1. Thermodynamics Chander Gupta and Matt Hagopian

2. Introduction into Thermo Thermodynamics is the study of energy and its transformations Thermochemistry deals with the transformations of energy (usually heat) during chemical reactions

3. Types of Energy Kinetic Energy- energy associated with the motion of an object KE=1/2mv 2 Potential Energy- energy an object possesses relative to others Energy is measured in Joules (J) or calories (cal) 1 cal = the energy required to increase the temperature of 1 gram of water by 1°C = 4.184 J 1 J = 1 (kg x m 2 )/s 2 Potential Energy is converted into Kinetic Energy

4. System and Surroundings Everything that is singled out in a reaction is called the system (reactants and products) Everything outside of this system is the surroundings Closed systems are the ones that are most readily studied - they can exchange energy but not matter with surroundings

5. Transferring Energy: Work and Heat Work is the energy used to move an object against force (w= F x d) Energy can be transferred back and forth between a system and its surroundings in the form of work and heat The measurement of work is usually Joules (J)

6. The First Law of Thermodynamics Energy is conserved - it cannot be created or destroyed Any energy lost by the system must be gained by the surroundings Internal energy- the sum of all kinetic and potential energy of a system ∆E = E final - E initial ∆E+ = energy gained from surroundings ∆E- = energy lost to surroundings ∆E = q + w (q = sum of heat transferred in/out of system, and w = work done on or by the system) Endothermic processes absorbs heat from the surroundings Exothermic processes release heat to the surroundings

7. Enthalpy (∆H) ∆H = the heat flow = enthalpy ∆H rxn = H products - H reactants ∆H- = exothermic reaction ∆H+ = endothermic reaction Ex. 2 H 2 (g) + O 2 --> 2H 2 O (g) ∆H = -483.6 kJ The enthalpy of the products is lower than the enthalpy of the reactants, therefore ∆H is negative

8. Properties of Enthalpy Enthalpy is an extensive property: the magnitude of ∆H is proportional to the amount of reactant used in the reaction Enthalpy change for a rxn is equal in magnitude, but opposite sign, for the reverse rxn Enthalpy change depends on the state of the reactants and products - 2H 2 O (l) --> 2H 2 O (g) ∆H=+88 kJ

9. Calorimetry The measurement of heat flow Calorimeter is a device used to measure heat flow The amount of heat a substance gains differs from substance to substance Heat capacity (C) = the amount of heat required to raise the temperature of a substance 1 K (or 1°C) The greater the heat capacity, the greater the amount of heat required to increase the temp.

10. Calorimetry Molar Heat Capacity = C molar = heat capacity of one mole of substance Specific heat= heat capacity of one gram of substance q= m x C x ∆T q= the change in heat of the substance m=the mass C=specific heat (J/g-K or J/g-°C) ∆T = change in temperature (T final - T initial )

11. Calorimetry ∆H = q p, if carried out under constant pressure We assume the calorimeter does not lose heat and doesn’t absorb it - the heat is fully confined in the system In an exothermic/endothermic reaction, the heat lost/gained by the system is gained/lost by the surroundings, so q soln =-q rxn Combustions take place in bomb calorimeters- heat from the combustion is measured by the ∆T if the surrounding water We must know the heat capacity of the calorimeter q rxn = -C cal x ∆T

12. Hess’s Law If a reaction is carried out in a series of steps, ∆H for the overall reaction will equal the sum of the enthalpy changes for the individual steps Example: CH 4(g) + 2O 2(g) --> CO 2(g) + 2 H 2 O (g) ∆H =-802kJ 2H 2 O (g) --> 2H 2 O (l) ∆H = -88kJ CH 4(g) + 2O 2(g) + 2H 2 O (g) --> CO 2(g) + 2 H 2 O (g) + 2H 2 O (l) ∆H = -890kJ Net equation : CH 4(g) + 2O 2(g) --> CO 2(g) + 2 H 2 O (l) ∆H=-890kJ

13. Enthalpy of Formation ∆H f = the enthalpy change for the reaction in which a substance is formed from its elements ∆H° = standard enthalpy change that takes place when all reactants and products are at 1 atm pressure and usually 298 K These combined to form standard enthalpy of formation ∆H° f, which forms one mole of substance from its elements in their most stable form under the standard conditions For any element in its most stable state, ∆H° f = 0. C, H 2, O 2, etc. ∆H° rxn = ∑n ∆H° f (products) - ∑m ∆H° f (reactants)

14. Covalent Bond Energies Strength of a covalent bond is measured by its bond enthalpy- molar enthalpy change upon breaking a bond Strength of covalent bonds increase with number of shared electron pairs (C=C > C-C). Bond length decreases with increasing number of bonds Breaking bonds is an endothermic process Forming bonds in an exothermic process Example: CH 4 + 2 O 2 --> CO 2 + H 2 O [4(C-H) + 2 (O=O)] - [2(C=O) + 2(O-H)] Bond energies are in Appendix 18

15. Changes of State As a substance changes phases, the molecules increase in vibrational frequency Fusion- melting -∆H fus - kJ/mol Vapor pressure increases with increasing temp. Boiling- ∆H vap ∆H vap > ∆H fus

16. Heating Curves Phase changes are endothermic processes The rise in the graph represents the rise in temperature The plateaus represent the phase changes First plateau- melting Second plateau- boiling Energy required for total curve is the sum of the individual parts (q 1 + q 2 + q 3 + q 4 + q 5 ) Plateau points- q=m∆H fus or q=m∆H vap Source: www.bbc.co.uk

17. Entropy (S) Molecules undergo 3 types of motion: Translational - entire molecule moves Vibrational - atoms of molecule move back and forth Rotational- entire molecule spins Entropy increases as these motions increase, and also with increases in volume and temperature Gases are the most random, and therefore have ∆S +

18. Entropy 2nd law of thermodynamics- ∆ S univ = ∆S sys + ∆S surr 3rd law of thermodynamics- entropy of a pure crystalline solid at 0 K = 0. There is no movement at absolute zero Standard molar entropy = S° ∆S+ means increasing randomness ∆S° = ∑n ∆S° f (products) - ∑m ∆S° f (reactants) ∆S is measured in J/K

19. Sample Problem Calculate ∆S° for: N 2 (g) + 3 H 2 (g) --> 2 NH 3 (g) ∆S° N 2 = 191.5 J/mol-K ∆S° H 2 = 130.6 J/mol-K ∆S° NH 3 = 192.5 J/mol-K

20. Answer N 2 (g) + 3 H 2 (g) --> 2 NH 3 (g) [2 mol (192.5 J/mol-K)] - [1 mol (191.5 J/mol-K) + (3 mol) (130.6 J/mol-K) ∆S° = -198.3 J/K

21. Gibbs Free Energy ∆G = ∆H - T ∆S ∆G- means spontaneous reaction At equilibrium the process is reversible and ∆G = 0 ∆G° f is analogous to ∆H° f. Equals 0 for substances in elemental state

22. Gibbs Free Energy Under nonstandard conditions, ∆G is related to ∆G° and the value of the reaction quotient, Q. ∆G = ∆G° + RT ln Q At equilibrium, (∆G=0 Q=K), ∆G° = -RT ln K Q = [products] x / [reactants] y ∆G = kJ/mol ∆G° = ∑n ∆G° f (products) - ∑m ∆G° f (reactants) K 1, products favored (highly spontaneous). K=1, rxn is in equilibrium