2-01 Equilibrium We have hinted that reactions can proceed in the forward or reverse direction • CH 3 -CH(OH)-CH 3 with an H 2 SO 4 catalyst CH 3 -CH=CH 2 • But CH 3 -CH=CH 2 with an H 2 SO 4 catalyst CH 3 -CH(OH)-CH 3 • This makes sense from our PE diagrams • The forward and reverse reactions just have different activation energies and opposite ∆H values A reversible reaction is said to be at equilibrium when the rate of the forward reaction is equal to the rate of the reverse reaction • If the products are able to escape from the reaction vessel they won’t be available to reform reactants o The reaction must be in a closed system Closed system • Nothing can enter or leave Assign 1-2 • The temperature affects equilibrium o A new equilibrium is attained at a new temperature
Transcript
2-01 Equilibrium We have hinted that reactions can proceed in the forward or reverse direction
• CH3-CH(OH)-CH3 with an H2SO4 catalyst CH3-CH=CH2 • But CH3-CH=CH2 with an H2SO4 catalyst CH3-CH(OH)-CH3 • This makes sense from our PE diagrams
• The forward and reverse reactions just have different activation energies and opposite ∆H values
A reversible reaction is said to be at equilibrium when the rate of the forward reaction is equal to the rate of the reverse reaction
• If the products are able to escape from the reaction vessel they won’t be available to reform reactants o The reaction must be in a closed system
Closed system
• Nothing can enter or leave Assign 1-2
• The temperature affects equilibrium o A new equilibrium is attained at a new temperature
• The same equilibrium will exist regardless of whether we start with excess reactants or products
• When a system is at equilibrium no macroscopic changes occur
Dynamic equilibrium (chemistry)
• Equilibrium in which microscopic changes but not macroscopic changes occur
• Constant microscopic back and forth between reactants and products but the rates cancel out and nothing is visible on the large scale
2-02 Characteristics of Equilibrium Since equilibrium reactions are reversible there is both a forward and reverse rate.
REACTANTS ↔ PRODUCTS
The rate of a reaction increases when the [reactants] increases • The rate must be proportional to the [reactants] • Rateforward = kforward [reactants] • Ratereverse = kreverse [products] • K is some constant for each reaction…a number Try 6 and 7 on graph paper
• Rate of consumption of reactants = rate of production of reactants
• [reactants] is not = [products] in general • [reactants] is constant and [products] is constant • forward and reverse rates do not change • a system which is not at equilibrium will tend to
move towards equilibrium once equilibrium is reached, no further change
occurs reversible-reactions.jnlp
For H2 + Cl2 <--> 2HCl + = The large boxes represent the total amounts at some equilibrium
• the number of moles of each gas present is not evident just from the equation
• the ratio of [product]/[reactant] is a constant at equilibrium more later…
H2
Cl2
HCl
The small boxes are the amounts that react every second at equilibrium
• total reactants changing = total products changing back= no net change
• 1H2 and 1Cl2 react to make 2HCl while at the same time 2HCl react to make 1H2 and 1Cl2
• The rate that something is reacted is equal to the rate that it is produced
Assign 8-13
2-03 Predicting if a reaction is spontaneous or not
Spontaneous change • Occurs by itself without outside assistance • Some conditions allow while other forbid spontaneous
reactions Exothermic reactions should be spontaneous provided the activation energy is not too high. PE Reaction proceeds If you give the reaction a little nudge it goes to completion
• Match to paper • Ignite explosives…
But endothermic reactions require constant energy input to proceed and are not expected to be spontaneous unless sufficient heat is available.
• Molecules on the downhill side would be expected to stay there unless energy is supplied to them.
• The side having minimum energy (minimum enthalpy) is favored
• lazy There is more to spontaneity
• Some endothermic reactions are spontaneous o Ammonium nitrate and water react spontaneously and
absorb energy from the surroundings • Some exothermic reactions don’t go to completion
o Equilibrium is established as the reverse endothermic reaction occurs
Randomness is a second factor that comes into play for spontaneity • Most arrangements in space are disordered • Only a few special arrangements are ordered
o Throw a deck of cards in the air • There are a few situations where order could
develop • Card house • Back in sequence • All Spades together
• Most situations from that random event are random (disordered)
• A disordered result from a random event is more probable than an ordered result.
As a crystal of salt dissolves water molecules strike the ions and they bounce out of the lattice
• To get back in another molecule of water will have to hit it perfectly to go back into its ordered position
o There are many other possible ways the ion could be hit that won’t take it back to its crystal
o Disorder is favored and the crystal dissolves Entropy is the amount of randomness in a system.
• Entropy tends to a maximum value • Things tend to get jumbled up over time • Slob
The reactants are highly ordered crystals and the products are very random ions so entropy favors the reaction even if enthalpy does not. The reactants absorb energy and then become so random that they can’t reorganize to reform.
• The reaction is spontaneous and can drop the surrounding temperature by 55o C
All reactions are driven towards maximum randomness (entropy) and minimum energy (enthalpy)
• The universe favors the sloppy and the lazy
2NH4SCN (s) + Ba(OH)2.8H2O(s)
Ba2+(aq) + 2SCN-
(aq) 2NH3(g) + 10H2O(l)
Tendency to minimum enthalpy Exothermic Endothermic AB + heat C + heat D The arrow points to the tendency for minimum enthalpy
• the down hill side • the side with heat as a reactant
Tendency to maximum entropy
• look at the relative randomness of phases • gases >> solutions > liquids >> solids
H2O (s) ↔ H2O(l)
• the liquid is more random and the products are favored CaC(s) + 2H2O(l) ↔ C2H2(g) + Ca(OH)2(aq)
• the most random phase present is gas and the products are favored
A(g) + B(s) ↔ 2C(g) + D(s) • if the reactants and products are in the same phase then the
side with more molecules is more random (more possible disordered arrangements)
Consider • C2H2(g) + 2Cl2(g) C2H2Cl4(l) + 386kJ
o Enthalpy favors the products o Entropy favors the reactants o Opposite tendencies lead to equilibrium
• CH4(g) + O2(g) CO2(g) + 2H2O(g) + 394 kJ o Enthalpy favors the products o Entropy favors the products o Reaction goes “100%”
• 4Au(s) + 3O2(g) + 162kJ 2Au2O3(s) o enthalpy favors the reactants o entropy favors the reactants o the reaction goes “0%”
no reaction o gold doesn’t oxidize!
Try 14-16
Notes 2-04 Le Chatelier’s Principle
Le Chatelier’s Principle • If a closed system at equilibrium is subjected to a
change, processes will occur that tend to counteract that change
• Whatever we do, nature tries to undo This principle is crucial to quickly predicting what will happen in equilibrium systems and will be used throughout the course. Fe3+ + SCN- FeSCN2+
• If we increase [Fe3+] or [SCN-] then the [FeSCN2+] increases.
• If we decrease [Fe3+] or [SCN-] then the [FeSCN2+] decreases.
Think of the water in a bathtub representing a system at equilibrium with the reactants at one side of the tub and the products at the other. • If you dump some water at one end of the tub the water
flows towards the other end of the tub to reestablish equilibrium
• If you scoop out some water at one end of the tub the water flows back towards the “hole” to reestablish equilibrium.
The effect of temperature changes: • If we decrease the temperature in a reaction like:
2NO(g) + Cl2 (g) ↔ 2NOCl(g) + 76kJ
• We removed energy (a product)
• It’s like we scooped out some water on the product side
• The reaction shifts to the right to compensate • More reactants become products
Time at which temp changed
[…]
Cl2
NO
NOCl
[Cl2] only changes half as much as [NO] since they react in a 1:2 ratio
Time
The effect of concentration changes: • If we increase the [Cl2] in a reaction like:
2NO(g) + Cl2 (g) ↔ 2NOCl(g) + 76kJ
• We added a reactant
• It’s like we dumped some water on the reactant side • The reaction shifts to the right to compensate
• More reactants become products
Time at which Cl2 was injected quickly
[…] Time
The effect of pressure changes: • A decrease in volume increases the pressure and the
concentration of all the gases.
2NO(g) + Cl2 (g) ↔ 2NOCl(g) + 76kJ
Cl2
NO
NOCl
Not all the Cl2 added is used up… the change can’t be cancelled completely because the other parts changed
• Figure out which side has the most gases (3 moles reactants and 2 moles products) • It’s like we dumped some water on the high gas side
(reactants) • The reaction shifts to the right to compensate (want fewer
gases at high pressure) • More reactants become products
Time at which pressure changed
[…]
Time
The effect of adding a catalyst • Since a catalyst speeds up the forward and reverse rate
the reaction remains at equilibrium • Just gets there faster if it wasn’t already there.
Graphing notes: • Temperature changes
• Concentrations change slowly to a new value • Changing the concentration of a species
• One species suddenly jumps up or down then all adjust slowly to compensate
• Changing the pressure • The concentration of all gasses simultaneously jump
up or down then slowly adjust.
Cl2
NO
NOCl
All gases increase in concentration at first and then the equilibrium shifts to make products
Assign 17-28 PS It helps to draw a little mound or hole over the substance that increases or decreases to visualize which way the equilibrium will shift.
Notes 02-05 Equilibrium in industry
In WWll Germany couldn’t get nitrates from the main global source, Chile. Nitrates are used to make Trinitrotoluene (TNT). Fritz Haber figured out how to make ammonia on a huge scale, which can easily be oxidized to nitrates. The Haber Process for making ammonia (important to remember)
• N2(g) + 3H2(g) ↔ 2NH3(g) + 92kJ Figure out what conditions will maximize the yield of ammonia
• Pressure…high • Temperature?
o Reaction rate = high vs yield = low • Other…iron oxide catalyst
http://www.youtube.com/watch?v=c4BmmcuXMu8 The Making of Cement from Limestone
• CaCO3(s) + 175kJ ↔ CaO(s) + CO2(g) • Best conditions are
o Pressure….low o Temperature…high
Is there a conflict between rate and yield?…no
Notes 2-06 The equilibrium expression and the equilibrium constant
Pull out the two graphs we drew for questions 6 and 7. At the first equilibrium there was 5 times as much B as A. At the second equilibrium there was 5 times as much B as A.
• The constant ratio is not a fluke! For the equilibrium equation:
D C B A +⇔+ it has been found that
[ ] [ ][ ] [ ] constant a=
××
=BADCKeq
the expression is called the equilibrium expression and Keq is a number called the equilibrium constant. Examples
• H2O(g) + CO(g) ↔ H2(g) + CO2(g) o [ ] [ ]
[ ] [ ](g)(g)2
2(g)2(g)
CO OHCO H
××
=eqK
• PCl5(g) ↔ PCl3(g) + Cl2(g)
o [ ] [ ][ ] PCl
Cl PCl5(g)
2(g)3(g) ×=eqK
• H2(g) + F2(g) ↔ HF(g) + HF(g) or H2(g) + F2(g) ↔ 2HF(g) o [ ] [ ]
[ ] [ ][ ]
[ ] [ ] F H HF
F HHF HF
2(g)2(g)
2(g)
2(g)2(g)
(g)(g)
×=
××
=eqK
o The coefficient becomes the exponent In general:
• but for a pure solid the concentration is a constant because its density is a constant o so [CaF2(s)] = 40.7M from 3.18 g/L and 78.1 g/mol o if the concentration of the solid never changes the
expression is really:
[ ] [ ] 132
)()(2
104.840.7 Ca −
−+
×=×
=aqaq
eqFK
Why bother having 2 constants in the same expression?
Combine the constants to get
[ ] [ ] 112
)()(2
)( 104.3 Ca −−+
×=×
=aqaq
neweqFK
Instead of having to calculate the concentration of a solid every time, just combine it into the Keq.
• Eliminate any concentrations that have a constant value in the equilibrium expression
o Solids…the concentration comes from the density which doesn’t change
o Pure liquids…the concentration can come from density
Only counts if it is the only liquid that exists in
the entire equilibrium expression
If there is more than one then dilution occurs and
Adding a reactant or a product which is a solid or pure liquid will have no effect on the equilibrium Assign 31-35
02-07 Le Chatelier’s Principle and the Equilibrium Constant When the concentration pressure or temperature are changed the reaction tends to counteract these changes.
• One shot change The system regains equilibrium
• Continuous change You have an open system Does not reach equilibrium
When temperature is decreased and held
2NO(g) + Cl2(g) ↔ 2NOCl(g) + 76kJ shifts to the product side. This results in an increase in [products] and a decrease in [reactants] so that increases.
To figure out the direction of the change in Keq draw an arrow in the Keq expression that shows the shift.
• An upward arrow indicates a higher Keq and vice versa •
Only changes in temperature change Keq Large Keq means that [products] > [reactants]
• lots of products at equilibrium Small Keq means that [products] < [reactants]
• lots of reactants at equilibrium
Assign 36-46
02-08 Equilibrium Calculations To determine the value of Keq for any reaction you need to accurately know the concentrations of all of the species present.
• Must be experimentally determined Example A For 2NO(g) + O2(g) ↔2NO2(g) A 2.0 L bulb contains 6.00 mol of NO2(g), 3.0 mol of NO(g) and 0.20 mol of O2(g) at equilibrium. What is the Keq? First determine the equilibrium expression:
Use the data to calculate each of the concentrations:
Substitute the values into the expression:
40[0.10][1.5]
[3.0]][O[NO]
][NOK 2
2
22
22
eq ===
Example B 4.00 mol of NO2 is introduced into a 2.00L bulb. After a while equilibrium is attained. At equilibrium 0.50 mol of NO is found. What is Keq? The question implies that time had to pass for equilibrium to pass. The initial concentration is not an equilibrium concentration. ∴ you need to figure out the equilibrium [ ]’s.
The Keq expression is ][O[NO]][NOK
22
22
eq =
Calculate the concentrations from the questions:
2.0M2.00L
4.00mol][NO start2 == M250.02.00L
0.500mol[NO]eq ==
Set up an “ice” box 2NO(g)
O2(g) ↔ 2NO2(g)
Initial 0 0 2.00 Change Equilibrium 0.250 Units are omitted but they must all be the same (usually all M but it can work with mol, contrary to what your book says) Fill in the rest of the box. The values in the change row obey the rules of stoichiometry. 2:1:2 ratio ∆ NO = + 0.250 (a shift left) ∆ O2 = ½ ∆ NO = + 0.125 (the shift added to the reactants) ∆ΝΟ2 = − ∆ NO = - 0.250 (the shift took away NO2) 2NO(g)
O2(g) ↔ 2NO2(g)
Initial 0 0 2.00 Change +0.250 +0.125 -0.250 Equilibrium 0.250 0.125 1.75 Substitute the equilibrium concentrations into the equilibrium expression:
392[0.125][0.250]
[1.75]][O[NO]
][NOK 2
2
22
22
eq ===
Example C An unknown amount of NO2 was introduced to a 5.00L bulb. When equilibrium was attained according to the equation
2NO(g) + O2(g) ↔2NO2(g)
the concentration of NO was 0.800M. If Keq is 24.0, how many moles of NO2 were originally put in the bulb? There is a start and then time passing before equilibrium. We know [NO]eq = 0.800M and we want [NO2]start = x. Set up the ice box 2NO(g)
O2(g) ↔ 2NO2(g)
Initial 0 0 x Change Equilibrium 0.800 [NO] went from 0 to 0.800 so the shift was left. Fill in the box with changes that match the stoichiometry of 2:1:2. ∆ NO = + 0.800 ∆ O2 = ½ ∆ NO = + 0.400 ∆ΝΟ2 = − ∆ NO = - 0.800 2NO(g)
O2(g) ↔ 2NO2(g)
Initial 0 0 x Change +0.800 +0.400 -0.800 Equilibrium 0.800 0.400 x-0.800 Substitute the values into the equilibrium expression:
0.24[0.400][0.800]
0.800]-[x][O[NO]
][NOK 2
2
22
22
eq ===
You won’t have to use the quadratic equation in chem. 12. There will always be a way around it. (x-0.800)2=(24.0)(0.800)2(0.400)=6.144 take the square root of both sides
x-0.800=2.479 x=3.279M = [NO2]start We needed moles at the start so:
# moles NO2 = 16.4mol5.00LL
mol3.279 =×
Example D Keq= 49 for 2NO(g) + O2(g) ↔2NO2(g). if 2.0mol of NO , 0.2mol O2 and 0.40mol of NO2 are put into a 2.0L bulb, which way will the reaction shift in order to reach equilibrium? This question is asking for a decision not a numerical answer. You need to compare two numbers
Q==][O[NO]
][NOK2
2
22
eq , where Q is a trial value for Keq
if Q = Keq then the system is at equilibrium if Q < Keq then
][reactants[products]
is too small and the reaction will shift to
products to raise Keq
if Q > Keq then ][reactants
[products] is too big and the reaction will shift to
reactants to lower Keq
First find the equilibrium expression:
][O[NO]][NO
][reactants[products]K
22
22
eq ==
Calculate the starting concentrations: 1.0M
2.0L2.0mol[NO]start == M20.0
2.0L0.40mol][NO start2 ==
Calculate the reaction quotient
∴ the reaction must shift right to the products
Example E Keq = 3.5 for SO2(g) + NO2(g) ↔ SO3(g) + NO(g). If 4.0 mol of SO2(g) and 4.0 mol of NO2(g) are placed in a 5.0 L bulb and allowed to come to equilibrium. What will be the equilibrium concentrations of all the species? First write the equilibrium expression
]][NO[SO][NO][SOK
22
3eq =
Set up an ice box SO2(g) NO2(g) SO3(g) NO(g) Initial 0.80 0.80 0 0 Change Equilibrium We will need to enter variables for the rest of the spaces. It’s easiest to assign x as the change for one of the species: SO2(g) NO2(g) SO3(g) NO(g) Initial 0.80 0.80 0 0 Change -x -x +x +x Equilibrium 0.80-x 0.80-x x x Substitute the equilibrium concentrations into the Keq expression:
5.3x)-(0.80
x]][NO[SO
][NO][SOK 2
2
22
3eq === take the square root of both
sides
∴[SO3]=[NO]= 0.52M; [SO2]=[NO2]=0.28M
Example F A 1.0 L reaction vessel contained 1.0 mol of SO2, 4.0 mol of NO2, 4.0 mol of SO3 and 4.0 mol of NO at equilibrium
according to SO2(g) + NO2(g) ↔ SO3(g) + NO(g). If 3.0 mol of SO2 is added to the mixture, what will the new concentration of NO be when equilibrium is re attained? You have all the information needed to calculate Keq.
Set up an ice box (any additions or removals are put in the initial line) SO2(g) NO2(g) SO3(g) NO(g) Initial 1.0 +3.0 4.0 4.0 4.0 Change -x -x +x +x Equilibrium 4.0-x 4.0-x 4.0+x 4.0+x
