Thermodynamics The Most Important Section

1. ThermodynamicsThe Most Important Section

2. Graph of Energy Added vs Time http://library.thinkquest.org/C006669/media/Chem/img/Graphs/HeatCool.gif

3. Explanation • On the slanted sections, the temperature and KE are increasing. • On the horizontal sections, the temperature is constant and the PE is increasing. • Molecules of a solid move in vibrational motion • Molecules of a liquid move in vibrational and rotational motion • Molecules of a gas move in vibrational, rotational, and translational motion

4. Graph of Temperature vs Pressure http://library.thinkquest.org/C006669/media/Chem/img/Graphs/Phase.gif

5. Phase Diagram • In this graph the line between solid and liquid is slanted towards the right. This is the typical trend, but water is different and slants to the left because water is more dense than ice. • All of the lines intersect at the triple point. • The end of the liquid/gas line is the critical point. Beyond the end of the critical point is an area in which the substance is considered a superficial fluid

6. Endothermic vs Exothermic • Endothermic Reaction • Energy is absorbed from the surroundings into the system • ΔH is positive • ΔH is considered a reactant • Exothermic Reaction • Energy is released from the system into the surroundings • ΔH is negative • ΔH is considered a product

7. Entropy • Entropy is randomness or disorder in a system • In nature substances are more likely to gain entropy and become more disordered • In general gasses have greater entropy than liquids which have greater entropy than solids, but they overlap • Substances with more molecules and atoms have higher entropies

8. Clausius-Clapeyron Equation • The Clausius-Clapeyron equation is used to find the pressure of a substance at a certain temperature or temperature at a certain pressure. • To find it, you must know the ΔH, the original pressure and temperature, and either the final pressure or final temperature

9. ΔH, ΔS,ΔG, and Spontaneity ΔH Change in Enthalpy kJ/mol ΔS Change in Entropy J/mol*K ΔG Change in Gibbs Free Energy

10. Finding ∆H Using Bond Energies • C2H5OH+3O2 -->3H2O+2CO2 • +[1(C-C)+5(C-H)+1(C-O)+1(O-H)+3(O=O)]-[6(O-H)+4(C=O)] Bond Energies • +[1(348)+5(413)+1(358)+1(463)+3(498)]-[6(463)+4(799)] • ∆H=+4728kJ-5974kJ=-1246kJ • The reaction is exothermic

11. Finding ∆H Given Standard Enthalpies of Formation • C2H5OH+3O2 -->3H2O+2CO2 • ∆Hrxn=∑∆Hp-∑∆Hr • [2(-393.5kJ/mol)+3(241.8kJ/mol)]-[1(-277.7kJ/mol)+3(0kJ/mol)] • =-1234.76 kJ

12. Finding ∆H Using Hess’s Law • 2C2H6+7O26H2O(g)+4CO2 • 2(C2H6 2C + 6H) -2(∆H)=-2(-84.7kJ/mol) • 4(C + O2CO2) 4(∆H)=4(-393.5kJ/mol) • 6(H2 + .5O2 H2O(g)) 6(∆H)=6(-242kJ/mol) +------------------------------------------- 2C2H6+4C+4O2+6H2+3O24C+12H+4CO2+6H2O(g) 2C2H6+7O24CO2+6H2O(g) ∆H=-2856.6 kJ/mol

13. Calorimetry • Determine the change in temperature of 2.00kg of water when 2.00g of C2H6 is burned and all heat is transferred. • 2C2H6+7O24CO2+6H2O • q=mcΔT=nΔH m=2.00kg • c=-4.184J/gK∆H=-1428300 J/mol n=2.00g/(30.264g/mol)=.001487mol ΔT=(nΔH)/mc=11.3K