The reaction was carried out
in
an evacuated 10.0L
container,
and was started by adding 1.00
mole
I2(g)
and 1.00 mole
H2(g).
After equilibrium was reached,
we analyzed the flask and found
1.60
moles of HI(g).
Kc = [HI(g)]2/[H2(g)][I2(g)]
H2(g)
I2(g)
HI(g)
Init
[]
0.100 M 0.100
M
0.0 M
D[]
-0.0800 M -0.0800
M
0.160 M
Equil
[]
0.0200 M 0.0200
M
0.160 M
Therefore,
Kc = [HI(g)]2/[H2(g)][I2(g)]
= (0.160M)2/(0.0200M)(0.0200M)
Kc =
64.0 OF
COURSE, Kc
FOR THE REVERSE
REACTION = 1/64.0 = 0.0156
***********************************************************************
Find Kc for
the reaction 2 SO2(g) + O2(g)
----> 2 SO3(g)
<----
Suppose 0.400 mole SO2(g)
and 0.200 mole O2(g) are
injected into a closed 1.00L
container. At equilibrium,
0.056 mole of SO3(g) are found. What is Kc ?
Kc = [SO3(g)]2/[SO2(g)][O2(g)]
SO2(g)
O2(g)
SO3(g)
Init
[]
0.400 M 0.200
M
0.00 M
D[]
-0.056 M -0.028
M
0.056 M
Equil
[]
0.344 M 0.172
M
0.056 M
Therefore,
Kc = [SO3(g)]2/[SO2(g)][O2(g)]
Kc
= (0.056M)2/(0.344M)(0.172M)
Kc = 0.154
*************************************************************************
15.3 What Does the Equilibrium Constant Tell Us?
Predicting
the Direction of Reaction from K:
For any given concentrations
or pressures, a value of
Q (the reaction quotient) can be calculated from:
Q = [C]c[D]d/[A]a[B]b
OR
Q = (Pc)c(Pd)d/(PA)a(PB)b
These equations are just
like the K equilibrium
expression, but Q denotes that we are NOT at
equilibrium.
If Q<
K, then the reaction will proceed toward
products
until Q
= K
If Q
> K, then the reaction will proceed toward
reactants
until Q
= K
If Q
= K, then the reaction is at equilibrium already.
***************************************************************************
Calculating
Equilibrium Concentrations
If you know Kc ,
you can find equilibrium concentration
of all
the species:
At 490oC,
Kc =54.00 for H2(g)
+ Br2(g) ----> 2 HBr(g)
<----
Starting with 1.000 mole H2
and 1.000 mole of Br2 in a
1.000L
flask,
find the equilibrium
concentration of all
species.
Kc = [HBr(g)]2/[H2(g)][Br2(g)]
H2(g)
Br2(g)
HBr(g)
Init
[]
1.000 M
1.000
0
D[]
-x
-x
2x
Equil
[]
1.000-x
1.000-x
0+2x
Therefore,
Kc
=
[HBr(g)]2/[H2(g)][Br2(g)]
= 54.00
Kc
= (2x)2/(1.000-x)(1.000-x)
54.00 = (2x)2/(1.000-x)2
NOTE THAT THIS IS A PERFECT SQUARE!!!!
Take square root of both
sides:
(54.00)1/2=(2x)/(1.000-x)
7.3485 = (2x)/(1.000-x)
7.3485(1.000-x) = 2x
7.3485 - 7.3485x
= 2x
7.3485 = 9.3485x
0.7861 = x
Therefore,
[H2] = [Br2]
= 1.000 - x
= 1.000 - 0.7861 = 0.214 M
and [HBr] = 2x = 2(0.7861) = 1.572 M
Check: (1.572)2/(0.214)(0.214)
= 53.96 ==> 54.00
(round-off errors)
*****************************************************************************
Consider the Haber
process:
2 NH3(g) ----> N2(g)
+ 3 H2(g)
<----
Kc =
0.104 at 300oC. If you start
with
2.00 moles of NH3(g)
in a 1.50L
container, what are the equilibrium
concentration
of all species?
H2(g)
N2(g)
NH3(g)
Init
[]
0
0 2.00moles/1.50L
= 1.33M
D[]
+3x
+x
-2x
Equil
[]
+3x
+x
1.33-2x
Kc =
[N2(g)][H2(g)]3/[NH3(g)]2
= 0.104
=(x)(3x)3/(1.33-2x)2
0.104 = 27x4/(1.33-2x)2
Take square root of both sides:
(0.104)1/2 = 5.196x2/1.33-2x
0.322(1.33-2x) = 5.196x2
0.428 - 0.644x - 5.196x2
= 0
5.196x2 + 0.644x -0.428 = 0
Solve this by quadratic
equation:
x = [-b (+/-)(b2-4ac)1/2]/2a
Two possible roots:
x1 = 0.232
**answer**
or
x2 = -0.356 (This cannot
be
correct, since x
is a concentration, & cannot be -)
Therefore, [H2]
= 3x = 3(0.232) = 0.696 M
[N2] = x = 0.232 M
[NH3] = 1.33 - 2x = 1.33 - 2(0.232) =
0.866
M
Check: (0.232)(0.696)3/(0.866)2
= 0.104
,,,,,,,,,,,OK!!!!!!
*****************************************************************************
15.4 Factors That Affect Chemical Equilibrium
A system at equilibrium will
remain
at equilibrium until it
is affected
by some change of conditions. The system
will shift
its equilibrium position so as to counteract the
effect of
the disturbance, and it will go to a NEW position
of
equilibrium.
Le
Chatelier's
Principle: If a stress is applied to
a system
at equilibrium, the system will shift in the direction that
reduces the stress to move toward a new state of
equilibrium.
Ex.: H2(g)
+ I2(g) ----> 2 HI(g)
<----
Changes
in concentrations
1) adding a reactant: will shift the equilibrium to
right ------>
2) adding
a product: will shift the equilibrium to
left <------
3) subtract
a reactant: will shift the equilibrium to
left <------
4) subtract
a product: will shift the equilibrium to
right ------>
Changes
in Pressure and Volume
5)
decrease
the volume: same as increasing the
pressure: will shift towards the direction that
reduces the # moles of gas; if same # moles of
gas on both sides (this example), then no shift
in equilibrium. (AND vice-versa for increase V
and decrease P!).
Changes
in Temperature
6) change
in T: this causes a change in the value of
the equilibrium constant Kc:
*for exothermic processes:
reactants ----> products + heat
<----
treat heat like a product, and therefore, in this
case,
the equilibrium will shift to
the left as T is
increased
or heat is added.
*for endothermic processes:
reactants + heat ----> products
<----
treat heat like a reactant, and therefore, in this
case,
the equilibrium will shift to
the right as T is
increased
or heat is added.
The
Effect
of a Catalyst
7) add
catalyst: this has no effect as it does not change
either Q or Kc
And
8) add
inert gas: this does not alter the partial pressure
of any of the reacting species (reactants or
products)
and therefore there is no
shift in equilibrium.
****************************************************************************
PROBLEM: A 1.00L flask
containing
the equilibrium mixture
CO(g) + Cl2(g) -----> COCl2(g)
was found to contain
0.40 mole COCl2, 0.10 mole CO and
0.50
mole of Cl2.
If 0.30 mole of CO(g) is added at constant volume and T,
calculate the new equilibrium concentration of all
species.
1)
Must
first find the equilibrium constant (not given!!):
Kc
= [COCl2]/[CO][Cl2]
= 0.40M/(0.10M)(0.50M) = 8.0
2)
Find concentrations of species (at new equilibrium):
CO(g)
Cl2(g)
COCl2(g)
Init. equil.
[]
0.10
M
0.50M
0.40M
D[]
+0.30-x
-x
+x
New equil.
[]
0.40-x
0.50-x
0.40+x
3) Plug into Kc
Kc = [COCl2]/[CO][Cl2] = (0.40+x)/(0.50-x)(0.40-x)
Kc = 8.0 = 0.40+x/(0.20 - 0.50x - 0.40x + x2)
1.6 - 7.2x + 8x2 = 0.40+x
8x2 - 8.2x + 1.2 = 0 (Solve using the quadratic)
x1 = 0.85 M (impossible answer, because it is too
large!)
x2 = 0.17 M (correct solution)
Therefore, the new equilibrium concentrations are:
[COCl2] = 0.40 + x = 0.40 + 0.177 = 0.577
M
[CO] = 0.40 - x = 0.40 - 0.177 = 0.223
M
[Cl2] = 0.50 - x = 0.50 - 0.177 = 0.323
M
Check: Kc =
[COCl2]/[CO][Cl2]
= 0.577/(0.223)(0.323)
Kc
= 8.01 ===>
8.0