Chern 12

1.8 Activation Energy ILl

Activation Energy

Recall that we have discussed a minimum energy that molecules must have inorder to collide successfully. We called this "min energy" the (lct\if Cth\O'il W-tr:Jj

(E~)

This means that there is an energy 'oav(\'e( that the reactant molecules have toovercome.

PEdiagram: rY'U~\-- ~ OV~( -\--he. h \\\ 1;>

PotentialEnergy(kJ)

Proare of Reaction

noK.." ~W~~\OO~

CA- \- p~ J.: a.eF(f\\-\or 14:

pt::r \..eR><.~

The "hill" is where the actual reaction occurs. Reactants without enough energywill climb part way up the hill (but not react). They may collide, but they just" InAne 'off of each other.

PE and KE of the molecules:

KEand PEare inversely related. ~= TE (total energy is constant)

1it ant fP~t)0-"",(,(~1~

Let's look at what is happening to the energy in a successful collision (REACTION):

100 • ~ Fo("rf\ \\90 - Ii \\Gltt\~~a~ LOm~\O<'

--sl~-aown80 - t(jt\~n Pc:70 \J

_1 t605040

3020-

REACTANTSLowPE) h0h Kt

hljh ~E:)low p~PRODUCTS

1 0 ----------------------------------~----------

Progress of Reaction

• At first the reactants are moving quickly (high KE, low PEl.• 2 reactant molecules approach each other:

o they begin to SLoW down

o The outer electrons of each molecule r-e .\ each other

negatively chargedelectron cloud

Repul lveForce negatively charged

electron cloud

o KE and PE If' (KE is converted to PEl

• A reaction (successful collision) occurs!o The molecules add the "gained" PEto the PEthey already possess. If

the PEis sufficient, bonds can be broken and they can form anOcti1t\kc\ Cornp'tX atthetopofthehilll

• Product molecules move apart:o The newly-formed product molecules (.e (l \ each other and move

apart. As they move away, PEis converted to KE (KE ±- and PE ~)

\\Ke COmr~eSSI·':.90.. SpYI nj

2

Comparing a KE distribution curve to a PE diagram:

Let's compare the 2 diagrams to understand why some reactions are SLOW andsome are FAST:

(m\r\ t. ) b~ hi' \ \ (t:~)

\<E. ~E:

Recall: the KEcurve shows us how much KEthe molecules have

If the PE hill (Ea) is large, {,w molecules have sufficient KE ($-L.e... sht1tl..t-d. a (etL0'(1 K E c« vv' e.)

• Very few molecules will be able to react in a given time interval so thereaction rate will be SL-.ovv IL-O\/J

If the PEhill (Ea) is small, O(t. molecules have sufficient KE

• more molecules will be able to react in a given time interval so the reactionrate will be fASt I t-t\G. rr

f~- -

3

The Activation Energy ~~l

o The Ea(or the size of the "hill") is fixed by thenct\y(e. of '( tOt cJ-C\ (It s . In other words, the r, is

determined by the # and strength of bO(\u\S in reactants.o Ex: when there are a lot of strong bonds, the energy hill will be

higher.o The Eais not affected by ilTemp or ileonc

Defn of Ea: the minimum PErequired to change the reactants into the activatedcomplex.

The Activated Complex:

The activated complex is the arrangement of atoms that occurs when the

reactants collide (they are in the process of rearranging to form products). r~~

The activated complex is: PE-- An Lht-e V mec\\' C\-tL molecule

-- t'l\'\St-a'o\e) \I\~E)~'nO ( \- - \\ \j ~C\

2 Requirements of an "Effective Collision":

1. SU ff \"C\".e n-\-\< S (\\e. ~ (>E)

2. to-'"0 (a~\f..- '2J~e -h~(C 0.\\3'("'1m.( Ii-\- 0 ~ (e.RG \- V\ (\ \-, )

(*if the reactants are not aligned correctly, the reaction requires even more

energy to occur. ..Eais even larger!)

Hebden Q's: p. 23 # 34 --40

4

EXOTHERMIC and ENDOTHERMIC Reactions

Forward and Reverse Reactions:

Reactions can go both ways!

REACTANTS PRODUCTS

Notice, however, that the "energy hill" (Ea)is different for the forward ANDthe reverse reactions.

120

100 -j------~806040

E~l (')

\1ev.t' I -the -eV\(~v8i pct({l'tr

-\-or ~~ v-e\l ~v¥-- rxn\ ~ mvtC'h \\\B~(\

'Rev~~~ ()en

\ S S\OvJ-tf

+hO\Yl tcrwo",A

Ex:

200180

'160

AClWATEDCOMPLEX

'140

REACTANTSForward R"~Cli n ••

fxf\ .

PRODUCTS20

Progress OTReaction

Eaf) = Eaof the forward reaction, Ea(r)= Eaof the reverse reaction

5

Relationship between Enthalpy Oi) and Eal

Consider an exothermic and Endothermic reaction PEdiagram. Let's draw in Eaand ~H:

EN\)()

-- -. -

~~.Reaction Coordinate

Reacri on Pa!lh

Ea(f)- 6-\-\ -= E "b-l()

Eb..l~).= E 1x-CY- ) i-lJt+Q

e "

(~~-==-)

You don't have to memorize that equation. You can simply draw yourself a quick

PEdiagram to calculate a value for ~H, Ea(f)or Ea(r).

Here is a question for you!

Label the diagram and calculate:

80E~~~~\/ \ E"bI..lY)

A+B I \

A t+ \,_ C+O2DI~ _

lOCIEa(f);o\ 00 - I..{ 0 ~ 6Q \(}"

Ea(r)= \ 00 - ~O:: so K:J~ ~Gt- \-\(~t-\-

~H = \"" -= ~~- 4 C) ':: - ~ C) ~J

Is the reaction EXOor ENDO?

EXO

P' so(klJ 40

Progr s of the reaction

6

I Do Hebden Q's: p. 25 #41 - 45