Equilibrium is the state where the rate of the forward reaction is _________ the rate of the reverse reaction. ◦ At these conditions, concentrations of all reactants and

products remain constant with time _________ equilibrium has been established at constant temperature.

For stoichiometry problems, we assumed chemical systems “_________ _________ _________”

In reality, the majority of chemical reactions do not go to completion; they just have very large K values

Large K value indicate a product favored reaction.

_________ _________ – occurs when the forward and reverse rxns happen at the same time

_________ _________ equilibrium – chemical rxnhappens continuously with no net change

Ex. saturated sugar in iced tea ~ the sugar is undergoing molecular solvation at the same time that its re-crystalizing so you can’t see a net change of sugar on the bottom

Ex. Ocean ~ water is evaporating and condensing instantaneously

equilibrium is _________ _________ _______

Whether the reaction lies far to the right (favors products) or to the left (favors reactants) depends on 3 main factors:

Initial concentrations ◦ (more collisions—faster reaction)

Relative energies of reactants and products ◦ (nature goes to minimum energy)

Degree of organization of reactants and products◦ (nature goes to maximum disorder)

the equilibrium _________ tells the tendency of the rxn to either _________ _________ or _________ _________ _________

if Kc is large, _________ _________ = rxn likes to form products (_________ _________)

if Kc is small, _________ _________ = rxn likes to stay as reactants (_________ ________)

if Kc = 1, then the rxn is at _________ ____(very rare)

THE EQUILIBRIUM EXPRESSION: A general description of the equilibrium condition proposed by Gudberg and Waage in 1864 is also known as the _________ _________ ______.

Equilibrium is temperature dependent, however, it _________ _________ _________ with concentration or pressure.

For aqueous activity = _________ _________gas activity = _________solids & liquids = ____

For the reaction A + B C + D rate =

After time, C & D react together to make A & B. (_________ _________)

C + D A + B rate =

So for the equilibrium reaction of

aA + bB cC + dD Kc = [C]c[D]d[A]a[B]b

Kc = conc. of products (or activities)conc. of reactants

The product concentrations appear in the numerator and the reactant concentrations in the denominator. Each concentration is raised to the power of its stoichiometriccoefficient in the balanced equation.

When finding K, don’t use liquids or solids, also K has no units

Ex. 1) N2(l) + 3H2(g) 2NH3(g)

If concentrations are given use [substance]:Kc =

If pressures are given use P (substance)Kp =

Ex. 2) MgCO3(s) Mg2+(aq) + CO3

2-(aq)

Kc =

Ex. 3) Write the equilibrium expression for the following reaction

a) for concentration b) for pressure

4 NH3(g) + 7 O2(g) 4 NO2(g) + 6 H2O(g)

Kp = Kc(RT)∆n or Kc = Kp(RT) -∆n

∆n = (moles gas products) – (moles gas reactants)

R = universal gas law constant 0.08206 L atm/mol K T = temperature in Kelvin

Kc = Kp _________ _________ the number of moles of gas products _________ the number of moles of gas reactants since (RT)Δn = (RT)0 = 1

Kp = Kc(RT)Δn is often referred to as the “politically correct” (pc) equation to help you remember the order of the K’s in the equation

Ex. 4) For the rxn 2S(g) + 3O2(g) 2SO3(g), Kp = 2.2 x 108 at 301oC. What is Kc?

Kreverse rxn = 1K

Ex) HF(aq) H+(aq) + F-

(aq) K = 7.2 x 10-4

Find K if H+(aq) + F-

(aq) HF(aq)

Kreverse rxn = 1 7.2 x 10-4

Kreverse rxn = 1400

For a rxn at a specific temp, Kc is always the same value when the concentrations are at equil.

Ex. 5) a) Determine the equilibrium constant for the for

the following reaction at 25oC b) What is the value of K for the reverse reaction?c) What is the value of K for SO3 SO2 + ½O2 ?

2SO2(g) + O2(g) 2 SO3(g)Exp. 1 = 0.344 M 0.172 M 0.056 MExp. 2 = 0.424 M 0.212 M 0.076 M

Ex. 5)2SO2(g) + O2(g) 2 SO3(g)

Exp. 1 = 0.344 M 0.172 M 0.056 MExp. 2 = 0.424 M 0.212 M 0.076 M

1st find the equilibrium expression

Ex. 5) 2SO2(g) + O2(g) 2 SO3(g)Exp. 1 = 0.344 M 0.172 M 0.056 MExp. 2 = 0.424 M 0.212 M 0.076 M

2nd – substitute the values into the expression

Kc1=

Kc2 =

Part b) - substitute the value for K into the inverse equation

2 SO3(g) 2SO2(g) + O2(g)

Kreverse rxn = 1K

Part c) - substitute the value for K into the inverse equation and take the square root b/c the coefficients are cut in half.

2 SO3(g) 2SO2(g) + O2(g) is changed toSO3(g) SO2(g) + ½O2(g)

Kreverse rxn = 1√K

For use when the system is NOT at equilibrium. For the general reaction

aA + bB cC + dD Qc = [C]c[D]d[A]a[B]b

Qc has the appearance of K but the concentrations are not necessarily at equilibrium.

Q measures the progress of the reaction. Quite useful for predicting what will happen under special conditions.

If…Q = Kc the rxn is at _________ _________

Q < Kc the rxn proceeds to the right (_________ _________ _________ _________ _______)

Q > Kc the rxn proceeds to the left (_________ _________ _________ _________ _______)

Ex. 6) At a high temp, Kc = 65.0 for this rxn: 2HI(g) H2(g) + I2(g)

Using the following conc. determine if the system is at equilibrium. If not, which direction will the rxn proceed to reach equilibrium?

[HI] = 0.500 M, [H2] = 2.80 M, [I2] = 3.40 M

Ex. 7) For the synthesis of ammonia at 500°C, the equilibrium constant is 6.0 × 10-2. Predict the direction in which the system will shift to reach equilibrium in each of the following cases:

a. [NH3] = 1.0×10-3 M; [N2] = 1.0×10-5 M; [H2] = 2.0×10-3 M

Ex. 7) For the synthesis of ammonia at 500°C, the equilibrium constant is 6.0 × 10-2. Predict the direction in which the system will shift to reach equilibrium in each of the following cases:

b. [NH3] = 2.00×10-4 M; [N2] = 1.50×10-5 M; [H2] = 3.54×10-1 M

Ex. 7) For the synthesis of ammonia at 500°C, the equilibrium constant is 6.0 × 10-2. Predict the direction in which the system will shift to reach equilibrium in each of the following cases:

c. [NH3] = 1.0×10-4 M; [N2] = 5.0 M; [H2] = 1.0×10-2 M

A rxn at equilibrium will tend to stay at equilibrium

If you disturb the rxn with a stress, the equil. will shift in a direction to _________ the stress.

3 types of stressors◦ Concentration◦ Pressure◦ Temp/heat

Adding or removing a reagent causes the equilibrium to shift to reestablish K.

Shifts occur to reestablish equilibrium positions. Think about K!

Generally, [products][reactants]

Adding a catalyst to a reaction has causes no shift and has NO EFFECT on K. It just causes equilibrium to be established faster!

– What happens when you add or remove reactants/products?

Ex. 8) H2(g) + CO(g) H2CO(aq)

a. Add more CO -

b. Remove H2CO -

c. Remove H2 -

for gases only. Why? s & l are only slightly compressible.

look at the number of moles to determine which way the rxn will shift◦ If P decreases, V increases; shift to the side with the larger

# of gas moles◦ If P increases, V decreases; shift to the side with the

smaller # of gas moles

Increasing pressure causes the equilibrium to shift to the side containing the fewest number of moles of gas. The reverse is also true.

Ex. 9) 3H2(g) + N2(g) 2 NH3(g) Increase Pressure

◦ If P decreases, V increases; rxn shifts to the side with more gas moles◦ If P increases, V decreases; rxn shifts to the side with less

gas moles

Ex. 10) C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)Increase pressure

Changing the temperature is a lot like adding or removing a “reactant or product”. Think of heat energy as a “reactant” or “product”.

Endothermic: +ΔH value (heat is added into the system \ heat is a reactant)

Exothermic: −ΔH value (heat is lost from the system \ heat is a product)

Reminder - heat is a part of the rxn.

Endothermic : ∆H is a reactant

Exothermic : ∆H is a product

Note: K of an exothermic rxn decreases with increasing temp, and K of an endothermic rxnincreases with increasing temp.

Ex. 11) The following is an endothermic reaction. What happens when?

2 SO3(g) 2SO2(g) + O2(g)

a. Remove heat

b. Add heat

Ex. 12) The following is an exothermic reaction. What happens when?

C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)

a. Remove heat

b. Add heat

The magnitude of Kc is a measure of the extent to which reactions occur. For any balanced eqn. the value of Kc

◦ is constant at a given temp

◦ changes if temp. changes

◦ does not depend on the initial concentrations.

There are basically six types of equilibrium problems in this chapter. All are approached in the same way except for the solving of x. All can be solved using RICE.

R = reaction I = Initial conc.C = Change in conc. E = Equilibrium or ending amount

The following are examples of each type of equilibrium problem. Work each of them on a separate sheet of paper.

Ex. 13) What is the value of Kc for N2 + 3 H2 ↔ 2NH3, if you start with 2.00M N2, 4.00M H2, and 5.00M NH3 and at equilibrium you have 2.00M H2.

R

I

C

E

Ex. 14) K for A + B ↔ C + D is 49. If 0.400 moles of each A and B are placed in a 2.00 L flask, what are the concentrations of all species at equilibrium?

Ex. 15) For A + B ↔ C + D, the equilibrium constant is 49 at a certain temperature. If 0.600 moles of A and 0.200 moles of B are placed in a 2.00L flask, what concentration of all species are present at equilibrium?

Ex. 16) A .200 M HF solution has a Ka of 7.2x10-4. What is the equilibrium concentration of H+? HF(aq) ↔ H+

(aq) + F-(aq)

*solids and pure liquids are not included in the equil expression

Ex. 17) A(s) ↔ B(g) + C(g) Solid A is placed in a vessel. After equilibrium is established, the total pressure is 0.75 atm. What is the value for Kp?

Ex. 18) H2 + I2 ↔ 2HI, K for this reaction is 50. The following concentrations are initial concentrations: H2 = 0.080 M, I2 = 0.060 M, HI =0 .79 M. What concentrations of H2, I2, and HI will be present at equilibrium?

Ex.19) Nitrosyl bromide, NOBr, is 34% dissociated by the following reaction at 25oC, in a vessel in which the total pressure is 0.25 atmosphere.

2NOBr(g) ↔ 2NO(g) + Br2(g)A) What is the value of Kp? B) What is Kc ?

Reminder:The change in the Gibbs Free Energy is a reliable

indicator of spontaneity of a physical process or chemical reaction.

When:∆G is < 0 (negative)

reaction is thermodynamically favored (spontaneous) product favored

∆G is = 0 system is at equilibrium

∆G is > 0 (positive) reaction is not thermodynamically favored

(nonspontaneous) reactant favored

Thermodynamic View of Equilibrium Equilibrium point occurs at the lowest value of free

energy available to the reaction system

∆Gorxn = -RTlnK use R = 8.314 J/mol K b/c of ∆Go

∆Gorxn < 0 K >1 products favored

(thermodynamically favored)

∆Gorxn = 0 K = 1 at equilibrium (very rare)

∆Gorxn > 0 K < 1 reactants favored

(not thermodynamically favored)

Ex. 20) Find the Kp for N2O4(g) 2NO2(g) for the temp at 25oC.

van’t Hoff eqn. allows us to find K at various tempsln KT2 = ∆Ho 1 - 1

KT1 R T1 T2

Ex. 21) Estimate Kp at 250. oC for the rxn of 2NO2(g) 2NO(g) + O2(g) Kp = 4.3 x 10-13 at 25oC and ∆Ho = 114kJ/mole.

Part II

Standard electrode potentials are determined at thermodynamic standard conditions.1 M solutions1 atm of pressure for gasesliquids and solids in their standard statestemperature of 250 C

Potentials change if conditions are nonstandard.

Nernst equation describes the electrode potentials at nonstandard conditions.

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(

E = E - 2.303RTnF

log Q

E = potential under condition of interest

E potential under standard conditions

R = universal gas constant = 8.314 Jmol K

T = temperature in Kn = number of electrons transferred

F = the faraday = (96,487 C / mol e 1 JC V

J / V mol eQ = reaction quotient

0

0

-.

-

=

×

=

)

,96 487

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Substitution of the values of the constants into the Nernst equation at 250 C gives:

2.303 RTF

so

E = E - 0.0592n

log Q0

= 0 0592.

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Ex. 22) Calculate the initial potential of a cell that consists of an Fe3+/Fe2+ electrode in which [Fe3+]=1.0 x 10-2 M and [Fe2+]=0.1 M connected to a Sn4+/Sn2+ electrode in which [Sn4+]=1.0 Mand [Sn2+]=0.10 M . A wire and salt bridge complete the circuit at 25oC.

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1st Calculate the E0 cell by the usual procedure. (Use your standard reduction potentials table)

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2nd – Write the expression for Q

55

3rd - Substitute the ion concentrations into Q and the number of electrons transferred to calculate Ecell.

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Ex. 23) Calculate the cell potential for the following electrochemical cell at room temerature, 25oC. If the [Sn2+] = 4.50 x 10-1 M and [Ag+] = 0.110 M

Sn(s) + 2Ag+ Sn2+ + 2Ag(s)

Spontaneous – Thermodynamically favorable rxns◦ E0

cell = positive◦ ∆G0 = negative◦ K > 1 (product favored)

Nonspontaneous – Not thermodynamically favored rxns◦ E0

cell = negative◦ ∆G0 = positive◦ K < 1 (reactant favored)

the relationship of ∆G0 and K (equilibrium constant) for a reaction.

K log RT 303.2Gor lnK -RTG 00 −=∆=∆

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The relationship between ∆G0 and E0cell is

also a simple one.

e ofnumber ne mol J/V 96,487 F where

E F-n G

-

-

0cell

0

=

=

=∆

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Combine these two relationships into a single relationship to relate E0

cell to K.

RT

E Fn K ln

orlnK RT -E Fn -

0cell

0cell

=

=

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Ex. 24) Calculate the thermodynamic equilibrium constant for the reaction at 250C. ∆G0 = -89.4 kJ/molrxn

Cu PbCuPb 22 +→+ ++

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Ex. 24) Calculate the thermodynamic equilibrium constant for the reaction at 250C. ∆G0 = -89.4 kJ/molrxn

63

Ex. 25) Calculate the equilibrium constant for the following reaction at standard conditions.

Sn(s) + 2Ag+ Sn2+ + 2Ag(s)

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