Chapter 21 and 14 Test

Tomorrow (Friday, February 11th, 2011) is the test covering nuclear chemistry (Chapter 21) and kinetics / reaction rates (Chapter 14). You have a total of three work packets covering all the material from the two chapters. Solutions to all three packets have been posted on the Events Calendar in the forms of PodCasts. The test will be 13 multiple choice questions (all but one from old AP Chemistry exams) and two free response questions (also from old AP Chemistry exams). You will be able to use a periodic table and equations sheet from the AP Chemistry exam and a calculator on all parts of the exam. Below are some highlights of what you should pay special attention to as you prepare for the exam.

• Make sure that you KNOW the nuclear decay products of alpha, beta, and gamma particles. Know how each would be represented.
• Be able to write a balanced nuclear decay emissions reaction. Make sure that you know how to balance the atomic numbers and atomic mass numbers of the parent and daughter isotopes.
• Know that all nuclear decay rates are first-order. A first-order integrated rate law applications can be used. Original isotope amounts, final isotope amounts, rate constants, and time (including half life time) can be solved for.
• Know how to determine rate order from concentration and time data. This is the integrated rate law which will require you to determine which mathematical relationship ([X] vs time OR ln[X] vs time OR 1/[X] vs time) gives the straightest line.
• The slope of the integrated rate law graph is the rate constant. All rate law constants are positive in value. Zero and first order integrated rate laws have a negative sign in front of the "k" value because of the negative slopes of each graph.
• Each order of reaction has its own units. Look at the packet given in class that summarizes rate law constant units.
• Know how to write an overall rate law.
• Remember, the sum of all of the orders of reactants is the overall rate order. Example: A reactant that is first order and another reactant that is second order will have an overall rate order of 3. An example for the units for the rate constant would be L^2/mol^2 x seconds.
• Know the differential method for determining orders of reactants and rate laws.
• **Be aware of the method for determining order of reactants that was presented in class with the 1997 Free Response question from work packet #2. Once one order is found, that order can be used in subsequent ratios of experiments. Thereby, cancelling concentrations of the reactant that has the known order is not necessary.
• Know how to determine the overall enthalpy (energy) of a reaction and the activation energy from a graph.
• Know that a catalyst lowers the activation energy needed to form an activated complex.
• A catalyst can be used in a reaction to form an intermediate, but is released in subsequent steps of a mechanism, therefore never consumed in the reation. A catalyst used in such a way is a homogeneous (same state of matter as reactant) catalyst. The intermediate is a substance that is later consumed in a later step of the mechanism.
• A heterogeneous catalyst (different state of matter than the reactants) usually controls the orientation of one of the reactants. This means that fewer collisions between molecules will result in the correct orientation for the reaction to occur. Fewer collisions needed for the reaction to occur results in a lower activation energy.
• For a reaction to occur, the particles must strike with: (1) sufficient energy and (2) the correct orientation.
• Know the theories behind why reaction rates change. A change in temperature changes the rate of collisions. Fewer collisions means a slower rate of reaction. More collisions means a faster rate of reaction. Temperature changes alter the value of the rate constant for a reaction.
• Changing the concentrations of reactants changes rate without changing the rate constant. A higher concentration means more collisions will happen between reactants, thus the rate of the reaction will go up. The reverse is true for a decrease in concentration. Fewer molecules, less chance of having a collision. The analogy of being in the hall at school during passing period (high probablility of bumping into someone) creating a high reaction rate OR being in the hall at school during a period (very low probability of bumping into someone) creating a low reaction rate is good when thinking about this concept.
• Be able to solve for any variable from any integrated rate law. (zero, 1st, 2nd)
• A change in concentration of a zero order reactant has NO EFFECT on the rate of reaction. A doubling of a concentration of a 1st order reaction doubles the rate of reaction, the halfing of a concentration of a 1st order reaction cuts the rate in half, etc. Doubling the concentration of a second order reactant will quadruple the reaction rate. Halfing the concentration of a second order reactant will cut the rate by 1/4th.
• The sum of all of the reactants and products of all of the steps of a mechanism must equal the balanced equation. Intermediates and catalysts must cancel out.
• The rate determining step of a mechanism is the slow step in a mechanism. The rate determining step must contain all or parts of the reactants that are part of the rate law.
• Reactants in a fast step can be part of the rate law if they are DIRECTLY responsible for making an intermediate that is consumed in a rate determining (slow) step of a mechanism.
• Know the concept of how the Arrhenius equation can be used to determine the activation energy of a reaction.

Remember, tomorrow morning (Friday, February 11th, 2011) is a review day for the AP Chemistry exam. Also, the rate lab performed today (Thursday, February 10th, 2011) will be a formal lab write-up that will be due Monday, February 14th, 2011. See me before the test if you have any final questions.