Lecture 16 – Chapter 16, Sections 1-2 Equilibriumkrieg/Chem121_2007/lectures/Lecture16...Lecture 16 – Chapter 16, Sections 1-2 Equilibrium • Reversible reactions • Dynamic

Lecture 16 – Chapter 16, Sections 1-2 Equilibrium

• Reversible reactions

• Dynamic Equilibrium

• Equilibrium Constant

• Pure materials, solvents

First, Exam summary

High 100Mean 73.0 (other two sections were 74)

90s 580s 770s 860s 5≤ 60 8

Reactions are never one-way

• Remember from chapter 14 if ∆G < 0 reaction is spontaneous as written

• But if ∆G > 0 reaction is spontaneous in opposite direction

• Thus, it is ALWAYS possible for reactions to go in either direction

• ALL REACTIONS ARE REVERSIBLE

Example

• Last chapter we saw 2 NO2 N2O4

– Rate = kf[NO2]2

• This can just as well be N2O4 2 NO2

– Rate = kb[N2O4]

If there is lots of NO2 and just a little N2O4, which reaction is probably faster?

54321

33%33%33% 1. Forward 2 NO2 N2O4

2. Back N2O4 2 NO2

3. They are the same

If there is lots of N2O4 and just a little NO2, which reaction is probably faster?

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33%33%33% 1. Forward 2 NO2 N2O4

2. Back N2O4 2 NO2

3. They are the same

Dynamic Equilibrium

• At some point the forward rate and back rate will be equal

• This is dynamic equilibrium– Product and reactant are

both being produced and consumed at equal rates.

– Concentrations are constant

Approaching Equilibrium

Dynam

ic Equilibrium

Equilibrium constant

• At equilibrium both reactant and product are present• Rates are equal• Ratio of rate constants called the Equilibrium Constant

[ ] [ ][ ][ ]

[ ][ ]22

42

22

42

422

2

eq

eq

b

feq

eq

eq

b

f

eqbeqf

NOON

kk

K

NO

ONkk

ONkNOk

=≡

=

=

Equilibrium constant

• In general for a reaction likeaA + bB dD + eE

[ ] [ ][ ] [ ]beqa

eq

eeq

deq

eq BAED

K =

Look familiar? Keq is just Q when at equilibrium

A few details…

• Remember that Q is unitless – so is Keq

– Each concentration (or pressure) is implicitly divided by the standard concentration (1 M) or pressure (1 bar or 1 atm)

2

bar 1

bar 1

2

42

=

NO

ON

eqp

p

K

A few more details…

• Pure substances (liquids or solids) by definition have a concentration of 1– Concentrations are the same

before and after reaction• Solvents also have a concentration

of 1.– Technically we use mole

fraction– Solvent mole fraction always

approx. 1– Again amount of solvent is

effectively unchanged

Keqs can be combined• Suppose we have a 2-step reaction

step 1: 2NO2 NO3 + NOstep 2: NO3 O2 + NO . Overall: 2NO2 2NO + O2

• Add reactions to get overall reaction… multiply Keqs

( )

( )( )

( )22

2,

2

21

2

2

3

2

2

3

3

2

2

3

NO

NOO

NO

NOO

NO

NONOtoteq

NO

NOOeq

NO

NONOeq

ppp

ppp

ppp

K

ppp

K

ppp

K

==

=

=

Compare Q and K

• Q and K are constructed the same way– Q is true at all concentrations/pressures– At equilibrium, Q is equal to the value of K

• Comparing Q and K tells us which direction a rxn will go– If Q < K (numerator too small) make products– If Q = K (just right) equilibrium– If Q > K (numerator too big) make reactants

So if Q > Keq, can we say anything about ∆Grxn at these conditions?

54321

25%25%25%25% 1. Yes, ∆Grxn must be < 0

2. Yes, ∆Grxn must be = 03. Yes, ∆Grxn must be > 04. No, don’t talk to me about ∆Grxn

Keq and ∆G are connected

• Recall from Chapter 14

• If we are equilibrium, then…

QRTGG rxnrxn ln+∆=∆ o

eqrxn

eqrxn

eq

rxn

KRTG

KRTG

KQG

ln

ln0

0

−=∆

+∆=

==∆

o

o

Since you know how to calculate ∆G°rxn, you can calculate Keq!

Today• CSI your exam

Wednesday• Read Chapt 16