Reaction Rates PDF

Reaction Rates Learning Outcomes: 1. To understand the nature and importance of reaction transition states. 2. To understand how this explains the effects of reactant concentration and temperature on reaction rates. http://www.sciencedemonstrations.com/demonstrations/ custard_powd...

Reaction Rates Learning Outcomes: 1. To understand the nature and importance of reaction transition states. 2. To understand how this explains the effects of reactant concentration and temperature on reaction rates. http://www.sciencedemonstrations.com/demonstrations/ custard_powder_flame_thrower.pdf http://www.wichitagov.org/CityOfWichita/Templates/TwoColumnImageGallery.aspx?NRMODE=Published&NRNODEGUID= %7bF1818B64-EF5E-47D7-AD5F-7A4652A21DBF%7d&NRORIGINALURL=%2fCityOffices%2fFire%2fPhotos %2fGallery1%2ehtm&NRCACHEHINT=Guest For this reaction, why don’t A and B immediately reach equilibrium with C and D? Energy Reaction Coordinate There is an “energy barrier”. The high energy intermediate is know as a “transition state”. Energy Reaction Coordinate  Transition States The transition state is highly unstable. Therefore, once it is formed it will rapidly break down either back to the reactants or to the products. The rate of reaction is therefore determined by the rate of formation of the transition state. Consider a nucleophilic substitution reaction (SN2 type). http://en.wikipedia.org/wiki/Image:BromoethaneSN2reaction- small.png The reaction takes place via this transition state. http://en.wikipedia.org/wiki/Image:BromoethaneSN2reaction- small.png Energy Reaction  The first factor in determining the rate of formation of a transition state is the concentration of the reactants involved in forming the transition state (“law of mass action”). For reactions with more than one reactant, this can be visualised as the probability of collisions between reactant molecules.  Collisions not very likely. Collisions more likely. It is likely that if a collision between a molecule of A and B is required for the reaction then (often): Rate  [A] and  [B] so  [A][B] Or Rate = k[A][B] where k is the “rate constant” Note that for the reaction 2A ⇌ C +D It is likely that the rate will often be given by: Rate = k[A]2 (i.e. = k[A][A]) For example, for the reaction we have been considering: CH3CH2Br + OH- ⇌ CH3CH2OH + Br- Both CH3CH2Br and OH- are required to form the transition state, Therefore the concentrations of both affect the reaction rate, which is given by: Rate = k[CH3CH2Br][OH-] In other cases, the transition state can be reached by stretching or bending bonds (molecules are dynamic, not like stick models or pictures on a page), The rate of formation of the transition state is proportional to the number of molecules that could reach the transition state, So the rate of reaction is proportional to concentration of A nucleophilic substitution reaction (SN1 type). Adapted from http://en.wikipedia. org/wiki/Image:Bro moethaneSN2react ion-small.png In this case, this is the transition state for the first reaction step. In cases like this when the transition state is formed from only one of the reactants, then generally only the concentration of this reactant affects the reaction rate. For example for this reaction: (CH3)3CBr + OH- ⇌ (CH3)3COH + Br- the concentration of OH- does not affect the reaction rate, which is given by: Rate = k[(CH3)3CBr]  It should be noted that as C and D accumulate and A and B are used up, the reverse reaction will speed up and the forward reaction slow down and so the conversion of A and B to C and D reaction will slow down and eventually stop. Rate A + B →AB‡ = k1[A][B] Energy Rate C + D →AB‡ = k-1[C][D] Reaction Dispersal of flour or custard particles increases the reaction rate by increasing their surface area. It increases contact between oxygen and the flour or custard so that they can react more efficiently. But what about the flame?  Not all molecules and collisions can form transition states. Formation of the high energy transition state additionally requires a minimum energy. This is called the “activation energy” and is often written as “Ea”. In many cases it can be thought of as the minimum collision energy for the molecules to “stick together”.  Ea Energy Reaction Coordinate  Proportion of molecules Ea Energy  At the lower temperature only these Proportion of molecules react molecules Energy  But at the higher temperature all Proportion of of these molecules react molecules Energy  The effect of this is described by the “Arrhenius equation”. This gives the relationship between the rate constant k and temperature. For example for the equation: Rate = k[A][B] k changes (increases) with temperature. The Arrhenius equation can be written as:  E a / RT k  Ae o r Ea ln k ln A  RT ln(e-E /RT) = -Ea/RT a The important thing to note at this stage is that the relationship is non- linear and that k increases as temperature increases.  A plot of k against temperature. 0.18 0.16 0.14 0.12 k (l 2 s -1 mol -1) 0.1 0.08 0.06 0.04 0.02 0 270 280 290 300 310 320 330 Te m perature (K)  At “normal” temperatures on Earth and for typical activation energies, k values (and therefore the reaction rates) tend to approximately double for a temperature increase of 10 K (or 10°C). k is less affected by temperature if Ea is low. You may have been taught that temperature increases the reaction rate by increasing the frequency of collisions. This does contribute, but only to a very minor extent, On increasing the temperature from 20°C to 30°C (from 293 K to 303K) the increase in collision frequency would be Ea Q5) What is the best description of how catalysts increase reaction rates? a) It reduces the activation energy b) It provides a reaction mechanism with a lower Ea c) It increases the concentration of the transition state d) It provides a source of energy for the reaction e) It makes the transition state more reactive