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CHEMICAL EQUILIBRIUM Chapter 16. equilibrium vs. completed reactions equilibrium constant expressions Reaction quotient computing positions of equilibria: examples Le Chatelier’s principle - effect on equilibria of: addition of reactant or product pressure temperature. - PowerPoint PPT Presentation
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3 Nov 97 Entropy & Free Energy (Ch 20) 1
CHEMICAL EQUILIBRIUMChapter 16
• equilibrium vs. completed reactions• equilibrium constant expressions• Reaction quotient• computing positions of equilibria: examples• Le Chatelier’s principle - effect on equilibria of:
• addition of reactant or product• pressure• temperature
YOU ARE NOT RESPONSIBLE for section 16.7 (relation to kinetics)
3 Nov 97 Entropy & Free Energy (Ch 20) 2
For any type of chemical equilibrium of the typeTHE EQUILIBRIUM CONSTANTTHE EQUILIBRIUM CONSTANT
K =[C]c [D]d
[A]a [B]b
conc. of products
conc. of reactantsequilibrium constant
If K is known, then we can predict concentrations of products or reactants.
a A + b B c C + d Dthe following is a CONSTANT (at a given T) :
3 Nov 97 Entropy & Free Energy (Ch 20) 3
All reacting chemical systems can be characterized by their REACTION QUOTIENT, Q.
Q = = 0.350.25
= 1.40
Q - the reaction quotient
If Q = K, then system is at equilibrium.
To reach EQUILIBRIUM[Iso] must INCREASE and [n] must DECREASE.
Since K =2.5, system NOT AT EQUIL.
[iso][n]
Q has the same form as K, . . . but uses existing concentrations
n-Butane iso-Butane 0.25 0.35
3 Nov 97 Entropy & Free Energy (Ch 20) 4
Q/K
Q
Typical EQUILIBRIUM Calculations2 general types: a. Given set of concentrations, is system at equilibrium ?
Calculate Q compare to K
1
Q = K
IF:Q > K or Q/K > 1 REACTANTS
Q < K or Q/K < 1 PRODUCTS
Q=K at EQUILIBRIUM
3 Nov 97 Entropy & Free Energy (Ch 20) 5
Step 2 Put equilibrium [ ] into Kc .
Kc = [2x]2
[1.00 - x][1.00 - x] = 55.3
Step 1 Define equilibrium condition in terms of initial condition and a change variable
[H2] [I2] [HI]At equilibrium 1.00-x 1.00-x 2x
H2(g) + I2(g) 2 HI(g) Kc = 55.3
Step 3. Solve for x. 55.3 = (2x)2/(1-x)2 Square root of both sides & solve gives: x = 0.79
[H2] = [I2] = 1.00 - x = 0.21 M[HI] = 2x = 1.58 M
At equilibrium
3 Nov 97 Entropy & Free Energy (Ch 20) 6
“...if a system at equilibrium is disturbed, the system tends to shift its equilibrium position to counter the effect of the disturbance.”
EQUILIBRIUM AND EXTERNAL EFFECTS
• The position of equilibrium is changed when there is a change in: – pressure– changes in concentration– temperature
• The outcome is governed by LE CHATELIER’S PRINCIPLE
Henri Le Chatelier1850-1936- Studied mining
engineering- specialized in glass
and ceramics.
3 Nov 97 Entropy & Free Energy (Ch 20) 7
• If concentration of one species changes, concentrations of other species CHANGESto keep the value of K the same (at constant T)
• no change in K - only position of equilibrium changes.
Shifts in EQUILIBRIUM : Concentration
ADDING REACTANTS- equilibrium shifts to PRODUCTS
ADDING PRODUCTS- equilibrium shifts to REACTANTS
REMOVING PRODUCTS - often used to DRIVE REACTION TO COMPLETION
- GAS-FORMING; PRECIPITATION
3 Nov 97 Entropy & Free Energy (Ch 20) 8
SolutionA. Calculate Q with extra 1.50 M n-butane.
INITIALLY: [n] = 0.50 M [iso] = 1.25 MCHANGE: ADD +1.50 M n-butane What happens ?What happens ?
Q < K . Therefore, reaction shifts to PRODUCT
Q = [iso] / [n] = 1.25 / (0.50 + 1.50) = 0.63
n-Butane Isobutane
Effect of changed [ ] on an equilibrium
16_butane.mov(16m13an1.mov)
K = [iso][n]
= 2.5
3 Nov 97 Entropy & Free Energy (Ch 20) 9
Solution
B. Solve for NEW EQUILIBRIUM - set up concentration table [n-butane] [isobutane]Initial 0.50 + 1.50 1.25Change - x + xEquilibrium 2.00 - x 1.25 + x
Butane/Isobutane
K = 2.50 = [isobutane][butane]
1.25 + x2.00 - x
x = 1.07 M. At new equilibrium position, [n-butane] = 0.93 M [isobutane] = 2.32 M.
Equilibrium has shifted toward isobutane.
A B
3 Nov 97 Entropy & Free Energy (Ch 20) 10
Effect of Pressure (gas equilibrium)
Increase P in the system by reducing the volume.
Kc = [NO2 ]2
[N2O4 ] = 0.0059 at 298 K
NN22OO44(g) (g) 2 NO 2 NO22(g)(g)
16_NO2.mov(16m14an1.mov)
Increasing P shifts equilibrium to side with fewer molecules (to try to reduce P). Here, reaction shifts LEFT PN2O4 increases
See Ass#2 - question #6 PNO2 decreases
3 Nov 97 Entropy & Free Energy (Ch 20) 11
EQUILIBRIUM AND EXTERNAL EFFECTS
• Temperature change change in K• Consider the fizz in a soft drinkCO2(g) + H2O(liq) CO2(aq) + heat
• Increase TEquilibrium shifts left: [CO2(g)] [CO2 (aq)] K decreases as T goes up.
•Decrease T[CO2 (aq)] increases and [CO2(g)] decreases.K increases as T goes down
Kc = [CO2(aq)]/[CO2(g)]HIGHER T
LOWER T
• Change T: New equilib. position? New value of K?
3 Nov 97 Entropy & Free Energy (Ch 20) 12
Temperature Effects on Chemical Equilibrium
Kc = 0.00077 at 273 KKc = 0.00590 at 298 K
Kc [NO2 ]2
[N2O4 ]
N2O4 + heat 2 NO2 (colorless) (brown)
Horxn = + 57.2 kJ
Increasing T changes K so as to shift equilibrium in ENDOTHERMIC direction
16_NO2RX.mov(16m14an1.mov)
3 Nov 97 Entropy & Free Energy (Ch 20) 13
EQUILIBRIUM AND EXTERNAL EFFECTS
• Add catalyst ---> no change in K• A catalyst only affects the RATE of
approach to equilibrium.
Catalytic exhaust system
3 Nov 97 Entropy & Free Energy (Ch 20) 14
CHEMICAL EQUILIBRIUMChapter 16
• equilibrium vs. completed reactions• equilibrium constant expressions• Reaction quotient• computing positions of equilibria: examples• Le Chatelier’s principle - effect on equilibria of:
• addition of reactant or product• pressure• temperature
YOU ARE NOT RESPONSIBLE for section 16.7 (relation to kinetics)
3 Nov 97 Entropy & Free Energy (Ch 20) 15
Entropy and Free Energy (Kotz Ch 20)
• Spontaneous vs. non-spontaneous• thermodynamics vs. kinetics• entropy = randomness (So)• Gibbs free energy (Go)• Go for reactions - predicting spontaneous direction• Grxn versus Go
rxn • predicting equilibrium constants from Go
rxn
3 Nov 97 Entropy & Free Energy (Ch 20) 16
Entropy and Free Energy ( Kotz Ch 20 )
• How can we predict if a reaction can occur, given enough time?
• Note: Thermodynamics DOES NOT say how quickly (or slowly) a reaction will occur.
• To predict if a reaction can occur at a reasonable rate, one needs to consider:
• some processes are spontaneous; others never occur. WHY ?
THERMODYNAMICS
KINETICS
9-paper.mov20m02vd1.mov
3 Nov 97 Entropy & Free Energy (Ch 20) 17
Thermodynamics• state of a chemical system (P, T, composition)• From a given state, would a chemical
reaction decrease the energy of the system? • If yes, system is favored to react — a
product-favored system which will have a spontaneous reaction.• Most product-favored reactions are
exothermic.• Spontaneous does not imply anything about
time for reaction to occur. (kinetics)
3 Nov 97 Entropy & Free Energy (Ch 20) 18
Thermodynamics versus Kinetics
• Diamond to Graphite– spontaneous from thermodynamics– but not kinetically favored.
• Paper burns.- product - favored reaction.
- Also kinetically favored once reaction is begun.
3 Nov 97 Entropy & Free Energy (Ch 20) 19
Product-Favored Reactions
E.g. thermite reactionFe2O3(s) + 2 Al(s)
2 Fe(s) + Al2O3(s)H = - 848 kJ
In general, product-favored reactions are exothermic.
3 Nov 97 Entropy & Free Energy (Ch 20) 20
Non-exothermic spontaneous reactions
But many spontaneous reactions or processes are endothermic . . .
NH4NO3(s) + heat NH4+ (aq) + NO3
- (aq)Hsol = +25.7 kJ/mol
or have H = 0 . . .
3 Nov 97 Entropy & Free Energy (Ch 20) 21
Entropy, SEntropy, SOne property common to
product-favored processes is that the final state is more DISORDERED or RANDOM than the original.
Spontaneity is related to an increase in randomness.
The thermodynamic property related to randomness is ENTROPY, S. Reaction of K
with water
3 Nov 97 Entropy & Free Energy (Ch 20) 22
PROBABILITY - predictor of most stable state
WHY DO PROCESSES with H = 0 occur ?Consider expansion of gases to equal pressure:
This is spontaneous because the final state,with equal # molecules in each flask, is much more probable than the initial state,with all molecules in flask 1, none in flask 2
SYSTEM CHANGES to state of HIGHER PROBABILITYTHIS IS USUALLY the more RANDOM state.
3 Nov 97 Entropy & Free Energy (Ch 20) 23
Gas expansion - spontaneity from greater probability
Consider distribution of 4 molecules in 2 flasks
P1 < P2 P1 > P2P1 = P2
With more molecules (>1020) P1=P2 is most probable by far
3 Nov 97 Entropy & Free Energy (Ch 20) 24
WHAT about EXOTHERMIC REACTIONS ?Consider 2 H2 (g) + O2 (g) 2 H2O (l)
• HIGHLY EXOTHERMIC• final state (liquid) is MUCH MORE ORDERED (less random arrangement) than initial state (2 gases) • BUT PROBABILITY of final state is higher when one considers change in the surroundings. WHY ?• Heat evolved increases motion of molecules in the surroundings
>Increased disorder of surroundings
decreased disorder of system
MUST consider change in disorder in BOTH SYSTEM and SURROUNDINGS to predict DIRECTION of SPONTANEITY
3 Nov 97 Entropy & Free Energy (Ch 20) 25
Directionality of Reactions
How probable is it that reactant molecules will react?
PROBABILITY suggests that a product-favored reaction will result in the dispersal of energy or dispersal of matter or both.
3 Nov 97 Entropy & Free Energy (Ch 20) 26
Probability suggests that a product-favored reaction will result in the dispersal of energy or of matter or both.
Directionality of Reactions
Energy DispersalMatter Dispersal9_gasmix.mov20m03an1.mov
9_exorxn.mov20m03an2.mov
3 Nov 97 Entropy & Free Energy (Ch 20) 27
Directionality of ReactionsEnergy Dispersal
• Exothermic reactions involve a release of stored chemical potential energy to the surroundings.
• The stored potential energy starts out in a few molecules but is finally dispersed over a great many molecules.
• The final state—with energy dispersed—is more probable and makes a reaction product-favored.
3 Nov 97 Entropy & Free Energy (Ch 20) 28
Standard Entropies, So
• Every substance at a given temperature and in a specific phase has a well-defined Entropy• At 298o the entropy of a substance is called
So - with UNITS of J.K-1.mol-1
• The larger the value of So, the greater the degree of disorder or randomness e.g. So (in J.K-1mol-1) : Br2 (liq) = 152.2
Br2 (gas) = 245.5
For any process: So = So(final) - So(initial)
So(vap., Br2) = (245.5-152.2) = 93.3 J.K-1mol-1
3 Nov 97 Entropy & Free Energy (Ch 20) 29
S (gases) > S (liquids) > S (solids)
So (J/K•mol)H2O(gas) 188.8
H2O(liq) 69.9
H2O (s) 47.9
Ice Water
Vapour
Entropy, S