1. ENTHALPY, HESS’ LAW, AND THERMOCHEMICAL EQUATIONS

2. Quick Review of Concepts  We have been introduced to heat producing (exothermic) reactions and heat using (endothermic) reactions  Heat is a measure of the transfer of energy from a system to the surroundings and from the surroundings to a system

3. Enthalpy  The heat given off or absorbed during a reaction (change in heat of a system) is called the change in enthalpy ( Δ H) when the pressure of the system in kept constant  H = H (products) – H (reactants)

4. H products < H reactants  H < 0 H products > H reactants  H > 0 6.3 More on Enthalpy

5. Calorimetry  We measure the transfer of heat (at a constant pressure) of a chemical reaction by a technique called calorimetry  In calorimetry:  The heat released by the system is equal to the heat absorbed by its surroundings  The heat absorbed by the system is equal to the heat released by its surroundings  The total heat of the system and the surroundings remains constant  First Law of Thermodynamics

6. Calorimetry  We use an insulated device called a calorimeter to measure this heat transfer  A typical device is a “coffee cup calorimeter”  Reaction is open to the atmosphere Therefore, constant pressure  q = ∆H at constant pressure

7. Calorimetry  To measure Δ H for a reaction using a coffee-cup calorimeter: 1. Dissolve the reacting chemicals in known volumes of water 2. Measure the initial temperatures of the solution 3. Mix the solutions 4. Measure the final temperature of the mixed solution

8. Calorimetry  The heat generated by the reactants is absorbed by the water  We know the mass of the water, m water  We know the change in temperature, ∆T water  We also know that water has a specific heat of C water = 4.18 J/°C-g.  We can calculate the heat of reaction by: q sys = ∆H = − q surr q rxn = ∆H = − q water q rxn = ∆H = -(m water × C water × ∆T water )

9. Practice Problem When 25.0 mL of water containing 0.025 mol of HCl at 25.0°C is added to 25.0 mL of water containing 0.025 mol of NaOH at 25.0°C in a coffee cup calorimeter, a reaction occurs. Calculate ∆H (in kJ) of one mole of water during this reaction if the highest temperature observed is 32.0°C. Note: Assume the densities of the solution are 1.00 g/mL Assume all heat was absorbed by water

10. Summary  Heat is a measure of the transfer of energy from a system to the surroundings and from the surroundings to a system  The change in heat of a system is called the change in enthalpy ( Δ H) when the pressure of the system in kept constant  We measure the transfer of heat (at a constant pressure) of a chemical reaction by a technique called calorimetry  We use an insulated device called a calorimeter to measure this heat transfer

11. H 2 O (s) H 2 O (l)  H = + 6.01 kJ Is  H negative or positive? System absorbs heat Endothermic  H > 0 6.3 Thermochemical Equations

12. Is  H negative or positive? System gives off heat Exothermic  H < 0 Thermochemical Equations CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (l)  H = -890.4 kJ

13. Thermochemical Stoichiometry  Enthalpy is STOICHIOMETRIC!  Recall that the stoichiometric coefficients always refer to the number of moles of a substance  You would read the above equation as: “6.01 kJ are absorbed for every 1 mole of ice that melts at 0 0 C and 1 atm” “890.4 kJ are released for every 1 mole of methane that is combusted at 25 0 C and 1 atm” H 2 O (s) H 2 O (l)  H = 6.01 kJ CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (l)  H = -890.4 kJ

14. 6.3 How much heat is evolved when 266 g of white phosphorus (P 4 ) burn in air? P 4 (s) + 5O 2 (g) P 4 O 10 (s)  H = -3013 kJ 266 g P 4 1 mol P 4 123.9 g P 4 x 3013 kJ 1 mol P 4 x = 6470 kJ Practice Problem

15. Let’s Try Some More!  Page 165

16. If you reverse a reaction, the sign of  H changes H 2 O (l) H 2 O (s)  H = - 6.01 kJ If you multiply both sides of the equation by a factor n, then  H must change by the same factor n 2H 2 O (s) 2H 2 O (l)  H = 2 x 6.01 = 12.0 kJ Manipulating ∆H