Chapter 14 Chemical Equilibrium
Forward reaction
H2(g) + I 2(g)
2HI(g)
Reverse reaction
2HI(g)
H2(g) + I 2(g)
At equilibrium
H2(g) + I 2(g)
2HI(g)
Chemical equilibrium is reached when reactants combine to form
products at the same rate as the products react to form reactants
forward
A
reverse
B
ratef =kf[A]
A
B
rater =kr[B]
at equilibrium, ratef = rater, k f [A] = k r [B]
Ch 14 Equilibrium 1
Reaction Quotient, Q, and Equilibrium Constant, K
In general,
aA + bB
[C ]c [ D]d
Q=
[ A]a [ B]b
cC + dD
Q=
Pr oducts
Re ac tan ts
Generally use Q when a reaction has not reached equilibrium.
At equilibrium, all [ ] do not change over time, and Q = K
3H2(g) + N 2(g)
2NH3(g)
Whatever effects rates, effects equilibrium -Temp, concentration
aA + bB
cC + dD
At equilibrium: ratef = rater
if ratef = k f [A][B] and rater = k r[C][D]
and kf[A][B] = kr[C][D]
and
[C ]c [ D]d k f
=
=K
a
b
k
[ A] [ B]
r
The first quotient is known at the equilibrium constant expression.
Ch 14 Equilibrium 2
[ C]c [ D]d
K=
[ A]a [ B]b
Law of Mass Action
• Products in the numerator and reactants in the denominator, "Products
over Reactants"
• Raise each concentration to the power of the stoichiometric coefficient
in the balanced equation
• K « 1 favors reactants (10–3)
• K » 1 favors products (103)
•
If 10–3 < K < 103 , products and reactants present
Note: For the reaction
PCl3(g) + Cl 2(g)
PCl5(g)
at 400K K = 2.89
For
PCl5(g)
K=1
PCl3(g) + Cl 2(g)
2.89 = 0.346
The equilibrium constant is dependent on the direction of the reaction
Ch 14 Equilibrium 3
If a chemical equation can be expressed as the sum of two or more
chemical equations, the equilibrium constant for the overall reaction is
the product of the equilibrium constants for the component reactions.
2
PCl3 ]
[
K1 =
3
[ P ]2 [Cl2 ]
2 P( g ) + 3 Cl2 ( g ) ⇔ 2 PCl 3 (g)
PCl3 ( g ) + Cl2 (g) ⇔ PCl5 (g )
K2 =
2 P( g ) + 5 Cl 2 (g ) ⇔ 2 PCl 5 (g)
K3 =
[PCl5 ]
[PCl3 ][Cl2 ]
[ PCl5 ]2
[ P ]2 [Cl2 ]5
2 P( g ) + 3 Cl2 (g ) ⇔ 2 PCl3 (g )
2 PCl3 (g ) + 2 Cl2 (g ) ⇔ 2 PCl5 ( g)
2 P( g ) + 5 Cl 2 (g ) ⇔ 2 PCl5 (g)
[ PCl5 ]2
2
PCl 3 ]
[
[PCl5 ]
K3 =
=
×
5
3
[ P ]2 [Cl2 ] [P ]2 [Cl2 ] [ PCl3 ][Cl2 ]
×
[PCl 5 ]
[PCl3 ][Cl2 ]
= K1 × K 2 × K 2 = K1 K 22
Relations Between Equilibrium Constants
Chemical Equation
Equilibrium Constant
aA + bB ¸ cC + dD
K1
cC + dD ¸ aA + bB
K2 =
naA + nbB ¸ ncC + ndD
1
= K1−1
K1
K3 = K1n
H 2S, Ka1 = 9.5x10 -8 , Ka2 = 1x10-19
Ch 14 Equilibrium 4
Be aware of K c and K p
Kc is used with concentration expressed in molarity or moles per liter.
Kp is used with pressures, if you are given partial pressures in
atmospheres.
 n
P =   RT
 V
 n 
V 
3H2(g) + N 2(g)
KP =
2
PNH
3
PH32 PN2
is a measure of concentration
2NH3(g)
(
[ NH3 ]RT )2
( RT )2
=
= Kc
= K c (RT )−2
3
3
(RT ) ( RT )
([ H2 ]RT ) ([ N 2 ]RT )
K p = Kc ( RT )∆n
∆n = moles products − molesreac tan ts
For above reaction, ∆n = 2 - (3 + 1) = -2
K p = Kc ( RT )−2
Ex: In the synthesis of ammonia from nitrogen and hydrogen, Kc = 9.60 at
300°C. Calculate Kp for this reaction at this temperature.
Ch 14 Equilibrium 5
Heterogeneous Equilibrium
CaCO3(s)
CaO(s) + CO2 (g)
The concentration of a solid is proportional to its density and remains
constant as long as the temperature does not change.
Density g cm3 mol
=
=
M
g mol cm3
Since concentrations of solids are constant, they are included in the
equilibrium constant and don't appear in the reaction quotient.
K=
[ CaO][CO2 ]
[CaCO3 ]
[CO2 ] = K
[CO2 ] = Kx
[ CaCO3 ]
[CaO ]
PCO2 = K p
Write the equilibrium-constant expression for Kc and Kp for each of the
following reactions.
CO2(g) + H2(g)
SnO2(s) + 2CO (g)
CO(g) + H2O(l)
Sn(s) + 2CO2(g)
Ch 14 Equilibrium 6
Calculating Equilibrium Constants
Substitute the equilibrium concentrations into the equilibrium
concentrations.
Ex: Nitryl chloride, NO2Cl, is in equilibrium with NO2 and Cl2,
2 NO2Cl(g)
2 NO2(g) + Cl 2(g)
At equilibrium the concentrations of the substances are [NO2Cl] =
0.00106 M, [NO 2] = 0.0108 M, and [Cl2] = 0.00538 M. From these data
calculate the equilibrium constant, Kc .
Predicting the Direction of a Reactions.
•
Q < K, not equilibrium, reactants will be converted to products
•
Q = K, equilibrium
•
Q > K, no equilibrium, products will be converted to reactants
Ex: At 448°C the equilibrium constant, Kc for the reaction,
H 2(g) + I 2(g)
2 HI(g), is 51. Predict how the reaction will proceed to
reach equilibrium at 448°C if we start with 2.0x10-2 mol of HI, 1.0x10–2
mol of H2, and 3.0x10–2 mol of I 2 in a 2.0 L container.
Ch 14 Equilibrium 7
Calculating Equilibrium Concentrations
Follow flow chart on page 580
•
Balanced Equation
•
Difference Table: include initial conc., change in conc. ∆ [ ], and
equilibrium conc.
•
Substitute equilibrium conc. into equilibrium equation and solve for x.
If you use quadratic equation, use the root that makes sense.
•
Calculate equilibrium concentrations by inserting value of x.
•
Check your results by substituting your answer into equilibrium
equation.
EX: The equilibrium constant, Kc , for the reaction of H2 with I2 is 57.0 at
700 K:
H 2(g) + I 2(g)
2 HI(g)
Kc = 57.0 at 700 K.
If 1.00 mol of H2 is allowed to react with 1.00 mol of I2 in a 10.0 L
reaction vessel at 700 K, what are the concentrations of H2, I2 and HI at
equilibrium? What is the equilibrium composition of the equilibrium
mixture in moles?
Ch 14 Equilibrium 8
EX: For the reaction
I2(g) + Br2(g)
2 IBr(g), K c = 280 at 150°C .
Suppose that 0.500 mol of IBr in a 1.00 L flask is allowed to reach
equilibrium at 150°C. What are the equilibrium concentrations of IBr, I2
and Br2.
Ch 14 Equilibrium 9
Factors that Effect Equilibrium
•
Concentrations change
•
Pressures change
•
Temperature changes
Le Châtelier's Principle
If a stress (such as a change in concentration, pressure, or temperature)
is applied to a system in equilibrium, the equilibrium shifts in a way that
tends to undo the effect of the stress.
Example
A change in concentration will effect the value of Q. The concentrations
will shift until Q again equals K.
3H2(g) + N 2(g)
2NH3(g)
Ch 14 Equilibrium 10
NH3 ]2
[
K=
= 1.2
3
[ H2 ] [ N2 ]
H2(g) + I 2(g)
K=
[ HI ]2
[ H2 ][ I2 ]
at 375°C
2HI(g)
= 50.0
If H2 is added, Q will decrease. The equilibrium will shift so that more HI
is produced by consuming H2 and I2 until Q = K
Endothermic:
Reactants + heat = products
Increasing T results in increase in K.
Exothermic:
Reactants = products + heat
Increasing T results in decrease in K.
Equilibrium shifts in the direction that absorbs heat.
Ch 14 Equilibrium 11
Effects of Disturbances on Equilibrium and K
Disturbances
Addition of
reactant
Addition of
Product
Decrease in
volume, increase
in pressure
Increase in
volume, decrease
in pressure
Rise in
temperature
Drop in
temperature
Change as
Mixture
Returns to
Equilibrium
Some of Added
reactant is
consumed
Some of added
product is
consumed
Pressure
decreases
Effect on
Equilibrium
Effect on K
Shift to right
No Change
Shift to left
No Change
Heat energy is
consumed
Shift in the
endothermic
direction
Shift in
exothermic
direction
Shift toward
No change
fewer gas
molecules
Pressure increases Shift toward more No change
gas molecules
Heat energy is
generated
Change
Change
The Effect of Catalyst
A catalyst increases the rate at which equilibrium is achieved, but it does
not change the composition of the equilibrium mixture.
Ch 14 Equilibrium 12
Le Châtelier's Examples
N2(g) + O2(g)
2NO(g)
∆H = +180.5 kJ
Equilibrium Constant, K
4.5 x 10
6.7 x 10
Temperature
-31
298 K
-10
900 K
-3
2300 K
1.7 x 10
2 NO2
N2O4 (g)
∆H = -57.2 kJ
Equilibrium Constant, K
Temperature
1300
273 K
70
298 K
3H2(g) + N 2(g)
2NH3(g)
Which way does equilibrium shift when:
• Add H2?
• Add NH3?
• Volume increased?
• Volume decreased?
Ch 14 Equilibrium 13
The Link Between Chemical Equilibrium and Chemical Kinetics
K=
kf
kr
k = Ae− Ea
RT
ln k = −
slope = –Ea/R
lnk
Ea
∆ ln k
=
−
∆ T1
R
Ea
RT
slope = –Ea/R
lnk
1/T (1/K)
1/T
Potential Energy
k1
k−1
Potential Energy
K=
Reaction Coordinate
Reaction Coordinate
Ch 14 Equilibrium 14