Chapter 16 - Principles of Chemical Equilibrium

- all reactions are "reversible" - principle of micro-reversibility

- the "committed step"- much theory

- not always obvious- for some the reverse reaction is very difficult

- a measure of reversibility is the "equilibrium constant"

16.1 Dynamic Equilibriumthe book provides a number of examples:

- vapour over a liquid- substance in a saturated solution- iodine partitioning (solid/vapour)

º all of these are “physical” equilibria

what about chemical equilibria?- protonation/deprotonation of an acid- special type of equilibrium - Chapter 17- CaO + CO2 º CaCO3

- 3O2 º 2O3

all equilibria are dynamic

that is, A º B

all equilibria are simply a competition between a forward and reverse reactionequilibrium is achieved when the rate of the forward reaction is exactly equal to the rate of the backwards (or reverse) reaction

- the net change is “zero” but this is not the same thing as saying there is “no reaction” - the reactions occur but balance each other out for no net change

note that the individual reactions (the forward and reverse reactions) for an equilibrium are “elementary steps” so the coefficients are the stoichiometricfactors

hence, if: 2A º B

then: kforward[A]2 = kreverse[B]

16.2 The Equilibrium Constant Expression

there are many ways to write the equilibrium constant expression but they all equate to:

Keq = [products] / [reactants]

that is, a numerical constant gives the proportion of products relative to the proportion of reactants

recall: by definition, products = right hand sidereactants = left had side

the equilibrium constant expresses a ratio

so, what happens if part of the equilibrium is missing?for example,

CO(g) + 2H2 (g) º CH3OH(g)

(this is the production of methanol from ‘syn gas’)

the equilibrium can be written as:

Keq = [CH3OH(g)] / [CO(g)][H2(g)]2 = 14.5 M-2

Keq = [CH3OH(g)] / [CO(g)][H2(g)]2 = 14.5 M-2

if we start with, say, [CO] = 1.0 M and [H2] = 1.0 M, then to balance this equation, we are going to have to have enough methanol present so that the ratio of methanol to carbon monoxide and hydrogen is 14.5 to 1.

Note that since we do not have any methanol to begin with, reducing either the concentration of carbon monoxide or hydrogen will not get us to a balanced equation. The only way to achieve the prescribed ratio is by increasing the amount of methanol present. (In this case, to 14.5 M.)

the ratio is called “the equilibrium constant expression” and Keq is the “equilibrium constant”

normally, when the concentrations are in molarity, then the equilibrium constant can be called ‘Kc’ but if pressure is used then it is also designated as ‘Kp’

We also have the terms ‘Ka’ for acid equilibrium constant and ‘Kb’ for base equilibrium constant and, of course, there is ‘Ksp’ for solubility equilibrium constant which is more frequently referred to as the “solubility product”

º all use a capital K

General expression for equilibria:

aA + bB º cC + dD

Kc = [C]c[D]d / [A]a[B]b

(wrong in note package)

thermodynamics

16.3 Relationships involving Equilibrium Constants

between Kc and the balanced equation:1) reversing the equation inverts Kc

i.e. A º B k1[A] = k2[B] k1/k2 = [B]/[A]

B º A k3[B] = k4[A] k3/k4 = [A]/[B]

since, [B]/[A] = ([A]/[B])-1, k1/k2 = (k3/k4)-1

and K12 = (K34)-1

2) multiplying the coefficients or stoichiometric factors by a constant does the same thing to the exponents

i.e. 2H2 + O2 º 2H2O Kc = [H2O]2/[H2]2[O2]

H2 + ½O2 º H2O Kc = [H2O]/[H2][O2]½

3) adding equilibria, we multiply the constants

i.e. A º B K1 = [B]/[A]B º C K2 = [C]/[B]

overall: A º C K = [C]/[A] = ([B]/[A])([C]/[B])= K1K2

Equilibria involving gases, Kp

for a gas, PV = nRT so, (n/V) = P/RTif we consider an equilibrium involving gases, then we have to account for the ‘RT’ term in the gas law equations

two examples:H2 + Cl2 º 2HCl

Kc = [PHCl/RT]2 / [PH2/RT][PO2/RT] = [PHCl]2 / [PH2][PO2] x (RT)2/(RT)2

2H2 + O2 º 2H2O

KC = [PH2O/RT]2 / [PH2/RT]2[PO2/RT]

= [PH2O]2 / [PH2]2[PO2] x (RT)3/(RT)2

= [PH2O]2 / [PH2]2[PO2] x (RT)3/(RT)2

Kc = Kp x RT

General expression:Kp = Kc(RT))n gas

where )ngas is the difference in the number of gas molecules between the products and the reactants

i.e. 2H2 (g) + O2 (g) º 2H2O(g)

where )ngas = (2) - (2+1) = -1

2H2 (g) + O2 (g) º 2H2O(l)

where )ngas = (0) - (2+1) = -3

Pure substances and Equilibrialiquids and solids - when pure - are not included in the Equilibrium Constant Expression

Why? Because their concentration is constant and just subsumed by the equilibrium constant

i.e. pure water - always 55.5 M - never changespure gold - density 19.3 g/cm3 and 1cm3 = 1 mL

therefore, 1 L of gold = 19.3 kgand gold is 98 M - a constant

16.4 Magnitude of Equilibrium Constants

As it is a ration, it gives the relationship between the products and the reactants. Hence,- large K - more products than reactants- small K - more reactants that products- K . 1 - balance between products and reactants

note that a large K tells us that the reaction favoursthe products but still have the activation barrier to consider - reaction may not reach equilibrium quickly

16.5 The Reaction Quotient Concept- allows you to determine the direction of the shift in an equilibrium when it has been disturbed- not really necessary!

Le Ch>telier’s PrincipleWhen a system at equilibrium is subjected to a change in temperature, pressure, or concentration of a reacting species, the system responds by attaining a new equilibrium that partially offsets the impact of the change.

much easier to say “if an equilibrium is disturbed, it will respond in a way so as to minimize the disturbance”

but which way will that be?that is what ‘Q’, the reaction quotient, is

supposed to tell us

‘Q’ is the ‘instantaneous’ equilibrium quotient

that is, it is the value of the equilibrium constant expression calculated based on the concentration of the species present at any instant in time

It is not really necessary to calculate ‘Q’. However, consider the following:

CO + 2H2 º CH3OHKeq = [CH3OH]/[CO][H2]2 = 14.5 M-2

if [CO] [H2] [CH3OH] then0.0 M 0.0 M 0.1 M Q = 40.1 M 0.1 M 0.0 M Q = 00.1 M 0.1 M 0.1 M Q = 100

according to the book,

if Q = 0 then the reaction proceeds to the right and produces more product

Q < Keq then the reaction proceeds to the right and produces more product

Q = Keq then the reaction is at equilibriumQ > Keq then the reaction reverses and shifts in

favour of the reactants (proceeds to the left)

Q = 4 then the reaction reverses and shifts in favour of the reactants (proceeds to the left)

16.6 Altering Equilibrium Conditions- applying Le Ch>telier’s Principle

effect of changing concentration- equilibrium shifts so as to remove the effect- finds a “new” equilibrium

effect of changing pressure- two ways to do this - adding an inert gas and adding one of the reactants

- both change the concentration terms, Pgas

- change in partial pressure is a change in concentration and the system responds accordingly

note: a decrease in overall pressure - equilibrium shifts in favour of the side with the most number of gas molecules

an increase in overall pressure - equilibrium shifts to the side with the fewest gas molecules

effect of temperaturebook says “raising the temperature or an equilibrium mixture shifts the equilibrium condition in the direction of the endothermic reaction. Lowering the temperature causes a shift in the direction of the exothermic reaction.”

- of course, this means that you have to know the thermicity of the reaction!!

more important is the Arrhenius equation as a change in temperature is the only thing that can change the value of the rate constant and the equilibrium constant is a function of both the forward and reverse rate constants!

changes in temperature are the only thing that can change the value of Keq

effect of catalyst- speeds up both the forward and reverse reaction (by changing the activation energy of both)

- changes how fast equilibrium is obtained- doesn’t change the position of the equilibrium

16.7 Equilibrium Calculationseight simple steps:

1) write out the equilibrium (most likely already done for you)

2) write out the equilibrium constant expression3) write out and fill in an ICE table

ICE stands for: InitialChangeEquilibrium

this table should contain values for each reactant and product in the equilibrium expressed mathematically

4) substitute the equilibrium expressions from the ICE table into the equilibrium constant expression

5) multiply out and manipulate the resulting algebraic expression to get a quadratic (or simpler) equation

6) substitute the coefficients into the quadratic formula and solve

7) substitute the resulting answer into the equilibrium expressions from the ICE table to get the equilibrium concentrations of all of the reaction species

8) CHECK YOUR ANSWER!

step 1) A + B º Cstep 2) Keq = [C]/[A][B]

step 3) [A] [B] [C]initial 0.100 0.100 0.0change - x - x x

equilibrium 0.100 - x 0.100 - x x

step 4) Keq = (x)/(0.100 - x)(0.100 - x) = 10.3

step 5) (x) / (0.100 - x)(0.100 - x) = 10.3

(x)/ (0.01 - 0.200 + x2) = 10.3

x = 10.3 (0.01 - 0.200 + x2)

x = 10.3x2 - 2.06x + 0.103

10.3x2 - 3.06x + 0.103 = 0

(quadratic: ax2 + bx + c = 0 )

step 6) x = [ -b ± (b2 - 4ac)½ ] / 2a

= [-(-3.06) ± ( -3.062 - 4(10.3)(0.103))½] / 2(10.3)

= [3.06 ± (9.3636 - 4.2436)½] / 20.6

= [3.06 ± (5.12)½] / 20.6

= [3.06 ± 2.263] / 20.6

x = 0.2583 or x = 0.03869

step 7) if x = 0.2583

[A]e = 0.100 - 0.2583 = -0.1583

not possible!! a concentration can not be negative!

hence, x = 0.03689

[A]e = 0.100 - 0.03869 = 0.06131[B]e = 0.100 - 0.03869 = 0.06131[C]e = 0.03869

step 8)Keq = (0.03869) / (0.06131)2 = 10.3 °

Note: What if I had chosen wrong? That is, if I had set up the ICE table with [A] = 0.100 + x, [B] = 0.100 + x, and [C] = -x? The answer would have given me a negative value for x and step 7 would still have given me the correct values!!You don’t have to work out the reaction quotient to determine the equilibrium values. If you correctly follow all eight steps, every time, then you will minimize the chance of getting it wrong. And you can check your answer!

ExamplesA flask initially contains only 0.048 M H2 and 0.063 M F2.These two compounds react to product HF. The concentration of HF at equilibrium is 0.088 M. What is the value of Keq?

step 1)

more commonly, questions are like:

At 25°C, the equilibrium constant for the reaction:Ca2+

(aq) + SO42-

(aq) º CaSO4 (aq)

is 156 M-1. If the initial concentrations are [Ca2+] = 0.26 M, [SO4

2-] = 0.21 M, and [CaSO4] = 0.83 M, calculate the concentration of the species at equilibrium.

step 2)

What if we set up the ICE table the other way around?

step 3) [Ca2+] [SO42-] [CaSO4]

initial 0.26 0.21 0.83change + x + x - x

equilibrium 0.26 + x 0.21 + x 0.83 - x

Another example,Br2 + I2 º 2IBr

if Keq = 1.36, [Br2]o = 0.146M, [I2]o = 0.233 M, and [IBr]o = 0.0M, then what are the concentrations at equilibrium?

step 2)

if [IBr]o = 0.078 M, then what would the concentrations be at equilibrium?

step 1)

An example of the most complicated type of question:for the equilibrium, HI + Br2 º HBr + IBrif Keq = 10.4 and [HI]o = 0.146 M, [Br2]o = 0.302 M, [HBr] = 0.267 M, and [IBr] = 0.0 M,then what are the concentrations at equilibrium?

step 2)