Dr. Williamson’s Equilibrium Notes Demo Fe [Fe(SCN)]
Transcript of Dr. Williamson’s Equilibrium Notes Demo Fe [Fe(SCN)]
Dr. Williamson’s Equilibrium Notes
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Chemical Equilibrium
Dr. V.M. WilliamsonStudent Version
DemoFe3+(aq) + SCN–(aq) -> [Fe(SCN)]2+(aq)Clear/yellow clear brick brown
Chemical Equilibrium
Fe3+(aq) + SCN–(aq) -> [Fe(SCN)]2+(aq)Copyright 1999 by HOLT Publishing Fe3+(aq) + SCN–(aq) -> [Fe(SCN)]2+(aq)
Establishment of Equilibrium2SO2(g) + O2(g) « 2SO3(g)
Conc
entra
tions
(M)
Timeteq
0.400
0.100
[SO2]
[SO3]
[O2]
0.344
0.056
0.172
A
[SO2]
teq
0.400
0.100
Time
[SO3]
[O2]
0.344
0.172
0.056
B
N2 + 3H2 <=> 2NH3
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The Equilibrium Constant -Obj #2● For the general reversible reaction:
aA + bB « cC + dDreactants products
● The equilibrium constant Kc at a given temperature is written as:
equilibrium concentrations
● For the general reversible reaction:aA + bB « cC + dD
reactants products
● If it is a 1 step mechanism:ratef = kf [A] [B] and rater = kr [C] [D]
● So: kf [A] [B] = kr [C] [D]● kf = [___] [___]
kr [ ] [ ] ● K = [C] [D] K is related to the ratio of the
[A] [B] ________________ .
Homogeneous vs HeterogeneousEquilibria
• Homogeneous: single-phase reaction2H2S(g) + 3O2(g) « 2H2O(g) + 2SO2(g)
• Heterogeneous: multi-phase reaction2HgO(s) « 2Hg(l) + O2(g)
What is K for the reverse?2Hg(l) + O2(g) « 2HgO(s)
Properties of Kc
● Dimensionless quantity● Reaction-specific● Constant for a given temperature● Independent of initial concentrations● Measured experimentally
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The Magnitude of Kc● A measure of the extent of reaction● Dependent on nature of reaction and
temperature● Consider: N2(g) + 3H2(g) « 2NH3(g)
☛ Kc >> 1: [ ] >> [ ]☛ Kc << 1: [Products] [Reactants]☛ Kc = 1: [Products] [Reactants]
Kc =NH3 2
N2 H2 3= 3.6 x 108
_________-FavoredReversible Reaction
2NO(g) + O2(g) « 2NO2(g)Copyright 1995 by Saunders College Publishing
_________-FavoredReversible Reaction
PbI2(s) « Pb2+(aq) + 2I–(aq)Copyright 1995 by Saunders College Publishing
Particle View● http://www.chem.arizona.edu/chemt/Flash/HI.
html
● H2 + I2 à 2HI
● K = [HI]2
[H2][I2]
● Equilibrium Arrows
Gas-phase Homogeneous Equilibria● Equilibrium constant can be written as Kp
since:P = (n/V)RT, and n/V º concentration
● When all in an equation are gases: 2H2S(g) + 3O2(g) « 2H2O(g) +2SO2(g)
Kp =
● KP = ∆n = mol gas products - mol gas reactants. So when the moles of gas products = moles of gas reactants, then KP = __________
● Consider the gas-phase reaction:H2 + I2 « 2HI
Equal amounts of gaseous H2 and I2 are injected into a 1.50-L reaction flask at a fixed temperature. At equilibrium, analysis shows that 1.80 mol H2, _____ mol I2, and ____ mol HI are present. Calculate Kc for the above reaction.
Obj. = Calculate K using equilibrium concentrations or pressures (vv)
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● Determine equilibrium concentrations and use these to calculate Kc:
• [H2] = =
• [I2] = 1.80 mol = 1.50L
• [HI] = mol = L
● Kc = [HI]2 = ( )2 = [H2] [I2 ] ( ) ( )
Obj. = Calculate K using equilibrium concentrations or pressures (vv)
● For the following reaction:N2 + 3H2 « 2NH3
Kc is 2.37 x 10 -3 at 1000K. An equilibrium mixture is found to contain ______ mol of N2 and _____ mol of H2 in a 1.0–L container. What equilibrium concentration of NH3 must be present?
• Set up the equilibrium constant expression and solve for the unknown concentration:
N2 + 3H2 « 2NH3
• Kc = [ ][ ] [ ]
• Kc [ ] [ ] = [ ]• (2.37 x 10 -3) ( M) ( M)3 = [NH3]2
• = [NH3]2
• M = [NH3]
● The value of Kc depends on the form of the balanced equation for the reaction.
● If the reaction is reversed, the new Kc is the reciprocal of the original Kc:
2HBr(g) + Cl2 (g) « 2HCl(g) + Br2(g)
2HCl(g) + Br2(g) « 2HBr(g) + Cl2(g)
Variations in Kc
Kc = HCl 2 Br2HBr 2 Cl2
= 4 x 104
Kc' = HBr 2 Cl2
HCl 2 Br2= 2.5 x 10-5 =
● If the reaction is multiplied by a factor n, the new Kc is obtained by raising the original Kc to the power n:
N2(g) + 3H2(g) « 2NH3(g)K1 = 3.5 x 108
1/2N2(g) + 3/2H2(g) « NH3(g)K2 = ( )1/2 = ( )1/2 =
Variations in Kc Variations in K● If reactions are summed up, overall K is the
product of individual K �s:N2 + 2O2 « 2NO2 Kp1= 6.9 x 10–19
2NO « N2 + O2 Kp2 = 2.3 x 1030
2NO + O2 « 2NO2 Kp3 = Kp1 x Kp2
Kp3 = x =
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Obj. = Calculate K given inital concentration and 1 equilibrium concentration or pressure (vv)
● Consider the gas-phase reaction:A « B + 2C
In a 1.00-L container 1.00 mol of A is placed. After equilibrium has been established, ____ mol of A are present in the container. Calculate Kc for the reaction.
• Determine given concentrations: Initial conc. of A = 1.00 mol = 1.00 M
1.00 LEquil. conc. of A = 0.700 mol = 0.70 M
1.00 L• Then set up a reaction summary:
A « B + 2CI 1.00 0.0 0.0CE 0.70
• Kc = [B] [C]2 = [ ] [ ]2 = [A] [ ]
● Consider the gas-phase reaction:2A + B « C + 3D
In a 2.00-L container are placed 2.00 mol each of A, B, C, and D. After equilibrium has been established, ________ mol of D are present in the container. Calculate Kc for the reaction.
Obj. = Calculate K given inital concentration and 1 equilibrium concentration or pressure (vv)
• Determine given concentrations: Initial conc. of each = 2.00 mol = 1.00 M
2.00 LEquil. conc. of D = 0.400 mol = 0.20 M
2.00 L• Then set up a reaction summary:
R 2A + B « C + 3DI 1.00 1.00 1.00 1.00C E 0.20• Kc = [C] [D]3 = [ ] [ ]3 =
[A]2 [B ] [ ]2 [ ]
The Reaction Quotient● For the general reaction:
aA + bB « cC + dD● A mass action expression or reaction
quotient Q can be set up at any time during a reaction:
not necessarily equilibrium concentrations
● Progress of a Reaction: Q vs Kc● Q Kc:System at equilibrium; no
further net reaction● Q Kc: Forward reaction proceeds
until equilibrium is established● Q Kc: Reverse reaction proceeds
until equilibrium is established
Obj. = Calculate Q and compare it to a given K to determine the direction of the equil.
K c =C c D d
A a B b
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Obj. = Using the Reaction Quotient● For the following reaction:
2HF(g) « H2(g) + F2(g) Kc = 1.0 x 10–13
the following concentrations were measured: [HF ] = 0.500 M, [H2] = 1.00 x 10–3 M, and [F2 ] = 4.00 x 10–3 M.
● Is this system at equilibrium? If not, what must occur for equilibrium to be established?
Using the Reaction Quotient• Substitute the concentrations into the
expression for the reaction quotient Q and compare Q with Kc:
• Since Q ___ Kc, the system is not at equilibrium; the __________ reaction must occur to ______________ [H2] and [F2] and _________________ [HF] until Q = Kc
Q vs. K● 2NO(g) + O2 (g) --à 2NO2(g)
● K = [NO2]2
[NO]2[O2]
Obj. = Use K and initial conc. to find the equil concentrations
● Express equilibrium concentrations of all species in terms of initial concentrations and a single unknown (x) that represents a change in concentration
● Set up the equilibrium constant expression in terms of equilibrium concentrations and solve for x
● Use x to calculate equilibrium concentrations● Solve by algebra, sq root, quadratic, or
estimation
● Given the following data at a certain temperature for this gas rxn:
A « C Kc = 9.0If 2.0 moles of A is put into a 1-liter container,
what is the equilbrium M of C?● Set up a reaction summary;
let x = mol/L of A that reacts:R A « C
K = [C]/[A]
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● Given the following data at a certain temperature:
A + B « C + D Kc = 49.0If _____ mol A and _____ mol B are placed in a 2.00-L container, what are the concentrations of all the species when the system reaches equilibrium?
● Determine initial concentrations:
[A] = 0.400 mol2.00 L
= 0.200 M
[B] = 0.400 mol2.00 L
= 0.200 M
● Then set up a reaction summary; let x = mol/L of A that reacts:
A + B « C + DI 0.200 0.200 0 0CE
Set up the equilibrium constant expression and solve for x:
[Kc] =[C][D][A][B]
= 49.0
(x)(x)(0.200 - x)(0.200 - x)
= 49.0
(x)2
(0.200- x)2 = 49.0
(x)(0.200- x)
= 7
x = 0.175
x = 7 (0.200-x) ; x = 1.400 - 7x 8x = 1.400 ; x = 0.175
● Given the following data at a certain temperature:
A + B C + D Kc = 49.0
● If 0.600 mol A and 0.200 mol B are placed in a ________-L container, what are the concentrations of all the species when the system reaches equilibrium?
● Determine initial concentrations, and set up a reaction summary; let x = mol/L of A that react:
R A + B « C + DICE
[A] = 0.600 mol2.00 L
= 0.300 M
[B] = 0.200 mol2.00 L
= 0.100 M
Set up the equilibrium constant expression and solve for x:
[Kc] = [C][D][A][B] = 49.0
(x)(x)(0.300� x)(0.100 � x) = 49.0
(x)2
(0.0300 � 0.400x + x2)= 49.0
On rearrangingthis reduces to:48.0x2 �19.6x + 1.47 = 0Solve for x using the quadratic formula :
x = �b ± b2 �4ac2a
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● The values obtained for x are:X = 0.309 or X = 0.099
● Select the answer that is physically significant, x = 0.099, since 0.309 yields a negative concentration for A:
● [A] = ● [B] = ● [C] = [D] =
● Given the following data at a certain temperature:
A 2C Kc = 1.0 x 10–8
● If 0.10 mol A is placed in a 1.00-L container, what are the concentrations of all the species when the system reaches equilibrium?
Problem
• Determine given concentrations: Initial conc. of A = 0.10 mol = 0.10 M
1.00 L• Then set up a reaction summary:
A « 2CI C E
• Kc = [C]2 = [A]
● Approximation using the 100x rule. If 100K = x is not significant when compared to the original concentration, drop the X.
● Otherwise keep the X, multiply, rearrange, and use the quadratic equation if necessary.
● 100 times the K in this problem is:100 (1.0 x 10–8 ) = 1.0 x 10–6
• Try this as x in 0.10-x Kc = [2X]2
[0.10-X ]• But [0.10 - ]= [ ] in 2 sig figs• So approx. with Kc = [2X]2 =
[0.10]
• Kc = [2X]2
[0.10] = X2
= X[C] at equil. = 2X = M
1.0 x 10–8 = 4X2
0.10
•proof: find all conc, plug into K=A2/B to see if you get the same K as given. (or do quad)• 1.0 x 10–8 = 4x2
0.10-x0.100 x 10–8 - 1.0 x 10–8 x = 4x2
0 = 4x2 + 1.0 x 10–8 x - 0.100 x 10–8
quadratic gives x = 1.58 x 10–5 = 1.6 x 10–5 M[C] at equil. = 2x = ___________M
● A system at equilibrium will remain in equilibrium until it is disturbed
● Disturbances (stresses applied) include changes in concentration, pressure, volume, and temperature.
● If a stress is applied to a system at equilibrium, it will proceed in the direction that will reduce the stress until a new equilibrium state is reestablished (Q = K)
● Reaction direction is _____________ the stress applied. Shift is ______from an increase, ____________a decrease.
Obj.= Le Chatelier�s Principle
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Le Chatelier�s Principle:Effect of Stress on Equilibrium
Increase in reactant concentrationshifts equilibrium to the _______
Le Chatelier�s Principle:Effect of Stress on Equilibrium
Decrease in reactant concentrationshifts equilibrium to the _______
Le Chatelier�s Principle:Effect of Stress on Equilibrium
Increase in product concentrationshifts equilibrium to the _____
Le Chatelier�s Principle:Effect of Stress on Equilibrium
Decrease in product concentrationshifts equilibrium to the _______
Effect of Changing Reactant and Product Concentrations
K = [products][reactants]
reactants ßà products K?
Le Chatelier�s Principle:Effect of Volume Change
2NO2(g) « N2O4(g)Copyright 1995 by Saunders College Publishing
Consider:•V •P •[ ] •Shift ________ from side with ______ gas moles; Shift _______side with ______ gas moles
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Effect of P-V Changes● In reactions for which ______________,
volume or pressure changes do not affect theposition of equilibrium
● In reactions for which ____________, volume or pressure changes affect the position of equilibrium
System (gases) Effect of P-V stress
2NO « N2 + O2N2O4 « 2NO2
Effect of Pressure-Volume ChangeStress Direction of Shiftvolume decrease, pressure increase
volume increase,pressure decrease
stress on side with more moles
�decrease� on side with more moles
Volume (Pressure) Changes
● Change in P or V caused by the addition of an inert gas _____________________ the equilibrium (total pressure increases, but partial pressures of reactants is constant)
● Changes in P, V, and amount of pure solid or liquid have ________________ on the equilibrium involving pure solid or liquid
● Temperature increase favors ____________ reactions; shift to right; Kc _________
● Temperature increase disfavors __________ reactions; shift to left; Kc ____________
Copyright 1995 by Saunders College Publishing
N2O4(g) + heat « 2NO2(g)colorless brown
Le Chatelier�s Principle: Temperature
Effect of Changing Temperature
K = [products][reactants]
Reactants + Heat ßà productsK?
N2O4(g) + heat « 2NO2(g)OR
N2O4(g) « 2NO2(g) ∆H = 58.0 kJ
● _____● _____● _____● _____● _____
● _____● _____
● Add N2O4
● Remove NO2● Increase Pressure● increase Volume● Decrease
Temperature● Decrease Volume● Decrease Pressure
[NO2] K shift_____ _________ _________ ____Inc. _________ ____
_____ _________ ____
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Effect of a Catalyst● Catalyst lowers both ____for
the forward reaction and ____for the reverse reaction to the same extent
● Catalyst _______________ the position of equilibrium
● Catalyst _________________ Kc
● Catalyst ________ _________ required to reach equilibrium
Le Chatelier�s Principle:An Application
● Ammonia, NH3, used in the manufacture of fertilizers, plastics, dyes, explosives, and synthetic fibers, is synthesized industrially using the Haber process:N2(g) + 3H2(g) « 2NH3(g) Kc = _________
∆H = ________ kJ/mol ● Although Kc is _______, the reaction occurs
very slowly at 25oC. How would you adjust the temperature and pressure so that the NH3 yield is optimized?
The Haber Process: Effectof T and P on Ammonia Yield
oC Kc 10 atm 100 atm 1000 atm
Mole %NH3 in Equilibrium Mixture
209 650 51 82 98467 0.5 4 25 80758 0.014 0.5 5 13
Obj. = Relationship Between DGo
and the Equilibrium Constant
DG� Product FormationK
DG� 0K 1
DG� 0 K 1
● DG = DG� + RT ln Q● At equilibrium: Q = K DG = 0● DG� =
DG� 0K 1
● Find the equilibrium constant for: CO (g) + NO (g) « CO2 (g) +1/2 N2 (g)
● ∆H�= -373.37kJ; ∆S�= -99.19 J/K∆G�= -343.81kJ
● DG� = –RT ln KDG� = ln K
-RT● ( ) = ln K
( )= ln K= K
The Equilibrium Constant as a Function of Temperature
● The van�t Hoff equation relates the equilibrium constant with temperature:
equilibrium constants at the two temperatures
temperatures in Kelvin
gas constant
standard enthalpychange for reaction
lnKT2KT1
= ΔHo
R1T1
– 1T2