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)

http://www.chem.arizona.edu/chemt/Flash/HI.html


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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


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


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] =


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


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 _______


Decrease in reactant concentrationshifts equilibrium to the _______


Increase in product concentrationshifts equilibrium to the _____


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

Dr. Williamson’s Equilibrium Notes Demo Fe [Fe(SCN)]

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