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2/5/10
1
A flask contains 1.000 atm H
2
(g) and 2.000 atm F
2
(g). If
they react to form HF(g), calculate the equilibrium
pressures for all gases at 25 °C given:
H
2
(g) +F
2
(g)
!
2HF(g) K = 115.0 at 25 °C
38
A) P
H2
= 0.000atm, P
F2
= 1.000atm, P
HF
= 2.000atm
B) P
H2
= 0.032atm, P
F2
= 1.032atm, P
HF
= 1.936atm
C) P
H2
= 0.017atm, P
F2
= 1.017atm, P
HF
= 1.966atm
D) P
H2
= 0.009atm, P
F2
= 1.009atm, P
HF
= 1.982atm
E) P
H2
= 0.008atm, P
F2
= 1.008atm, P
HF
= 1.984atm
K
=
P
HF
2
P
H2
P
F2
!
115
Shortcut:
=
(1.936)
2
(0.032)(1.032)
=
113
Step 1
. Calculate Q at initial conditions to determine
direction of reaction
39
How to solve an Intermediate K problem?
Step 2
. Make up reaction I.C.E. table
Step 3
. Substitute into K.
Is stoichiometry correct?
Be Careful with C!
Is a species produced (+) or consumed ()?
Solve exactly to find P
equilibrium
or [A]
equilibrium
Solution to previous problem in online notes
Step 1
. Calculate Q at initial conditions
Q
=
P
HF
2
P
H2
P
F2
40
=
0.00
( )
2
1.00
( )
(2.00)
=
0
< K
Not suprisingly, rxn proceeds to right
(no products initially)
41
Step 2
. Make up reaction I.C.E. table
I
C
E
nitial
hange
quilibrium
1.00 atm
2.00 atm
 x
+ 2x
1.00  x
2.00  x
Step 3
. Substitute into K. Solve for x
!
P
equilibrium
K
=
115.0
=
2x
( )
2
1.00  x
( )
2.00
!
x
( )
"
#
$
%
’
=
2x
( )
2
2.00
!
3.0x
+
x
2
Rearrange to get :
111.0 x
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This note was uploaded on 04/22/2010 for the course CHEM 102 taught by Professor Brown during the Winter '10 term at University of Alabama at Birmingham.
 Winter '10
 Brown
 Chemistry, Equilibrium, pH

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