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.

For better or for worse, nature 'likes' chaos, disorder, high entropy... In fact, much of our life consists in fighting this disorder! A system (such as a room) is in a state of high entropy when its degree of disorder is high.

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 ways to arrange?

 Gases have higher entropies than liquids.  Liquids have higher entropies than solids.

 Entropy is greater for larger atoms.  Entropy is greater for molecules with more atoms.

 Ar has higher entropy than Ne as Ar molecules are larger.  C 8 H 18(l) has higher entropy than C 5 H 12(l) as complex molecules have higher entropy than simple ones.  Br 2(g) has higher entropy than Br 2(l) as gases have higher entropies than liquids since gases have more ways of being arranged

 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

 We use specific heat capacity (C p ) 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.

 Translation – movement of the whole molecule from one place to another.  Rotation – Spinning around.  Vibration – Stretching and compressing bonds.

 Adding further energy may increase the electronic energy of the molecules, breaking bonds between or within them.  However these changes don’t affect the temperature as electronic and bonding energy don’t affect motion energy.

 Insert Fig 2 CI 4.4

 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.

Molecule 1Molecule ways

Molecule 1Molecule 2 More quanta 5  7 more ways

Molecule 1Molecule ways

Molecule 1Molecule 2Molecule 3 More molecules 5  15 More ways

 Add the Qualitative Summary from page 72 to your notes!

 H 2 O (l)  H 2 O (s)  ΔS = JK -1 mol -1  ΔH = 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.

 ΔS surr = -ΔH T The entropy change in the surroundings is equal to the energy transferred (enthalpy change) divided by the temperature.

 ΔS total = ΔS sys + Δs surr  ΔS sys =  ΔS surr =  ΔS total =

 What about melting an ice cube at 10°C?  ΔS total = ΔS sys + Δs surr  ΔS sys =  ΔS surr =  ΔS total =

 If ΔS total is POSITIVE the reaction will be spontaneous.  This is the second law of thermodynamics!  Insert CI 4.4 Fig 3

 Salt water will have a higher entropy than pure water, i.e. More negative than -22JK -1 mol -1.  This means ΔS surr must be greater than + 22JK -1 mol -1.  As ΔS surr = -ΔH T  This can only happen with a smaller value of T, therefore salt water freezes at a lower temperature.

 CaCO3 (s)  CaO (s) + CO2 (g)  ΔS = +159 JK -1 mol -1  ΔH = +179 KJ mol -1  What are the total entropy changes at 298K (25°C) and 1273K (1000°C)?

 CaCO3 (s)  CaO (s) + CO2 (g)  ΔS = +159 JK -1 mol -1  ΔH = +179 KJ mol -1  298K (25°C) = -442 JK -1 mol -1  1273K (1000°C) = +18 JK -1 mol -1

 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 ΔS surr is constant.  However ΔS sys becomes more positive the greater the volume of liquid as there are more particles of water and therefore more ways of arranging these particles.

 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.

 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.

 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

 Chemical Ideas 4.4  Problems 1-4