1. Chapter 5 – Thermochemistry

2. 2 Energy - capacity to do work. Heat is the transfer of thermal energy between two bodies that are at different temperatures. Temperature is a measure of thermal energy. Law of conservation of energy – energy can be converted from one form to another but can be neither created nor destroyed. Thermochemistry is the study of heat change in chemical reactions. Thermochemistry

3. 3 Potential energy – energy due to position or composition. Kinetic energy – energy due to motion of the object and depends on the mass of the object and its velocity. Kinetic Energy (KE) = ½ mv 2 Note: 1 L∙atm = 101.3 J 1 kgm 2 /s 2 = 1 J Ex: An object with mass of 125 g travels at 75 m/s. Calculate the kinetic energy in Joules. Energy

4. 4 System – part of the universe on which we wish to focus attention. Surroundings – include everything else in the universe. Work – force acting over a distance. Energy is a state function; ( a property that does not depend on the system’s past) (only depends on present state). Energy

5. 5 Endothermic Reaction: Absorbs energy from the surroundings.(Heat flows into the system from surroundings) Exothermic Reaction: Releases energy from the system to the surroundings. Energy gained by the surroundings = energy lost by the system. Heat flows from warmer objects to cooler objects. Energy

6. 6 ΔE = q + w ΔE is the change in internal energy of a system. q is the heat exchange between the system and the surroundings. W is the work done on (or by) the system. W= -P ΔV when a gas expands against a constant external pressure Enthalpy and the First Law of Thermodynamics

7. 7 Sign reflects the system’s point of view. Work done by the system on the surroundings: w is negative (-) Work done on the system by the surroundings: w is positive (+) Sign conventions for work

8. Enthalpy Practice Calculate the total energy of the system that absorbs 454 kJ heat and releases 565 kJ of work.

9. Enthalpy Practice Calculate the ΔE for a system undergoing an endothermic process in which 15.6 kJ of heat flows and where 1.4 kJ of work is done on the system.

10. Q = ΔH = m x ΔT x Cp Now that you know a little bit about heat (Q), let’s talk about how it applies to chemical reactions. This is true as long as pressure does NOT change (the pressure does not change in our lab, so lets start using the word enthalpy change ) ΔHΔHQ enthalpy changeheat =

11. Enthalpy Practice A rare stone has a specific heat of 2.45 cal/g˚C. The mass of the stone is 4390 g and the amount of thermal energy the stone absorbed is 5403 calories. What is the change in temperature?

12. Calorimeter: An insulated device used to measure heat changes Calorimeter Bomb Calorimeter

13. Notation H 2 O (s) H 2 O (l) H 2 O (g) Do you think it matters which state a chemical is in when we are talking about heat (enthalpy)?

14. Notation CaO(s) + H 2 O(l) → Ca(OH) 2 (s) + 65.2 kJ productsreactants “yields” “produces” “equals” From now on: the CHEMICAL REACTION is our SYSTEM

15. 2NaHCO 3 (s) + 129 kJ →Na 2 CO 3 (s) + H 2 O(g) + CO 2 (g) Do you think this is an endothermic or exothermic reaction? CaO(s) + H 2 O(l) → Ca(OH) 2 (s) + 65.2 kJ Do you think this is an endothermic or exothermic reaction?

16. ΔH = -65.2 kJ or Notation CaO(s) + H 2 O(l) → Ca(OH) 2 (s) + 65.2 kJ CaO(s) + H 2 O(l) → Ca(OH) 2 (s) 2NaHCO 3 (s) + 129 kJ →Na 2 CO 3 (s) + H 2 O(g) + CO 2 (g) or 2NaHCO 3 (s) → Na 2 CO 3 (s) + H 2 O(g) + CO 2 (g)ΔH = 129 kJ

17. ΔH = -65.2 kJ ΔH = 129 kJ Notation -ΔH = exothermic reaction(rxn) ΔH = endothermic reaction(rxn) Exothermic Endothermic

18. Hess’s Law Hess’s law: 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. Enthalpy is a state function, where the path is independent of the history of the system.

19. diamond graphite Both are made of pure carbon (C) Diamond will eventually break down into graphite C (diamond) → C (graphite) =

20. So for reactions that: -take a long time -are expensive/rare -are explosive -toxic We can use two or more equations that we already know to give us the final equation. Reactions Continued

21. Characteristics of Enthalpy Changes If a reaction is reversed, the sign of ΔH is also reversed. Multiply or divide by a coefficient Reduce or cancel out the common reactant or product at opposite sides of the reaction.

22. Example Consider the following data: Calculate ΔH for the reaction 22

23. Problem-Solving Strategy Work backward from the required reaction, using the reactants and products to decide how to manipulate the other given reactions at your disposal. Reverse any reactions as needed to give the required reactants and products. Multiply reactions to give the correct numbers of reactants and products. 23

24. Example Reverse the two reactions: Desired reaction: 24

25. 1)C (s, graphite) + O 2 (g) → CO 2 (g)ΔH = -393.5 kJ 2)C (s, diamond) + O 2 (g) → CO 2 (g) ΔH = -395.4 kJ Hess’s Law: The enthalpy change of a reaction is the same whether it occurs by one step or a series of steps We don’t have millions of years to wait, but it is possible to burn these items to lead to the production of carbon dioxide

26. Example Multiply reactions to give the correct numbers of reactants and products: 4[ ] 4( ) 3[ ] 3( ) Desired reaction:

27. Example Final reactions: Desired reaction: ΔH = +1268 kJ 27

28. Calculation of  H 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. Ex: Calculate  H for the following reaction: C 6 H 12 O 6 (s) + 6O 2 (g) → 6 CO 2 (g) + 6H 2 O(l) Given the following information:  H f ° (kJ/mol) H 2 O(l) –285.83 C 6 H 12 O 6 (s) –2812 CO 2 (g) - 393.5 29 Standard Enthalpies of Formation

30. Consider the combustion of propane: C 3 H 8 (g) + 5O 2 (g) → 3CO 2 (g) + 4H 2 O(l) ΔH = –2221 kJ Assume that all of the heat comes from the combustion of propane. Calculate ΔH in which 6.50 g of propane is burned in excess oxygen at constant pressure.