1. Aka the Law of conservation of energy, Gibbs in 1873 stated energy cannot be created or destroyed, only transferred by any process The net change in energy is equal to the heat that flows across a boundary minus the work done BY the system  U = q + w –Where q is heat and w is work –Some heat flowing into a system is converted to work and therefore does not augment the internal energy 1 st Law of Thermodynamics

2. Directionality from the 2 nd Law For any spontaneous irreversible process, entropy is always increasing How can a reaction ever proceed if order increases?? Why are minerals in the earth not falling apart as we speak??

3. NEED FOR THE SECOND LAW The First Law of Thermodynamics tells us that during any process, energy must be conserved. However, the First Law tells us nothing about in which direction a process will proceed spontaneously. It would not contradict the First Law if a book suddenly jumped off the table and maintained itself at some height above the table. It would not contradict the First Law if all the oxygen molecules in the air in this room suddenly entered a gas cylinder and stayed there while the valve was open.

4. MEANING OF ENTROPY AND THE SECOND LAW Entropy is a measure of the disorder (randomness) of a system. The higher the entropy of the system, the more disordered it is. The second law states that the universe always becomes more disordered in any real process. The entropy (order) of a system can decrease, but in order for this to happen, the entropy (disorder) of the surroundings must increase to a greater extent, so that the total entropy of the universe always increases.

5. 3 rd Law of Thermodynamics The heat capacities of pure crystalline substances become zero at absolute zero Because dq = CdT and dS = dq / T We can therefore determine entropies of formation from the heat capacities (which are measureable) at very low temps

6. ‘Free’ Energy Still need a function that describes reaction which occurs at constant T, P G = U + PV – TS = H – TS (dH = dU + PdV + VdP) The total differential is: dG = dU + PdV + VdP – TdS – SdT G is therefore the energy that can run a process at constant P, T (though it can be affected by changing P and T) Reactions that have potential energy in a system independent of T, P  aqueous species, minerals, gases that can react…

7. Can start to evaluate G by defining total differential as a function of P and T dG = dU + PdV + VdP – TdS – SdT Besides knowing volume changes, need to figure out how S changes with T For internal energy of a thing: dU = dq tot – PdV; determining this at constant volume  dU = C V dT where C V is the heat required to raise T by 1°C

8. 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…

9. 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

10. 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)

11. Does water behave like this? Water exists as liquid, solids, gas, and supercritical fluid (boundary between gas and liquid disappears – where this happens is the critical point) C p is a complex function of T and P (H-bond affinities), does not ascribe to Maier- Kelley forms…

12. 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

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

15. 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

16. 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).

17. 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.