Rate Versus … From Experiment Lecture

TP For CH3 3CBr ‒OCH3  CH3 3COCH3 what is the order in ‒OCH3? 25% 25% 25% 25% 1. 2. 3. 4. 1 2 Neither of the above More info needed Br‒ ‒OCH 3 Rate M/s 0.0001 0.0001 2.8 x 10‒5 0.0002 0.0001 5.6 x 10‒5 0.0001 0.0002 2.8 x 10‒5 CH3 3CBr • Rate versus … from experiment • Making sense of rate versus … : Reaction mechanism • Making sense of rate constants: the Arrhenius relation Final lecture: Rate versus temperature: Ea and A. Putting it all together: First law, second law, equilibrium, and kinetics. Differential rate laws: rate vs […] Rate versus … from experiment CH3 3CBr We know how to define rate. We also know that rate In liquid methanol, CH3OH, the following reaction occurs: concentration ‒OCH 3  CH3 3COCH3 Br‒ Write a rate law for this reaction. So, we need to learn just how rate depends on concentration and what determines the rate constant, . We’ll use data to learn how rate depends on concentration, the so‐called differential rate law. We’ll see that each reaction will have it own particular dependence 1 (CH3)3CBr + ‒OCH3  (CH3)3COCH3 + Br‒ Differential rate laws: rate vs […] In liquid methanol, CH3OH, the following reaction occurs: CH3 3CBr ‒OCH 3  CH3 3COCH3 How can we know whether the rate law is correct? ratefor 1. 2. 3. 4. TP For CH3 3CBr ‒OCH3  CH3 3COCH3 what is the order in CH3 3CBr ? 25% 25% 25% 25% 1. 2. 3. 4. 1 2 Neither of the above More info needed Rate M/s 0.0001 2.8 x 10‒5 0.0002 0.0001 5.6 x 10‒5 0.0001 0.0002 2.8 x 10‒5 CH3 3CBr a, ratefor ‒OCH 3 b What is the order, , in CH3 3CBr? What is the order, , in ‒OCH3? What is the full differential rate law? What is the value of kfor? TP For CH3 3CBr ‒OCH3  CH3 3COCH3 what is the order in ‒OCH3? Br‒ ‒OCH 3 Rate M/s 0.0001 0.0001 2.8 x 10‒5 0.0002 0.0001 5.6 x 10‒5 0.0001 0.0002 2.8 x 10‒5 CH3 3CBr ‒OCH 3 0.0001 CH3 3CBr Br‒ 25% 25% 25% 25% 1. 2. 3. 4. 1 2 Neither of the above More info needed Br‒ ‒OCH 3 Rate M/s 0.0001 0.0001 2.8 x 10‒5 0.0002 0.0001 5.6 x 10‒5 0.0001 0.0002 2.8 x 10‒5 CH3 3CBr 2 TP For CH3 3CBr ‒OCH3  CH3 3COCH3 what is the full differential rate law? 25% 25% 25% 25% 1. 2. 3. 4. rate k CH3 3CBr rate k CH3 3CBr rate k ‒OCH3 Neither of the above ‒OCH 3 TP For CH3 3CBr ‒OCH3  CH3 3COCH3 what is the value of k ? Br‒ CH3 3CBr 0.0001 ‒OCH 3 Rate M/s 0.0001 2.8 x 10‒5 10‒5 0.0002 0.0001 5.6 x 0.0001 0.0002 2.8 x 10‒5 25% 25% 25% 25% 1. 2. 3. 4. 2.8 x 10‒5 M s‒1 2.8 x 10‒1 s‒1 2.8 x 10‒1 M s‒1 None of the above Br‒ ‒OCH 3 Rate M/s 0.0001 0.0001 2.8 x 10‒5 0.0002 0.0001 5.6 x 10‒5 0.0001 0.0002 2.8 x 10‒5 CH3 3CBr Making sense of rate versus […] Making sense of rate versus … : Reaction mechanism From data, we have found that the differential rate law for CH3 3CBr is rate ‒OCH 3  CH3 3COCH3 Br‒ k CH3 3CBr Shouldn’t the rate depend on ‒OCH3 too? 3 (CH3)3CBr + ‒OCH3  (CH3)3COCH3 + Br‒ A possible mechanism that accounts for rate not depending on 1 CH3 3CBr  CH3 3C 2 CH3 3C ‒OCH 3 ‒OCH Br‒ , slow  CH3 3COCH3 , fast Show that this accounts for the observed rate law. 3 Making sense of rate constants is … The mechanism 1 CH3 3CBr  CH3 3C 2 CH3 3C ‒OCH 3 Br‒ , slow  CH3 3COCH3 , fast presumes reaction 1 is much slower than reaction 2. Let’s see if we can understand how could this be. Making sense of rate constants Making sense of rate constants: Arrhenius equation The mechanism 1 2 CH3 3CBr  CH3 3C CH3 3C ‒OCH 3 Br‒ , slow  CH3 3COCH3 , fast Reaction 1 is much slower than reaction 2 because … k1,for is much smaller than k2,for What determines the relative size of rate constants? 4 Making sense of rate constants NO2(g) + CO(g)  NO(g) + CO2(g) To understand what determines the relative size of rate constants, let’s begin by imagining how enthalpy changes as an exothermic reaction, R  P, such as Because reactions involve both bond breaking in reactants and bond making in products, energy must be supplied for both R  P and P  R. NO2 g CO g  NO g CO2 g evolves as reactants, R, are transformed into products, P. Sketch a diagram showing how enthalpy changes vertical axis as time passes horizontal axis , starting with all reactants, R, and ending with all products, P. NO2(g) + CO(g)  NO(g) + CO2(g) NO2(g) + CO(g)  NO(g) + CO2(g) Where does the energy come from for R  P? Kinetic energy of reactant molecules. Only molecules with kinetic energy greater than the activation energy, Ea, can react. 5 NO2(g) + CO(g)  NO(g) + CO2(g) NO2(g) + CO(g)  NO(g) + CO2(g) Only molecules with kinetic energy greater than Ea can react Raising temperature increases the number with at least kinetic energy Ea Only molecules with kinetic energy greater than the activation energy, Ea, can react. exp Ea/ R T … Fraction of molecules with minimum kinetic energy Ea for reaction always less than 1 NO2(g) + CO(g)  NO(g) + CO2(g) NO2(g) + CO(g)  NO(g) + CO2(g) What role does the relative orientation of the colliding molecules collision geometry play? Collisions must have the correct geometry : A 6 Putting geometry and temperature together k A A exp Ea/ R T … frequency of collisions … probability that collisions have an appropriate collision geometry exp Ea/ R T … Fraction of molecules with minimum kinetic energy Ea for reaction always less than 1 7