1. Energy Review

2. Things to know How is heat transferred? Exo and endo Specific heat Heat Specific Heat of lead lab Hess’s law Stoich with enthalpy

3. How is heat transferred? Remember heat always goes from hot to cold. Temperature – Thermal energy - Motion of particles - friction Size does affect the amount of thermal energy

4. Exo and endo Exothermic – Endothermic – Signs of q and  H Exo : -q and –  H (lost heat) Endo: +q and +  H (gained heat) Releasing of heat to surroundings Feels warm to the touch (usually) Taking in heat from the surroundings Feels cold to the touch (usually)

5. Specific heat The amount of heat needed to raise 1 gram of substance by 1 o C. Large value – Small value – Which substance will have the greatest change in temp: S water = 4.180 J/g o C or S rock = 35 J/g o C? Water – smallest specific heat Stores a lot of heat, slow temp changes Stores very little heat, fast temp changes

6. Heat problems q = ms  T Calculate heat needed to raise 25.0 g of lead by 30.0 o C. S lead = 0.13 J/g o C q = ? m = 25.0 g s = 0.13 J/g o C  T = 30.0 o C q = ms  T q = (25.0 g)(0.13 J/g o C)(30.0 o C) q = + 97.5 J Endothermic: positive q and raised the temp

7. Calculate the specific heat(J/g o C) of a 100.0 g object that loses 12.0 kJ of heat as it cools from 98.2 o C to 22.3 o C. q = m = s =  T = T f – T i -12 kJ 100.0 g ? J/g o C q = ms  T algebra S = q m  T 22.3 o C – 98.2 o C = -75.9 o C S = 1.58 J/g o C

8. Heat gained = - heat lost Cold water and hot lead The heat gained by the cold is equal to the heat lost by the hot lead. Why the temp very close to the initial cold water? Water has much, much greater specific heat

9. Hess’s law Enthalpy change(heat change) of a series of steps in chemical process is equal the enthalpy change of the overall reaction. (2 step process) A + B  C  H = +40 kJ C  D  H = - 25 kJ A + B  D  H = +15 kJ Overall (1 step process) A + B  D  H = +15 kJ

10. Calculate  H for NO(g) + O(g)  NO 2 (g) Step 1) (2 O 3 (g)  3 O 2 (g)  H = -427 kJ) Step 2) (O 2 (g)  2 O(g)  H = +495 kJ) O (g)  ½ O 2 (g)  H = - 247.5 kJ Step 3) NO(g) + O 3 (g)  NO 2 (g) + O 2 (g)  H = -199 kJ NO(g) + O(g)  NO 2 (g)  H = -660 kJ Are all 3 chemicals from the overall equation on the correct side of the arrow? Need to flip step 2 to get O on correct side and times by ½ Times step 1 by ½ and flip to get rid of O 3 (g) Now add up the reactions ½ 3/2 O 2 (g)  O 3 (g)  H = - 213.5 kJ ½ NO

11. Stoich with enthalpy 2H 2 (g) + O 2 (g)  2H 2 O(l)  H = -482 kJ/mol a)If 4.00 mols of H 2 was used calculate  H. b)If 15.0 g of O 2 was used calculate  H. = -968 kJ = -227 kJ