1. 11 Spontaneous Process A process is spontaneous if it occurs without outside intervention, it happens on its own. Spontaneous processes can be fast or slow! Examples of Spontaneous Processes: –a ball rolls downhill, but the ball never spontaneously rolls uphill –steel rusts, but the rust never spontaneously forms iron and oxygen –Water spontaneously freezes at temperatures below 0 o C

2. 22 Entropy, S ENTROPY (  S) is a measurement of the change in RANDOMNESS or DISORDER of the particles in a chemical reaction. Spontaneous reactions will always move toward an increase in Entropy (Increase in randomness) Why is this? Because things are more likely to occur in a random order than in an organized arranged order.

3. 33 2nd Law of Thermodynamics The second law states that the disorder in the universe is always increasing. As disorder increasing, energy is transferredThe second law states that the disorder in the universe is always increasing. As disorder increasing, energy is transferred to less usable forms. So… If ∆S > 0, entropy is increasing, the particles are becoming more chaotic. If ∆S < 0, entropy is decreasing, the particles are becoming more organized.

4. 44 Which is more likely? Which kind of card hand are you more likely to be dealt?  Random cards Or a Royal Flush  It is more likely that you will get a collection of random cards rather than an ordered hand.

5. 55 Probability of Disorder If you never clean your room, will your stuff become more organized or more chaotic? This is the natural progression in the universe! If there is no outside force, Entropy will increase!

6. 66 Factors that affect Entropy There are several factors that determine the randomness or disorder of compounds and molecules. 1) States of Matter / Temperature 2) Atom and Molecular Complexity 3) Amount of a compound 4) Mixing Compounds Think about situations that have the greatest amount of particle movement. An increase in movement is an indicator for increased entropy and more randomness!!!

7. 77 State of Matter Solids have the least entropy due to a more organized and structured shape. Gases have the greatest entropy due to an undefined shape and lots of kinetic energy. S solid < S liquid << S gas Increase temperature, Increase Entropy.

8. 88 Entropy increases when you increase atomic complexity Increases in the number of protons, electrons, neutrons, increases entropy. -Increase in molecular complexity generally leads to increase in S. -The more complex the molecule, the more likely it is to be disorganized and random. Factors affecting Entropy, S

9. 99 -Increasing the amount of a compound will cause the entropy to increase. -Larger molar quantities have more disorder than smaller molar quantities. Look for the coefficients! Entropy increases when a pure liquid or solid is dissolved in a solvent. Breaking a solid up into smaller pieces leads to a less organized structure, and more kinetic energy. Factors affecting Entropy, S

10. 1010 Practice Problems #1 For each, circle which has a greater entropy (  S) A) Hg (l) or Hg (s) B) C 2 H 4 (g) or C 2 H 6 (g) C) 1.0 mol of H 2 (g) or 2.0 mol of H 2 (g) D) NaCl (s) or NaCl (aq) E) NH 3 (g) or Ne(g) Predict if entropy is increasing or decreasing: 1) solid sugar is added to water 2) iodine vapor condenses onto a cold surface forming crystals

11. 1111 Consider the reaction: ___H 2 (g) + ___O 2 (g) ---> ___H 2 O(l) Would we expect entropy to increase or decrease? Be sure to balance the reaction ∆S o = 2(H 2 O) - [2 (H 2 ) + (O 2 )] *Get ∆S values from your chart!* ∆S o = 2(69.91) - [2(130.6) + (205.0)] ∆S o = - 326.38 J/mol Calculating ∆S for a Reaction ∆S o = S o (products) -  S o (reactants)

12. 1212 ∆S o = - 326.38 J/mol. K – Negative ∆S means the entropy is decreasing, and the particles are becoming more organized. 2 H 2 (g) + 1 O 2 (g) ---> 2 H 2 O(l) Entropy decreases, it becomes more organized because 3 moles of gases produced 2 mol of liquid. Lots of Motion (Disorder)  More organized. Calculating ∆S for a Reaction ∆S o = S o (products) -  S o (reactants)

13. 1313 Practice Problem #2 Calculate  S at standard conditions __NiS(s) + __O 2 (g) --> __SO 2 (g) + __NiO(s) * Predict if you think  S will be positive or negative SubstanceS o (J/mol. K) SO 2 248 NiO38 O 2 205 NiS53

14. 1414 Practice Problem #3 Calculate  S for __Al 2 O 3 (s)+__H 2 (g)  __Al(s) + __ H 2 O(g) * Predict if you think  S will be positive or negative SubstanceS o (J/K. mol) Al 2 O 3 (s)?? H 2 (g)??? Al(s)?? H 2 O(g)???

15. 1515 1 Al 2 O 3 (s)+ 3 H 2 (g)  2 Al(s) + 3 H 2 O(g) Why is enthalpy large and positive? –H 2 O is nonlinear and triatomic »H 2 O is larger and is more complex –H 2 is linear and diatomic »H 2 is smaller and less complex The more complex the molecule, the higher the  S

16. 1616 Entropy Changes for Phase Changes For a phase change, ∆S = q/T where q = heat transferred in phase change For H 2 O (liq) ---> H 2 O(g) ∆H = q = +40,700 J/mol