let: rate constant at sea level be ks and that on mountain be km ks/km = 100 ( 3mins as opposed to 300 mins)

Transcription

1 homework solution : "egg question" let: rate constant at sea level be ks and that on mountain be km ks/km = 100 ( 3mins as opposed to 300 mins) ln ks/km = Ea x 10 / 373 x 363 x x = 10Ea/ Ea = 518 kj mol

2 Kinetics and Mechanism: Theories of Reaction Rates: Elementary Reactions:- These are reactions that occur in a SINGLE step i.e. there are no other reactions. Once you know that a reaction is an elementary reaction (and this must be found out by experiment), the rate law follows from the stoichiometry of the reaction. one step - we need some examples to understand this Thus for an elementary reaction, but only for an elementary reaction, the rate equation can be written down just by looking at the balanced chemical equation. (example coming) in an elementary reaction there is one step. in simple words the molecules do not form one intermediate then another...etc on the way to products. 32-2

3 elementary reactions can be thought of as steps taken during the overall reaction a reaction can occur via one elementary step or by coupled elementary steps example Unimolecular Reactions: One molecule reacts/decomposes to give products O 3 6 O 2 + O N 2 O 4 6 2NO 2 2 elementary steps Bimolecular Reactions: Two molecules react to give products NO + O 3 6 NO 2 + O 2 2NO 2 6 N 2 O 4 O + O 2 O 3 Termolecular Reactions are rare - why?? 32-3

4 For Elementary Reactions (mechanisms) the overall order is equal to the molecularity of the Elementary Reaction. example: 2A B Rate = k [A] 2 : molecularity is 2 (2 A's involved) [Recall that the overall rate Law can NOT be predicted from the stoichiometric (reaction) equation]: here it is --but only because this is an elementary step We can think of a mechanism as being a sequence of Elementary Reactions which convert the reactants into products. example A B C The sum of the Elementary Reactions will simply give the overall Stoichiometric Equation. Also if you have a postulated Mechanism, the overall rate law derived from it MUST agree with that which is found by Experiment - or the postulated Mechanism is wrong! we need an example to understand

5 example The reaction of NO 2 in the laboratory CO(g) + NO 2 (g) v NO(g) + O 2 (g) follows the following rate experimental law. Rate = k[no 2 ] 2 (i) Is this an elementary reaction? (ii)does it involve a bimolecular reaction between NO 2 and CO? Answer: The answer to each question is "no". This cannot be an elementary reaction, because the rate law would have been given by: rate = k[no 2 ][CO] Furthermore, only the rate equation just given would be compatible with a bimolecular reaction between NO 2 and CO. The reaction actually IS second order, and we will consider its mechanism in a later lecture. we don t know quite enough to see why yet

6 before we go on ---- note well. the kinetic order of a reaction can only be determined experimentally from the rate equation. You can talk about the order of the reaction, or the order with respect to any reactant. the molecularity can only be applied to an elementary reaction. It describes the number of chemical species participating in the elementary reaction. For elementary processes, the molecularity is the same as the overall kinetic order. more later

7 Collision Theory of Gas Phase Bimolecular Reactions Consider the gas phase bimolecular elementary reaction A + B 6 C + D Rate = -d[a]/dt = -d[b]/dt = k[a][b] collisions between molecules cause reactions + OH - H 3 O + COLLIDE + 2H2 O rate depends on number of successful collisions 32-7

8 Assumptions: 1. A and B must collide to react Rate is proportional to the frequency of collisions per litre (e.g., collisions L -1 s -1 ) The frequency of collisions can be calculated from the average velocity of A and B (from the Kinetic Theory of Gases) and their cross-sectional areas. This frequency is proportional to [A] and [B] i.e. moles of collisions between A and B per litre per second = Z[A][B] {Z = moles of collisions L -1 s -1 : it is the collision frequency when [A] = [B] = 1 M and Z will have units mol -1 Ls -1 } Rate = Z[A][B] units: mol L -1 s -1 mol-1 Ls -1 mol L-1 mol L

9 Assumption #2 Collisions must have greater than some minimum K.E. to result in a reaction. Rate is proportional to the fraction of collisions having K.E. greater than or equal to some value E, where E depends on the reaction. For gases at STP (1 atm, 273 K) the molar volume is 22.4 L which means that the concentration is (1/22.4) M, there are typically around 10 8 moles of collisions L -1 s -1. If every collision resulted in a reaction, the initial rate would be approximately 10 8 mol L -1 s -1 and the reaction would be 99.9% complete in around 10-7 s. So: only a small fraction of the collisions can lead to product in most reactions. Z increases with temperature (Z is proportional to T 0.5 ). 32-9

10 The fraction of collisions having K.E. $ E can also be calculated as e -E/RT. e.g. if E = 100 kj, then the fraction at 298 K is e -E/RT = e-(100 x 1000)/(8.314 x 298) = 3.0 x This fraction increases sharply with temperature and is the main reason that k increases with T. e.g., at 400K and for E =100kJ... e -E/RT = 8.7 x up by over 10,000 times A if E<E a molecules "bounce" apart A B 32-10

11 Assumption # 3.Only collisions having the proper relative orientation of the two reactants can result in a reaction. Rate is proportional to the fraction (p) of the collisions having the proper relative orientation. (steric factor) Putting this all together Rate = p x Z[A][B] x e -E/RT = k[a][b] where: p is the fraction of collisions having the proper orientation for reaction. Z[A][B] is the moles of collisions L -1 s -1 e -E/RT $ E i.e. k = Rate = pze -E/RT [A][B] (Z = moles of collisions L -1 s -1 when [A]=[B]=1 M) Rate = k[a][b] = pz[a][b] e -Ea/RT 32-11

12 Now cast your mind back to the Arrhenius Equation: k = Ae -Ea/RT compare k = pze -Ea/RT i.e. E = E a and pz = A E a is the minimum K.E. for reaction steric factors A collisions 32-12