1. Entropy Explanation and theory

2. What is Entropy? Entropy is the measure of the disorder of a system. Ex, When a liquid turns into gas by heating it, the intermolecular forces decrease and the speed of the particles increase. The particles become more disordered.

3. Numerical Values for Entropy Entropy is given the symbol S ∆S (Entropy change) is as important as ∆H (Enthalpy change) ∆S is defined in a system as: ∆S = q / T q: Heat added to the system from the surrounding T: Thermodynamic temperature (in Kelvin) at which heat is transferred (from the surrounding to the system)

4. Ex, heating a system that is under going a change of matter (e.g, ice melting to liquid). This liquid, it’s heat does not increase, it only makes the matter (water molecules) in the system more disordered.

5. Entropy changes in system and surroundings Types of entropy changes: ◦ ∆S (System) ◦ ∆S (Surrounding) In terms of entropy changes ∆S, the significance of the heat released by an exothermic reaction into the surrounding depends on how hot the surroundings already are.

6. The size of entropy change is proportional to the temperature. ∆S (surr) = q/T = - ∆H/T ∆S (tot) = ∆S (sys)+ ∆S (surr) ∆S (tot) = ∆S (sys) – (∆H/T) ∆S (tot) = ∆S – (∆H/T) Becomes By conversion

7. Laws of Thermodynamics First law of Thermodynamics states: Energy cannot be created or destroyed. Second law of Thermodynamics states: The total entropy of the universe [∆S (tot)] tends to increase when any change takes place, it never goes down.

8. As a result to the second law of Thermodynamics, ∆S (tot) for any change is always greater than (or equal to) zero. A negative entropy change isn’t possible.

9. For a spontaneous change to occur: ∆S (tot) > 0 So substituting the equation above gives: ∆S – (∆H/T) > 0 If we multiply through by T, the criterion for spontaneous change becomes: T∆S - ∆H > 0 For a spontaneous reaction

10. Predicting entropy changes Change∆S Solid  LiquidIncrease (+) Solid  GasIncrease (+) Liquid  GasIncrease (+) Liquid  SolidDecrease (-) Gas  SolidDecrease (-) Gas  LiquidDecrease (-) When predicting entropy changes, the change due to a change in the number of particles in the gaseous state is usually greater than any other possible factor.

11. What is a spontaneous process? Processes that proceed in a definite direction when left to themselves and in the absence of any attempt to drive them in reverse — are known as natural processes or spontaneous changes. Is a process in which matter moves from a less stable state to a more stable state.

12. Spontaneity: Examples Drop a teabag into a pot of hot water, and you will see the tea diffuse into the water until it is uniformly distributed throughout the water. What you will never see is the reverse of this process, in which the tea would be sucked up and re-absorbed by the teabag. The making of tea, like all changes that take place in the world, possesses a “natural” direction.

13. A closer look at disorder: microstates and macrostates How can we express disorder quantitatively? From the example of the coins, you can probably see that simple statistics plays a role: the probability of obtaining three heads and seven tails after tossing ten coins is just the ratio of the number of ways that ten different coins can be arranged in this way, to the number of all possible arrangements of ten coins. The greater the number of microstates that correspond to a given macrostate, the greater the probability of that macrostate.

14. All of the changes described above take place spontaneously, meaning that: Once they are allowed to commence, they will proceed to the finish without any outside intervention. It would be inconceivable that any of these changes could occur in the reverse direction (that is, be undone) without changing the conditions or actively disturbing the system in some way.

15. Disorder is more probable than order because there are so many more ways of achieving it.

16. What determines the direction in which spontaneous change will occur? It is clearly not a fall in the energy, since in many cases the energy of the system does not change. If there is no net loss of energy when these processes operate in the forward or natural direction, it would not require any expenditure of energy for them to operate in reverse. In other words, contrary to the common sense, The direction of a spontaneous process is not governed by the energy change...... and thus the First Law of Thermodynamics cannot predict the direction of a natural process.

17. EXAMPLE

18. NA OVERVIEW WORD DOCUMENT

20. Defining Standard Gibbs Energy Change

23. THE EFFECT OF TEMPERATURE ON THE SPONTANEOUS REACTIONS FOR DIFFERENT REACTIONS IS SUMMARIZED IN THE TABLE BELLOW

24. http://www.chem.ufl.edu/~itl/2045/lecture s/lec_u.html http://www.chem.ufl.edu/~itl/2045/lecture s/lec_u.html http://www.chem1.com/acad/webtext/ther meq/TE1.html#PageTop

25. Calculus EX: Calculate ∆G reaction for the reaction 2Al (s) + fe 2 O 3 (s) → 2fe(s) + Al 2 O 3 (s) from the following data. Compound ∆Gf/kJmol -1 fe2O3 (s) -742 Al 2O3(s) -1582 Comment on the significance of the value obtained

26. Solution 2Al (s) + fe 2 O 3 (s) → 2fe(s) + Al 2 O 3 (s) 2 (0) -742 2 (0) -1582 ∆G reaction =∑∆G f (products) - ∑∆G f (reactant) = -1582 – (- 742) = -840 KJ mol -1 The reaction is spontaneous under stander conditions.

27. Using ∆S reaction and ∆H reaction values to calculate ∆G reaction at all temperatures Stander value = 298 K The expression = ∆G reaction = ∆H sys - T∆S sys

28. Worked example Calculate ∆G reaction at 298 K for the thermal decomposition of calcium carbonate from the following data Compound ∆H f /KJ mol -1 S/Jk -1 mol -1 CaCo3 (s) -1207 92.9 Cao (s) -635 39.7 Co2 (g) -394 214

29. Solution First, calculate ∆H reaction CaCo3 (s) → Cao (s) + Co2 (g) -1207 -635 -394 ∆H reaction = ∑∆H f (products) - ∑∆H f (reactants) = (-635 + -394) – (-1207) = +178 KJ mol -1 Second, calculate ∆S reaction = ∑S (products) - ∑S (reactant) = (39.7 + 214) – 92.9 = 160.8 KJ -1 mol -1

30. Now calculate the change in Gibbs’ free energy of the reaction. ∆G reaction = ∆H reaction - T ∆S reaction = - 178 – (298) ( 160.8 * 10 -3 ) = + 130 KJ mol -1