1. Increasing energy with temp? The added energy in a substance that occurs as temperature increases is stored in modes of motion in the substance For any molecule – modes are vibration, translation, and rotation –Solid  bond vibrations –Gases  translation –Liquid water – complex function…

2. Heat Capacity When heat is added to a phase it’s temperature increases (No, really…) Not all materials behave the same though! dq=C V dT  where C V is a constant (heat capacity for a particular material) Or at constant P: dq=C p dT Recall that dq p =dH then: dH=C p dT Relationship between C V and C p : Where a and b are coefficients of isobaric thermal expansion and isothermal compression, respectively

3. Enthalpy at different temps… HOWEVER  C isn’t really constant…. C also varies with temperature, so to really describe enthalpy of formation at any temperature, we need to define C as a function of temperature Maier-Kelley empirical determination: C p =a+(bx10 -3 )T+(cx10 -6 )T 2 –Where this is a fit to experimental data and a, b, and c are from the fit line (non-linear)

4. Heats of Formation,  H f Enthalpies, H, are found by calorimetry Enthalpies of formation are heats associated with formation of any molecule/mineral from it’s constituent elements

5. Calorimetry Measurement of heat flow (through temperature) associated with a reaction Because dH = q / dT, measuring Temperature change at constant P yields enthalpy

7. Problem When 50.mL of 1.0M HCl and 50.mL of 1.0M NaOH are mixed in a calorimeter, the temperature of the resultant solution increases from 21.0 o C to 27.5 o C. Calculate the enthalpy change per mole of HCl for the reaction carried out at constant pressure, assuming that the calorimeter absorbs only a negligible quantity of heat, the total volume of the solution is 100. mL, the density of the solution is 1.0g/mL and its specific heat is 4.18 J/g-K. q rxn = - (c s solution J/g-K) (mass of solution g) (  T K) = - (4.18 J/g-K) [(1.0g/mL)(100 mL)] (6.5 K) = - 2700 J or 2.7 kJ  H = 2.7 kJ Enthalpy change per mole of HCl = (-2.7 kJ)/(0.050 mol) = - 54 kJ/mol

8. Hess’s Law Known values of  H for reactions can be used to determine  H’s for other reactions.  H is a state function, and hence depends only on the amount of matter undergoing a change and on the initial state of the reactants and final state of the products. If a reaction can be carried out in a single step or multiple steps, the  H of the reaction will be the same regardless of the details of the process (single vs multi- step).

9. CH 4 (g) + O 2 (g) --> CO 2 (g) + 2H 2 O(l)  H = -890 kJ If the same reaction was carried out in two steps: CH 4 (g) + O 2 (g) --> CO 2 (g) + 2H 2 O(g)  H = -802 kJ 2H 2 O(g) --> 2H 2 O(l)  H = -88 kJ CH 4 (g) + O 2 (g) --> CO 2 (g) + 2H 2 O(l)  H = -890 kJ Net equation Hess’s law : if a reaction is carried out in a series of steps,  H for the reaction will be equal to the sum of the enthalpy change for the individual steps.