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# Is this your room? - PowerPoint PPT Presentation

Is this your room?. Then you already know about entropy. Using a simple approach, Define entropy as a measure of disorder. A system (such as a room) is in a state of high entropy when its degree of disorder is high. As the order within a system increases, its entropy decreases. .

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Define entropy as a measure of disorder.

A system (such as a room) is in a state of high entropy when its degree of disorder is high.

As the order within a system increases, its entropy decreases.

A system (such as a room) is in a state of high entropy when its degree of disorder is high.

For better or for worse, nature 'likes' chaos, disorder, high entropy... In fact, much of our life consists in fighting this disorder!

2 its degree of disorder is high. 20 ways to arrange?

This can be explained in terms of probabilities. Disordered states are simply more likely to exist (or emerge) than ordered states. The spontaneous direction of change is from a less probable to a more probable state, as illustrated above.

In general its degree of disorder is high.

• Gases have higher entropies than liquids.

• Liquids have higher entropies than solids.

Also … its degree of disorder is high.

• Entropy is greater for larger atoms.

• Entropy is greater for molecules with more atoms.

Examples its degree of disorder is high.

• Ar has higher entropy than Ne as Ar molecules are larger.

• C8H18(l) has higher entropy than C5H12(l) as complex molecules have higher entropy than simple ones.

• Br2(g) has higher entropy than Br2(l) as gases have higher entropies than liquids since gases have more ways of being arranged

On to CI 4.4 its degree of disorder is high.

• AS – entropy is a measure of the number of ways particles can be arranged.

• A2 – entropy is a measure of the number of ways quanta (packages of energy) can be arranged.

• Entropy greater if more quanta (heat it)

• Entropy greater if more molecules

Stack of Energy its degree of disorder is high.

• We use specific heat capacity (Cp) as a measure of how much energy is required to warm something up.

• Temperature is related to kinetic energy of the molecules – when they move more energetically they feel hotter.

Types of kinetic energy its degree of disorder is high.

• Translation – movement of the whole molecule from one place to another.

• Rotation – Spinning around.

• Vibration – Stretching and compressing bonds.

Relationship between the levels... the molecules, breaking bonds between or within them.

• Insert Fig 2 CI 4.4

Bigger atoms/molecules have more entropy? the molecules, breaking bonds between or within them.

• Energy levels for heavier atoms are generally closer together.

• Number of energy levels also increases with number of atoms in the molecule, thus making adjacent levels closer together.

5 ways the molecules, breaking bonds between or within them.

4 quanta – 2 molecules

5 the molecules, breaking bonds between or within them.  7

more ways

6 quanta – 2 molecules?

More

quanta

5 ways the molecules, breaking bonds between or within them.

4 quanta – 2 molecules

5 the molecules, breaking bonds between or within them.  15

More ways

4 quanta – 3 molecules

More

molecules

Summary the molecules, breaking bonds between or within them.

Making ice cubes... the molecules, breaking bonds between or within them.

• H2O (l)  H2O (s)

• ΔS = -22.0 JK-1mol-1

• ΔH = -6.01 kJ mol-1

• ΔS is negative because entropy decreases moving from a liquid to a solid.

• ΔH is also negative as the process is exothermic.

• The release of heat to the surroundings increases the entropy of the surroundings.

Calculating the molecules, breaking bonds between or within them. ΔSsurr

• ΔSsurr = -ΔH

T

The entropy change in the surroundings is equal to the energy transferred (enthalpy change) divided by the temperature.

Spontaneity the molecules, breaking bonds between or within them.

• ΔStotal = ΔSsys + Δssurr

• ΔSsys =

• ΔSsurr =

• ΔStotal =

Spontanaeity the molecules, breaking bonds between or within them.

• What about melting an ice cube at 10°C?

• ΔStotal = ΔSsys + Δssurr

• ΔSsys =

• ΔSsurr =

• ΔStotal =

Spontanaeity the molecules, breaking bonds between or within them.

• If ΔStotal is POSITIVE the reaction will be spontaneous.

• This is the second law of thermodynamics!

• Insert CI 4.4 Fig 3

Science of de-icing the roads... the molecules, breaking bonds between or within them.

• Salt water will have a higher entropy than pure water, i.e. More negative than

-22JK-1mol-1.

• This means ΔSsurr must be greater than

+ 22JK-1mol-1 .

• As ΔSsurr = -ΔH

T

• This can only happen with a smaller value of T, therefore salt water freezes at a lower temperature.

Your turn... the molecules, breaking bonds between or within them.

• CaCO3 (s)  CaO (s) + CO2 (g)

• ΔS = +159 JK-1mol-1

• ΔH = +179 KJ mol-1

• What are the total entropy changes at 298K (25°C) and 1273K (1000°C)?

Your turn... the molecules, breaking bonds between or within them.

• CaCO3 (s)  CaO (s) + CO2 (g)

• ΔS = +159 JK-1mol-1

• ΔH = +179 KJ mol-1

• 298K (25°C) = -442 JK-1mol-1

• 1273K (1000°C) = +18 JK-1mol-1

What about dissolving and crystallisation? the molecules, breaking bonds between or within them.

• Dissolving the same amount of solid in two different volumes of water, e.g. 100ml and 1000ml.

• Changes in ΔH so small we assume constant.

• Temperature remains constant.

• Therefore presume ΔSsurr is constant.

• However ΔSsys becomes more positive the greater the volume of liquid as there are more particles of water and therefore more ways of arranging these particles.

What about dissolving and crystallisation? the molecules, breaking bonds between or within them.

• By making the volume of water smaller, such as by evaporation, the entropy of the system becomes less positive.

• Eventually this leads to the total entropy becoming negative and therefore dissolving becoming unfavourable.

• Hence crystallisation takes place.

The magic number... the molecules, breaking bonds between or within them.

• What about if ΔS total = 0?

• This means that there is no net change in either direction and the products and reactants are at equilibrium.

And finally... the molecules, breaking bonds between or within them.

• You may need to calculate the entropy change for a chemical reaction given the entropies of reactants and products.

• You can do this easily using the following equation:

• ΔS reaction = ΔS products - ΔS reactants

Practise Time... the molecules, breaking bonds between or within them.

• Chemical Ideas 4.4

• Problems 1-4