Lecture 16 – Chapter 16, Sections 1-2 Equilibriumkrieg/Chem121_2007/lectures/Lecture16...Lecture...
Transcript of Lecture 16 – Chapter 16, Sections 1-2 Equilibriumkrieg/Chem121_2007/lectures/Lecture16...Lecture...
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?
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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?
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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