the position of equilibrium section 7.2. the equilibrium constant the position of equilibrium for a...
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THE POSITION OF EQUILIBRIUM
•Section 7.2
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THE EQUILIBRIUM CONSTANT
The position of equilibrium for a particular reversible reaction can be defined by a constant
This constant has a numerical value found by relating the equilibrium concentrations of the products to those of the reactants
Denoted by Kc the equilibrium constant
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DEDUCING THE EQUILIBRIUM CONSTANT
Consider this standard reaction: aA + bB cC + dD Where a, b, c and d represent
the stoichiometry of each substance, then:
c d
a bc
C D
A BK
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CONTINUED
The concentrations must be those that occur at equilibrium, not the starting values
Kc is constant for a given temperature
Change the temperature, you change Kc
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EQUILIBRIUM LAW
Equilibrium law: in any reversible reaction at a state of equilibrium, the rate of the forward reaction equals the rate of the reverse reaction
•
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PROBLEM #1
Consider the following reaction: N2(g) + 3H2(g) 2NH3(g)
How is the Kc calculated?
2
33
2 2
c
NH
N HK
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PROBLEM #1 CONTINUED
An experiment has this data at equilibrium at 500°C:
[N2] = 0.922 mol dm-3
[H2] = 0.763 mol dm-3
[NH3] = 0.157 mol dm-3
Calculate the Kc for this experiment
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WORKING OUT
0.0602 A note about units: the
concentrations calculated would give you varying units, Kc is unitless because you are using the “activity” which is close to concentration
2
3
0 .157
0 .922 0 .763cK
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ANOTHER CONSIDERATION
H2(g) + I2(g) 2HI(g)
Kc for the forward reaction:
For the reverse reaction:
2
2 2c
HI
H IK
2 2
2c
H I
HIK
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CONTINUED
The value of the equilibrium constant for the reverse reaction is the reciprocal of that for the forward reaction
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HOW FAR WILL A REACTION GO? Consider the following reactions and Kc
H2(g) + I2(g) 2HI(g) Kc = 2 at 277°C
H2(g) + Cl2(g) 2HCl(g) Kc = 1018 at 277°C
The large difference illustrates how stable HCl is at this temperature compared to HI
Virtually all of the reaction between H2 and Cl2 has produced HCl
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CONTINUED
The magnitude of Kc is a good indicator of how far a reaction goes toward completion (under given conditions)
If Kc>>1, then the reaction has gone virtually to completion
If Kc<<1, then the reaction has hardly taken place at all
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CONTINUED
For Kc = 1, there are equal amounts of reactants and products
The Kc for a given reversible reaction at equilibrium only changes if the temperature changes
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OTHER CONSIDERATIONS
The concentrations of certain substances remain constant, so they are omitted from the equilibrium constant expression
All solids have a fixed density, so these are omitted
The concentration of pure liquid water is also constant (important to consider when solutions are aqueous, water omitted)
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CONTINUED
If water is used as a liquid with other liquids ( not aqueous solutions), then it is included
Water must be included if it is in the gas phase
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LE CHÂTELIER'S PRINCIPLE
This principle is used to predict the qualitative effects of changes on the position of equilibrium
When a system at equilibrium is disturbed, the equilibrium position will shift in the direction which tends to minimize, or counteract, the effect of the disturbance
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MORE Possible changes in conditions that
we need to consider are: Changes in the concentrations of
either the reactants or the products Changes in pressure for gas phase
reactions Changes in temperature The presence of a catalyst
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CHANGES IN CONCENTRATION
Increasing the concentration of a reactant will move the position of equilibrium to the right
This favors the forward reaction and will increase the equilibrium concentrations of the products
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MORE The addition of more product to an
equilibrium mixture would shift the position of equilibrium to the left
This would favor the reverse reaction When new equilibrium
concentrations are substituted into the equilibrium expression, the value of Kc remains unchanged
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EXAMPLE
Fe(H2O)63+
(aq) + SCN-(aq) [Fe(H2O)5SCN]2+ + H2O(l)
yellow-brown colorless blood-red
If the concentration of either the thiocyanate ion, or the iron (III) ion is increased, then the intensity of the color increases
The shift in equilibrium to the right causes the concentration of the added reactant to fall again
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CONTINUED If the concentration of iron(III) ions is decreased by
adding fluoride ions, which form a stable complex with the iron (III) ions, then the intensity of the coloration decreases
The shift of the equilibrium to the left produces more aqueous iron (III) ions to counteract the reduction caused by the F-
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CHANGES IN PRESSURE
Changing the pressure only affects reactions that involve gases
Gas phase reactions will only be affected by a change of pressure if the reaction involves a change in the number of moles on the two sides of the equation
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INCREASE IN PRESSURE
Increase the pressure and the equilibrium shifts to the side with the least moles of gas
2SO2(g) + O2(g) 2SO3(g) Increase P: equilibrium → , to side with
fewer moles of gas
C(s) + H2O(g) CO(g) + H2(g)
Increase P: equilibrium ←, to side with fewer moles of gas (
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DECREASE IN PRESSURE
Decrease pressure and the equilibrium shifts towards the side with the most moles of gas
2SO2(g) + O2(g) 2SO3(g) Decrease P: equilibrium ←, to side with
more moles of gas
C(s) + H2O(g) CO(g) + H2(g) Decrease P: equilibrium →, to side with
more moles of gas
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MORE ABOUT PRESSURE
H2(g) + I2(g) 2HI(g)
Changing pressure has no effect as 2 moles of gas are converted to 2 moles of gas
The changes in concentrations that result from the changes in pressure are such that the value of Kc remains unchanged
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TEMPERATURE CHANGE
If the temperature of a system is increased, then the equilibrium shifts in the direction of the endothermic change, this will tend to lower the temperature
N2(g) + O2(g) 2NO(g) ∆H = +180 kJ mol-1
Increased T: Kc increases, equilibrium →
Decreased T: Kc decreases, equilibrium ←
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TEMPERATURE CHANGE
If the temperature of a system is decreased the equilibrium shifts in the direction of the exothermic change, this is to generate heat and raise the temperature
2SO2(g) + O2(g) 2SO3(g) ∆H = -197 kJ mol-1
Decreased T, Kc increases, equilibrium →
Increased T, Kc decreases, equilibrium ←
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MORE ABOUT TEMPERATURE
Changes in temperature affect the rate constants of the forward and reverse reactions to different extents
This means that the actual value of Kc changes also
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PRESENCE OF A CATALYST
The presence of a catalyst reduces the activation energy of both the forward and reverse reactions by the same amount
Both reactions are sped up by the same factor
Equilibrium is established more rapidly, but the position of equilibrium and the value of Kc are not affected
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SUMMARY
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INDUSTRIAL PROCESSES
Many industrial processes involve equilibria
The goal is to produce the desired product efficiently and rapidly, but with the minimum amount of waste and the minimum input of energy
Kinetics and equilibria have to be considered
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HABER PROCESS
Produces ammonia from nitrogen and hydrogen gases
A heated iron catalyst is used N2(g)+ 3H2(g) 2NH3(g) ∆H = -92 kJ mol-
1 4 moles of gas are converted to 2
moles of gas A high pressure will favor the product
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CONTINUED The forward reaction is exothermic,
so a low temperature would favor the products
Issues with the process: Low temperature causes a low
reaction rate High pressure is expensive to
provide A compromise is chosen to maximize
product
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COMPROMISE Typical conditions chosen for the Haber
process are pressures in the range of 200-1000 atm and temperatures about 700K
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COLLECTING AMMONIA
The reaction is not allowed to reach equilibrium because the reaction rate slows as equilibrium is approached
Usually only about 20% of the reactants are converted to ammonia
The ammonia is condensed and the reactants are recycled to form more ammonia
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USES FOR AMMONIA
The manufacture of fertilizers The manufacture of nitrogen-
containing polymers such as nylon Can be oxidized to produce nitric acid Nitric acid is used to produce
explosives such as TNT and dynamite Nitric acid is also used in making
dyes
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CONTACT PROCESS
The Contact process is the production of sulfuric acid by the oxidation of sulfur
First, pure sulfur is burned in air S(s) + O2(g) SO2(g)
Then sulfur dioxide is mixed with air and passed over a vanadium (V) oxide catalyst
2SO2(g) + O2(g) 2SO3(g) ∆H = -196 kJ mol-1
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CONTINUED
High pressure would favor the formation of sulfur trioxide, but low pressure actually does a good job
Low temperature favors the product, but not too low (slows the reaction)
Compromise: 700-800K, the use of a catalyst (finely divided V2O5) and about 2 atm
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CONTINUED
The result is over 90% conversion to sulfur trioxide
The sulfur trioxide must be reacted with water to produce sulfuric acid
SO3(g) + H2O(l) → H2SO4(l)
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USES OF SULFURIC ACID
Manufacture of: Fertilizers Polymers Detergents Paints and pigments
Used in the petrochemical industry and in the processing of metals
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ANOTHER USE OF SULFURIC ACID
Used as an electrolyte in automobile batteries (battery acid)