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Chapter 13. Exothermic release energy into the surroundings temp of surroundings increases products have less energy - Δ E R  P. - PowerPoint PPT Presentation

### Transcript of Chapter 13

Chapter 13

Exothermic release energy into the surroundings temp of surroundings increases products have less energy - E R P Chapter 13Endothermic absorb energy from the surroundings temp of surroundings decreases products have more energy + E P R

R 1 A + B ----> C + D + 30 joules A + B ----> C + D E = -30 joulesForward reaction is exothermic

Reverse reaction is endothermic30 joules + C + D ----> A + BC + D ----> A + B E = + 30 joules

2Reactions involveBond breaking requires energy +E

Bond formation releases energy -E

If more energy is needed to break bonds than is released +E ( endothermic)

If more energy is released in bond formation than is absorbed -E (exothermic)

3Reaction mechanism- steps by which a reaction occursEach step of a reaction mechanism involves the collision of two molecules.

4Colliding molecules need1. enough energy to break the bonds Activation Energy ( A.E) or threshold energy2. the proper geometry or orientation (correct angle)

5 Energy (P.E.) reaction coordinatec6Factors affecting rate of reactionNature of reactants (number of bonds)Surface area (solids and liquids)TemperatureConcentration of reactants (solutions and gases)Catalyst7Rate =(# of coll/time) (fract with A.E.) (fract with orientation)

If orientation factor is 1 (orientation does not matter)TEMPERATURE affects fraction with A.E.

8Temp - measure of the average K.E.At a given temp all molecules do not have the same K.E.

# of molecules

Kinetic energy threshold energy (A.E)9

Rate =(# of coll/time) (fract with A.E.) (fract with orientation)

Concentrationaffects the # of collisions At a given temp Rate = k (# of collisions /time)

10Rate = k (# of collisions /time)

k large - fast rxnk small - slow rxnConcentration of reactants raised to some power (order)Rate = (k ) [A]x [B]yX and Y (orders)Orders found only by experiment

11Rate = (k ) [A]x [B]y

If concentration of a reactant is doubled and the rate doubles the order is 1If concentration of a reactant is doubled and the rate quadruples the order is 2If changing concentration of a reactant has no effect on the rate of reaction it is not included in the rate lawSum of all orders is the order of the reaction

12

Reaction coordinatePotential energyCatalysts and rate13A catalyst does not change the K.E of the molecules

# of molecules

Kinetic energy

catalyst no catalyst14Catalyst -changes the pathway (steps) Steps require less energyChanges the orientation requirement- more molecules have required orientation Not consumed in the reaction

Reaction mechanism and Rate LawEach step involves the collision of two moleculesadding steps gives net reactionEach step has its own rate law.In the steps the coefficients are the orders of the rate lawSlowest step determines the overall rate -rate determining stepThe rate law for the slowest step is the rate law for the reaction.

16Reaction MechanismA + B- C + 2DD+ B DBDB + D F

A + 2B - C + F17Determining Reaction Mechanism1. Do an experiment to determine the rate law. Experimental rate law

2. Postulate possible steps by which the reaction could take place. Reaction Mechanism

18A2+ B2 2ABrate = k [A2]2 (from experiment)Possible mechanism 1 A2+ B2 2AB rate = k [ A2] [ B2]Predicted does not match experimentalNot the reaction mechanism

19A2+ B2 -> 2ABrate = k [A2]2 (from experiment)Possible mechanism 2 Step 1 A 2 + A2 2A + A2 Step 2 A + B2 AB2 Step 3 AB2 + A 2AB

Step 1 rate = k [A2][A2] or rate =k [A2]2Step 2 rate = k [A][B2]Step 3 rate = k [AB2] [A]Do any rate laws match the experiment rate law? rate = k [ A2]2

202NO + O2 2NO2ExperimentInitial concentration [NO]Initial concentration [O2]

Rate of formation of NO2 (M/s) 1 0 .015 M 0 .015 M 0.048

2 0 .030 M

0 .015 M

0.192 3 0 .015 M

0 .030 M

0.096 4 0 .030 M

0 .030 M

0.38421