Unit 5 Chemistry- Thermodynamics

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  • Created by: Hannah
  • Created on: 23-06-11 08:22

Definitions

Enthalpy change is the heat transferred in a reaction at constant pressure

Lattice formation enthalpy:

The enthalpy change when 1 mole of solid ionic compound is formed from its gaseous ions under standard conditions (always negative/exo): Na+ (g) + Cl- (g) ---> NaCl (s)

Lattice dissociation enthalpy:

the enthalpy change when 1 mole of solid ionic compound is completely dissociated into its gaseous ions under standard conditions (always positive/ endo): NaCl (s) ---> Na + (g) + Cl- (g)

Enthalpy of formation:

The enthalpy change when 1 mole of a compound is formed when its elements in their standard states under standard conditions: 2C (s) + 3H2 (g) + 1/2 O2 (g) --> C2H5OH (l)

Bond dissociation enthalpy:

The enthalpy change when all the bonds of the same type in 1 mole of gaseous molecules are broken: Cl2 (g) ---> 2Cl (g)

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More definitions

Enthalpy of atomisation of an element:

The enthalpy change when 1 mole of gaseous atoms is formed from an element in its standard state: 1/2 Cl2 (g) ---> Cl (g)

Enthalpy of atomisation of a compound:

The enthalpy change when 1 mole of gaseous atoms is formed from a compound in its standard state: NaCl (s) ---> Na (g) + Cl (g)

First ionisation enthalpy:

The enthalpy change when 1 mole of gaseous 1+ ions is formed from 1 mole of gaseous atoms: Mg (g) --> Mg + (g) + e-

Second ionisation enthalpy:

The enthalpy change when 1 mole of gaseous 2+ ions is formed from 1 mole of gaseous 1+ ion: Mg + (g) ---> Mg 2+ (g) + e-

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And more definitions

First electron affinity

The enthalpy change when 1 mole of gaseous 1- ions is made from 1 mole of gaseous atoms: O (g) + e- ---> O- (g)

Second electron affinity

The enthalpy change when 1 mole of gaseous 2- ions is made from 1 mole of gaseous 1- ions: O- (g) + e- ---> O2- (g)

Enthalpy of hydration

The enthalpy change when 1 mole of aqueous ions is formed from gaseous ions: Na + (g) ---> Na + (aq)

Enthalpy of soluion

The enthalpy change when 1 mole of solute is dissolved in sufficient solvent that no further enthalpy change occurs on further dilution: NaCl (s) ---> NaCl (aq)

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Born-Haber Cycles

Hess' Law states that the total enthalpy change of a reaction is always the same, no matter which route is taken. ( Delta H f = All the others added)

Calculations involving group 2 elements are a bit different:

e.g. A haber cycle of MgCl2 - Magnesium chloride

1. There's two moles of chlorine ions in each mole of MgCl2- so you need to double the atomisation enthalpy of chlorine

2. Group 2 elements from 2+ ions so you need to include the second ionisation enthalpy of magnesium

3. You are forming 2 moles of Cl- ions so you need to double the first electron affinity of chlorine.

Order is something like - formation one way, and then atomisation, ionisation, affinity and lattice enthalpy the other way 

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Theoretical values

Can work out a theoretical lattice enthalpy by doing some calculations based on the purely ionic model of a lattice. 

The model assumes that all the ions are spherical, and have their charge evenly distributed around them.

But the experimental lattice enthalpy from the Born-Haber cycle is usually different. This is evidence that ionic compounds usually have some covalent character.

The positive and negative ions in a lattice aren't usually exactly spherical. Positive ions polarise neighbouring negative ions to different extents, and the more polarisation there is, the more covalent the bonding will be, and so the negative ion distorts and is not spherical. 

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Comparing lattice enthalpies tell how ionic a latt

1. If the experimental lattice energies are bigger than the theoretical values by quite a lot...

It tells you that the bonding is stronger than the calculations from the ionic model predict. 

The difference shows that the ionic bonds are quite strongly polarised and so they have quite a lot of covalent character.

2. If the experimental and theoretical values are a pretty close match, you can say that the compounds fit the 'purely ionic" model very well...

This indicates that the structure of the lattice for these compounds is quite close to being purely ionic. There's almost no polarisation so they don't have much covalent character. 

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Enthalpies of solution

When a solid ionic lattice dissolves in water these two things happen:

1. The bonds between the ions break - this is endothermic. This enthalpy change is the lattice enthalpy of dissociation.

2. Bonds between the ions and the water are made- this is exothermic. The enthalpy change here is called the enthalpy change of hydration. 

(Oxygen is more electronegative than hydrogen, so it draws the bonding electrons towards itself, creating a dipole).

3. The enthalpy change of solution is the overall effect on the enthalpy of these two things.

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Enthalpy Cycles

You can work out the enthalpy change of solution using an enthalpy cycle, You just need to know the lattice dissociation enthalpy of the compound and the enthalpies of hydration of the ions. 

How to draw (looks like a triangle):

1. Put the ionic lattice and the dissolved ions on top and connect them by the enthalpy change of solution. This is the direction route. e.g.:

NaCl (s) ----- Enthalpy change of solution ----->  Na + (aq) + Cl- (aq)

2. Connect the ionic lattice to the gaseous ions by the lattice enthalpy of dissociation. This will be a positive number:

NaCl (s) ---- Lattice dissociation enthalpy -----> Na + (g) + Cl- (g)

3. Connect the gaseous ions to the dissolved ions by the hydration enthalpies of each ion. This completes the indirect route:

Na +(g) + Cl-(g) --Enthalpy of hydration for Na + & then Cl- -> Na +(g) + Cl-(aq) 

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Mean bond enthalpies

Energy is taken in when bonds are broken

Energy is released when bonds are formed so....

Enthalpy change for reaction =

sum of enthalpies of bonds broken - sum of enthalpies of bonds formed

If you need more energy is break bonds than is released when bonds are made, its an endothermic reaction. The enthalpy change, delta H, is positive.

If more energy is released than is taken in, its exothermic and delta H is negative.

A given type of bond will vary in strength from compound to compound. It can even vary within a compound. 

Mean bond enthalpies are the averages of these bond enthalpies. Only the bond enthalpies of diatomic molecules, such as H2 and HCl will always be the same. 

So calculations using mean bond enthalpies will never be perfectly accurate. You get more exact results from experimental data obtained from specific compounds. 

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Entropy

Entropy tells you how much disorder there is. It is a measure of the number of ways that particles can be arranged and the number of ways that the energy can be shared out between the particles. 

Entropy is represented by the symbol S

Substances really like disorder, so move to try and increase the entropy. 

Physical state affects entropy:

Solid particles wobble about a fixed point, there's hardy any randomness so they have the lowest entropy. Gas particles move where they like, have the most random arrangement and so the highest entropy. 

Dissolving affects entropy:

Dissolving a solid increases entropy -they can move more freely- not held in one place

More particles means more entropy:

The more particles you have, the more ways they and their energy can be arranged. Entropy increases because the number of moles increases.

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More about entropy

A spontaneous (or feasible) change is one that will just happen by itself- you don't need to give it energy. 

You need to supply energy to endothermic reactions, but some are spontaneous.

In some reactions, the entropy increases so much that the reaction will happen by itself. 

Reactions won't happen unless the total entropy change is positive:

During a reaction, there's an entropy change between reactants and products- the entropy change of the system and also a entropy change of the surroundings.

Total entropy change= entropy change of system+entropy change of surroundings

Delta S total = Delta S system + Delta S surroundings

Delta S system = S products - S reactants

Delta S surroundings = - (Delta H/ T)  Delta H in Jmol-1 and T in K

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Free energy change

Free energy change, Delta G, is a measure used to predict whether a reaction is feasible. 

If Delta G is negative or equal to zero, the reaction might happen by itself. 

( Even if Delta G shows that a theoretically feasible, it might have a really high activation energy and be so slow that you wouldn't notice it happening at all).

Free energy change takes into account the changes in enthalpy and entropy.

Delta G = Delta H - T x Delta S system 

Delta G is in Jmol-1. 

Delta H is the enthalpy change in Jmol-1

T is temperature in K

Delta S system is the entropy of the system in JK-1mol-1

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Feasibility depends on temperature

If a reaction is exothermic (negative delta H) and has a positive entropy change, then Delta G is always negative. 

Since Delta G = Delta H - T x Delta S system, these reactions are feasible at any temperature. 

If the reaction is endothermic (positive delta H) and has a negative entropy change, then Delta G is always positive. These reactions are not feasible at any temperature. 

But for other combinations, temperature has an effect:

If Delta H is positive (endothermic) and Delta S system is positive, then the reaction won't be feasible at some temperatures but will be at a higher temp.

If Delta H is negative (exothermic) and Delta S system is negative, then the reaction will be feasible at some temperatures but won't be feasible at higher temperature. 

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Just feasible...

When Delta G is zero, a reaction is just feasible.

You can find the temperature when Delta G is zero by free arranging the free energy equation:

Delta G = Delta H - Tx Delta S system 

So when Delta G = 0....

Then T x Delta S system = Delta H

So T = Delta H/ Delta S system 

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Comments

Chloe Norton

This is brilliant! Just what I needed! :) **

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