Rate Law and Mechanics A + 2B C 2 B E (slow) E + A C (fast)

OUTCOME QUESTION(S): C12-3-10 RATE MECHANICS Vocabulary & Concepts Explain the concept of a reaction mechanism. Include: rate determining step, intermediates, and catalysts Determine the reaction orders and rate law of a chemical reaction from experimental data. Explain the scientific process connecting a chemical reaction to its experimental rate law, and to the prediction of an appropriate reaction mechanism. Include: connecting the rate law to the RDS Vocabulary & Concepts Elementary reaction

aA + bB cC + dD rate = k[A]a[B]b Remember from before - Elementary reaction: A simple reaction that takes place in one step. For reactions that occur in a single step (elementary): The order of each reactant in the rate law is equal to the coefficient in the balanced equation. aA + bB cC + dD rate = k[A]a[B]b

2 H2O(l) → H3O+(aq) + OH-(aq) O3(g) + NO(g) → NO2(g) + O2(g) Determine a rate law for the following elementary reactions: The self-ionization of water: 2 H2O(l) → H3O+(aq) + OH-(aq) Smog – a reaction of ozone and nitrogen monoxide: O3(g) + NO(g) → NO2(g) + O2(g) rate = k[H2O]2 rate = k[O3][NO]

With its own energy requirements and elementary rate law Each step in a mechanism is elementary. With its own energy requirements and elementary rate law A + 2B C 1. 2 B E (slow) 2. E + A C (fast) Rate1 = k[B]2 Rate2 = k[E][A] RateOverall = ? But the overall rate law for a complex reaction must be determined experimentally…

A + 2B C 1. 2 B E (slow) 2. E + A C (fast) RateOverall = k[B]2 Rate1 = k[B]2 Rate2 = k[E][A] Trial [A] mol/L [B] mol/L Rate (mol/Lmin) 1 1.0 0.50 2 3.0 3 2.0 2 RateOverall = k[B]2 The overall rate law matches the information of the rate determining step.

A + 2B C 1. E (slow) 2 B 2. E + A C (fast) Rate = k[B]2 WHY? We already determined: RDS has the greatest effect on the overall rate. Changes to the reactants in the other steps will have no effect. 1. E (slow) 2 B 2. E + A C (fast) A + 2B C Rate = k[B]2 Reactants not found in the RDS (E and A) - Have no measurable effect on rate - Would be zero order (and not appear in the rate law) The RDS is the only step that affects rate; the rate law is an equation of the only reactants that affect rate… – they have to match –

Putting it all together Chemists first determine the rate law experimentally though trials ... THEN deduce a RDS and a hypothetical mechanism of how the reaction could occur. A chemist creates a reaction by putting reactants in a beaker…after some time all that is left is nitryl fluoride –microscopic molecular collisions can’t be viewed, so how might the reaction proceed? 2 NO2 + F2 2 NO2F

2 NO2 + F2 2 NO2F Rate = k[NO2][F2] RDS: NO2 + F2  Analysis of trial data has shown the rate law to be: Rate = k[NO2][F2] The RDS must have NO2 and F2 – using coefficients (1) that match the first orders. RDS: NO2 + F2  The RDS and the rate law must display the same message of what is important

2 NO2 + F2 2 NO2F The proposed mechanism is then: Rate = k[NO2][F2] Of course, the proposed steps must add up to the net equation for the mechanism to be acceptable Step 1: NO2 + F2 NO2F + F- (RDS) Step 2: NO2 + F- NO2F 2 NO2 + F2 2 NO2F Chemists would now stop the reaction early to look for the intermediate (F-) to prove that this mechanism is correct – if not found, a new mechanism is proposed.

NO2 + CO  NO + CO2 Reaction rate data has shown the rate law to be: Rate = k[NO2]2 The proposed mechanism is might be: Step 1: 2 NO2  NO3 + NO (RDS) Step 2: NO3 + CO  NO2 + CO2 You will be asked to make the connection between rate law and the RDS – but not asked to create a mechanism NO2 + CO  NO + CO2

3 M + N P + 2 Q If the experimental rate law was determined to be: Rate = k[M]2 a) What would be the effect of doubling the [N]? b) What would be the effect of tripling the [M]? c) What is the stoichiometry of the RDS? a) No significant effect (zero order – not in RDS) b) 9x increase (2nd order – proportional squared) c) RDS: 2M 

A + B C Rate = k[A][B] If the rate law for this reaction is: What does this evidence suggest? If the rate law already matches the coefficients of the overall reaction – then this is the ONLY step and it is an elementary reaction.

CAN YOU / HAVE YOU? C12-3-10 RATE MECHANICS Vocabulary & Concepts Explain the concept of a reaction mechanism. Include: rate determining step, intermediates, and catalysts Determine the reaction orders and rate law of a chemical reaction from experimental data. Explain the scientific process connecting a chemical reaction to its experimental rate law, and to the prediction of an appropriate reaction mechanism. Include: connecting the rate law to the RDS Vocabulary & Concepts Elementary reaction