1. Thermodynamics Is it hot in here or what?

3. Energy Many forms and sources Thermochemistry is interested in heat exchanges Breaking bonds takes energy (endothermic) Making bonds produces energy (exothermic) SI Unit is the Joule (J), common unit calorie 4.18J = 1 cal 4.18 kJ = 1 kcal

4. Temperature and Energy Change Every substance has a unique ability to absorb and release energy. Called heat capacity. Water has a very large value. Specific Heat (c p ) is the measure of how much energy a substance will absorb or release to change its temp. 4.18J/g°C for water

5. Q = m c p ΔT Energy = mass x spec. heat x temp. change Ex: If 20 g of water changes its temp from 25.0C to 30.0C, what amount of energy is absorbed? Q = (20.0g) x (4.18 J/g C) x (5.0 C) Q = 418 J Practice!!!!

6. Heat of Reaction Energy change that occurs during a chemical reaction Called enthalpy Symbol (ΔH) change in enthalpy Negative ΔH value = exothermic reaction Positive ΔH value = endothermic reaction

8. Energy Reaction coordinate Reactants Products Overall energy change Exothermic reaction

9. Endothermic reaction Energy Reaction coordinate Overall energy change

10. Heat of Formation ∆H f ° The energy change associated with the formation of one mole of a compound from its elements. Sometimes uses fractional coefficients Ex. H 2 (g) + ½ O 2 (g) → H 2 O (l) ∆H f ° = -286 kJ/mol exothermic (comp. is stable)

12. Thermochemical Equation An equation which includes the heat energy as a reactant (endothermic) or product (exothermic) H 2 (g) + ½ O 2 (g) → H 2 O (l) + 286kJ – Notice that the energy is written as a + even though the ∆H f ° = -286 kJ/mol

13. Calculation of ΔH reaction Two methods – both work Hess’ Law and a shortcut Learn both (use the shortcut more often)

14. Hess’ Law - finding ΔH Says that if you can write a reaction in steps that add up to the original reaction, the ΔH of the overall reaction is the sum of the ΔH of the steps. (route independent) Guidelines 1. If the reaction is reversed, so is the sign of ΔH. 2. If coefficients are multiplied, so is ΔH.

15. Example: find ΔH CH 4(g) + 2O 2(g) → CO 2(g) + 2H 2 O (l) Solution 1. Write ΔH f eq. for each compound C(s) + 2H 2 (g) → CH 4 (g) ΔH = -74.8 kJ C(s) + O 2 (g) → CO 2 (g) ΔH = -393.5 kJ H 2 (g) + ½ O 2 (g) → H 2 O(l) ΔH = -285.8 kJ

16. solution for: CH 4(g) + 2O 2(g) → CO 2(g) + 2H 2 O (l) 2. Multiply by any coefficients in the orig. eq. 2H 2 (g) + O 2 (g) → 2H 2 O(l) ΔH = -571.6 kJ 3. Reverse reactions to get reactants on left CH 4 (g) → C(s) + 2H 2 (g) ΔH = +74.8 kJ 4. Add the resulting equations together……

17. Orig. eq. CH 4(g) + 2O 2(g) → CO 2(g) + 2H 2 O (l) CH 4 (g) → C(s) + 2H 2 (g) ΔH = +74.8 kJ 2H 2 (g) + O 2 (g) → 2H 2 O(l) ΔH = -571.6 kJ C(s) + O 2 (g) → CO 2 (g) ΔH = -393.5 kJ _____________________________________ CH 4 (g) + 2O 2 (g) → CO 2 (g) + 2H 2 O(l) (same as orig.) so add up ΔH = -890.3 kJ

18. Hess’ Law Tedious, but it works……. Practice…….

19. ALTERNATIVE TO HESS’ LAW Handy shortcut to use for calculations. ΔH = Σ ΔH f (products) - Σ ΔH f (reactants) Remember to multiply values by any coefficients in the balanced equation. ΔH = (-393.5 kJ + 2 x -285.8 kJ) – (-74.8 kJ) ΔH = -890.3 kJ

20. SPONTANEOUS REACTIONS Proceed without outside assistance (beyond the initial Ea) -- they just happen. Spontaneous chemical reactions generally occur if the products have lower PE than the reactants. -sometimes this is not the case ( ex--- H 2 O(s) only forms if the temp is less than 0 C) -reason for this is the other driving force in nature...........

21. Entropy A measure of the disorder or randomness of a system. -represented by S (change in..... ΔS) ΔS = Σ S f (products) - Σ S i (reactants)

22. All entropies (Sf and Si) are positive, but …ΔS can be negative or positive. if ΔS is negative ===> Less disorder (lower entropy) Example: crystallization of a liquid if ΔS is positive ===> More disorder (more entropy) – Example: evaporation of a liquid….. favored in nature!

23. Prediction of entropy change synthesis of compounds decreases entropy (due to bonding) decomposition of compounds increases entropy mixing a solute and solvent increases entropy Solid→ Liquid → Gas increases entropy Ie. evaporation of a liquid increases entropy

25. So>>>>>> Two forces influence the direction of a spontaneous reaction: ΔH and ΔS when ΔH decreases; lower energy results. when ΔS increases; higher disorder results

26. These two factors compete Like a tug of war. Usually the enthalpy prevails.

27. 4 Possible Cases ΔH ΔS Result: - + reaction favored + - no reaction - - rx if ΔH is large + + rx if ΔS is large

28. Gibbs Free Energy ΔG An equation called the “Gibbs equation” compares the values of ΔH and Δ S. ΔG = ΔH - T ΔS if ΔG is negative; the reaction is spontaneous if ΔG is positive: the reaction is not spontaneous, and will not occur.

29. Alternative to the Gibbs equation: ΔG = ΣΔG f (products) - ΣΔG f (reactants)

30. Collision Theory In order to react molecules and atoms must touch each other. They must hit each other hard enough to react. Anything that increase these things will make the reaction faster.

31. Things that Effect Rate l TEMPERATURE l Higher temperature = faster particles. l More and harder collisions. l Faster Reactions. l CONCENTRATION l More concentrated = molecules closer together. l Collide more often. l Faster reaction.

32. Things that Effect Rate l PARTICLE SIZE l Molecules can only collide at the surface. l Smaller particles = bigger surface area. l Smaller particles = faster reaction. l Smallest possible is atoms or ions. l Dissolving speeds up reactions. l Getting two solids to react with each other is slow.

33. Things that Effect Rate l CATALYSTS- substances that speed up a reaction without being used up.(enzyme). l Speeds up reaction by giving the reaction a new path. l The new path has a lower activation energy. l More molecules have this energy. l The reaction goes faster. l Inhibitor- a substance that blocks a catalyst.