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York University CHEM 1001 3.0 Chemical Equilibrium - 1
Chemical Equilibrium Basics
Reading: Chapter 16 of Petrucci, Harwood and Herring (8thedition)
Problem Set: Chapter 16 questions 25, 27, 31, 33, 35, 43, 71
York University CHEM 1001 3.0 Chemical Equilibrium - 2
Chemical Equilibrium
All chemical reactions are reversible; that is, if a reactioncan occur in one direction:
CO(g) + 2H2(g) 6 CH3OH(g)
then the reverse reaction can also occur:
CH3OH(g) 6 CO(g) + 2H2(g)
This applies to both elementary and overall reactions.
As a reaction proceeds, the rate of the forward reactiondecreases and the reverse reaction increases until they areequal. This balance is called equilibrium.
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Approach to Equilibrium - Example
Elementary reactions:
(1) N2O4(g) 6 2NO2(g) (forward)
(2) 2NO2(g) 6 N2O4(g) (reverse)
rate(1) = k1[N2O4] rate(2) = k2[NO2]2
At t=0 let [N2O4] > 0 and [NO2] = 0, then rate(1) > rate(2).
C As time passes, [N2O4] decreases and [NO2] increases.
C Rate(1) decreases and rate(2) increases until they are equal.
C When rate(1) = rate(2), we have equilibrium.
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Approach to Equilibrium - continued
When rate(1) = rate(2), we have equilibrium. Therefore:
k1[N2O4] = k2[NO2]2
= equilibrium constant
At equilibrium:
C Concentrations remain constant.
C Both reactions continue to occur.
C This is a dynamic equilibrium.
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Constant Conditions
In a static system, nothing happens. Example: Neon in theEarth's atmosphere is neither produced nor consumed.
At steady-state, the rate of production is balanced by therate of consumption. Example: Oxygen in the atmosphere isproduced by photosynthesis and consumed by respiration.
Dynamic equilibrium is a steady-state in which theconsumption process is the exact opposite of the productionprocess. Example: N2O4(g) W 2NO2(g).
People sometimes use equilibrium for all three of these.This is unacceptable in chemistry.
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Examples of Equilibrium
In a closed container, H2O(l) willcome into equilibrium with H2O(g).
(But H2O(g) and H2O(l) are not inequilibrium in the atmosphere.)
(a) I2 in H2O(l) (yellow).
(b) I2 equilibrium between H2O(l)(top) and CCl4(l) (bottom, purple).Most I2 is in the CCl4.
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Evidence of Dynamic Equilibrium
AgI(s) W Ag+(aq) + I-(aq)
Make up a saturatedsolution with excess solidAgI.
Add a small amount ofradioactive 131I-(aq) to thesolution.
The radioactivity willgradually appear in thesolid AgI.
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The Equilibrium Constant
Elementary reaction: aA + bB + ... W gG + hH + ...
At equilibrium: kf[A]a[B]b... = kb[G]g[H]h...
so = equilibrium constant
This applies to every step in a reaction mechanism.For the overall reaction, KC has the same form:
C products in the numerator, reactants in the denominator
C multiply concentrations raised to powers
C powers are the stoichiometric coefficients
York University CHEM 1001 3.0 Chemical Equilibrium - 9
Equilibrium Constant vs. Rate Laws
These are similar, but different.
C Both involve products of concentrations.
C The rate law may or may not include reactants,products, and other species.
C The equilibrium constant includes all reactants andproducts, but only reactants and products.
C In the equilibrium constant, the powers are thestoichiometric coefficients. Not so for rate laws(except elementary reactions).
York University CHEM 1001 3.0 Chemical Equilibrium - 10
Equilibrium Constant - Examples
CO(g) + 2H2(g) W CH3OH(g)
½CO(g) + H2(g) W ½CH3OH(g)
CH3OH(g) W CO(g) + 2H2(g)
The correct expression for the equilibrium constant dependson how the reaction is written.
York University CHEM 1001 3.0 Chemical Equilibrium - 11
Approach to Equilibrium
CO(g) + 2H2(g) W CH3OH(g)
C Three differentinitial conditions.
C Equilibriumreached in eachcase.
C Different finalconditions.
C All satisfy theequilibriumcondition.
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Final State - Equilibrium
Y 14.5
Y 14.4
Y 14.5
8
KC
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Beyond Concentrations
C In reality, rate laws do not always give the exactdependence of reaction rate on concentration.
C Also, equilibrium constants using concentrations, KC,are not constant (sometimes not nearly so).
C The expressions we use here are approximations.
C The thermodynamic equilibrium constant, Keq, isdefined using activities instead of concentrations.
C We use concentrations as approximations to activities.
York University CHEM 1001 3.0 Chemical Equilibrium - 14
Activities
C Thermodynamic equilibrium constants use activities.
C Activities are usually defined to be dimensionless.
C Thermodynamic equilibrium constants are dimensionless.
C Activities of pure solids and pure liquids are unity (1.0...).
C For dilute solutions:
solute activities approximately equal concentrations,
the solvent activity is approximately unity.
C For ideal gases, activities equal partial pressures.
C We will treat solutions as dilute and gases as ideal.
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Equilibrium Constant - More Examples
2H2(g) + O2(g) W 2H2O(l)
CaCO3(s) W CaO(s) + CO2(g)
2H2O(l) W H3O+(aq) + OH-(aq) Keq . [H3O
+][OH-]
HCl(aq) + H2O(l) W H3O+(aq) + Cl-(aq)
York University CHEM 1001 3.0 Chemical Equilibrium - 16
Equilibrium Constants Involving Gases
CO(g) + 2H2(g) W CH3OH(g)
For this reaction at 483 K, we found that:
The thermodynamic equilibrium constant is:
These are not the same.
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Converting between KC and KP
From the ideal gas law: PV = nRT
So, for CO: PCO = (nCO/V)RT = [CO]RT
So:
and
KC = 14.5 at T = 483 K. R = 0.08206 L atm mol-1 K-1
So RT = 39.6 L atm mol-1 and KP = 9.23×10-3
York University CHEM 1001 3.0 Chemical Equilibrium - 18
Magnitudes of Equilibrium Constants
C Equilibrium constants have a very wide range of values.
C Equilibrium constants can change enormously withchanges in temperature.
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Significance of Magnitude of K2H2(g) + O2(g) W 2H2O(l)
= 1.4×1083 at 298 K
If pO2 = 0.21 bar, what is the equilibrium pH2?
Y
Y PH2 = 5.8×10-42 bar Y < 1 molecule in the entireatmosphere
York University CHEM 1001 3.0 Chemical Equilibrium - 20
Significance of K - continued
C All reactions are reversible. So there are always somereactants left at equilibrium.
C A small K means that a reaction can not proceed in thedirection written (to the right). It is unfavorable.
C A reasonably large K means that the reaction may proceed.It is favorable. But it might be very slow.
C If K is so large that the equilibrium concentration of areactant is negligible, we say that the reaction "goes tocompletion".
C If K is very small, the reverse reaction goes to completion.
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Significance of K - example
C N2(g) + O2(g) W 2NO(g) KP = 4.7×10-31 at 298 K
The reaction can not proceed in the direction written.At 298 K, no significant NO can form from air.
C N2(g) + O2(g) W 2NO(g) KP = 1.3×10-4 at 1800 K
At 1800 K, significant NO may form. It turns out thatthe rate is fast at high temperatures.
C 2NO(g) W N2(g) + O2(g) KP = 2.1×1030 at 298 K
At 298 K, the NO may convert completely back to N2
and O2. But the rate is very slow.
York University CHEM 1001 3.0 Chemical Equilibrium - 22
The Equilibrium Constant - Summary
C All reactions are reversible. When the rates of the forwardand reverse reactions are equal, we have equilibrium.
C Equilibrium is a dynamic state, both forward and reversereactions continue to occur.
C The equilibrium constant, K, gives the condition thatmust obtain at equilibrium.
C K can be used to determine the direction in which a reactionis allowed to proceed. But the reaction might be very slow.
C K is expressed in terms of activities. Using concentrationsor partial pressures is an approximation.
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Predicting the Direction of Change
CO(g) + 2H2(g) W CH3OH(g)
C If we start with only CO and H2, the reaction will initiallyproceed to the right.
C If we start with only CH3OH, the reaction will initiallyproceed to the left.
C What if we start with a mixture of reactants and products?In what direction will the reaction proceed?
The answer depends on where the reaction stops - so itdepends on the equilibrium constant. But how?
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The Reaction Quotient
C The reaction quotient, Q, is defined to be the same asthe equilibrium constant, but using initial conditionsinstead of equilibrium conditions.
Q is used to decide the direction of change from aspecific set of initial conditions.
C If Q < K, the reaction will proceed to the right.
If Q > K, the reaction will proceed to the left.
If Q = K, the reaction is at equilibrium.
C QC is defined using concentrations; QP is defined usingpartial pressures. Must be consistent with KC or KP.
York University CHEM 1001 3.0 Chemical Equilibrium - 25
Using the Reaction Quotient
CO(g) + 2H2(g) W CH3OH(g) KC = 14.5 at 483 K
Initial conditions: [CO] = 1.0 M,
[H2] = 0.30 M,
[CH3OH] = 2.5 M
? In which direction will the reaction proceed?
QC > KC Y The reaction proceeds to the left.
York University CHEM 1001 3.0 Chemical Equilibrium - 26
Using the Reaction Quotient - continued
What happens as the reaction proceeds?
CO(g) + 2H2(g) W CH3OH(g)
We have QC > KC, so the reaction proceeds to the left.
In other words, the rate of the reverse reaction is greaterthan the rate of the forward reaction.
Thus: [CH3OH] decreases, [CO] increases, [H2] increases
Y QC decreases until QC = KC
When QC = KC, equilibrium isreached. The forward and reverserates are equal.
York University CHEM 1001 3.0 Chemical Equilibrium - 27
More on Using the Reaction Quotient
What if conditions change?
CO(g) + 2H2(g) W CH3OH(g)
At equilibrium, QC = KC. What if we then add more CO?
C Adding CO makes QC smaller.
C Then QC < KC.
C The reaction proceeds to the right.
As the reaction proceeds [CH3OH] 8, [CO] 9, [H2] 9, QC 8.
Eventually QC = KC and we have a new equilibrium.All concentrations are different from our first equilibrium.
York University CHEM 1001 3.0 Chemical Equilibrium - 28
Le Châtelier’s Principle
When an equilibrium system is subjected to a change, itresponds by attaining a new equilibrium that partiallyoffsets the impact of the change.
Input (Change) Response
increase inside pressure increase volume
increase outside pressure reduce volume
increase in temperature absorb heat
increase in concentration shift away from that species(as directed by Q)
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Le Châtelier’s Principle - Example
Closed container with liquid water andwater vapor in equilibrium:
H2O(l) W H2O(g)
C Increase external pressure:Volume decreases, some H2O(g)changes to H2O(l) (shift to left).
C Add He(g) to increase internal pressure: Volume increases,some H2O(l) changes to H2O(g) (shift to right).
C Increase temperature: Some H2O(l) changes to H2O(g)(shift to right absorbs heat).
York University CHEM 1001 3.0 Chemical Equilibrium - 30
Adding/Removing Reactants or Products
The equilibrium will shift to partially offset the change.
CO(g) + 2H2(g) W CH3OH(g)
C Change: adding CH3OH
Response: equilibrium shifts to left
C Change: adding CO
Response: equilibrium shifts to right
C Change: removing CH3OH
Response: equilibrium shifts to right
York University CHEM 1001 3.0 Chemical Equilibrium - 31
Effect of Changing the Volume
The equilibrium will shift to partially offset the change.
CO(g) + 2H2(g) W CH3OH(g)
C A shift to the right reduces the pressure (replaces 3 molesof gas with 1 mole of gas).
C A shift to the left increases the pressure.
C Change: decreased volume (increased pressure)
Response: equilibrium shifts to right
C Change: increased volume (decreased pressure)
Response: equilibrium shifts to left
York University CHEM 1001 3.0 Chemical Equilibrium - 32
Effect of Inert Gas
CO(g) + 2H2(g) W CH3OH(g)
(1) Add He(g) while keeping the volume fixed.
C No effect on the reactants or products.
C No effect on the equilibrium.
(2) Add He(g) while keeping the total pressure fixed.
C Volume increases.
C Partial pressures of reactants and products decrease.
C Equilibrium shifts to left.
York University CHEM 1001 3.0 Chemical Equilibrium - 33
Exothermic and Endothermic Reactions
C Breaking chemical bonds requires energy. Formingbonds releases energy.
C Chemical reactions may absorb or release energy. Thisenergy is often transferred as heat.
C Exothermic reactions release energy as heat:
CH4 + 2O2 6 CO2 + 2H2O
H2SO4(l) + 2H2O(l) 6 2H3O+(aq) + SO4
2-(aq)
C Endothermic reactions absorb energy as heat:
H2O(l) 6 H2O(g)
NaOH(s) 6 Na+(aq) + OH-(aq)
York University CHEM 1001 3.0 Chemical Equilibrium - 34
Effect of Change in Temperature
The equilibrium will shift to partially offset the change.
C For endothermic reactions, the equilibrium constantincreases when the temperature goes up.
C For exothermic reactions, the equilibrium constantdecreases when the temperature goes up.
Examples: T (K) KP
C CaCO3(s) W CaO(s) + CO2(g) 298 1.0×10-23
This reaction is endothermic. 1200 1.0
C CH4 + 2O2 6 CO2 + 2H2O
The equilibrium constant will be smaller at higher T.
York University CHEM 1001 3.0 Chemical Equilibrium - 35
Effect of a Catalyst
A catalyst is a substance that increases the rate of achemical reaction without being consumed by the reaction.
A catalyst:
C Does not change the position of equilibrium.
C Can decrease the time it takes to reach equilibrium.
C Will not make an unfavorable reaction occur.
C Can help a favorable reaction to proceed faster.
York University CHEM 1001 3.0 Chemical Equilibrium - 36
The Direction of Change - Summary
C Equilibrium is not achieved for just one set ofconcentrations. It is achieved for any concentrations thatsatisfy the equilibrium constant.
C The reaction quotient, Q, gives the direction of changefrom a specific set of initial conditions.
C Le Châtelier’s Principle: An equilibrium system respondsto a change in conditions by partially offsetting the change.
C The effect of changes in amount, pressure, or volume canalso be determined by using Q.
C The effect of temperature on K depends on whether thereaction is endothermic or exothermic.
York University CHEM 1001 3.0 Chemical Equilibrium - 37
Equilibrium Calculations
Many types of reactions rapidly reach equilibrium:
C acid/base reactions
C salts entering solution
C formation of complex ions
We need to learn how to calculate:
C equilibrium constants from composition data
C composition from equilibrium constants
Some of these calculations require the use ofstoichiometry and mass balance.
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Determining KC from Equilibrium Data
Example: N2O4(g) W 2NO2(g)
At equilibrium at 25 C, a 3.00 L vessel contains 7.64 gof N2O4 and 1.56 g of NO2. Find KC.
Solution:
Calculate the concentrations of reactants and products:
[N2O4] = 0.0277 M [NO2] = 0.0113 M
Evaluate the equilibrium constant:
KC = [NO2]2/ [N2O4] = 4.61×10-3
York University CHEM 1001 3.0 Chemical Equilibrium - 39
KP from Initial and Final Amounts of a Substance
Example: 2SO3(g) W 2SO2(g) + O2(g)
Initially, 0.0200 mol of SO3 are put into an evacuated 1.52 Lvessel. At equilibrium at 900 K, there are 0.0142 mol SO3.Find KP.
Solution:
(1) Calculate equilibrium moles of all reactants andproducts.
(2) Calculate partial pressures of all species.
(3) Evaluate the equilibrium constant: KP = P2SO2PO2/P
2SO3
York University CHEM 1001 3.0 Chemical Equilibrium - 40
KP from Initial and Final Amounts - continued
(1) The ICE Table
Reaction 2SO3(g) W 2SO2(g) + O2(g)
Initial 0.0200 mol 0.0000 mol 0.0000 mol
Change -0.0058 mol 0.0058 mol 0.0029 mol
Equilibrium 0.0142 mol 0.0058 mol 0.0029 mol
(2) Partial pressures: 0.690 atm 0.282 atm 0.141 atm
This used the ideal gas law: P = nRT/V = (48.6 atm mol-1)×n
(3) Evaluate KP = P2SO2PO2/P
2SO3
KP = 0.0235 = 0.024 (2 significant figures)
York University CHEM 1001 3.0 Chemical Equilibrium - 41
Variations in the ICE Table
C The ICE table is based on moles.
C Other quantities (concentrations, partial pressures) maybe used if they are strictly proportional to moles.
Examples:
Concentration = may be used if V is fixed.
Partial pressure may be used if T and V are fixed:
York University CHEM 1001 3.0 Chemical Equilibrium - 42
Equilibrium Partial Pressures from KP
Example: NH4HS(s) W NH3(g) + H2S(g)
At 25 °C, KP = 0.108
Excess solid NH4HS is placed in an evacuated flask. Findthe partial pressures of NH3 and H2S at equilibrium.
Solution: KP = PNH3 PH2S
Reaction NH4HS(s) W NH3(g) + H2S(g)
Initial - 0.0000 atm 0.0000 atm
Change - x atm x atm
Equilibrium - x atm x atm
x2 = KP = 0.108 Y PNH3 = 0.329 atm and PH2S = 0.329 atm
York University CHEM 1001 3.0 Chemical Equilibrium - 43
Equilibrium from Initial Conditions and KC
Example: N2O4(g) W 2NO2(g)
At 25 °C, KC = 4.61×10-3
0.0240 mole of N2O4 is placed in a 0.372 L flask.Find the moles of N2O4 at equilibrium.
Solution: KC = [NO2]2/ [N2O4]
Reaction N2O4(g) W 2NO2(g)
Initial 0.0240 mol 0.0000 mol
Change -x mol +2x mol
Equilibrium (0.0240-x) mol +2x mol
KC = (2x/V)2 / ((0.0240-x)/V)
York University CHEM 1001 3.0 Chemical Equilibrium - 44
Solving for the Unknown Concentration
From the previous slide: VKC (0.0240-x) = 4x2
Method 1: If only a little N2O4 dissociates, then
xn0.0240 Y 0.0240-x . 0.0240 Y 4x2 . 0.0240VKC
So x.0.00321 and nN2O4 = 0.0240-x = 0.0208 mol|
Check: x/0.0240 = 0.13, so assumption may be OK.
Method 2: Use the result from Method 1 to improve theapproximation: 0.0240-x . 0.0240-0.0032 = 0.0208
So 4x2 . 0.0208VKC Y x.0.00299 and nN2O4 = 0.0210 mol|
Can repeat until a constant value is reached.
York University CHEM 1001 3.0 Chemical Equilibrium - 45
Solving for Unknown Concentration - continued
Method 3: Use quadratic equation.
If 0 = ax2 + bx + c the
0 = 4x2 + VKCx - 0.0240VKC
a = 4 b = VKC c = -0.0240VKC
Y x = 0.00300 and nN2O4 = 0.0210 mol|
Check result: KC = [NO2]2/ [N2O4]
[N2O4] = nN2O4/V = 0.0564 M, [NO2] = 2x/V = 0.016 M
KC = 4.61×10-3 which is what it should be
York University CHEM 1001 3.0 Chemical Equilibrium - 46
Initial Concentrations Given for Multiple Species
Example: V3+(aq) + Cr2+(aq) W V2+(aq) + Cr3+(aq)
[V3+] = [Cr2+] = 0.0100 M [V2+] = [Cr3+] = 0.150 M
KC = 720 Find the equilibrium concentrations.
Solution:
Decide in which direction the reaction will proceed:
Since QC < KC, the reaction will proceed to the right.
York University CHEM 1001 3.0 Chemical Equilibrium - 47
Initial Concen. for Multiple Species - continued
Reaction V3+(aq) + Cr2+(aq) W V2+(aq) + Cr3+(aq)
Initial 0.0100 M 0.0100 M 0.150 M 0.150 M
Change -x M -x M x M x M
Equil. (0.0100-x) (0.0100-x) (0.150+x) (0.150+x)
Y
Y Y x = 4.25×10--3
Y [V3+] = [Cr2+] = 0.0057 M [V2+] = [Cr3+] = 0.154 M
York University CHEM 1001 3.0 Chemical Equilibrium - 48
Equilibrium Calculations - Summary
To calculate equilibrium constants from composition data:
C Calculate equilibrium amounts all species.
C Evaluate the equilibrium constant.
To calculate equilibrium composition from equilibriumconstants and initial composition:
C Construct the ICE table to get the all the equilibriumamounts in terms of a single unknown, x.
C Substitute into the expression for K, then solve for x.
C Evaluate all the equilibrium amounts.
Amounts can be moles, concentrations, or partial pressures.