Based on CaCO3(s)
CaO(s) + CO2(g), which
of the following can attain equilibrium? Start with...
(a) ...pure CaCO3 “YEP.”
(Kp = PCO2)
(it breaks down, forming products until the amt. of CO2 is “right”)
(b) ...CaO and some CO2 at P > Kp
“YEP.”
(too much CO2; it reacts w/available CaO until amt. of CO2 is “right”)
(c) ...CaCO3 and some CO2 at P > Kp
“NOPE.”
(too much CO2, but no CaO for it to react with in order to attain the
“right” amt. of CO2)
(d) ...CaCO3 and CaO “YEP.”
The Magnitude of the Equilibrium Constant
If K >> 1... products are favored.
Eq. “lies to the right.”
If K << 1... reactants are favored.
Eq. “lies to the left.”
The K for the forward and reverse reactions
are NOT the same.
-- they are reciprocals
-- You must write out the equation
and specify the temperature
when reporting a K.
Calculating Equilibrium Constants
(1) If the concentrations of all substances at
equilibrium are known, plug and chug.
Find Kc @ 472oC for N2(g) + 3 H2(g)
2 NH3(g)
At equilibruim @ 472oC…[NH3] = 0.00272 M
[N2] = 0.0402 M
[H2] = 0.1207 M.
2
[NH 3 ]
(0.00272) 2
Kc 
3 
3 = 0.105
[N2 ][H 2 ]
(0.0402)(0 .1207)
(2) If you know the concentrations
of only some substances at
equilibrium, make a chart and
use reaction stoichiometry to
figure out the other concentrations
at equilibrium. THEN plug and chug.
“Mr. B, what kind of chart?”
“Ice,
ice,
baby…”
I = “initial”
C = “change”
E = “equilibrium”
At 1000 K, the amounts shown below are known.
Find Kc @ 1000 K.
initial
2 SO3(g)
6.09 x 10–3 M
2 SO2(g)
0M
+
O2(g)
0M
D
–3.65 x 10–3 M
+3.65 x 10–3 M +1.825 x 10–3 M
at eq.
2.44 x 10–3 M
+3.65 x 10–3 M +1.825 x 10–3 M
Put “at eq.” #s into Kc expression…
(3.65 x 103 )2 (1.825 x 103 )
–3
=
4.08
x
10
Kc 
(2.44 x 103 )2
By knowing K, we can...
(1) predict the direction of a reaction
(2) calc. amts. of R and P once eq. has been
reached, if we know init. amts. of everything
reaction quotient, Q: what you get when you plug the
R and P amts. at any given time
into the eq.-constant expression
P
P
K  at eq., and Q  at any time.
R
R
If Q > K...rxn. proceeds
If Q < K...rxn. proceeds
If Q = K...we are at eq.
(Be sure
you know
WHY.)
For H2(g) + I2(g)
2 HI(g), Kc = 51 at a temp.
of 488oC. If you start with 0.020 mol HI, 0.010 mol H2,
and 0.030 mol I2 in a 2.0-L container, which way will
the reaction proceed?
0.02


[HI]
2
Q

= 1.3
[H ][I ] 0.01 0.03 
2
2
2
2
2
2
rxn. proceeds
to eq.
(need more
< 51 product)
HI!
From an MSDS for HI(aq)… (i.e., hydroiodic acid)
“May cause congenital malformation in the
fetus. Corrosive - causes burns. Harmful if
swallowed, inhaled, or in contact with skin.
Very destructive of mucous membranes. May
affect functioning of thyroid. May increase size of skull.”
For PCl5(g)
PCl3(g) + Cl2(g), Kp = 0.497 at
500 K. At equilibrium, the partial pressures of PCl5
and PCl3 are 0.860 atm and 0.350 atm, respectively.
Find the partial pressure of Cl2.
Kp 
PPCl3  PCl2
PPCl5
 0.497 
0.350 (PCl2 )
0.860
(Did this rxn start with only PCl5?)
1.22 atm
For the reaction in the previous problem, if a gas cylinder at 500 K is charged with PCl5 at 1.66 atm, find the
partial pressures of all three substances at equilibrium.
PCl5(g)
PCl3(g) + Cl2(g)
initial
1.66 atm
0 atm
0 atm
D
–x
x
x
at eq.
1.66 – x
x
x
Kp 
PPCl3  PCl2
PPCl5
2
x
0.497 
1.66  x
 b  b  4ac
x
2a
2
x = 0.693 atm or –1.190 atm
a x  0.497x  0.82502  0
c
b
2
PPCl3  0.693 atm
PCl2  0.693 atm
PPCl5  0.967 atm