CHEM 102 CLASS NOTES
Prof. Upali Siriwardane,
Chemistry Program, Louisiana Tech University, Ruston, LA 71272
CHAPTER 14, Chemical Equilibrium
14.1
14.2
14.3
14.4
14.5
14.6
14.7
14.8
14.9
TABLE OF CONTENT
Fetal Hemoglobin and Equilibrium 61 3
The Concept of Dynamic Equilibrium 61 5
The Equilibrium Constant (K) 61 8
Expressing the Equilibrium Constant in Terms of Pressure 622
Heterogeneous Equilibria: Reactions Involving Solids and Liquids 625
Calculating the Equilibrium Constant from Measured Equilibrium Concentrations 626
The Reaction Quotient: Predicting the Direction of Change 629
Finding Equilibrium Concentrations 631
Le Châtelier’s Principle: How a System at Equilibrium Responds to Disturbances 641
KEY SKILLS
Expressing Equilibrium Constants for Chemical Equations (14.3)
• Example 14.1 • For Practice 14.1 • Exercises 21, 22
Manipulating the Equilibrium Constant to Reflect Changes in the Chemical
Equation (14.3)
• Example 14.2 • For Practice 14.2 • For More Practice 14.2 • Exercises 27–30
Relating Kp and Kc (14.4)
• Example 14.3 • For Practice 14.3 • Exercises 31, 32
Writing Equilibrium Expressions for Reactions Involving a Solid or a Liquid (14.5)
• Example 14.4 • For Practice 14.4 • Exercises 33, 34
Finding Equilibrium Constants from Experimental Concentration Measurements (14.6)
• Examples 14.5, 14.6 • For Practice 14.5, 14.6 • Exercises 35, 36, 41, 42
Predicting the Direction of a Reaction by Comparing Q and K (14.7)
• Example 14.7 • For Practice 14.7 • Exercises 45–48
Calculating Equilibrium Concentrations from the Equilibrium Constant and One or
More Equilibrium Concentrations (14.8)
• Example 14.8 • For Practice 14.8 • Exercises 37–44
Finding Equilibrium Concentrations from Initial Concentrations and the
Equilibrium Constant (14.8)
• Examples 14.9, 14.10 • For Practice 14.9, 14.10 • Exercises 51–56
Calculating Equilibrium Partial Pressures from the Equilibrium Constant and
Initial Partial Pressures (14.8)
• Example 14.11 • For Practice 14.11 • Exercises 57, 58
Finding Equilibrium Concentrations from Initial Concentrations in Cases with a
Small Equilibrium Constant (14.8)
• Examples 14.12, 14.13 • For Practice 14.12, 14.13 • Exercises 59, 60
Determining the Effect of a Concentration Change on Equilibrium (14.9)
• Example 14.14 • For Practice 14.14 • Exercises 61–64
Determining the Effect of a Volume Change on Equilibrium (14.9)
• Example 14.15 • For Practice 14.15 • Exercises 65, 66
Determining the Effect of a Temperature Change on Equilibrium (14.9)
• Example 14.16 • For Practice 14.16 • Exercises 67, 68
KEY CONCEPTS
dynamic nature of equilibrium:
Law of mass action
initial concentration
equilibrium constant expression Kp and
Kc
equilibrium concentration
calculate Kp from Kc and vice versa
heterogeneous equilibria
reaction quotient (Q)
Le Châtelier's Principle
calculate equilibrium
concentrations from Kp and Kc
equilibrium constants from experiments
CHAChapter
forward and reverse
reaction
calculate equilibrium
constant
homogeneous equilibria
predicting equilibrium
shifts
14. Chemical Equilibrium
Characteristics of Chemical
Equilibrium
In doing stoichiometry calculations we assumed that reactions proceed to
completion, that is, until one of the reactants runs out. Many reactions do proceed
essentially to completion: complete reactions are indicated by
. For such reactions
it can be assumed that the reactants are quantitatively converted to products and that the
amount of limiting reactant that remains is negligible.
Irreversible or complete reactions: Chemical reactions can be considered to have
forward and backward reactions. Forward reaction is when reactants combine to form
products whereas products are converted back to reactants in the backward reaction. In
most chemical reactions, the rate of backward reaction is so small all reactants are
completely converted to products. This condition is usually represented in a chemical
equation by an arrow pointing to right
E.g. H2 + O2
H2O
Equilibrium Chemical Reactions
On the other hand, there are many chemical reactions that stop far short of completion.
An example is the dimerization of nitrogen dioxide:
NO2(g) + NO2(g)
N2O4(g)
The reactant, NO2, is a dark brown gas, and the product, N2O4, is a colorless gas.
When NO2 is placed in an evacuated, sealed glass vessel at 25°C, the initial dark brown
color decreases in intensity as it is converted to colorless N2O4. However, even over a
long period of time, the contents of the reaction vessel do not become colorless. Instead,
the intensity of the brown color eventually becomes constant, which means that the
concentration of NO2 is no longer changing. Making the container colder makes
equilibrium to shift to left and warming shift the equilibrium to right.
Other examples:
3H2(g) + N2(g)
2NH3(g)
Any chemical reaction could be considered as a forward and backward reactions
occurring at the same time(
) as described previously. If the rates of backward and
forward reactions chemical reactions are comparable both reactants and products can
coexist leading to a condition called chemical equilibrium reaction.
Dynamic Equilibrium:
Equilibrium chemical reaction could be considered as a forward and backward
reaction occurring at the same time(
). The reactants and products will interchange
constantly, however maintaining same concentrations of reactants and products. Dynamic
equilibrium means a constantly changing system. In an equilibrium reaction reactants and
products change continuously among each other (dynamic) as opposed to a static
equilibrium. This change maintaining a constant concentration of reactants and products
is called a dynamic equilibrium in equilibrium chemical reactions. Double arrows are
used in the equation to indicate this.
3H2(g) + N2(g)
The Equilibrium Constant
2NH3(g)
Equilibrium constant comes from a law that applies to chemical equilibrium
called Law of Mass action.
Law of Mass Action:
Law of mass action describes an equilibrium process by quantifying the
equilibrium concentration of reactants and products. It uses the ratio of backward and
forward reactions and express it terms of an equilibrium constant (K). For example
consider a hypothetical equation:
jA+kB
[C]l[D]m
K = ----------------[A]j[B]k
lC+mD
[A]… are equilibrium concentration of A, B, C, D etc.
j, k, l, m are stoichiometric coefficients
K = equilibrium constant
Equilibrium Expression:
Equilibrium expression is the Law of mass action equation with the equilibrium
concentration of reactants and products and the equilibrium constant (K).
Example: Equilibrium expression for the formation of NH3 gas:
N2(g) + 3H2(g)
2NH3(g)
[NH3]2
K = ------------ = 6.02 x 10-2 L2/mol2 at 127oC
[N2][H2]3
Equilibrium Constant (K): The constant in the equilibrium expression is called
equilibrium constant (K).
[H2O]2
[NH 3]2
K = ------------and K = ------------ are
[H2] 2[O2]
[N2][H2] 3
Examples of equilibrium expressions containing K.
From the numerical value of equilibrium constant following could be said about the
reactions.
K = (infinity) - Irreversible reactions
K = 0 - No reaction
K = between 0 and 1 - Equilibrium reactions
Write the equilibrium constant for the following reactions:
a) 3H2(g) + N2(g)
2NH3(g)
b) CO(g) + Cl2(g)
COCl2(g)
c) N2O4(g)
2NO2(g)
d) MgCO3(s)
MgO(s) + CO2(g)
e) NaCl(s)
Na+(aq) + Cl-(aq)
f) 3/2H2(g) + 1/2N2(g)
NH3(g)
g) C(s) + CO2(g)
2CO(g)
h) NH4Cl(s)
NH3(g) + HCl(g)
i) N2O(g) + N2H4(g)
3N2(g) + 2H2O(g)
j) NOBr(g)
2NO(g) + Br2(g)
k) N2(g) + O2(g)
2NO(g)
l) CaCl2(s) + 2H2O(g)
CaCl2.2H2O(s)
m) H2(g) + CO2(g)
H2O(g) + CO(g)
a) 3H2(g) + N2(g)
2NH3(g)
[NH3]2
K = ----------[H2]3[N2]
b) CO(g) + Cl2(g)
COCl2(g)
[COCl2]
K = ----------[CO][Cl2]
c) N2O4(g)
2NO2(g)
[NO2]2
K = ----------[N2O4]
d) MgCO3(s)
MgO(s) + CO2(g)
K = [CO2]
Pure liquid or solid concentrations
are not written in the expression.
e) NaCl(s)
Na+(aq) + Cl-(aq)
K = [Na+(aq)] [Cl-(aq)]
f) 3/2 H2(g) + 1/2 N2(g)
NH3(g)
[NH3]
K = --------------[H2]3/2[N2] 1/2
g) C(s) + CO2(g)
2CO(g)
[CO]2
K = -----[CO2]
h) NH4Cl(s) NH3(g) + HCl(g)
K = [NH3][HCl] i) 2N2O(g) + N2H4(g)
[N2]2[H2O]2
2N2(g) + 2H2O(g)
K = ---------------[N2O]2[N2H4]
j) 2NOBr(g)
2NO(g) + Br2(g)
[NO]2[Br2]
K = ------------[NOBr]2
k) N2(g) + O2(g)
2NO(g)
[NO]2
K = ---------[N2][O2]
Which of the above chemical reactions, a-k, are examples of homogeneous
equilibrium and which ones are examples of heterogeneous equilibrium?
Homogenous equilibrium: Chemical equilibrium where reactants and products are in
same phase. Usually homogenous equilibrium is where all reactants and products are
gases. E.g. a, b, c, f, i, j, k in problem the problem above.
Heterogeneous equilibrium: Chemical Equilibrium where at least one phase of a
reactant or product is different from the rest. E.g. d, e, g, h.
The Meaning of Equilibrium Constant
jA+kB
lC+mD
rate constant forward reaction k+
K = ----------------------------------------rate constant backward reaction k Equilibrium occurs when k+ = k -.
[C]l[D]m
etc.
K = ----------------[A]j[B]k
[A]…. are equilibrium concentration of A, B, C, D
j, k, l, m are stoichiometric coefficients
K = equilibrium constant
E.g. Formation of water
2H2 + O2
2H2O
[H2O]2
since the backward reaction rate is almost 0 or very small
K = ------------ = a equal to 0.
K is equal to (infinity- ) i.e. only
[H2]2[O2]
H2O is found after equilibrium is reached.
rate constant of forward Reaction
k+
K = --------------------------------------------- = --=  (very large)
rate constant of backward Reaction
0
Keq >> 1
Keq ~ 1
Keq << 1
reaction will go mainly to products
reaction will produce roughly equal amounts of product and reactant
reaction will go mainly to reactants
Consider, for example, the equilibrium between N2O4(g) and NO2(g):
N2O4(g)
2NO2(g)
[N
]2
O
2
K
e
q
[N
]
O
2
4
Listed below is experimental data giving initial concentrations for N2O4(g) and NO2(g).
After some time the reaction reaches equilibrium and the concentrations listed.
Initial
@ Equilibrium
N2O4
NO2
N2O4
NO2
Keq
0.00
0.02
0.0014
0.017
0.21
0.00
0.03
0.0028
0.024
0.21
0.00
0.04
0.0045
0.031
0.21
0.02
0.00
0.0045
0.031
0.21
E.g. Formation of NH3 gas.
N2(g) + 3H2(g)
2NH3(g)
[NH3]2
K = ------------ = 6.02 x 10-2 L2/mole2 at 127oC
[N2][H2]3
Forward reaction is 100 times slower than the backward reaction at this temperature.
Taking reaction a in the above list, the formation of ammonia (32 billion pounds per
year produced in US alone) describes the importance of studying the equilibrium in
a chemical reaction.
Anhydrous ammonia is produced using process called Haber Process:
3H2(g) + N2(g)
2NH3(g) ; K = 6.0 x 10-2 L2/mol2
At 25o K = small, no or very little product is formed. However, at 500oC ,
K = 6.0 x 10-2 L2/mol2 some ammonia is formed NH3 is removed continuously from the
mixture using a "cold trap" to obtain liquid ammonia. Removing ammonia shift the
equilibrium to right (As discussed under Le Chatelier's Principle later in detail).
Determining Equilibrium Constants
Following definitions are needed to do an equilibrium constant calculation. These
ICE (Initial-Change-Equilibrium) calculations are based on the idea that initial,
change and equilibrium concentrations along with equilibrium constant are needed.
Initial concentration:
The concentrations in moles per liter (M) or partial pressure (P) of reactants and products
before the equilibrium is reached is called initial concentration.
E.g. 3H2(g) + N2(g)
2NH3(g)
[H2] or PH2 = Concentration of H2 before the equilibrium is reached.
[ N2] or PN2 = Concentration of N2 before the equilibrium is reached.
[NH3] or PNH3 = Concentration of NH3 before the equilibrium is reached.
Equilibrium concentration:
The concentrations in moles per liter (M) of reactants and products after the equilibrium
is reached is called equilibrium concentration.
E.g. 3H2(g) + N2(g)
2NH3(g)
[H2] or PH2 = Concentration of H2 after the equilibrium is reached.
[ N2] or PN2 = Concentration of N2 after the equilibrium is reached.
[NH3] or PNH3 = Concentration of NH3 after the equilibrium is reached.
K (Kc)
K (Kc) is when the mole/L (molarity which normally indicated by [A] ) units are
used for equilibrium concentration of reactants and products in the equilibrium
expression
Kp
Kp is the equlibrium constant of gaseous reactions when the concentrations are expressed
in terms of individual partial pressures (p) of a mixture of gases at equilibrium.
Consider the following reactions:
3H2(g) + N2(g)
2NH3(g)
CO(g) + Cl2(g)
COCl2(g)
NaCl(s)
Na+(aq) + Cl-(aq)
2N2O(g) + N2H4(g)
3N2(g) + 2H2O(g)
K (Kc) in the equilibrium expression is usually the constant calculated based on
moles/Liters concentration of reactants and products. This is also called Kc - equilibrium
constant based on M, [A] or moles/L -Concentrations.
However, many chemical reactions occur in gas phase and concentration of products and
reactants is easier to give K in partial pressures (P). The equilibrium constant calculated
based on partial pressure is called Kp - P - partial pressure.
Relationship between K (Kc) and Kp
Kp = K(RT)n
Kp and K (Kc) are related by the following equation:
Kp = K(RT)n
K = constant based on concentration in mole/Liters.
Kp = constant based on partial pressures.
R = universal gas constant
T = Kelvin Temperature,
n = (sum of stoichiometric coefficients of gaseous products) - (sum of the
stoichiometric coefficients of gaseous reactants)
E.g.
N2(g) + 3H2(g)
2NH3(g)
n
CO(g) + Cl2(g)
= 2 - 4 = -2
P2NH3
Kp = ----------Kp = K(RT)-2
3
PN2P NH3
Kp
K = --------- = Kp(RT)2
(RT)-2
COCl2(g)
n
= 1 - 2 = -1
PCOCl2
Kp = -------PCOPCl2
Kp = K(RT)-1 ;
K
Kp = ---- ;
RT
K = KpRT
NaCl(s)
Na+(aq) + Cl-(aq)
Kp = 1;
Kp = K(RT)n ;
n = 0
Kp = K
2N2O(g) + N2H4(g)
3N2(g) + 2H2O(g)
n = 5 - 3 = 2
P3N2 P2H2O
Kp
2
Kp = ---------------------- ; Kp = K(RT) ; K = ------P2N2O PN2H4
(RT)2
For example in the equilibrium system:
H2 (g) + I2(g) = 2HI(g); Kc = 49.7 at 458 C
What would be the Kp of this system if R = 0.0821 liter-atm/mole K?
1. Check to make sure the equation is balanced
2. Calculate  n
n = Total gas moles on the right - Total gas moles on the left
n=2-2=0
3. Calculate T in Kelvin from given Celsius temperature
T = 273 + 458 = 731 K
4. Using the relationship plug values in and solve for Kp
Kp = Kc (RT) n
Kp = 49.7 [(0.0821)(731)]0 = 49.7
Equilibrium Constant KpCalculation
At 500oC, Consider the following reaction (assume an IDEAL gas mixture):
PCl5(g)
PCl3(g) + Cl2(g)
A 1.0-liter vessel was initially filled with pure PCl5, at a pressure of 2.0 atm, at
25oC. After equilibrium was established, the partial pressure of PCl3 was 3.16 x 10-2
atm. What is Kp for the reaction?
PPCl3 PCl2
Kp = ---------PPCl5
PCl5
PCl3
Cl2
Initial partial pressure:
2.0 atm
__
__
Change:
-x
x
x
if x amount of PCl5 decomposed
(2.0 - x)atm
x atm
x atm
-2
-2
to reach equilibrium
(2.0 - 3.16 x 10 )atm 3.16 x 10 atm 3.16 x10-2
atm
At equilibrium
x = 3.16 x 10-2 atm
3.16 x 10-2 atm x 3.16 x 10-2 atm
9.99 x 10-4
Kp = --------------------------------------= ------------ = 5.07 x 10-4atm
(2.0- 3.16 x 10-2)atm
1.9684
A 5.0-g sample of solid NH4Cl is heated in a 2.5-L container to 900oC. At
equilibrium the pressure of NH3(g) (reaction 2h) is 0.60 atm. Calculate the
equilibrium constant, Kp, for this reaction.
NH4Cl(s) A 5.0-g sample of solid NH4Cl is heated in a 2.5-L
container to 900oC. At equilibrium the pressure of NH3(g) (reaction 2h) is 0.60 atm.
Calculate the equilibrium constant, Kp, for this reaction.
NH4Cl(s)
NH3(g) + HCl(g)
Kp = PNH3PHCl
;note that NH4Cl(s) doesn't appear in the
expression since it's a solid.
PNH3
PHCl
Initial concentration
0.0
0.0
Change
x
x
Equilibrium concentration 0.6 atm
x atm
0.6 atm
0.6 atm
PHCl should be equal to that of PNH3 since for each NH3 formed a HCl is formed.
Therefore, Kp = 0.6 atm x 0.6 atm = 0.36 atm2
NH3(g) + HCl(g)
Kp = PNH3PHCl
;note that NH4Cl(s) doesn't appear in the
expression since it's a solid.
PNH3
PHCl
Initial concentration
0.0
0.0
Change
x
x
Equilibrium concentration 0.6 atm
x atm
0.6 atm
0.6 atm
PHCl should be equal to that of PNH3 since for each NH3 formed a HCl is formed.
Therefore, Kp = 0.6 atm x 0.6 atm = 0.36 atm2
Using Equilibrium Constants
Knowing the equilibrium constant for a reaction allows us to predict several
important features of the reaction: the tendency of the reaction to occur (but not the speed
of the reaction which is the area of chemical kinetics), whether or not a given set of
concentrations represents an equilibrium condition, and the equilibrium position that will
be achieved from a given set of initial concentrations. The reaction quotient is used for
predicting the net direction of equilibrium reactions.
Reaction Quotient (Q)
When the reactants and products of a given chemical reaction are mixed, it is
useful to know whether the mixture is at equilibrium or, if not, the direction in which the
system must shift to reach equilibrium. If the concentration of one of the reactants or
products is zero, the system will shift in the direction that produces the missing
component. However, if all the initial concentrations are nonzero, it is more difficult to
determine the direction of the move toward equilibrium. To determine the shift in such
cases, we use the reaction quotient Q. The reaction quotient is obtained by applying the
law of mass action using initial concentrations instead of equilibrium concentrations.
jA+kB
lC+mD
[A]…. are initial concentration of A, B, C, D etc.
j, k, l, m are stoichiometric coefficients
Q = reaction quotient
[C]l[D]m
Q = ----------------[A]j[B]k
Q > Keq
Q = Keq
Q < Keq
reverse reaction will be spontaneous
reaction @ equilibrium
forward reaction will be spontaneous
Consider the following reaction:
SO2(g) + NO2(g)
NO(g) + SO3(g) (Kc = 85.0 at 460oC)
Given 0.040 mole of SO2(g), 0.0.500 mole of NO2(g), 0.30 mole of NO(g),and 0.020
mole of SO3(g) are mixed in a 5.00 L flask, determine:
a) The net the reaction quotient, Q .
b) Direction to achieve equilibrium.
The net the reaction quotient, Q .
The reaction quotient (Q) is constant in the equilibrium expression when initial
concentration of reactants and products are used.
[NO][SO3]
Q = ------------- Q = equilibrium constant calculated based on initial
concentrations.
[SO2][NO2]
0.040 mole
0.500 mole
0.30 mole
0.020 mole
[SO2] = -------------; [NO2] = ----------- ; [NO] = ------------; [SO3] = ----------5.00 L
5.00L
5.00L
5.00 L
[SO2] = 8 x 10-3mole/L ; [NO2] =0.1mole/L; [NO] = 0.06 mole/L; [SO3] = 4 x 103
mole/L
0.06 (4 x 10-3 )
Q = ------------------ = 0.3
8.0 x 10-3 x 0.1
The net direction to achieve equilibrium.
Since the equilibrium constant K = 85.0, at 460oC and Q = 0.3, to reach equilibrium Q
should increase to 85.0. To do that top term or product concentrations should be
increased. Therefore the reaction goes to right (towards products) to achieve equilibrium.
Calculation of unknown concentration of reactants or products in an equilibrium
mixture
At 100o C the equilibrium constant (K) for the reaction:
H2(g) + I2(g)
2HI(g)
is 1.15 x 102. If 0.400 moles of H2 and 0.400 moles of I2 are placed into a 12.0-liter
container and allowed to react at this temperature, what is the HI concentration
(moles/liter) at equilibrium?
[HI]2
K = ------------; K =1.5 x 102 .
[H2][I2]
Initial moles:
[H2]= 0.400 mol;
[I2] = 0.400 mol ;
[HI] = 0.00 mol
Change in moles:
[H2]= -x mol
[I2] = -x mol
[HI] = 2x mol
Equilibrium concentration in M (mole/L):
If x of H2 and x of I2 moles reacts to achieve equilibrium
0. 400-x mol
0.400-x mol
[H2]= ---------------;
[I2] = --------------;
12.0 L
12.0L
[HI]2
K = ------------ =1.5 x 102 .
[H2][I2]
2x mole
[HI] = -------------;
12.0L
(2x/12)2
2x2
K = ----------------------------- = --------------(0.400-x/12)(0.400-x/12)
(0.400-x)2
4x2
2x
K =1.5 x 102= ----------- ; Sq.Rt (1.5 x 102) = 10.72 = --------(0.4-x)2
(0.4-x)
2x = 10.72(0.4 -x) = 4.289 - 10.72x
2x + 10.72x = 4.289
12.72x = 4.289
x = 4.289/12.72 = 0.337 mole
[HI] = 2x/12 = 0.056 mole/L
At a certain temperature the value of the equilibrium constant is 3.24 for the
reaction:
H2(g) + CO2(g)
H2O(g) + CO(g)
If 0.400 mol H2 and 0.400 mol CO2 are placed in a 1.00 L vessel, what is the
concentration of CO at equilibrium?
K = 3.24
Here we assume mole/L concentration is same as moles since volume is 1L.
[H2 ]
[ CO2]
[H2O ]
[CO ]
Initial concentration:
0.4
0.4
0.00
0.00
Change in concentration: -x
-x
x
x
Equilibrium concentration: 0.40-x
0.40-x
x
x
If x of H2 and x of CO2 reacted to produce x moles of H2O and CO
[ H2O][CO]
x2
K = -------------- =
---------------[H2][CO2]
(0.40-x)(0.40-x)
x2
x
K = 3.24 = ------------ = ---------(0.40-X)2 (0.40-X)
Equilibrium:
Concentration
[H2]
0.40-x
0.143
x
sq.rt . 3.24 =1.8 = ---------(0.40-x)
x = 1.8 (0.400-x) = 0.72 - 1.8 x
x + 1.8x = 0.72
2.8x = 0.72
x = 0.72/2.8 = 0.257 mol/L
[CO2]
[H2O ]
0.40-x
x
0.143
0.257 mol/L
Shifting a Chemical Equilibrium: Le Chatelier's Principle
[CO ]
x
0.257 mol/L
Le Chatelier's principle is used to predict the shift of an equilibrium. Le
Chatelier's Principle is one of the general principles applicable to any equilibrium
process. It simply states that: If a change is imposed on a system at Equilibrium, the
position of the equilibrium will shift in a direction that tends to reduce that change.
Listed below are how various "changes" that affect equilibria:
1) Adding products (unless one of the products is a solid!) to a reaction will cause the
equilibrium to shift back to produce more reactants.
2) Adding reactants (unless one of the reactants is a solid!) to a reaction will cause the
equilibrium to shift forward to produce more products.
3) Removing reactants (unless one of the reactants is a solid and as long as there is some
left) will cause the equilibrium to shift back to produce more reactants.
4) Removing products (unless one of the products is a solid and as long as there is some
left) will cause the equilibrium to shift forward to produce more products.
5) The effect of temperature on a reaction is dependent on whether the reaction is
exothermic (Hrxn = negative) or endothermic (Hrxn = positive).
6) Pressure changes could affect a reaction if there is net change in gaseous reactants and
products
Temperature changes could be either heating or cooling and also due to heat
absorbed/released in the reaction.
Pressure changes could be either by increasing or decreasing the pressure of the vessel.
In addition changes in number of total particles in gaseous reaction mixtures going from
reactant to products can create pressure changes
E.g.
Increasing T of the equilibrium should shift equilibrium to left and vice versa.
Increasing P will shift equilibrium to right and vice versa.
Note: volume changes can be considered as pressure changes. Increased volume have
the same effect as a decrease in pressure.
For the following equilibrium reactions:
H2(g) + CO2(g)
Predict the equilibrium shift if:
a) The temperature is increased
b) The pressure is decreased
H2O(g) + CO(g) ?H = 40 kJ
a) Temperature increased for the above equilibrium: Reaction is endothermic (H = +
value). Equilibrium should shift to a direction to absorb heat. Forward reaction absorbs
heat. Backward reaction should release heat. Therefore, forward reaction occurs or
equilibrium shift to right to absorb heat.
b) Pressure is decreased: In this equilibrium there is no change in number of particles in
going from reactant to products. Equilibrium cannot respond to pressure changes directly.
There is no change in the position of equilibrium.
Equilibrium at the Nano-scale
At the nano or sub-microscopic level (i.e. looking at the molecular species), the state of
equilibrium does not mean that there is no change. However, the equilibrium constant K
remain essentially constant. The following point will be illustrated.
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Equilibrium is a dynamic process at the molecular level. There is no net change,
but two opposing processes are taking place.
The equilibrium constant fluctuates slightly due to unequal reaction rates in
opposite directions.
A system moves spontaneously toward a state of equilibrium.
The driving forces for equilibrium are:
(a) molecules assume the state of lowest energy,
(b) molecules tend to reach a maximum disorder or entropy.
Controlling Chemical Reactions: The Haber-Bosch Process
The Haber-Bosch Process (aka Haber process) is the reaction of nitrogen and hydrogen
to produce ammonia. The nitrogen and hydrogen are reacted over an iron catalyst under
conditions of 200 atmospheres, 450°C:
N2(g) + 3H2(g) <--> 2NH3(g)
The process was developed by Fritz Haber and Carl Bosch in 1909 and patented in 1910.
It was first used on an industrial scale by the Germans during World War I: Germany had
previously imported nitrates from Chile, but the demand for munitions and the
uncertainty of this supply in the war prompted the adoption of the process. The ammonia
produced was oxidized for the production of nitric acid in the Ostwald process, and the
nitric acid for the production of various explosive nitro compounds used in munitions.
The nitrogen is obtained from the air, and the hydrogen is obtained from water natural
gas in the reaction:
CH4(g) + H2O(g) → CO(g) + 3H2(g)
Equilibrium and the Haber Process
The reaction of nitrogen and hydrogen is reversible, meaning the reaction can proceed in
either the forward or the reverse direction depending on conditions. The forward reaction
is exothermic, meaning it produces heat and is favored at low temperatures. Increasing
the temperature tends to drive the reaction in the reverse direction, which is undesirable if
the goal is to produce ammonia. However, reducing the temperature reduces the rate of
the reaction, which is also undesirable. Therefore, an intermediate temperature high
enough to allow the reaction to proceed at a reasonable rate, yet not so high as to drive
the reaction in the reverse direction, is required.
The forward reaction favors high pressures because there are fewer molecules on the
right side. So the only compromise in pressure is the economical situation trying to
increase the pressure as much as possible.
The iron catalyst has no effect on the position of equilibrium, however it does increase
the reaction rate. This allows the process to be operated at lower temperatures, which as
mentioned before favors the forward reaction. Other catalysts are also active for this
reaction, in fact the first Haber-Bosch reaction chambers, used uranium catalysts.
The ammonia is formed as a gas but on cooling in the condenser liquefies at the high
pressures used, and so is removed as a liquid. Unreacted nitrogen and hydrogen is fed
back in to the reaction.
Notwithstanding its original adoption as a military necessity, the Haber process now produces about
half of all the nitrogen used in agriculture: billions of people are alive and fed from its use.
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This page was last modified on April 4, 2013