11.07.2015 Views

Harpers

Harpers

Harpers

SHOW MORE
SHOW LESS
  • No tags were found...

Create successful ePaper yourself

Turn your PDF publications into a flip-book with our unique Google optimized e-Paper software.

62 / CHAPTER 8that anything which increases the frequency or energy ofcollision between substrates will increase the rate of thereaction in which they participate.TemperatureRaising the temperature increases the kinetic energy ofmolecules. As illustrated in Figure 8–1, the total numberof molecules whose kinetic energy exceeds the energybarrier E act (vertical bar) for formation of productsincreases from low (A), through intermediate (B), tohigh (C) temperatures. Increasing the kinetic energy ofmolecules also increases their motion and therefore thefrequency with which they collide. This combination ofmore frequent and more highly energetic and productivecollisions increases the reaction rate.Reactant ConcentrationThe frequency with which molecules collide is directlyproportionate to their concentrations. For two differentmolecules A and B, the frequency with which they collidewill double if the concentration of either A or B isdoubled. If the concentrations of both A and B are doubled,the probability of collision will increase fourfold.For a chemical reaction proceeding at constant temperaturethat involves one molecule each of A and B,Number ofmolecules∞0Figure 8–1.reactions.AA+ B→PEnergy barrierBKinetic energyThe energy barrier for chemicalC∞(12)the number of molecules that possess kinetic energysufficient to overcome the activation energy barrier willbe a constant. The number of collisions with sufficientenergy to produce product P therefore will be directlyproportionate to the number of collisions between Aand B and thus to their molar concentrations, denotedby square brackets.Rate ∝ [ A][ B]Similarly, for the reaction represented byA+ 2B→P(13)(14)which can also be written asthe corresponding rate expression isor(15)(16)(17)For the general case when n molecules of A react withm molecules of B,the rate expression is(18)(19)Replacing the proportionality constant with an equalsign by introducing a proportionality or rate constantk characteristic of the reaction under study gives equations(20) and (21), in which the subscripts 1 and −1refer to the rate constants for the forward and reversereactions, respectively.(20)(21)K eq Is a Ratio of Rate ConstantsWhile all chemical reactions are to some extent reversible,at equilibrium the overall concentrations of reactantsand products remain constant. At equilibrium,the rate of conversion of substrates to products thereforeequals the rate at which products are converted tosubstrates.Therefore,andA+ B+ B→PRate ∝ [ A][ B][ B]Rate ∝ [ A][ B] 2nA + mB →Pn mRate ∝ [ A] [ B]n mRate1= k 1[ A] [ B]Rate−1= k −1[ P]Rate1= Rate−1n mk1[ A] [ B] = k − 1[ P]k 1 [ P]=kn m−1 [ A] [ B](22)(23)(24)The ratio of k 1 to k −1 is termed the equilibrium constant,K eq . The following important properties of a systemat equilibrium must be kept in mind:(1) The equilibrium constant is a ratio of the reactionrate constants (not the reaction rates).

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!