CHEM 1332 1st Edition Lecture 9 Outline of Last Lecture - Arrhenias equation- Example of a problem- Submicroscopic scale Outline of Current Lecture - Temperature and collisions- Effects of concentration - Reaction mechanism- Slow step Current Lecture- An increase in temperature results in particles moving more quickly- Two things happen the number of collisions increase and the energy of average collision increases therefore the rate increases because the number of collisions that are greater than the activation energy increase- Activation energy stays the same and is independent of temperature- In order for a reaction to happen particles need energy and they get that from collisions- Collisions have to have the correct orientation and relative position of reactant molecules These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.- Rate law is rate=k[CO2][ NO]- If you double concentration of CO2 you double the rate this doubles the number of collisions - If you double NO you double the rate and you double the number of NO possible present to a reaction- If you are given rate=k [ A ]2- If you double A you double the collisions we quadruple the rate- Reaction mechanisms is how does a reaction actually happen at submiscroscopic level -CO2+NO is one step reaction is - A lot of reactions are multi-step - In a mechanism the sum of the steps equals the overall reaction- Each step must be describable by a balance equation and it must be elementary and be unimolecular or bimolecular- The rate is soley determined by the slowest step - Rate law corresponds to reactants of slowest step - For example if the first second and fourth step last less than five minutes and the third step is 30 minutes the whole reaction will take thirty minutes because the reaction depends on the slowest step-(CH3)3CCl+H2O →(CH3)3COH + HCl-rate=k [(CH3)3CCl] in this reaction H2O is the zero order - From this we know that H2O is not in the slow step and the mechanism is more thanone step - Step One −¿+¿+Cl¿(CH3)3CCl→(CH3)C¿ this the slowest step - Step Two +¿+¿+H2O→(CH3)3COH +H¿(CH3)C¿- Step Three −¿→ HCl+¿+Cl¿H¿ - From the steps you can see that there are intermediates that are and then used -2 NO2+H2→ N2O3+H2O- The rate law is given and it is rate=k [ NO2]2[ H2]- It cannot be one step because it would have to be termolecular - The slow step involves NO2, NO2+H2- 1st step NO2+NO2→ NO2O4 This is fast- 2nd step NO2O4+H2→ NO2O3+H2O This is the slow step -rate=k[NO2O4][ H2]-NO2O4 is an intermediate - But it is equal to 2 NO2- Which is equal to rate=k [ NO2]2[ H2]- A catalyst speeds up a reaction by providing an alternate mechanism that has a lower activation energyKey TermsIntermediates-produced by reactants and reacts further to provide productCatalyst- speeds up a reaction by providing an alternate mechanism that has a lower activation