EXAM 3 REVIEW LBS 172 REACTION MECHANISMS GENERAL Step by step process of bond making and breaking by which reactants become products Summation of steps must be equal to overall reaction Example NO2 g F2 g FNO2 F g NO2 g F g FNO2 g Overall 2NO2 g F2 g 2FNO2 g TERMS Elementary reaction individual step in a mechanism one chemical event Ex from above rxn NO2 g F2 g FNO2 F g Intermediate species that is formed and subsequently used up in the reaction Ex F g Molecularity number of particles colliding in an elementary step 1 unimolecular 2 bimolecular 3 termolecular Ex NO g F g FNO2 g is a bimolecular step Catalyst species that is used up by the mechanism and is subsequently reformed RATES FOR ELEMENTARY STEPS Rate equation of an elementary step is based on reaction stoichiometry Rate product of rate constant and the concentrations of each reacting particle Neither solids nor pure liquid reactants are used in rate equations For any elementary step A B C D rate k A B Ex Rate for step NO2 g F2 g FNO2 F g k NO2 F2 RATE EQUATIONS BASED ON REACTION MECHANISMS The rate of any reaction is limited by its slowest step the rate of which is often nearly the same as the overall reaction rate Rate determining step RDS the slowest elementary step in a reaction mechanism The RDS involves the highest total energy of all elementary steps in a given mechanism The rate of a rate determining step is assumed to be equal to the rate of its overall reaction 1 A B X M Slow large Ea 2 M A Y Fast small Ea Overall 2A B X Y Step 1 is RDS so rate of overall reaction kstep 1 A B In a laboratory setting proposed mechanisms can be disproved by examining rate dependence on reactants in RDS Mechanisms cannot however be proved correct Reaction mechanisms involving an equilibrium step Common case a fast equilibrium step followed by a slow step Ex 1 NO2 g NO g O g fast equilibrium where k1 is constant for forward rxn and k 1 is constant for reverse reaction 2 O g NO2 g NO g O2 g slow k2 is constant Overall 2NO2 g 2NO g O2 g Rate equation may be written as k2 O NO2 O is an immeasurable intermediate and should not appear in the rate equation Since in equilibrium k1 NO2 k 1 NO O O can be written in terms of measurable species O k1 NO2 k 1 NO Rate equation can be written as k2 k1 NO2 k 1 NO NO2 COLLISION THEORY AND ACTIVATION ENERGY Collision theory states that for any reaction to occur three conditions must be met 1 Reacting molecules must collide 2 Collision must have enough energy to break bonds 3 Molecules must collide in the appropriate orientation Activation energy EA the minimum amount of energy that must be absorbed by a system to cause it to react The proportion of molecules in a sample that have energy at or above EA is directly related to the speed of their reaction Reaction coordinate diagram Shows chemical potential energy over time during a reaction Ex P O T E N T I A L E N E R G Y EA Reactants EA Products TIME The energy of the reactants at left is higher than the energy of the products at right therefore the reaction is exothermic reactants would have lower E than products if endothermic EA is the difference from the beginning E of the sample and the energy required for the sample to react When the sample has reached EA there is enough energy to break the appropriate bond s this occurs at the peak A single peak represents one step New bonds are created to form products To begin a reverse reaction EA must be added to the products to reach the peak Effects of Temperature Heating a sample has the effect of increasing the fraction of molecules in a sample having higher energies This in turn increases the fraction of molecules at or above the activation energy barrier and therefore increases the rate of reaction Heat does not change concentration of reactants so for the rate to increase an increase in temperature must increase the rate constant Arrhenius equation rate constant k Ae Ea RT where A is a parameter called the frequency factor R is the gas constant EA is activation energy and T is temperature EQUILIBRIUM EQUILIBRIUM CONSTANT AND REACTION QUOTIENT Equilibrium constant Reaction aA bB cC dD at equilibrium Equilibrium constant Keq C c D d A a B b Keq may only be expressed at equilibrium Keq depends only on the reaction and temperature Keq has no units Solid and liquid reactants and products are omitted in Keq calculations Keq may be written as Kc as it is based on concentrations From equation for Keq if Keq 1 then reaction is reactant favored if Keq 1 then reaction is product favored Equilibrium constant can be written in terms of partial pressures for reactions involving gasses Reaction aA g bB g cC g dD g at equilibrium Kp PCc PDd PAa PBb By the ideal gas law Kp Kc RT n where n change in moles of gas Reaction quotient Q Reaction aA bB cC dD Q C c D d A a B b Q is the same expression as Keq but it may be expressed when a system is not in equilibrium System is at equilibrium if Q Keq Q can be used to predict concentration changes as systems approach equilibrium If Q K reaction will shift right converting more reactants into products to reach equilibrium If Q K reaction will shift left converting more products into reactants Q can be applied similarly to partial pressures and compared to Kp LeChatelier s Principle and rates Equilibrium will react to a stress to reach equilibrium This can be explained with rates Ex Co2 4Cl CoCl42 If Cl is added Q will decrease rxn will make more CoCl42 to reach equilibrium Why At equilibrium k Cl 4 Co2 k CoCl42 if Cl is added the rate of the forward reaction increases increasing CoCl42 and then increasing the reverse reaction to stabilize at equilibrium In endothermic reactions the addition of heat has the same effect on rate as the addition of reactant In exothermic reactions the addition of heat has the same effect on rate as the addition of product EQUILIBRIUM CALCULATIONS General steps to solve equilibrium problems 1 Write out equation 2 Use Q to determine shift 3 Set up ICE table use balanced equation to find ratios 4 Plug in Eq values into K expression 5 Solve for variable change 6 Use change to find equilibrium concentrations 7 Check answer To avoid using the quadratic formula the addition or subtraction of a variable change x in a K expression can be neglected only if the addition subtraction of x does not significantly affect the number that is being added to subtracted from How to use an ICE table Ex C graphite CO2 g 2CO …