Jeff Notes 1

Information from the Rate Law Expression Monday, October 27, 2008 8:19 AM

Consider the rate law, r=k[A]2[B]1 What is the rate determining step? Must be consistent with the rate law 2A + 1B -> Product

Things to Remember 1) 2) 3) 4) 5)

Given the following rate laws, which reaction is slower? R1=k[A]3[B]2

R2=k[A]2[B]2

Actual work:

R1 is going to be slower: 5 particles are needed top collide versus only 4 for R2.

2a) r=k[A] 2b) A2 = 2A (fast) 2A+B (fast) 2AB+B (slow)(rate determining step)

Order of Reaction Gives us information about how changes in concentration of a reactant will affect the rate Example: r=k[A]2[B]1 The order of the reaction with respect to A is second order. The means if we double the concentration of A the rate will quadruple (22= 4) The order of reaction with respect to B is first order This means if we triple the concentration of B the rate will triple (31=3) The overall order predicts how a change in the concentration of all reactants will affect the rate To find the overall order, add the powers of each of the reactants. For our example the overal order is 2+1=3 If we double the concentration of A and B the rate will increase 8 fold (23= 8)

1) An elementary step is just one step involving intermediary reactants, the rate-determining step is the slowest out of these elementary step. a) 2A + B + E -> F b) Step 1 (2A + B = C) because it has the most particles and therefore is most likely to be the rate determining step. 3a) -35kJ 3b) X, because it has the highest potential energy 3c) Exothermic because the product has lower energy than the reactants 3d) V, X, and Z because they are 'hills' (ie increases in energy). 3e) W and Y because the energy decreases in these steps

2) 3)

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5) 6)

Chemistry Page 1

Finish Thurday's Note Questions: 2 pg 390, 1-3 pg 391 Unit Review 1-13 pg 408-9 Rates Worksheet H2O2 Lab (Due Tomorrow)

1a) Funnel with tube of water 1b) Method of initial concentration c) Colour absorption over time Surface area, catalysts, concentration, nature of the reactants This would cause the surface area to become 24cm2 (4 sides would have 0.1cm2, 2 would have 1.00cm2, and multiply by 10 pieces.) Therefore the surface area would increase by a factor of 4 (24/6) and the reaction would be 80mL/s (20mL/s * 4) Increase of rate is relative to the ratio of the increase in concentration squared. b) Rate of reaction decreases c) Rate of reaction increases d) Increase (or decrease depending on nature of reactants) in rate 1.47333mL/s b) Higher rate ii) Slower rate r=k[ClO2]2[OH-]1 b) 3 c) Increases by a factor of 4 d) Increases by a factor of 2