1. TO LIVE IS THE RAREST THING IN THE WORLD. MOST JUST EXIST. THAT IS ALL.

2. THERMOCHEMISTRY THE STUDY OF ENERGY CHANGES THAT ACCOMPANY CHEMICAL AND PHYSICAL CHANGES

3. Heat a form of energy. can be transferred between samples heat flows from matter at a higher temperature to matter at a lower temperature Temperature a measure of the average kinetic energy of the particles in a sample.

4. Units of Heat Joule (SI unit) calorie cal the amount of energy required to raise the temperature of one gram of water one degree Celsius. Calorie Cal a dietary calorie. kilocalorie, kcal (1,000 calories) 1 cal = 4.184 Joules

5. Enthalpy the heat content of a system represented by H only changes in enthalpy can be measured ∴ ΔH is used

6. Specific heat or heat capacity, c p the amount of energy required to raise the temperature of one gram of a substance one degree Celsius used in equation q = m x c p x ΔT

7. Note that in the Metric System, Joules are the units of measure for heat. Energy, or heat (J) mass (g) specific heat (J/g°C) Δ temp (°C)

8. Heat capacity, c p

9. For water, C = 4.18 J/(g o C), and also C = 1.00 cal/(g o C) Thus, for water: it takes a long time to heat up, and it takes a long time to cool off! Water is used as a coolant!

10. q = m x c p x ΔT 1.A 45.0-gram sample of iron is heated from 25.0°C to 50.0°C. How much energy is required? (c p iron = 0.449 J/g°C) q = ? m = 45.0 g c p = 0.449 J/g°C ΔT = 50.0°C – 25.0°C = 25.0°C q = m c p ΔT q = 45.0g (0.449 J/g°C) (25.0°C) = 505.125 J q = 505 J

11. What is the specific heat capacity of an object if a 12.5-gram sample is heated from 12.0°C to 28.0°C using 100.0 joules? q = 100.0 J m = 12.5 g c p = ? ΔT = 28.0 – 12.0 = 16.0°C q = m c p ΔT

12. Heat - represented by “q”, is energy that transfers from one object to another, because of a temperature difference between them. In studying heat changes, think of defining these two parts: the system – the part of the universe you focus your attention on the surroundings – everything else If heat flows into a system from the surroundings, the system gains energy, and the change is said to be endothermic. Heat has a positive value.

13. If heat flows out of a system to the surroundings, the system loses heat, and the change is said to be exothermic. Heat has a negative value. Every reaction has an energy change associated with it. Exothermic reactions release energy, usually in the form of heat. Endothermic reactions absorb energy. Energy is stored in bonds between atoms.

14. The Law of Conservation of Energy states that in any chemical or physical process, energy is neither created nor destroyed. All the energy is accounted for as work, stored energy, or heat.

15. Calorimetry - the accurate and precise measurement of heat change for chemical and physical processes. For systems at constant pressure, the heat content is the same as a property called Enthalpy (H) of the system.

16. Changes in enthalpy =  H q =  H These terms will be used interchangeably. Thus, q =  H = m x C x  T  H is negative for an exothermic reaction.  H is positive for an endothermic reaction.

17. 17 C + O 2  CO 2 Energy ReactantsProducts  C + O 2 C O 2 395kJ + 395 kJ

18. THIS IS AN EXOTHERMIC REACTION. THE CHEMICAL BONDS OF THE PRODUCTS CONTAIN LESS CHEMICAL POTENTIAL ENERGY THAN THE BONDS OF THE REACTANTS. THE SYSTEM GIVES OFF ENERGY TO THE SURROUNDINGS.  H IS NEGATIVE. ANOTHER WAY OF SHOWING THIS IS THE ENERGY CHANGE IS SHOWN AS A PRODUCT.

19. 19 CaCO 3  CaO + CO 2 Energy ReactantsProducts  CaCO 3 CaO + CO 2 176 kJ CaCO 3 + 176 kJ  CaO + CO 2

20. THIS IS AN ENDOTHERMIC REACTIION. THE CHEMICAL BONDS IN THE PRODUCTS HAVE MORE CHEMICAL POTENTIAL ENERGY THAN THE CHEMICAL BONDS IN THE REACTANTS. THE SYSTEM GAINS ENERGY FROM THE SURROUNDINGS.  H IS POSITIVE.

21. Chemistry Happens in MOLES u An equation that includes energy is called a thermochemical equation u CH 4 + 2O 2  CO 2 + 2H 2 O + 802.2 kJ u 1 mole of CH 4 releases 802.2 kJ of energy. u When you make 802.2 kJ you also make 2 moles of water

22. 22 Thermochemical Equations u A heat of reaction is the heat change for the equation, exactly as written The physical state of reactants and products must also be given. Standard conditions for the reaction is 101.3 kPa (1 atm.) and 25 o C

23. CH 4 + 2 O 2  CO 2 + 2 H 2 O + 802.2 kJ If 10. 3 grams of CH 4 are burned completely, how much heat will be produced? [(10.3 g) / (16.05 g/mole)] x (802.2 kJ/mole) = 514 kJ

24. HEAT OF REACTION u The heat that is released or absorbed in a chemical reaction  Equivalent to  H u C + O 2 (g)  CO 2 (g) + 393.5 kJ  C + O 2 (g)  CO 2 (g)  H = -393.5 kJ u In a thermochemical equation, it is important to indicate the physical state a) H 2 (g) + 1/2O 2 (g)  H 2 O(g)  H = -241.8 kJ b) H 2 (g) + 1/2O 2 (g)  H 2 O(l)  H = -285.8 kJ

25. HEAT OF COMBUSTION u The heat from the reaction that completely burns 1 mole of a substance:  C + O 2 (g)  CO 2 (g) + 393.5 kJ  C + O 2 (g)  CO 2 (g)  H = -393.5 kJ

26. HEAT OF SOLUTION u Heat changes can also occur when a solute dissolves in a solvent. Molar Heat of Solution (  H soln. ) = heat change caused by dissolution of one mole of substance q = mol x  H soln. (no temperature change) u Sodium hydroxide provides a good example of an exothermic molar heat of solution (next slide)

27. HEAT OF SOLUTION NaOH (s)  Na 1+ (aq) + OH 1- (aq)  H soln. = - 445.1 kJ/mol u The heat is released as the ions separate (by dissolving) and interact with water, releasing 445.1 kJ of heat as  H soln. thus becoming so hot it steams! H2OH2O

28. SAMPLE PROBLEM HOW MUCH HEAT IS RELEASED WHEN 2.5 MOLES OF NaOH ARE DISSOLVED IN WATER? NaOH (s)  Na + (aq) + OH - (aq)  H SOL = -445.1 kJ/mol  H = -445.1 kJ/mol X 2.5 mol = -1112.75 kJ

29. HESS’S LAW u If you add two or more thermochemical equations to give a final equation, then you can also add the heats of reaction to give the final heat of reaction. Called Hess’s Law of Heat Summation

30. HOW DOES IT WORK? 1) If you turn an equation around, you change the sign: If H 2 (g) + 1/2 O 2 (g)  H 2 O(g)  H=-285.5 kJ then the reverse is: H 2 O(g)  H 2 (g) + 1/2 O 2 (g)  H =+285.5 kJ 2) If you multiply the equation by a number, you multiply the heat by that number: 2 H 2 O(g)  2 H 2 (g) + O 2 (g)  H =+571.0 kJ 3) Or, you can just leave the equation “as is”

31. STANDARD HEATS OF FORMATION  The  H for a reaction that produces 1 mol of a compound from its elements at standard conditions u Standard conditions: 25°C and 1 atm. u Symbol is u The standard heat of formation of an element = 0 u This includes the diatomics

32. WHAT GOOD ARE THEY? u Tables are available for standard heats of formation u The heat of a reaction can be calculated by: subtracting the heats of formation of the reactants from the products  H o = (  H f o Products) – (  H f o Reactants)

33. EXAMPLE CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O(g)  H f o CH 4 (g) = - 74.86 kJ/mol  H f o O 2 (g) = 0 kJ/mol  H f o CO 2 (g) = - 393.5 kJ/mol  H f o H 2 O(g) = - 241.8 kJ/mol   H= [-393.5 + 2(-241.8)] - [-74.68 +2 (0)]  H= - 802.4 kJ