Reaction RateReaction Rate

The rate of appearance of a productThe rate of appearance of a product

or disappearance of a reactant or disappearance of a reactant

units: M/sunits: M/sCollision Theory- reaction rate depends Collision Theory- reaction rate depends

on the rate of “effective” collisions on the rate of “effective” collisions between reactants. between reactants. See “collision Model” videoSee “collision Model” video

Factors that affect reaction Factors that affect reaction raterate Nature of reactantsNature of reactants

Ionic reactions tend to be fast (double Ionic reactions tend to be fast (double displacement)displacement)

Molecular reactions are slower (bonds rearrange)Molecular reactions are slower (bonds rearrange) ConcentrationConcentration

Rate = Rate = kk[A][A]mm[B][B]nn (rate constant x product of concentration (rate constant x product of concentration

of reactants)of reactants) Increase pressure on a gasIncrease pressure on a gas

TemperatureTemperature Directly proportionalDirectly proportional

Surface areaSurface area Catalyst presentCatalyst present

activated complex- a transitional activated complex- a transitional structure created from an effective structure created from an effective collision.collision.

activation energy- the energy activation energy- the energy required to form the activated required to form the activated complexcomplex

EEff = activation energy for forward reaction = 30kJ = activation energy for forward reaction = 30kJ EErr = activation energy for the reverse reaction = = activation energy for the reverse reaction =

60kJ60kJ

catalystcatalyst

a substance that speeds up a chemical reaction without being changed itself.

It lowers the activation energy needed for a reaction to occur.

Exothermic- gives off heatExothermic- gives off heat Products have less energy than Products have less energy than

reactantsreactants Tend to be spontaneousTend to be spontaneous

Endothermic- absorbs heatEndothermic- absorbs heat Products have more energy than Products have more energy than

reactantsreactants

NN2(g) 2(g) + 3H+ 3H2(g) 2(g) 2NH 2NH3(g)3(g)

If the rate of disappearance of HIf the rate of disappearance of H2 2 is .048 is .048 M/s, what would be the rate of formation M/s, what would be the rate of formation of ammonia?of ammonia?

.048M/s x .048M/s x 2 mol NH2 mol NH3 3 = .032 M/s= .032 M/s

3 mol H3 mol H22Use stoichiometry to relate the rate of one species to the rate of another species in the same reaction (under the same conditions)

PredictPredict

Draw a graph of concentration vs. time forDraw a graph of concentration vs. time for

A + 2B A + 2B 3C 3C

http://www.chm.davidson.edu/vce/kinetics/http://www.chm.davidson.edu/vce/kinetics/

ReactionRates.htmlReactionRates.html

Time

Conce

ntr

ati

on (

M)

legend----- [A]…….[B]___ [C]

Reaction rate varies directly as the Reaction rate varies directly as the product of the concentrations of product of the concentrations of

the reactants in the rate the reactants in the rate determining step.determining step. Rate LawRate Law

Rate = Rate = kk[A][A]nn[B][B]mm

KK is the rate constant, increases with is the rate constant, increases with temp.temp.

[A], [B] are molar concentrations of reactants[A], [B] are molar concentrations of reactants

n, m – reaction order, determined n, m – reaction order, determined experimentallyexperimentally

Overall reaction order = n + mOverall reaction order = n + m

0 order: Rate = k0 order: Rate = k

11stst order: Rate = k [A] order: Rate = k [A]

If [A] doubles, rate doublesIf [A] doubles, rate doubles

If [A] triples, rate triplesIf [A] triples, rate triples

22ndnd order: rate = k [A] order: rate = k [A]22

If [A] doubles, rate quadruples (2If [A] doubles, rate quadruples (222))

If [A] triples, rate is 9x (3If [A] triples, rate is 9x (322))

33rdrd order: rate = k [A] order: rate = k [A]33

If [A] doubles, rate is 8x (2If [A] doubles, rate is 8x (233))

If [A] triples, rate is 27x (3If [A] triples, rate is 27x (333))

A + B A + B C CTrialTrial [A][A] [B][B] RateRate

(M/s)(M/s)

11 0.1M0.1M 0.1M0.1M 0.00760.0076

22 0.1M0.1M 0.2M0.2M 0.00760.0076

33 0.2M0.2M 0.1M0.1M 0.03040.0304

Determine the rate expression for the above reaction.

Rate = k[A]2

Determine the rate constant, k.

K = .0076/(.1)2 = .76 M-1s-1

CC22HH44BrBr22(l) + 3I(l) + 3I--(aq) (aq) C C22HH44(l) + 2Br(l) + 2Br--(aq) + I(aq) + I33--

(aq)(aq) Three particles colliding with another is Three particles colliding with another is

unlikely to occur in one step.unlikely to occur in one step. Actual reaction mechanism:Actual reaction mechanism:CC22HH44BrBr22 + I + I- - C C22HH44BrBr-- + IBr (slow) + IBr (slow)

CC22HH44BrBr- - C C22HH44 + Br + Br-- (fast) (fast)

IBr + IIBr + I-- Br Br-- + I + I22 (fast)(fast)

II22 + I + I-- I I33- - (fast)(fast)

Rate = k[CRate = k[C22HH44BrBr22][I][I--]]

overall order of a reaction = twooverall order of a reaction = two

Reaction MechanismsReaction MechanismsNONO22 + F + F22 NONO22F + F slowF + F slow

F + NOF + NO22 NO NO22FF fast fast

2NO2NO22 + F + F22 2NO2NO22F overallF overall The rate law should have the reactants from the The rate law should have the reactants from the

slow (rate determining) step in it.slow (rate determining) step in it. Rate = k [NORate = k [NO22][F][F22]]

Catalyst- appears as a reactant and a product Catalyst- appears as a reactant and a product Intermediate- is created then used up (not a Intermediate- is created then used up (not a

reactant or product.reactant or product.

molecularity Elementary Reaction or slow step

Rate Law

unimolecular A products Rate = k[A]

bimolecular A + A products Rate = k[A]2

bimolecular A + B products Rate = k[A][B]

Termolecular A + A + A products Rate = k[A]3

termolecular A + A + B products Rate = k[A]2 [B]

termolecular A + B + C products Rate = k[A][B][C]

You can determine the rate law from a balanced equation IF it is an elementary reaction (one step) or if you know the slow step.

Write the rate law for the single elementary reaction below.H2(g) + Br2(g) 2HBr(g)Rate = k [H2][Br2]

Half lifeHalf life

TT1/2 1/2 is the time required for the is the time required for the concentration of a reactant to drop by concentration of a reactant to drop by one half.one half.

The longer the half life, the more stable The longer the half life, the more stable the compound.the compound.

NN22OO5(sol)5(sol) 2NO 2NO2(sol)2(sol) + ½O + ½O2(g)2(g)

What is the half life in this reaction? 2.5 hours

What concentration of N2O5 will be left after two half lives (5 hour)?

.37M

order Rate equation

K units Straight line graph

Half life

0 Rate = k M/s [A] vs time

1 Rate = k[A] S-1 ln[A] vs time .693/k

2 Rate = k[A][B] = k[A]2

M-1 s-1 1/[A]t vs time

1/k[A]0

Summary

Based on these graphs of concentration of hydrogen Based on these graphs of concentration of hydrogen peroxide vs. time, what order reaction is this?peroxide vs. time, what order reaction is this?

Which graph is most linear?

Activation energyActivation energy k = Aek = Ae(-Ea/RT)(-Ea/RT) Arrhenius equation Arrhenius equation A is a constant A is a constant related to the number, related to the number,

orientation and frequency of collisions occurring orientation and frequency of collisions occurring between the particles in the reaction, called between the particles in the reaction, called frequency factorfrequency factor

k is rate constant k is rate constant R is universal gas constant R is universal gas constant T is absolute temperature T is absolute temperature EEa a is activation energyis activation energy (J/mol) (J/mol)

e is 2.71828 . . . e is 2.71828 . . .

As activation energy increases, reaction rate (and k) decreases- a smaller fraction of the molecules will have enough energy to react

As seen on the AP Chem equations sheet

A graph of lnk vs 1/T will be a line with slope –Ea /R and a y-intercept of lnA.

Y = mx + b

Sample problemSample problemTemp (°C) K (s-1)

189.7 2.52 x 10-5

198.9 5.25x 10-5

230.3 6.30x 10-4

251.2 3.16x 10-3

From this data, calculate the activation energy.

Temp (K)

1/T (K-1) ln k

462.9 2.160 x 10-3 -10.589

472.1 2.118 x 10-3 -9.855

503.5 1.986 x 10-3 -7.370

524.4 1.907 x 10-3 -5.757

Slope = Slope = -6 – (-10.4)-6 – (-10.4)

.00195-.00215.00195-.00215

= -1.9 x 10= -1.9 x 1044

Slope = - Slope = - EEaa

RR

EEa a == -(slope)(R) -(slope)(R)

= -(-1.9x10= -(-1.9x104 4

K)(8.31J/mol∙K)(1kJ/1000J)K)(8.31J/mol∙K)(1kJ/1000J)

=1.6 x 10=1.6 x 102 2 kJ/molkJ/mol