1. Plan for Mon, 20 Oct 08 Lecture –Constant P and Constant V Calorimetry (6.2) –Characteristics of enthalpy changes and Hess’s Law (6.3) Q3, Ex1, Exp1 lab report returned Don’t forget there is a quiz on Wednesday!

2. Heat Flow System: 1 L of water T = 50 o C Surroundings: T = 25 o C 1 L of water T = 25 o C How much heat did the water lose? Well, first we need to know how much heat it had. The liter of water is allowed to cool, until it reaches room temperature.

3. System: 1 L of water T = 50 o C How much heat did the water lose? The liter of water is allowed to cool, until it reaches room temperature, 25 o C.

4. NaOH + HCl  H 2 O + NaCl;  H = -58 kJ The heat evolved in this reaction is trapped in the water…this heat increases the average E k of the water molecules, leading to an increase in the temperature of the water. Formally, q is the amount of heat that must be exchanged with the surroundings to return the system to its original temperature. In calorimetry we don’t let it escape… q goes into raising the temperature of the water. q system = - q surroundings Constant P Calorimetry

5. Example 5.00 g of ammonium nitrate is dissolved in 500. mL of water at 25.00 o C. What is the final temperature of the water? NH 4 NO 3 (s)  NH 4 + (aq) + NO 3 - (aq)  H soln = +82.93 kJ/mol Think of the reaction as the system and the water as the surroundings… Then the heat gained by the reaction is exactly equal to that lost by the water:

6. Constant V Calorimetry For reactions involving gases, such as combustion. The heat produced in the reaction is transferred to the steel bomb, and then to the surrounding water bath.

7. Constant V Calorimetry The “calorimeter constant” C cal must be obtained experimentally, using a reference compound for which the  E of combustion is well-known. q rxn = -C cal  T -q rxn = q bomb + q H2O + q O2 + q products + … -q rxn = q calorimeter

8. Example: Combustion of Sucrose A 1.010-g sample of sucrose (C 12 H 22 O 11 ) is combusted in a bomb calorimeter with C cal = 4.90 kJ/ o C. The initial temperature was 24.92 o C, and the final temperature was 28.33 o C. How much energy was evolved as heat during this reaction? How much energy is evolved per mole of sucrose combusted?

9. Enthalpy Changes Enthalpy is a measure of the potential energy stored in a chemical system. A given molecule will always have the same kind of bonds in it, no matter where or when or how it was made. This means a given molecule at a given temperature and physical state will always have the same enthalpy content, no matter where or when or how it was made. Therefore, enthalpy changes associated with chemical or physical processes are state functions… They depend only on the initial and final states of the system…not on the steps taken to get from the initial state to the final state. Methane, CH 4, always has four C-H bonds. Elemental oxygen, O 2, always consists of two doubly bonded O atoms. 2 + 2 +

10. + Consider the synthesis of nitrogen dioxide, NO 2, from its elements: N 2 and O 2.

11. Hess’s Law In going from a particular set of reactants to a particular set of products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps. This means we can use enthalpy changes for reactions that we can easily measure to determine the enthalpy changes for reactions that are not so easy to measure. Two characteristics of enthalpy will be useful in this endeavor: Enthalpy is a state function. If a reaction is reversed, the value of  H is also reversed. Enthalpy is an extensive property. If the coefficients in a balanced equation are multiplied by an integer,  H must be multiplied by the same integer.

12. NOTE: The value of  H included with balanced chemical equations is for the reaction as it is written. The processes below represent an enthalpy change only for the molar amounts present. “Heat of Fusion” - enthalpy change associated with melting H 2 O(s)  H 2 O(l)  H fus = +0.334 kJ/mol “Heat of Vaporization” - enthalpy change associated with boiling H 2 O(l)  H 2 O(g)  H vap = +2.26 kJ/mol “Heat of Reaction” - enthalpy change associated with a chemical reaction CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(g)  H rxn = -50.1 kJ/mol “Heat of Solution” – enthalpy change associated with the dissolution of ionic solids in water NH 4 NO 3 (s)  NH 4 + (aq) + NO 3 - (aq)  H soln = +82.93 kJ/mol CaCl 2 (s)  Ca 2+ (aq) + 2Cl - (aq)  H soln = -26.2 kJ/mol

13. What is the enthalpy change associated with changing graphite into diamond? C graphite  C diamond  H = ??

14. Using Enthalpy We can use the  H for various reactions to determine  H for a composite reaction. Example: C(s, diamond) + O 2 (g) CO 2 (g);  H = -396 kJ/mol C(s, graphite) + O 2 (g) CO 2 (g);  H = -394 kJ/mol

15. Using  H (cont.) C(s, graphite) + O 2 (g) CO 2 (g)  H = -394 kJ/mol CO 2 (g) C(s, diamond) + O 2 (g)  H = +396 kJ/mol C(s, graphite) C(s, diamond)  H = +2 kJ  H rxn > 0…..rxn is endothermic + C(s, diamond) C(s, graphite)  H = -2 kJ  H rxn < 0…..rxn is exothermic

16. Example Use the information below to determine  H for the reaction: 3 C(graphite) + 4 H 2 (g)  C 3 H 8 (g) C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4H 2 O(l)  H = -2219.9 kJ C (graphite) + O 2 (g)  CO 2 (g)  H = -393.5 kJ H 2 (g) + 1/2 O 2 (g)  H 2 O(l)  H = -285.8 kJ ANS:  H = -104 kJ

17. Hess Hints Using Hess’s Law involves some degree of trial and error when you are manipulating the given chemical equations. Some tips for success: –Work backward from the required reaction, using the reactants and products to guide you in manipulating the other given reactions. –Reverse any reactions as needed to give the required reactants and products in your final equation. Don’t forget to reverse the sign on  H too also!! –Multiply reactions to give the correct number of reactants and products in your final equation. Don’t forget to multiply the  H though, also!!