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THERMODYNAMICS Moses Kool and Miguel Grizzly Bear CONSERVATION LAWS Energy and mass are interchangable Slopes: q=ms∆T or q=mC∆T q=quantity of energy m=mass of material C=s=constant=specific heat capacity ∆T=change in temperature Slopes are kinetic energy and plateaus are potential energy THERMO LAWS 1st law= energy cannot be created nor destroyed 2nd law= Universal randomness is always on the rise. Irreversible processes result in increase in randomness. BRIEF INTRO TO HEATING CURVES Plateaus: q=m∆H q=quantity of potential energy ∆H= enthalpy of fusion or vaporization Enthalpies of Water Szzzzzzzzz ∆Hf= +6.02 kJ/mol ∆Hv= +40.7 kJ/mol INTERMOLECULAR BONDS -When energy is applied and physical states are altered, it is the intermolecular bonds that are being broken -Breaking requires input. Bonds that form release energy http://employees.csbsju.edu/hjakubowski/classes/ch111/olsg-ch111/statesmat/hbonds.gif ∆E WITH WORK AND HEAT ∆E = q + w q is energy liberated or added to the system while w is the work done on or by the system The change in energy depends on the amount of heat added or released from the system and the amount of work done on or from the system When heat is added or work is done to the system the energy increases Heat is hot!!!!! A -25oC 5g block of ice is heated to the point where it is 150oC. What is the quantity of the energy involved in this process? H2O: Csolid=2.06 J/goC Cliquid= 4.18 J/goC ΔHfusion= 6.02 kJ/mol Melting Point=0oC ΔHvap= 40.7 kJ/mol Boiling Point= 100oC (5g)(2.06 J/goC )(25oC)=257.5 J= .2575 kJ (.277 mols)(6.02 kJ/mol)=1.667 kJ (5g)(4.18 J/goC)(100oC)= 2090 J= 2.09 kJ (.227 mols)(40.7 kJ/mol)= 9.2389 kJ (5g)(2.03 J/goC)(50oC)= 507.5 J= .5075kJ ΔH=13.76 kJ Cgas= 2.03 J/goC CALCULATING ∆H AND ∆S IN REACTIONS For a particular equation: CH4 + O2 CO2 + 2H2O the ∆Hreaction is ∆Hreaction=∑ ∆Hproduct - ∑ ∆Hreactant ∆Hreaction= [1 mol(-395.5 kJ/mol) + 2mol (-285.83)] –[1mol(-74.8) + 2 mol (0)]= -890.36 kJ ∆S IS ENTROPY ∆Sreaction=∑ ∆Sproduct - ∑ ∆Sreactant ∆S = [1mol(213.6 J/K) + 2mol(69.91)] – [1mol(186.3) + 2 mol(205)] = -242.85 J/K Any change in energy increases the amount of energy in the universe SPONTANEOUS Spontaneous = natural, happens normally in nature + ∆H -∆H + ∆S Entropy driven. Can be spontaneous at high temp. Always spontaneous - ∆S Never spontaneous Enthalpy Driven. Can be spontaneous at a low temp. GIBBS FREE ENERGY ∆G= ∆H - T∆S ∆H = enthalpy ∆S = entropy ∆G = free energy ∆G= ∆Gº + RTlnQ Q = rxn quotient From that we derive ∆Gº = -RT ln K Equilibrium Productfavored Reactantfavored ∆G=0 Large - ∆G Large +∆G k= 1 k>> 1 k<< 1 MORE GIBBS • 0 = ∆Gº + RT lnKa • Q = e- ∆G/RT • R is a constant •T is the absolute temp •Ka is the equilibrium constant •Which is the ratio of the concentration of the products raised to their respective powers all divided by the concentrations of the reactants raised to their respective powers We’re All Free!!!! **Calculate ΔH, ΔS, and ΔG at 298K for the reaction: 2PCl3 + O2 2POCl3 ΔH= 2(-542.2) - 2(-288.07)= -508.26 kJ/mol ΔG= 2(-502.5) – 2(-269.6)= -465.8 ΔS= 2(325) – [2(311.7) + 205]= -178.4 J/mol K **At equilibrium, what will be the temperature of the system? ΔG= ΔH - T ΔS 0= ΔH - T ΔS T ΔS= ΔH T= (-508.26 kJ/mol)/(-.1784 J/ mol K) T= 2848.99 K **What is the K for this reaction occurring at 2009 K? ΔG2009= -508.26 kJ/mol – 2009(-.1784 J/mol K) -149.85 = (-8.314 J/ mol K)(2009) lnK .00897 = lnK e.00897= K K=1.00 ΔG2009= -149.85 HESS’ LAW In Hess’ law reactions that are done in one or more steps can have the enthalpies changes added up from each individual step to calculate the total ∆H CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g) 2H2O(g) 2 H2O(l) ΔH = −802 kJ ΔH = − 88 kJ --------------------------------------------------------------------CH4(g) + 2 O2(g) + 2 H2O(g) CO2(g) + 2 H2O(l) + 2 H2O(g) ΔH = − 890 kJ Example From Chemistry: The Central Science, Ch.5 Sc. 6 Sometimes you may have to flip the equation, then you change the sign of the enthalpy. Later on in this marvolous presentation you will be honored with a example Calculate the change in enthalpy for the reaction: P4O6 + 2O2 P4O10 Using these enthalpies for reactions: P4 + 3O2 P4O6 ΔH = -1640.1 kJ P4 + 5O2 P4O10 ΔH = -2940.1 kJ **Flip the first equation, as the product of that reaction is the reactant in our problem reaction. At the same time, the sign in front of the ΔH needs to be reversed as well. P4O6 P4 + 3O2 ΔH = 1640.1 kJ ** The P4’s cancel on each side and 3O2’s also cancel. Thus the net reaction will look like the problem reaction. Now that the reactions are equal, add the enthalpies together to get the total change in enthalpy. 1640.1 kJ + -2940.1 kJ = -1300 kJ LAWS OF THE THERMO-MAN First law of thermodynamics states that any energy lost by a system must be absorbed by its surroundings and vice versa Internal energy is the sum of the kinetic and potential energy in a system The change in energy … ∆E = Efinal - Einitial If the ∆E is positive then energy is gained from the surroundings. If the ∆E is negative then energy is released into the surroundings **Calculate the ΔH for this reaction in kJ/mol. HHH H 2 H-C-C-C-C-H + 13 O=O 8 O=C=O + 10 H-O-H HH HH 20(413 kJ/mol) + 3(348) + 13(498) = 15778 kJ/ mol 16(799) + 20(463)= 22044 kJ/ mol 15778-22044= -6266 kJ/mol http://www.justosphere.com/files/image/007logo-1.jpg ENTHALPY The word enthalpy come from the Greek word enthalpein, which means hot in this famous romance language It is the amount of heat that is absorbed or released by a reaction Enthalpy is defined by H = E + PV Enthalpy is a state function means that it can change. Thus the change in enthalpy equation is ∆H = ∆E + P ∆V Basically it is the thermal internal (it rhymes) energy of a system (no it doesn’t) ENTHALPIES OF FORMATION The enthalpy of the reaction is the heat of a reaction A negative ∆H would cause a exothermic reaction A positive ∆H would cause a endothermic reaction The change in enthalpy in a reaction is equal to the magnitude, but for a opposite sign of the enthalpy the reaction is reversed. CALORIMETRY Calorimetry is the measure of heat flow C is heat capacity The heat capacity for an object is the amount of heat energy required to raise the objects temperature by 1 ºC 20g of water is heated up from 20oC to 85oC. Calculate the ΔH in kJ/mol for this process. q=mc(ΔT) (20g)(65oC)(4.18 J/goC)= 5434 J/g (5434 J/g)(18 g/mol)(.001 kJ/J)= 97.812 kJ/mol --------------------------------------------------------------------------------------------------------------- “Random Mike Thing” -qfuel=qnet · qwater (-773261J)(12.25g/mol)=(1250J/K)(23.1-14.7) + m(23.1-14.7)(4.18J/g°C) m= 270677.2 g that is a lot of agua! The point of that was, -qrxn= qcal THERMODYNAMICS DONE RIGHT WITH MOSES AND MIKE