2/1/10
1
8
Consider the following reaction:
H
2
(g) + F
2
(g)
!
2HF (g)
What is the expression for K?
A)
B)
C)
D)
K
=
HF
[
]
H
2
[
]
F
2
[
]
K
=
2
HF
[
]
F
2
[
]
H
2
[
]
K
=
HF
[
]
2
H
2
[
]
F
2
[
]
K
=
H
2
[
]
F
2
[
]
HF
[
]
2
Form:
Products (RHS)
Reactants (LHS)
Exponents given by
stoichiometric coefficients
1.A. Equilibrium Constants using Pressures of Gases
For gases, use pressure instead of molarity (mol/L
)
** Given concentrations, use them to calculate K
Given pressures, use them to calculate K
All equilibrium constants K
9
H
2
(g) + F
2
(g)
!
2HF (g)
K
=
P
HF
2
P
H
2
P
F
2
Consider the previous reaction:
K is unitless
Equilibrium Constant:
P = pressure in bar
K is constant at a given temperature
(even if pressures do not seem to cancel)
(or atm since 1 bar
!
1 atm)
(K
c
)
(K
p
)
Book uses:
K
c
, K
p
, K
a
, K
b
, K
w
, K
eq
, K
sp
….
KP
=
P
NO
2
2
PNO
2
PO
2
=
KCRT
!
1
If,
(for above reaction)
Generally,
K
P
=
K
c
RT
(
)
!
n
gas
Where
"
n
gas
= mols of gas product  mols of gas reactant
For above reaction,
"
n
gas
= 2  3 = 1
K = K
c
is overall equilibrium constant for gases or solutions
1.
Equilibrium Constants using Pressures of Gases
(Section 17.3)
Ideal Gas Law :
PV = nRT
P = pressure (atm)
V = volume (L)
n = mols
R = 0.08206 L atm mol
1
K
1
T = Temperature (Kelvin)
UNITS !!
data sheet
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 Winter '10
 Brown
 Chemistry, Equilibrium, Molarity, Mole, Reaction, Trigraph

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