1. REACTION ENERGY CHAPTER 16 PAGE 500

2. A. THERMOCHEMISTRY 1. Introduction a. Every chemical reaction causes a change in energy b. Endothermic or exothermic c. Thermochemistry – study of the transfer of energy as heat that happens during chemical reaction and physical changes

3. 2. Temperature and Heat are Related a. Calorimeter – tool which measures energy absorbed or released as heat in a chemical or physical change b. Temperature – measure of the average kinetic energy of the particles 1) Typically use °C or K 2) K = 273.15 + °C 3) Measured by energy transfer in joules (J)

5. 3) Unit is either joules or calories a) A calorie – amount of heat it takes to raise the temperature of 1 gram of water 1°C b) A dietary Calorie = 1000 calories c) 1 Cal = 1 kcal = 1000 calories d) 1 cal = 4.184 J

6. 3. Energy transfers changes with the reaction a. Energy transfer as heat depends on: 1) Mass of material 2) Type of material 3) Size of the temperature change 4) Example: 1 gram of iron heated to 100°C and then cooled to 50°C transfers 22.5 J in a calorimeter. Silver under the same conditions only transfers 11.8 J

7. b. Specific heat 1) Amount of energy required to raise the temperature of one gram of substance by one Celsius degree (1°C) or one kelvin (1 K) 2) Unit: J/(g·K) or J/(g·°C) 3) Fig 1.2 (pg 503) 4) Water 4.18 J/(g·K) a) Very high for a common substance b) Allows for temperature regulation SubstanceSpecific Heat J/(g·K) Water (l)4.18 Water (s)2.06 Water (g)1.87 Ammonia (g)2.09 Ethanol (g)2.44 Aluminum (s)0.897 Iron (s)0.449 Lead (s)0.129

9. CALORIMETRY Two kinds https://www.youtube.com/watch?v=EAgbknIDKNo

10. 7) Examples: a) A 4.0 g sample of glass was heated from 274 K to 314 K, a temperature increase of 40 K, and was found to have absorbed 32 J of energy as heat. What is the specific heat of this type of glass? b) How much energy will the same glass sample gain when it is heated from 314 K to 344 K?

11. c) Determine the specific heat of a material if a 35 g sample absorbed 96 J as it was heated from 293 K to 313 K. Answer: 0.14 J/(gK) d) If 980 kJ of energy are added to 6.2 L of water at 291 K, what will the final temperature of the water be? Answer: 329 K

12. e) A piece of copper with a mass of 95.4 g increases in temperature from 25.0°C to 48.0°C when the copper absorbs 849 J of heat. What is the specific heat of copper? Answer: 0.387 J/(g·°C)

13. 4. Heat energy transferred a. Enthalpy Change 1) Amount of energy absorbed by a system as heat during a process at constant pressure. 2) ∆H – means “change in enthalpy” 3) ∆H = H products – H reactants 4) q = Δ H

14. b. Enthalpy of Reaction 1) Quantity of energy transferred as heat during a chemical reaction 2) Difference b/w stored energy of reactants and products 3) AKA: “heat of reaction”

15. c. Enthalpy of Reaction – Exothermic Reactions 1) Chemical equations don’t typically show energy release 2) Thermochemical equation a) Equation that includes the quantity of energy released or absorbed as heat during the reaction b) 2 H 2 (g) + O 2 (g) → 2 H 2 O + 483.6 kJ 3) Reactants have larger enthalpy than products

16. d. Enthalpy of Reaction – Endothermic Reactions 1) Products have larger enthalpy than reactants 2) 2 H 2 O (g) + 483.6 kJ → 2 H 2 (g) + O 2 (g)

17. e. Thermochemical Equations 1) Typically not written with energy as product or reactant 2) Written with Δ H a) Δ H is negative = exothermic (system loses energy) b) Δ H is positive = endothermic (system gains energy) 3) 2 H 2 (g) + O 2 (g) → 2 H 2 O (g) Δ H = -483.6 kJ 4) 2 H 2 O (g) → 2 H 2 (g) + O 2 (g) Δ H = 483.6 kJ

18. 5) Endothermic or Exothermic? a) CaO + H 2 O → Ca(OH) 2 + 65.2 kJ b) 2 NaHCO 3 + 129 kJ → Na 2 CO 3 + H 2 O + CO 2 c) What is Δ H for a) and b)? a) Δ H = -65.2 kJ b) Δ H = 129 kJ

19. 5. Enthalpy of Reaction Examples a. 25.0 mL of HCl and 25.0 mL of NaOH, both at 25.0°C are mixed in a calorimeter. The final temperature is 32.0°C. Calculate Δ H in kJ. Is this reaction exothermic or endothermic? Negative or positive value?

20. What we know Δ H = - (m x c q x Δ T) C q of water = 4.18 J/(g·°C) M = 50.0 g Δ T = 7.0°C Δ H = -(50.0g) 4.18 J/(g·°C) (7.0°C)