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advertisement ```Chemical Equilibrium:
“Big K”
kinetics: rate constant “little k”
kinetics “little k” told us how fast a reaction proceeds and
is used to indicate a possible mechanism.
Eq. tells us to what extent a RXN proceeds to completion.
react.  prod. @ Eq: rate forward = rate reverse
H2CO3(l)  H2O(l) + CO2(g)
1
The Equilibrium Constant: K (Temperature Dependent)
(mechanism independent)
aA + bB  dD + eE
Kc 
[products]
[reactants
Kc 
raised to their coeff .
]
raised to their coeff .
[D]
d
a
[E]
[A] [B]
e
b
Law of Mass Action: Values of Kc are constant for a RXN
at a given temperature. Any equilibrium mixture of the
above system at that temperature should give the SAME
Kc value.
2
We also have Kp which is sometimes used when dealing with
gases with the “P” referring to the pressure of the gases.
aA(g) + bB(g)  dD(g) + eE(g)
Kp 
(P D ) (P E )
d
e
a
b
(P A ) (P B )
Since PV = nRT (ideal gas law)
n
V

P
RT

moles
 concentrat
Liters
 n 
ion and .. P  
 RT
V 
so pressure is proportional to concentration.
K P  K c  RT
 n
0.0821L•atm
mol•K
moles of GAS
temp in K
3
Problem:
PCl3(g) + Cl2(g)  PCl5(g)
In a 5.00 L vessel (@ 230oC) an equilibrium mixture is found
to contain: 0.0185 mol PCl3, 0.0158 mol PCl5 and 0.0870 mol
of Cl2.
Question: determine Kc and Kp
Solution:
Kc 
[PC l5 ]
[PC l3 ][C l2 ]

 0.0158 m ol 


 5.00 L

 0.0185 m ol   0.0870 m ol 



 5.00 L
  5.00 L

Kc = 49.08
K P  K c (RT)
n
 49.08(0.08
21  503)
-1
one mol less gas
on product side.
 1.19
Question: what does the value of K mean?
4
Using I.C.E. (Initial, Change, Equilibrium)
Problem: Manufacture of Wood Alcohol. A 1.500 L Vessel
was filled with 0.1500 mol of CO and 0.300 mol of H2. @
Eq. @500K, 0.1187 mol of CO were present. How many
moles of each species were present @ Eq. and what is the
value of Kc?
CO(g) + 2H2(g)  CH3OH(g)
0.3000
0
I.
0.1500
C.
-x
-2x
+x
x
0.3000 - 2x
0.1500 - x
E.
Since @Eq. there were 0.1187 mol CO present, 0.1500 - x = 0.1187
Therefore x = 0.0313
We can now solve for each of the other Eq. terms.
5
Problem: Manufacture of Wood Alcohol. A 1.500 L Vessel was filled with
0.1500 mol of CO and and 0.300 mol of H2. @ Eq. @500K, 0.1187 mol of CO
were present. How many moles of each species were present @ Eq. and what is
the value of Kc?
I.
C.
CO(g) + 2H2(g)  CH3OH(g)
0.3000
0
0.1500
-x
-2x
+x
x
0.3000 - 2x
0.1500 - x
E.
Since @Eq. there were 0.1187 mol CO present, 0.1500 - x = 0.1187
Therefore x = 0.0313
We can now solve for each of the other Eq. terms.
H2: 0.3000 - 2x = 0.2374 moles
CH3OH: x = 0.0313 moles
CO(g) + 2H2(g)  CH3OH(g)
Therefore E. 0.1187
0.2374
0.0313
6
Problem: Manufacture of Wood Alcohol. A 1.500 L Vessel was filled with
0.1500 mol of CO and and 0.300 mol of H2. @ Eq. @500K, 0.1187 mol of CO
were present. How many moles of each species were present @ Eq. and what is
the value of Kc?
CO(g) + 2H2(g)  CH3OH(g)
0.3000
0
0.1500
I.
C.
-x
E.
0.1500 - x
+x
0.3000 - 2x
x
CO(g) + 2H2(g)  CH3OH(g)
0.1187
0.2374
0.0313
Therefore E.
Now find Kc:
-2x
Kc 
[C H 3 O H]
[C O ][H 2 ]
2

 0.0313 


 1.500 
 0.1187   0.2374 



 1.500   1.500 
2
Kc = 10.52
7
More I.C.E.
Problem: @ 77oC, 2.00 mol of NOBr (nitrosyl bromide) placed
in a 1.000L flask dissociates to the extent of 9.4%. Find Kc.
C.
2NOBr(g) 
2.00
-2x
E.
2.00 - 2x
I.
2NO(g) +
0
+2x
2x
Br2(g)
0
+x
x
Now what?
Since 9.4% dissociates, the Change in NOBr, -2x = 2.00(0.094)
and.... |x| = 0.0940
so substituting the x value into the “E. term” gives:
2NOBr(g) 
E.
1.812
2NO(g) +
0.188
Br2(g)
0.0940
8
More I.C.E.
Problem: @ 77oC, 2.00 mol of NOBr (nitrosyl bromide) placed
in a 1.000L flask dissociates to the extent of 9.4%. Find Kc.
I.
C.
E.
2NOBr(g) 
2.00
-2x
2NO(g) +
0
+2x
Br2(g)
0
+x
2.00 - 2x
2NOBr(g) 
2x
2NO(g) +
x
Br2(g)
E.
1.812
0.188
0.0940
2
Kc 
[NO]
2
[Br 2 ]
[N OBr]
2

 0.188   0.0940 

 

1
 1  

 1.812 


 1 
2
What does the value of Kc tell us?
= 1.01 x 10-3
9
Treatment of Pure Solids and Liquids (as solvents) in K expressions.
S(s) + O2(g)  SO2(g)
would expect Kc =
[SO2]
[S(s)][O2]
but since M is meaningless for solids, solids are dropped out.
and...
Kc 
[S O2 ]
[O2 ]
Experimentally Kc is found to be 4.2 x 1052 @ 25oC and
independent of S.
10
AgCl(s)  Ag+(aq) + Cl-(aq)
Ksp = [Ag+(aq)][Cl-(aq)] = 1.8 x 10-10 @ 25oC
This is an EQUILIBRIUM value independent of the amount
of solid AgCl left sitting on the bottom of the container.
11
PURE LIQUIDS (SOLVENTS)
NH3(g) + H2O(l)  NH4+(aq) + OH-(aq)
K 


[NH 4 ][O H ]
 1.8 x 10
-5
[NH 3 ]
Note: by convention the water is ignored.
HCOOH(aq) + H2O(l)  HCOO-(aq) + H3O+(aq)
K 
[HC O O


][H 3 O ]
 1.8 x 10
4
[HC O O H]
12
CaCO3(s)  CaO(s) + CO2(g)
What is the Kc expression?
Kc = [CO2]
What is the Kp expression?
Kp = PCO2
13
Reversing Equations: Reactants become Products and
Products become Reactants.
What is the relationship between the two K values?
1. 2H2(g) + O2(g)  2H2O(g)
K1 
[H 2 O ]
2
2
 3.4 x 10
81
[H 2 ] [O 2 ]
2. 2H2O(g)  2H2(g) + O2(g)
2
[H 2 ] [O 2 ]
 82
K2 
 2.9x10
2
[H 2 O ]
Relationship:
K2 = 1/K1 = K1-1
14
Knet for summing RXN’s:
If a RXN can be obtained from the sum of RXN’s,
KRXN = K1K2
RXN 1:
S(s) + O2 (g)  SO2(g) K 1 
[SO 2 ]
[SO 3 ]
1/2
K net 
[SO 3 ]
[O 2 ]
K 1K 2 
[O 2 ]
x
 2.6 x 10
[SO 2 ][O 2 ]
Net RXN: S(s) + 3/2O2(g)  SO3(g)
[SO 2 ]
52
[O 2 ]
K2 
RXN 2: SO2(g) + 1/2O2(g)  SO3(g)
 4.2 x 10
[SO 3 ]
1/2
[SO 2 ][O 2 ]
KRXN = K1K2

3/2
 1.1 x 10
[SO 3 ]
[O 2 ]
3/2
15
65
12
The RXN Quotient: Qc
Consider a system that may not yet be @ Equilibrium.
aA + bB  dD + eE
Qc 
[D]
d
a
[E]
[A] [B]
If Qc = Kc ?
If Qc &lt; Kc ?
If Qc &gt; Kc ?
e
b
@ Equilibrium
ratio
ratio
Prod.
React.
Prod.
is too small so RXN 
is too large so RXN 
React.
16
Example: PCl5(g)  PCl3(g) + Cl2(g) @ 250oC Kc= 4.0 x 10-2
If: [Cl2] and [PCl3] = 0.30M and [PCl5] = 3.0M, is the system
at Equilibrium? If not, which direction will it proceed?
Qc =
[PC l 3 ][C l 2 ]
[0.30][0.3 0]
-2

=
3.0
x
10
Find Qc and compare to Kc to decide.
[PC l 5 ]
[3.0]
Qc &lt; Kc (not @ Eq.)
Which way must the RXN go to achieve Equilbrium?
Remember ratio is prod./React
more products makes the number bigger
RXN goes
17
Calculations Using Kc: (1st case....Perfect Square)
H2(g) + I2(g)  2HI(g)
@ 699K
Kc = 55.17
Experiment: 1.00 mol of each H2 and I2 in a 0.500 L flask.
Find [ ] of products and reactants @ Equilibrium.
[HI]
C.
H2(g) + I2(g)  2HI(g)
[ ]
[ ]
[ ]
0
2.00
2.00
2x
-x
-x
[H2 ][I2 ]
E.
2.00 -x

[2x]
conc. in mol/L
I.
2
Kc 
2
Kc 

2.00 - x
2x
2
[2x]
[2.00- x][2.00- x]
2
= 55.17
[2.00- x]
“perfect square”
18
Calculations Using Kc: (1st case....Perfect Square--continued)
H2(g) + I2(g)  2HI(g)
@ 699K
H2(g) + I2(g)  2HI(g)
[ ]
[ ]
[ ]
2x
2.00 -x 2.00 - x
2
Kc 
[HI]
[H2 ][I2 ]
E.
2
Kc 
Kc = 55.17
2
[2x]
[2x]

[2.00- x][2.00- x]
2
= 55.17
[2.00- x]
“perfect square”
55.17 
[2x]
[2.00- x]
7.428(2.00 - x) = 2x
1.58 = x
19
Calculations Using Kc: (1st case....Perfect Square)
H2(g) + I2(g)  2HI(g)
@ 699K
2
Kc 
[HI]
[H2 ][I2 ]
E.
Kc = 55.17
H2(g) + I2(g)  2HI(g)
[ ]
[ ]
[ ]
2x
2.00 -x 2.00 - x
7.428(2.00 - x) = 2x
1.58 = x
[H2] = 2.00 - 1.58 = 0.42M
[I2] = 0.42M
[HI] = 2(1.58) = 3.16M
20
Kc Problems with Quadratic Equation
ax2 + bx +c = 0
x
b
2
b  4ac
2a
If equilibrium expression is not a perfect square must use
quadratic equation.
21
Problem:
H2(g) + I2(g)  2HI(g) @ 458 oC Kc = 49.7
Experiment: 1.00 mol H2, 2.00 mol I2 in a 1.00 L flask.
Find: conc. of the equilibrium mixture.
H2(g) + I2(g)  2HI(g)
[ ]
[ ]
[ ]
2.00
0
I.
1.00
-x
2x
-x
C.
2x
2.00 - x
E. 1.00 - x
K
2
Kc 
[HI]
[H2 ][I2 ]
2
K c  49.7 
[2x]
[1.00 - x][2.00 - x]
0.920x2 - 3.00x + 2.00 = 0
22
Problem:
H2(g) + I2(g)  2HI(g) @ 458 oC Kc = 49.7
Experiment: 1.00 mol H2, 2.00 mol I2 in a 1.00 L flask.
Find: conc. of the equilibrium mixture.
2
Kc 
[HI]
[H2 ][I2 ]
I.
C.
2
K c  49.7 
[2x]
[1.00 - x][2.00 - x] E.
H2(g) +
[ ]
1.00
-x
1.00 - x
I2(g)  2HI(g)
[ ]
[ ]
2.00
0
2x
-x
2.00 - x
2x
0.920x2 - 3.00x + 2.00 = 0
2 solutions for x: 1.63  0.70
x = 2.33 or 0.93
will give positive solution for Eq. Conc.
23
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