Lecture 4 - Nassau Community College 152/152 lecture 4... · 2012-05-23 · Lecture 4 Professor...
Transcript of Lecture 4 - Nassau Community College 152/152 lecture 4... · 2012-05-23 · Lecture 4 Professor...
Lecture 4Professor Hicks
Inorganic Chemistry (CHE152)
Add the following homework problems
Chapter 14: 61, 63, 69, 71
Equilibrium for a Multi-
step Mechanism
A + 2B + D E
A + 2B C k1F
k1R
C + D E k2F
k2R
At equilibrium forward and backward
rates for the first and second steps
k1F[A][B]2=k1R[C]
k2F[C][D] = k2R[E]
and
Multiply equation 1 by equation 2
= K1K2 = Keq (for the overall reaction)
[C]
[A][B]2
[E]
[C][D]x
[E]
[A][B]2[D]= Keq
=[C]
[A][B]2
k1F
k1R
(1)= K1
=[E]
[C][D]
k2F
k2R
(2)= K2
The equilibrium constant (Keq) for a multistep process can
be obtained from the overall reaction
It is equal to the ratio of products to reactants raised to
their stoichiometric numbers
Reaction quotient (Q)
• Q monitors progress of a reaction
• Q = 0 at moment reactants mixed
• increases as products form
amount products
amount reactantsQ =
• is equal to Keq at equilibrium
Reaction quotient for reactions in solution
Pb2+(aq) +2Cl-(aq) PbCl2 (s)
1
Q = ____________
[Pb2+][Cl-]2
For dissolved substances:
1) molarity appears in Q
2) pure substances (solids and liquids that are not
dissolved) never appear in Q
3) molarities raised to power of n in the overall
reaction
Reaction quotient (Q) for gases
2C8H18(g) +25O2(g) 16CO2(g) + 18H2O (g)
(PCO2 )16(PH2O)18
Q = ____________________
(PC8H18)2(PO2 )25
• for gases the partial pressures appear in Q
• each is raised to power of n
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Uses of Q
• calculating Q is like taking a snapshot of
the reaction
• Tells how far it is from equilibrium
• if Q less than Keq reaction will be moving
forward (towards products)
• if Q = Keq then reaction is at equilibrium
• if Q greater than Keq the reaction will be
moving backwards (towards reactants)
Keq
Equilibrium constant (Keq)
[C][D]
[A][B] Kc =
c stands for concentration in [Molarity]
PCPD
PAPB
Kp =p stands for partial pressure
equilibrium constant ALWAYS CAPITAL K
if it applies to solutions
if it applies to gases
c
Keqp
Meaning of the size of Keq
• if Keq is large the reaction tends to have
more products than reactants at
equilibrium
• if Keq is small the reaction tends to have
more reactants at equilibrium
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Manipulating Keq
• Keq for the reverse of a reaction
Keq = Keq2
Keq (reverse) =1
Keq (forward)
double reaction
Keq = Keq3
triple reaction
etc.
(old )(new) (new) (old )
• when reactions are added Keq for overall
reaction is the product of each of the steps
Keq values
Keq for an overall reaction
A + B C K1 = 10
C D + E K2 = 5
A + B D + E Koverall = K1 K2 = 50
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Le Chateliers’ Principle
• If a system at equilibrium is disturbed it will
move in a direction to counteract the
disturbance
• LCP is used to predict the direction a
reaction will move in response to changes
in temperature, pressure, or amounts of
reactants/products
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disturbance = Cl increased
response = system decreases
[Cl] by moving towards products
Le Chateliers’ Principle
What effect will adding Cl ions have?
Ag+ (aq) + Cl (aq) AgCl (s)
(changes in reactants/products)
Le Chateliers’ Principle - If a system
at equilibrium is disturbed it will respond
by moving in a direction to counteract the
disturbance
Le Chateliers’ Principle
Hb (aq) + O2 (aq) HbO2 (aq) hemoglobin = Hb
oxygenated hemoglobin = HbO2
disturbance = O2 decreased
response = system responds
to raise [O2 ] by moving towards
reactants
this occurs when the hemoglobin
reaches a cell that has a lower O2
concentration due to using it in
metabolism. The shift in equilibrium
is the “release” of oxygen to the cell
(changes in reactants/products)
Le Chateliers’ Principle - If a system
at equilibrium is disturbed it will respond
by moving in a direction to counteract the
disturbance
What will be the effect of reducing the partial pressure of O2 ?
Hemoglobin, O2 and equilibrium
Hb (aq) + (aq) (aq)
lungs [O2] high
cell
Hb (aq) + O2 (aq)
cell [O2] lowlungs
[O2]
high [O2] shifts reaction towards HbO2
hemoglobin = Hb
oxygenated hemoglobin = HbO2
low [O2] shifts reaction towards Hb + O2
O2HbO2
HbO2 (aq)
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Le Chateliers’ Principle (changes in applied
pressure or volume)
• If pressure is increased the system will shift towards the side that has smaller volume to reduce the pressure
• C(s, graphite) C(s, diamond)
larger volume smaller volume
putting graphite under large pressures reduces
the volume and causes it to turn into diamond
Le Chateliers’ Principle (changes in applied pressure or volume)
• If pressure is decreased (or V is increased) the system will respond by shifting towards the side that has larger volume (more moles of gas) attempting to increase the pressure
H2CO3 (aq) CO2 (g) + H2O (l)
less moles gas more moles gas
pressure is decreased when champagne is uncorked the system responds by trying to increase the pressure. This means shifting the equilibrium towards the side with more moles gas releasing CO2 gas
Le Chateliers’ Principle
(changes in temperature)
• raising temperature can be thought of as adding
heat in order to “remove the heat” the reaction will
move in the direction that consumes heat
NH4Cl (s) + heat NH3 (g) + H+ (aq) + Cl- (aq)
heating will drive reaction towards products
reactions that absorb heat (H positive) are said to be endothermic
reactions that release heat (H negative) are said to be exothermic
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LCP and partial pressuresN2 (g) + 3H2 (g) 2NH3 (g)
If the applied pressure is increased on this system it would tend to
decrease the volume – the LCP response is to move to right
The reaction quotient is disturbed by this - that is why it responds to
find a new equilibrium position
Kp = pNH3
2
PN2PH23
Say applied P was doubled
V will be halved
Each partial pressure will be doubled
Q =
22
2 23
= ¼ of previous value
If the total pressure is changed by adding a gas not involved in the reaction at
constant volume none of the partial pressures will be changed so there will be no
need to re-equilibrate
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How far will it go?
• Reactant and products concentrations
stop changing when Q = Keq
Q: How do we figure out [products] and
[reactants] at equilibrium?
A: The algebra to calculate [products] and
[reactants] is usually organized in a table
• Initial-Change-Equilibrium (ICE) are the
three concentrations involved
Initial-Change-Equilibrium (ICE) table
A + 2B D + 3E
initial 0.055 0.28 0 0
(M)
change
(M)
Do the initial values have to add up to or
multiply out to any particular quantity?
Do the changes in concentration
have add up to, or multiply out to
etc. any particular quantity?
No, they can be whatever we
decide to make them.
Yes, they must be in the same
ratio as the balanced equation
0.025 moles A x 2 moles B
1 mole A= 0.05 moles B
react
x moles A x2 moles B
1 mole A= 2x moles B
react
-x -2x
say for instance 0.025 moles A react then if x moles A react
1 2 1 2
1 2 1 3
+x +3x
1 2
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Initial-Change-Equilibrium table
A + 2B D + 3E
initial 0.055 0.28 0 0
(M)
change
(M)
equilibrium 0.055-x 0.28-2x x 3x
(M)
-x -2x +x +3x
the initial values plus changes are the equilibrium values
+ + + +
the equilibrium values if plugged into the reaction quotient will be equal to Keq
plug into Q
Keq
Q = [D][E]3
[A][B]2
Initial-Change-Equilibrium table
A + 2B D + 3E
initial 0.055 0.28 0 0
(M)
change
(M)
equilibrium 0.055-x 0.28-2x x 3x
(M)
-x -2x +x +3x
+ + + +
If the initial concentrations and Keq are known the equilibrium
concentrations can be calculated
The problem is getting x by itself
Keq =[D][E]3
[A][B]2 =x*3x
(.055-x)(.28-2x)
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CO + 2H2 CH3OHinitial (M)
change (M)
equilibrium (M)
Fe3+ + SCN- FeSCN2+
initial (M)
change (M)
equilibrium (M)
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H2 (g) + I2 (g) 2HI (g) initial (M)
change (M)
equilibrium (M)
2NO + Br2 2NOBrinitial (atm)
change (atm)
equilibrium (atm)