Lecture 1 - Equillibria, Transition States & Rates

Question 1

Change In Free Energy

Answer 1

The dG for the reaction is the difference in free energy between the intial and final states

• The free energy of the initial state is the sum of standard molar free energies of the reactants
• The free energy of the final state is the sum of the standard molar free energies of the products

dG = G_Products - G_Reactants

Question 2

Relation Of dG To Equillibrium Constant (K)

Answer 2

dG = -RTln(K)

Where K is defined as the concentration of the products over the concentration of the reactants

Question 3

Exergonic Reaction

Answer 3

If Change In Standard Reaction Gibbs Energy Is < 0 Then K > 1 Which Means That PRODUCTS Are Dominant At Equillibrium. This Type Of Reaction IS Thermodynamically Stable And Is Termed EXERGONIC

Question 4

An Endergonic Reaction

Answer 4

If Change In Standard Reaction Gibbs Energy Is > 0 Then K < 1 Which Means That REACTANTS Are Dominant At Equillibrium. This Type Of Reaction Is NOT Thermodynamically Stable And Is Termed ENDERGONIC

Question 5

Equillibrium Constant At 298K & Percentage Of Product At Equillibrium

Answer 5

Question 6

If 0 < K < 1 Then Where Does The Equillibrium Lie

Answer 6

To The Left

i.e Towards Reactants

Question 7

If K > 1 Where Does The Equillibrium Lie

Answer 7

To The Right

i.e Towards The Products

Question 8

The Transition State

Answer 8

The point of highest energy on the reaction co-ordinate / energy diagram

The minimum amount of energy needed to reach the transition state is refered to as the activation energy

Question 9

Convention For Charges When Drawing Transition States

Answer 9

The convention is to draw partal +ve or -ve on atoms with the understanding that the partial charges add up to th total charge of the products or reactants

Question 10

Rate Equation For A Bimolecular Reaction Of A + B ⇒ Products

Answer 10

Rate = k[A][B]

k = Rate constant in dm³ mol^-1 s^-1

Rate Is in mol dm^-3 s^-1

Question 11

Relation Of Rate Constant To Temperature

i.e Arrhenius Equation

Answer 11

k = Ae^(-E_act/RT)

Where A is The Pre-Exponential Factor And Has the Same Units As The Rate Constant

What this equation shows is that the rate of reaction is very sensitive to the height of the energy barrier

• The bigger the barrier the slower the reaction
• The smaller the barrier the quicker the reaction

Question 12

Definition Of Equillibrium

Answer 12

The definition of equillibrium means that the forward rate is equal to the backwards rate:

k₁[F^-][CH₃I] = k_-1[I^-][CH₃F]

Which if we rearrange then we can associate the rate constants (k₁ & k_-1) with the equillibrium constant (K)

k₁/k_-1 = [I^-][CH₃F] /[F^-][CH₃I] = K

Question 13

Principle Of Microscopic Reversability

Answer 13

This means that the transition state for the back reaction will be exactly the same as the forward reaction

Question 14

Initial & Final States As Vibrational Energy Diagrams

Answer 14

most energy diagrams are represented as single enrgy levels for the inital and final states however in reality the molecules have vibrational energy and will populate the lowest few energy levels of the vibrational potential energy well

The transition state occurs at a distance along the reaction co-ordinate where the parabolas for vibration along the reaction co-ordinate of reactants and products intersect

The energy of the transition point will be lower than the intersection point because a bond is being made at the same time as one is being broken

Question 15

Hammond’s Postulate

(Definition)

Answer 15

If two states occur consecutively during a reaction and have nearly the same energy content, their interconversion will involve only a small reorganisation of molecular structure.

Question 16

Hammonds Postulate Basis

Answer 16

As reactions increase in exothermicity, the potential energy wells for reactants and products will intersect increasingly near to the reactants and the activation energy will get smaller

• This implies that EXOTHERMIC reactions will have EARLY REACTANT LIKE transition states
• Implies that ENDOTHERMIC reactions will have LATE PRODUCT LIKE transition states

Question 17

Early Reactant Like Transition State & Late Product Like Transition State

Answer 17

Question 18

Early Reactant Like Transition State

(Addition Of Grignard Reagent To A Ketone)

Answer 18

Since this is an exothermic reaction then hammonds postulate states that the transition state will look more like that of the reactants (the ketone) than the product alkoxide

The reaction co-ordinate for this reaction is taken to be the reciprocal (reciprocal so that it increases during the reaction) of the R- - -C=O bond

Question 19

Late Product Like Transition State

(Enolisation Of A Ketone)

Answer 19

The free energy diagram can be drawn by using the reciprocal of the C—-H bond distance as a reaction co-ordinate

In the transition state the C-H and C=O bond will be mostly broken and the O-H and C=C π bond will be mostly formed

Question 20

Hammond’s Postulate With Aromatic Electrophillic Substitution

Answer 20

There are two transition states for this reaction TS₁ on the way to the intermediate and TS₂ on the way to the product.

The rate of the overall reaction is determined by the energy of TS₁ relative to benzene i.e the height of the first barrier

Question 21

Kinetic Control

Answer 21

This means that the product obtained is the one that is formed the fastest

In the digram below B is less stable than C, but is formed faster because the energy barrier leading to it is smaller

Question 22

The Ratio Of Concentrations Of B & C At Any Point During The Reaction For Two Competing Reactions:

A ⇒ B

A ⇒ C

C is More Stable Than B, But The Energy Barrier (Activation Energy) Leading To B Is A Lot Smaller Than That Leading To C

Answer 22