Download Lecture 02-08-12

Lecture February 6
Chap. 16 - All except the last two (Sec.
16.9 and 16.10) sections - will pick those
up after we do Thermodynamics
Chap. 16 SmartWork HW and Quiz
due Monday Feb. 13
Already Posted
CHEM271 - Spring 12 - Feb 06
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Quiz of the Day
Consider the following reaction:
2 SO2(g) + O2(g) → 2SO3 (g)
If you start with a closed vessel containing 1.00 atm of SO2 and
1.00 atm of O2, what is the final pressure in the vessel after the
reaction has gone to completion. Assume the temperature has
not changed.
2 SO2(g) + O2(g) → 2SO3 (g)
Initial 1.00atm
React -1.00
Final
0
1.00 atm
-0.50
0.50
0
+1.00
1.00
Final pressure = 1.50 atm
CHEM271 - Spring 12 - Feb 06
Remember P
is
proportional
to n
2
Examples of Equilibria
H2O(l)
H2O(g)
KP = PH2O
Remember no
condensed phases
Vapor pressure of water = 0.031 atm at 25oC
(increases with temperature) = 1.00 atm at 100oC
N2(g) + O2(g)
2NO(g)
KP = (PNO2)2/(PN2 PO2) = 4.1 x 10-31 at 25oC
AgCl(s)
Ag+(aq) + Cl-(aq)
KC = [Ag+] [Cl-] = 1.77 x 10-10 at 25oC
K << 1 mostly reactants
CHEM271 - Spring 12 - Feb 06
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Manipulating K
N2(g) + 3H2(g)
2NH3(g)
KP = (PNH32)/(PN2)(PH23)
1) Reverse the reaction:
2NH3(g)
N2(g) + 3H2(g)
KP’ = 1/KP
=(PN2)(PH23)/(PNH32)
2) Multiply by a constant:
2N2(g) + 6H2(g)
1/2N2(g) + 3/2H2(g)
KP’ =KP2
4NH3(g) (PNH34)/(PN22)(PH26)
NH3(g)
KP’ = (KP)1/2
=(PNH3)/(PN21/2)(PH23/2)
CHEM271 - Spring 12 - Feb 06
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3)Combining Chemical Equations
Combine two reactions:
Step 1: 3H2(g) +N2(g)
N2H4(g) + H2(g)
Step 2: N2H4(g) + H2(g)
Net reaction: 3H2(g) + N2(g)
2NH3(g)
K1
K2
2NH3(g)
Knet = K1 K2
Note: If subtracted (2) same as adding the
reverse equation and Knet = K1/K2
CHEM271 - Spring 12 - Feb 06
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Conversions Between KP and KC
CaCO3(s)
CaO(s) + CO2(g)
KP = PCO2 = 3.7x10-4
Sec. 16.3
KP =KC(RT)Δn
Note: CO2 only 0.037% by volume in atmosphere
KC = [CO2]
= 1.5x10-5
n/V = [ ] = P/RT = 1.5x10-5 M at T=298K
Fe2O3(s) + 3CO(g)
2Fe(l) + 2CO2(g)
KP = (PCO2)2/(PCO)3
KC = [CO2]2/[CO]3
Here KC = KP(RT)
Remember: Solids and liquids are not included!
CHEM271 - Spring 12 - Feb 06
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Reaction Quotient Q
Same form as for Keq but using the [ ] (or P) that
you have.
Q < K reaction will move to the product side
(right) to achieve equilibrium
Q > K reaction will move to the reactants
side (left) to achieve equilibrium
Q = K reaction is at equilibrium
CHEM271 - Spring 12 - Feb 06
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Q Example
H2O(g) + CO(g)
H2(g) + CO2(g)
KP = 1.5 at a given T
KP = PH2PCO2/PH20PCO
If you have a gas mixture that is 0.60 atm H2O
(g), 0.80 atm CO(g), 1.0 atm H2(g) and 0.9 atm
CO2, which direction will the reaction go to
achieve equilibrium?
QP = (1.0)(0.9)/(0.6)(0.8) = 1.875
Q>K
reaction will go to the LEFT or REACTANTS
CHEM271 - Spring 12 - Feb 06
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Keq from Reaction Data
H2O(g) + CO(g)
H2(g) + CO2(g)
Given the following:
KC =
1.50x10-2 moles H2O and 1.50x10-2
[H2][CO2]/[H2O][CO]
moles CO in a sealed vessel.
At equilibrium there is 8.3x10-3 moles
of CO2.
-2)2/(.67x10-2)2
=
(.83x10
What is the value of Kc?
NOT
EKC = 1.5
T
ICE table for theh
reaction
er e
are H (g) + CO (g)
H O(g) + CO(g)
n
o
Initial
1.50x10 1.50x10
0 un 0
kno
Change -.83x10 -.83x10 +.83x10 +.83x10 wn
s
2
Equil
2
-2
-2
-2
-2
2
-2
-2
0.67x10-2 0.67x10-2 0.83x10-2 0.83x10-2
CHEM271 - Spring 12 - Feb 06
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Quiz of the Day
The equilibrium constant KP for the thermal
decomposition of NO2 is 6.5x10-6 at
450.oC. What is KC at this temperature?
2 NO2(g)
ANS:
2 NO(g) + O2(g)
KP = PNO2PO2 /PNO22
P = [ ]RT
KP ={[NO]2[O2]/[NO2]2}(RT)3/RT2
KP=KC(RT)
KC
T = 450+273 = 723
KC = (6.5x10-6)/(0.08206)(723) = 1.1x10-7
CHEM271 - Spring 12 - Feb 06
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