1. 1 Thermodynamic Equations (a)Heat, Work and the 1 st Law PV=nRT equation of state for ideal (perfect) gas work done against external pressure work done by ideal gas in isothermal reversible expansion/compression

2. 2 First Law of Thermodynamics definition of enthalpy change in enthalpy at constant pressure molar heat capacity at constant pressure molar heat capacity at constant volume

3. 3 Molar internal increment from T 1 and T 2 Molar enthalpy increment from T 1 to T 2 Q: How to calculate enthalpy change for ideal gas expansion?

4. 4 (b) Thermochemistry Standard enthalpy of reaction from standard enthalpies of formation Kirchhoff equation For H 2 +1/2O 2 =H 2 O

5. 5 © 2 nd Law and Entropy Changes 2 nd law where entropy change for isothermal adsorption of heat Q: How to calculate the entropy of phase transformation?

6. 6 Entropy change on heating from T 1 to T 2 at constant pressure. if C p is constant from T 1 to T 2.

7. 7 Molar entropy change for isothermal expansion/compression of an ideal gas e.g. 10 moles ideal gas expanded from 1 to 10 m 3

8. 8 Entropy change for random mixing of n A moles of A and n B moles of B Entropy change for formation of one mole of an binary ideal gas mixture, or an ideal liquid solution or an ideal solid solution.

9. 9 Calculate the change in entropy when a block of iron (heat capacity=0.48J.g -1.K -1 ) of 500g mass at 473K is brought into contact 1000 g of water at 293K after thermal equilibrium is reached. T f = 302.7K  S=29.7JK -1

10. 10 Entropy and Probability Background  ‘Classical approach’ to thermodynamics.  What entropy ‘means’  What is happening in terms of the atoms and molecules when we change the state of the system   S>0, there is always an increase in ‘randomness’ or ‘disorder’. ·

11. 11 1. Expansion of a gas at constant T e.g. 10 moles ideal gas expanded from 1 to 10 m 3 2. Compression of gas at constant T

12. 12 1. Heat up a solid, liquid or gas at constant P If Cp=const e.g. Heat 2 moles of Al (~54g) from 298K to 373K, Cp(Al,s)=24.36JK -1 mol -1.

13. 13 1. Mixing gases e.g. mix 1 mole of O 2 with 3 moles of N 2

14. 14. ==Enthalpy of Fusion e.g. melt 10 moles of Pb at its melting point 600K, =4812 J.mole -1 Melting solid

15. 15 1. Oxidise a metal to its oxide e.g. oxidise 2 moles of Al(s) to Al 2 O 3 (s) at 298K =50.99-2(28.33)-3/2(205.0)=-313.2JK -1.

16. 16 Now examine these in terms of changes in degree of ‘disorder’. 1. Gas molecules occupy large volume. Each molecule has more space. Increase in disorder. 2. Gas molecules occupy less volume. Each molecule as less space. Decrease in disorder, ie. Increase in order. 3. For crystals, atoms vibrate about fixed lattice points. When the crystal is heated up, amplitude of vibrations has increased, ie. More disordered. 4. When gases have formed a mixture, there is an increase in disorder. 5. When a crystal is melted, ordered array of atoms on crystal lattice become free movement of atoms within liquid phase. 6. When gas reacts with solid to form another solid, the net result is removal of gas phase. Therefore, it leads to decrease in disorder, ie. Increase in order.

17. 17 What do we means by ‘disorder’ and how do we quantify this? By ‘probability’ A ‘disordered’ state is one where there are a large number of possible equally probable arrangements, ie a high probability state. The more disordered state has a higher probability  f (larger number of different arrangements). The less disordered state has a lower probability  i (smaller number of different arrangements). Example: two colour balls mixing

18. 18 What do we mean by ‘probability’? Let’s consider throwing a dice. What are the chances (ie the probability) of throwing a six? For two dice, what are chances of throwing 2 sixes? For three dice -3 sixes? The chance is 1 in 216.

19. 19 Therefore, the probability of several events occurring simultaneously is the product of the probabilities of the individual events 1/6  1/6  1/6=1/216 n dice-probability of throwing n sixes =1/6 ­ n Thus a ‘disordered’ state is one where there are a large number of possible equally probable arrangements, ie a high probability state.

20. 20  Thus entropy changes can be regarded as a measure of the increase in the ‘degree of mixing’ such that for Less disordered state  more disordered state  S=S final -S initial >0 The more disordered state has a higher probability  f (larger number of different arrangements). The less disordered state has a lower probability  i (smaller number of different arrangements).

21. 21 These are related by the Boltzman Equation

22. 22 When we use the term ‘disorder’, we are referring to a situation which can be realised in many different ways whereas ‘ordered’ states refer to those which can only be obtained in a few ways.

23. 23 The equation proposed by Boltzman is as follows: Where  final = Number of ways of achieving the final state (final probability)  initial = number of ways of achieving the initial state (initial probability).  final >  initial, then  S>0 then process will occur spontaneously in an isolated system.

24. 24 Example: expansion of gas into a vaccum We consider the situation after opening the valve and before the gas has expanded. What is the probability of finding a particular molecule in V 1 as opposed to finding it in V 2. Probability must be related to volumes of V 1 and V 2, ie. Chance of finding a molecule in V 1 is V 1 /(V 1 +V 2 ). Now add a second molecule, chance of finding this in V 1 is also V 1 /(V 1 +V 2 ). Probability of finding them both in V 1 =

25. 25 For N molecules, chance of all N molecules being in V 1 =, which is the probability of initial state  initial. Probability of final state=1 since all of the molecules must be somewhere inside the volume (V 1 +V 2 ) i.e.  final =1

26. 26 Apply Boltzman Equation to obtain  S for 1 moles i.e. N=6.023  10 23. Thus we get the same result as before.

27. 27 Conclusion: There are two ways to calculate entropy changes

28. 28 a)What is the reversible process? [5 marks] b)Give an example of the reversible process. [5 marks] c)Explain why a gas expansion to vacuum is not reversible process? [5 marks] d)For the reaction Calculate the heat of reaction at 1000 K if the heat capacities in J.mol -1 K -1 are given by Cp= 25.7 for CO(g), Cp=24.8 for CO 2 (g) Cp=24.6 for O 2 (g) Enthalpies of formation for CO(g) –105.6 kJ.mol -1, for CO 2 (g) -376 kJ mol -1 at 298 K.