Ch. 14

Question 1

Dynamic Equilibrium

Answer 1

To reach an equilibrium state, a reaction must be reversible
* Reversible reactions can proceed in both the forward and reverse directions

When the rate of forward and reverse reactions are equal they establish a dynamic equilibrium
* At equilibrium, the concentrations of reactants and products no longer change with time

• This does not mean that the concentrations of reactants and products are equal!

–> Rate in which they go from prods to reacts and vice versa

The equilibrium constant will tell us what the final concentrations will be

Question 2

The Equilibrium Constant

Answer 2

During equilibrium, the forward and reverse reactions proceed at equal rates

• Recall that rate = k[A] for the unimolecular (1st order) forward reaction, and rate = k[B]2 for the bimolecular reverse reaction(2nd order)
• In other words, k(forward)[A] = k(reverse)[B]² at equilibrium
  We can rearrange this equation to:

k(forward) = [B]²
———————————- = K
k(reverse) = [A]
* K (the equilibrium constant) is equal to the ratio of the rate constants for the forward
and reverse reactions

The same general process applies regardless of the actual rate law
- prod over reactants
- Exponents equal to stoichiometric coefficients

Question 3

equilibrium constant, Kc

Answer 3

For reactions occurring in solution, the ratio of concentrations of the products and reactants at equilibrium is the equilibrium constant.

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

• dont care about rxn order
• [] in mols per L

Question 4

equilibrium constant Kp

Answer 4

For gas-phase reactions we express the equilibrium constant Kp in terms of partial pressure

kp = p(N)2)^4p(O2) / p(N2O5)^5

Question 5

Relating Kc and Kp

Answer 5

From the ideal gas equation the concentration of a gas can be determined from its pressure

aA(g) + bB(g) ⇌ cC(g) + dD(g)

For any of the reactants [A] = nA/V
l
divide by V
l
v
PA = nART/V
l
B/c n/v = []
l
v
PA = [A]RT
l
pressure is equal to [] multiped by RT
l
v
[A] = PA/RT
[] = to pressure dived by RT

eventually

kc = kp(RT)^Dn

Important – The units of R must match the pressure units. Normally 0.08314 Lbar/molK

Question 6

If K is large (K»1)

Answer 6

The equilibrium lies far to the right
(favs products)

Equilibrium concentrations of products (numerator in K expression) are much greater
than concentrations of reactants (denominator in K expression)
– at eq have a lot of prod and not a lot of reactants
– b/c k = prod/react, numerator will be large and dominator will be small, resulting in a large number

Reaction goes almost to completion

a lot of time large k rxns are considered/treated as 1 way rxns b/c when rxn reaches eq theres so little of limiting reactant left, you can treat it has if its all been consumed

Question 7

If K is small (K«1)

Answer 7

The equilibrium lies far to the left
(favs reactants)

Equilibrium Concentrations of products (numerator in K expression) are much smaller
than concentrations of reactants (denominator in K expression)
– at eq [] of product will be small & [] of reactant will be large, so num is a small number and dem is a large number, resulting in a small number (small K)

Reaction barely proceeds at all

Question 8

If K is close to unity (K ~ 1)

Answer 8

There will be substantial []s of both reactants and products.

Note that while K is related to reaction rate, it is a ratio of rates

The value of K tells us nothing about how quickly equilibrium will be achieved!
* The ratio of two small rates may be the same as the ratio of two large rates, and either ratio may be small or large.

Question 9

Units of Kc, Kp, K

Answer 9

If any units other than the standard units are used, they must be included in the reported K

• Units must be included in Kc or a Kp (unless Dn = 0)
• Units are not included in K calculated using standard units
• When converting between Kp and Kc:
• Because pressure is in bar and concentration in mol/L, you must use the value of R in Lbar
  / molK (0.08314472)
• K is the dimensionless thermodynamic equilibrium constant
• Because we assume activities are equivalent to concentrations or pressures:
• For a gas phase reaction, K is equivalent to Kp (with pressures in bar)
• For a solution phase reaction, K is equivalent to Kc (with concentrations in mol/L)

Question 10

Manipulating Reactions: Reversing

Answer 10

• Reversing the way you write a reaction turns reactants into products and products into
  reactants
• K is always products over reactants, so K for the reverse reaction is the reciprocal of K for
  the forward reaction.

rxn that favs reactants in 1 direction, favs products when reverse it
rxn that favs products in 1 direction, favs reactants when reverse it

Question 11

Manipulating Reactions: coefficients of an equation are multiplied by a factor

Answer 11

• When the coefficients of an equation are multiplied by a factor, the equilibrium constant is raised to that exponent.

multiply coeff by 2 –> mult exponents in eq expression by 2 = square the entire eq expression –> double the reactant is going to square its eq constant

Question 12

Manipulating Reactions: adding equations to make a new equation

Answer 12

• When you add equations to get a new equation, the equilibrium constant of the new equation is the product of the equilibrium constants of the old equations.

add rxns together –> multiply their coefficients

Question 13

Heterogenous Equilibria

Answer 13

• Some reactions include reactants and / or products in different phases
• The activity of any substance that participates in a reaction but is not in the same phase as the other reactants is always 1.
• Typically, these will be solids participating in gas phase or solution phase reactants
  – Immiscible liquids participating in solution or gas phase reactions can also be heterogeneous equilibria
• The concentration or partial pressure doesn’t change during the course of the reaction b/c the substance is not in solution
• Because an equilibrium constant is a product, an activity of 1 just disappears

Question 14

Heterogenous Equilibria Ex:

CaCO3(s) ⇌ CaO(s) + CO2(g)
Kp = p(CO2)

Question 15

Reaction Quotient

Answer 15

• If we know the equilibrium constant for the reaction, we can use it to predict what will happen.
• First, we find the reaction quotient Q

* Q is calculated the same way as K, except the reaction need not be at equilibrium

• If the reaction is at equilibrium Q = K
• If the reaction is not at equilibrium Q ≠ K –> b/c if system not at eq, theres going to be a spontaneous rxn that’ss convert reacts into prods, which’ll change [], and if change [], change Q.

* If the reaction is not at equilibrium, net reaction will occur in the direction that makes the
value of Q closer to the value of K

• Net reaction will continue until Q = K and the system is at equilibrium
• Note that while a particular reaction at a given temperature has only one value of K, it may have any value of Q

diff between k and Q, K has to be calculated at with measurements of a system at eq, while Q does not

Question 16

If Q < K

Answer 16

• Reaction proceeds to the right (makes more products)
• Q increases
• not a lot of reactants, too mucg prod
• numerator too small, dominator too big

Question 17

If Q > K

Answer 17

• Reaction proceeds to the left (makes more reactants)
• Q decreases
• too much prod, not enough reactants
• num too big, dem too small

Question 18

If Q = K

Answer 18

• Reaction is at equilibrium (forward and reverse reactions proceed at equal rates
• Q does not change

Question 19

Le Châtelier’s Principle

Answer 19

When a chemical system at equilibrium is disturbed, the system shifts in a direction that minimizes the disturbance.

There are several ways we can disturb the equilibrium, including:
* Changing the amount of reactant or product
* Changing the volume (sometimes)
* Increasing or decreasing the temperature

Some changes will not affect the equilibrium, such as:
* Increasing the pressure by adding an inert gas
* Changing the volume (sometimes)
* Adding a catalyst

Question 20

Le Châtelier’s Principle: Changing Amount of Reactant or Product

Answer 20

N2O4(g) ⇌ 2 NO2(g) kp = p(NO2)² / p(N2O4)

• If the reaction above is initially at equilibrium and:
• More N2O4 is added, Q will be reduced (Q < K) and the reaction will proceed to the right until Q = K again
  - _{increase dem, goes from k to Q b/c no longer at eq. partial pressure of N2O4 increase, pp of NO2 doesn’t change, proceed in forward direction producing prods, causing num of Q to increase and the dem of Q to decrease}
• Some N2O4 is removed, Q will be increased (Q > K), and the reaction will proceed the left until Q = K again
  -_{decrease dem, making Q larger, rxn proceed in reverse. remove stuff and system will move to make more of what you removed}
• More NO2 is added, Q will be increased (Q > K), and the reaction shifts to the left
  -_{increase num of Q, make Q larger, prods being turned back into reactants, but when eq reestablished, you’ll have more N2O4 and NO2 then what initial had}
• Some NO2 is removed, Q will be decreased (Q < K), and the reaction shifts to the right
  -_{shift rxn forward, decrease num of Q, make Q smaller. pp of NO2 when system reestablishes eq will be lower than what it was before, b/c when rxn occurs, it also lowers the pp of N2O4}
• Be sure to check whether the reactant/product added/removed is part of the equilibrium constant expression (heterogeneous equilibria)

Question 21

Le Châtelier’s Principle: Changing the Volume

Answer 21

N2O4(g) ⇌ 2 NO2(g) kp = p(NO2)² / p(N2O4)

• From the ideal gas law there is an inverse relationship between pressure and volume
  -_{taking the same # of moles and putting it into a smaller vol or if reduce vol by half the orignal vol, will double all pp b/c same #of moles but half the vol, meaning pressure is twice as large}
• Compressing a gas mixture increases the partial pressure of every component by the same
  ratio
• The effect on Q depends on the exponents (the stoichiometric coefficients)
• Decreasing the volume causes the reaction to shift in the direction that has the fewer moles of gas (to the left for N2O4(g) ⇌ 2 NO2(g))
  -_{when a rxn does not have Dn = 0}
  -_{b/c want to reduce the pressure by reducing the # of moles of gas}
• Increasing the volume causes the reaction to shift in the direction that has the greater number of moles of gas (to the right for N2O4(g) ⇌ 2 NO2(g))
• If the moles of gas is the same on both sides then there will be no effect on the equilibrium

Question 22

Le Châtelier’s Principle: Changing the Temperature

Answer 22

• Equilibrium constants change with temperature
• Equilibrium constants of exothermic reactions decrease with increasing temperature, and those of endothermic reactions increase with increasing temperature
• We can rationalize this by including ‘heat’ as a reactant or product in our chemical
  equation:
• Exothermic: N2(g) + 3H2(g) ⇌ 2NH3(g) + heat (heat is a product)
  -_{shift towards reactants, b/c add product expect to shift that way}
• Endothermic: N2O4(g) + heat ⇌ 2NO2(g) (heat is a reactant)
  -_{shifts toward products, b/c add reactant expect to shift that way}
• Heating an endothermic reaction will increase K (favors products)
  -_{increase eq constant}
• Cooling an endothermic reaction will decrease K (favors reactants)
  -_{decrease eq constant}
• Heating an exothermic reaction will decrease K (favors reactants)
  -_{decrease eq constant}
• Cooling an exothermic reaction will increase K (favors products)
  -_{increase eq constant}

Question 23

Le Châtelier’s Principle: Changing the Pressure by Adding Inert Gas or Adding a Catalyst

Answer 23

• Adding an inert gas (any gas that does not participate in the reaction, not necessarily a group 18 element) does not affect the partial pressures (but does change total pressure) of the reactants (Dalton’s law
  of partial pressures)
• Thus, adding an inert gas does not change the equilibrium constant.

Adding a Catalyst
* A catalyst reduces the activation energy for both the forward and reverse reactions equally
* A catalyst thus cannot change the ratio of forward and reverse reaction rates, and cannot change the equilibrium constant

-Ea for reverse is the diff in energy between reactants for the reverse rxn (which is the prods for the forward rxn) and the transition state
- b/c exo, Ea for reverse rxn is higher, b/c the Ea for the reverse rxn is the Ea for the forward rxn + DH.
-rxn energy of endo rxn will always be higher than the rxn energy for exo ran
– endo Ea generally leads to slower rxns, smaller rate constant
–exo tends to form prods b/c forward rxn tends to be fav overed reverse rxn.

Question 24

Le Châtelier’s Principle: Practice

The reaction 2 SO2(g) + O2(g) ⇌ 2 SO3(g) with ΔH° = −198 kJ/mol is at equilibrium.

In which direction will the net reaction shift after each of the following disturbances to the
equilibrium?

What happens to K?

- adding more O2 to the container

Question 25

Le Châtelier’s Principle: Practice

The reaction 2 SO2(g) + O2(g) ⇌ 2 SO3(g) with ΔH° = −198 kJ/mol is at equilibrium.

In which direction will the net reaction shift after each of the following disturbances to the
equilibrium?

What happens to K?

- condensing and removing SO3

Question 26

Le Châtelier’s Principle: Practice

The reaction 2 SO2(g) + O2(g) ⇌ 2 SO3(g) with ΔH° = −198 kJ/mol is at equilibrium.

In which direction will the net reaction shift after each of the following disturbances to the
equilibrium?

What happens to K?

- compressing the gases

Question 27

Le Châtelier’s Principle: Practice

The reaction 2 SO2(g) + O2(g) ⇌ 2 SO3(g) with ΔH° = −198 kJ/mol is at equilibrium.

In which direction will the net reaction shift after each of the following disturbances to the
equilibrium?

What happens to K?

- cooling the container

Question 28

Le Châtelier’s Principle: Practice

The reaction 2 SO2(g) + O2(g) ⇌ 2 SO3(g) with ΔH° = −198 kJ/mol is at equilibrium.

In which direction will the net reaction shift after each of the following disturbances to the
equilibrium?

What happens to K?

- doubling the volume of the container

Question 29

Le Châtelier’s Principle: Practice

The reaction 2 SO2(g) + O2(g) ⇌ 2 SO3(g) with ΔH° = −198 kJ/mol is at equilibrium.

In which direction will the net reaction shift after each of the following disturbances to the
equilibrium?

What happens to K?

- warming the mixture

Question 30

Le Châtelier’s Principle: Practice

The reaction 2 SO2(g) + O2(g) ⇌ 2 SO3(g) with ΔH° = −198 kJ/mol is at equilibrium.

In which direction will the net reaction shift after each of the following disturbances to the
equilibrium?

What happens to K?

- adding the inert gas helium to the container

Question 31

Le Châtelier’s Principle: Practice

The reaction 2 SO2(g) + O2(g) ⇌ 2 SO3(g) with ΔH° = −198 kJ/mol is at equilibrium.

In which direction will the net reaction shift after each of the following disturbances to the
equilibrium?

What happens to K?

- adding a catalyst to the mixture

Question 32

Calculating Equilibrium Constant

Answer 32

• Easiest scenario to calculate equilibrium constant is when we know the concentrations of
  reactants and products at equilibrium
• If we can write the expression for the equilibrium constant we can substitute the values given
• Equilibrium constant for a reaction will always be the same at a given temperature - regardless of the starting concentration of reactants and products
• How can we calculate K, if we only know the starting concentration of reactants/products and only 1 equilibrium concentration?
• Use an ICE table!

Question 33

Calculating Equilibrium Concentrations

Answer 33

• A more complex problem occurs when we know the equilibrium constant, K and only the initial concentrations. How do we calculate equilibrium concentrations?
• This generally involves finding roots of polynominal functions, but in some cases simplifying
  assumptions can be made
• There are three scenarios that most equilibrium questions fall into:
• Intermediate K
• Small K
• Large K

Question 34

Intermediate K

Answer 34

• When K is close to 1 (between about 0.001 and 1000), there will be significant concentrations of all species in the reaction
• You will need to set up an ICE table, and use it to derive a polynomial expression for K in
  terms of the concentrations of reactants and products
• Finding the roots of the polynomial will allow you to find all of the concentrations
• Quadratic and higher order polynomials have multiple roots - use common sense to pick
  the right one
  – Negative concentrations or pressures are physically impossible

Question 35

Small and Large K Problems

Answer 35

• Sometimes the equilibrium expression can be simplified so that we do not have to solve a
  complex polynomial

* If K is very small, the reaction strongly favors reactants and very little product will be produced

* If K is very large, the reaction strongly favors products and very little of the limiting reactant will
remain

• For these reactions, one can approximate the final concentrations by first running the reaction all the way to the favored side, until one of the concentrations (that of the limiting reagent) reaches zero
  – A small K problem is already in this state if at least one of the product concentrations is zero
• Because this approximation isn’t far from the actual equilibrium, one can then assume that all of
  the non-zero approximate concentrations will be essentially the same as the equilibrium
  concentrations

Question 36

Small and Large K Problems - The 5% Rule

Answer 36

• How do you decide if K is very large or very small?
• Compare the reactant and/or product concentrations to K

* If Ci/K > ~1000, you can consider K to be very small

* If K/Ci > ~1000, you can consider K to be very large

• After using the approximation we need to verify that the differences we assumed would be small really are small
• If the change in concentration (x) is less than about 5% of the approximated concentration,
  we can consider the assumption valid

Question 37

What if your assumption of small K or large K is invalid?

Answer 37

• Your problem has just become an intermediate K problem
• For simple reactions (no more than 2 moles of reactants or products), you can use the
  quadratic equation
• For more complex reactions that produce cubic or higher order polynomials, you can:
  – Solve the polynomial with your fancy calculator or
  – Use the method of successive approximations
• For the method of successive approximations, proceed as for a large/small K problem, then use your calculated “equilibrium” concentrations as initial concentrations and repeat
  until the value of x no longer changes

Question 38

Large K Problems

Answer 38

• A large K problem (the ratio of Ci/K or Pi/K is less than 0.001, K>1000) is solved just like a small K problem, except that at least one of the initial reactant (not product) concentrations should be zero
• If your initial conditions include a non-zero amount of reactant, you will need a 5-line “ICICE” table
  – This also applies to small K problems where the initial amount of product is not zero
• The first change line calculates a new set of initial concentrations, with the concentration of limiting reactant (or product for a small K problem) now zero
• fav products
• will want to run rxn all the way to products until run out of 1 of the reactants, allowing us to get close to our equilibrium situation in which for large k problem = want as much product as possible and 0 amount of 1 reactant.

Question 39

Mixtures of reactants and products

Answer 39

• In the examples so far, the reactions could only proceed in one direction, as no products
  were initially present
• Problems involving initial conditions with both reactants and products present can be solved using the same methods

* For a small K problem, use a 5-line ICICE table, and ensure that at least one product concentration is zero in the second I line

* For a large K problem, use a 5-line ICICE table, and ensure that at least one reactant concentration is zero in the second I line

• For an intermediate K problem, you can proceed just like you would if the initial product concentration was zero
  – If you don’t want x to be negative, start by calculating Q, then decide which way the reaction is going to go so you can assign +x and –x appropriately

Question 40

Disturbing the Equilibrium

Answer 40

• What happens when you start with a system at equilibrium then do something to disturb the equilibrium?
• Le Chatelier’s principle will predict which way the equilibrium will shift
• Changing the volume or adding or removing a reactant or product will change the concentration(s), but will not affect K
• Changing the temperature will change K
• Either way, set up an ICE table and find the new equilibrium concentrations