   =   −  k1  R  T1  R  T2 -  R  T1 T2  T  Three variations of the Arrhenius Equation − Ea / RT k = Ae  − Ea  1 ln k =   + ln A  R T k 2  Ea  1 1  ln =   −  k1  R  T1 T2  Slide 18 of 56 General Chemistry: Chapter 14 Copyright © 2011 Pearson Canada Inc.  Outline: Kinetics First order Second order Second order Rate Laws Integrate d Rate complicated Laws Half-life complicated k(T) k = Ae − Ea / RT G k 2  Ea  1 1  ln =   −  k1  R  T1 Chemical T2  Kinetics Consider the reaction between nitrogen dioxide and carbon monoxide: NO2(g) + CO(g) → NO(g) + CO2(g) The rate constant at 701 K is measured as 2.57 L mol−1 s−1 and that at 895 K is > - - measured as 567 L mol-1 s−1. Find the activation energy for the reaction in kJ mol−1 hi 2 57 hz 567 h) = = (t () T 701T2 =. - , = = 395 - ↓ (i) E East= 15 Fal - Tak 5 3916 : E (3 09x10-7t) 314].. E1462k An32K 8 Chemical Ea. =. Kinetics Use the results from the previous example to predict the rate constant at 525 K. Ea= 1. 5 -05 To = ht Et -) = h (2) h(k) - = E( - b(b) - 0 944 = - 8 63 - 7 67. la.. = e -10-1 Chemical Kinetics  More Practice The half-life for the decay of radium is 1620 years. What is the rate constant for this first-order process? Chemical Kinetics Chemical Kinetics  More Practice The first-order decomposition of cyclopropane has a rate constant of 6.7 × 10-4 s-1. If the initial concentration of cyclopropane is 1.33 mol L-1, what is the concentration of cyclopropane after 644 s? Chemical Kinetics Chemical Kinetics  More Practice A 1st order rxn has a rate constant of 3.61×10-15 s-1 at 298 K and a rate constant of 8.66×10-7 s-1 at 425 K. Determine the activation energy for this reaction. Chemical Kinetics Chemical Kinetics