2. Chemical reactions involves the breaking and the making of bonds. Energy is needed to break down a bond. Energy is released when a bond is formed. If more energy is released than is absorbed, the reaction will be Exothermic. If more energy is needed to break the bonds than is given when new bonds are formed, the reaction will be endothermic.

5. Energy level Reaction progress Activation energy Energy given out by reaction Using a catalyst might lower the activation energy

6. Enthalpy The energy contained in a chemical bond that can be converted into heat is known as enthalpy. Enthalpy is given the symbol H. Enthalpy can not be measured directly, but we can measure the enthalpy change in a reaction, ΔH

7. A process is endothermic when  H is positive.

8. A process is exothermic when  H is negative.

9. EXOTHERMIC – more energy is given out than is taken in (burning, respiration)

10. ENDOTHERMIC – energy is taken in but not necessarily given out. (photosynthesis)

11.  H changes sign when a process is reversed. Therefore, a cyclic process has the value  H = 0. Chapter 6: Thermochemistry11 Same magnitude; different signs.

12. Using  H 12 Values are measured experimentally Negative values indicate exothermic reactions Positive values indicate endothermic reactions Changes sign when a process is reversed. Therefore, a cyclic process has the value  H = 0 For problem-solving, one can view heat being absorbed in an endothermic reaction as being like a reactant and heat being evolved in an exothermic reaction as being like a product

13. We measure heat flow using calorimetry. A calorimeter is a device used to make this measurement. A “coffee cup” calorimeter may be used for measuring heat involving solutions. 13 A “bomb” calorimeter is used to find heat of combustion; the “bomb” contains oxygen and a sample of the material to be burned.

14. 14 Specific heat of a substance is the amount of heat required to raise the temperature of one gram by 1 o C or by 1 Kelvin. Specific heat = C = q/m  T units of C: J g –1 o C –1 or J g –1 K –1

15. Chapter 6: Thermochemistry15 EOS

16. 16 The standard enthalpy of reaction (  H o ) is the enthalpy change for a reaction in which the reactants in their standard states yield products in their standard states

17. The standard enthalpy of formation (  H o f ) of a substance is the enthalpy change that occurs in the formation of 1 mol of the substance from its elements when both products and reactants are in their standard states

18. Is the enthalpy change that occurs during the complete combustion of one mole of a substance. The enthalpy of combustion is defined in terms one mole of reactants, where the enthalpy of formation is defined in terms of one mole of products. the symbol  H o C is used to represent standard enthalpy change os combustion.

20. Chapter 6: Thermochemistry20 EOS

21. The heat of a reaction is constant, regardless of the number of steps in the process  H overall =  H’s of individual reactions When it is necessary to reverse a chemical equation, change the sign of  H for that reaction When multiplying equation coefficients, multiply values of  H for that reaction

22.   H is well known for many reactions, and it is inconvenient to measure  H for every reaction in which we are interested. However, we can estimate  H using  H values that are published and the properties of enthalpy.

24. The change in enthalpy,  H, is the enthalpy of the products minus the enthalpy of the reactants:  H = H products − H reactants

25. Hess’s law states that “If a reaction is carried out in a series of steps,  H for the overall reaction will be equal to the sum of the enthalpy changes for the individual steps.”

26. Imagine this as occurring in 3 steps: C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O (l) C 3 H 8 (g)  3 C (graphite) + 4 H 2 (g) 3 C (graphite) + 3 O 2 (g)  3 CO 2 (g) 4 H 2 (g) + 2 O 2 (g)  4 H 2 O (l)

27. C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O (l) C 3 H 8 (g)  3 C (graphite) + 4 H 2 (g) 3 C (graphite) + 3 O 2 (g)  3 CO 2 (g) 4 H 2 (g) + 2 O 2 (g)  4 H 2 O (l) C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O (l) The sum of these equations is:

28. We can use Hess’s law in this way:  H =  n  H f(products) -  m  H f(reactants) where n and m are the stoichiometric coefficients. 

29. C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O (l)  H= [3(-393.5 kJ) + 4(-285.8 kJ)] - [1(-103.85 kJ) + 5(0 kJ)] = [(-1180.5 kJ) + (-1143.2 kJ)] - [(-103.85 kJ) + (0 kJ)] = (-2323.7 kJ) - (-103.85 kJ) = -2219.9 kJ