T HERMO - DYNAMICS Moses Kool and Miguel Grizzly Bear

C ONSERVATION 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

T HERMO L AWS 1 st law= energy cannot be created nor destroyed 2 nd law= Universal randomness is always on the rise. Irreversible processes result in increase in randomness.

B RIEF INTRO TO HEATING CURVES Plateaus: q=m∆H q=quantity of potential energy ∆H= enthalpy of fusion or vaporization Enthalpies of Water Szzzzzzzzz ∆H f = kJ/mol ∆H v = kJ/mol

I NTERMOLECULAR B ONDS -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

∆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 -25 o C 5g block of ice is heated to the point where it is 150 o C. What is the quantity of the energy involved in this process? H 2 O: C solid =2.06 J/g o C C liquid = 4.18 J/g o C C gas = 2.03 J/g o C ΔH fusion = 6.02 kJ/mol ΔH vap = 40.7 kJ/mol Melting Point=0 o C Boiling Point= 100 o C (5g)(2.06 J/g o C )(25 o C)=257.5 J=.2575 kJ (.277 mols)(6.02 kJ/mol)=1.667 kJ (5g)(4.18 J/g o C)(100 o C)= 2090 J= 2.09 kJ (.227 mols)(40.7 kJ/mol)= kJ (5g)(2.03 J/g o C)(50 o C)= J=.5075kJ ΔH=13.76 kJ

C ALCULATING ∆H AND ∆S I N R EACTIONS For a particular equation: CH 4 + O 2  CO 2 + 2H 2 O the ∆H reaction is ∆H reaction =∑ ∆H product - ∑ ∆H reactant ∆H reaction = [1 mol( kJ/mol) + 2mol ( )] –[1mol(-74.8) + 2 mol (0)]= kJ

∆S IS E NTROPY ∆S reaction =∑ ∆S product - ∑ ∆S reactant ∆S = [1mol(213.6 J/K) + 2mol(69.91)] – [1mol(186.3) + 2 mol(205)] = J/K Any change in energy increases the amount of energy in the universe

S PONTANEOUS Spontaneous = natural, happens normally in nature + ∆H - ∆H + ∆SEntropy driven. Can be spontaneous at high temp. Always spontaneous - ∆SNever spontaneous Enthalpy Driven. Can be spontaneous at a low temp.

G IBBS F REE E NERGY ∆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 EquilibriumProduct- favored Reactant- favored ∆G=0Large - ∆GLarge +∆G k= 1k>> 1k<< 1

M ORE G IBBS 0 = ∆Gº + RT lnK a Q = e - ∆G/RT R is a constant T is the absolute temp K a 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: 2PCl 3 + O 2  2POCl 3 ΔH= 2(-542.2) - 2( )= kJ/mol ΔG= 2(-502.5) – 2(-269.6)= ΔS= 2(325) – [2(311.7) + 205]= 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= ( kJ/mol)/( J/ mol K) T= K **What is the K for this reaction occurring at 2009 K? ΔG 2009 = kJ/mol – 2009( J/mol K) ΔG 2009 = = ( J/ mol K)(2009) lnK = lnK e = K K=1.00

H ESS ’ L AW 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 CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O(g) ΔH = −802 kJ 2H 2 O(g)  2 H 2 O(l) ΔH = − 88 kJ CH 4 (g) + 2 O 2 (g) + 2 H 2 O(g)  CO 2 (g) + 2 H 2 O(l) + 2 H 2 O(g) ΔH = − 890 kJ 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 Example From Chemistry: The Central Science, Ch.5 Sc. 6

Calculate the change in enthalpy for the reaction: P 4 O 6 + 2O 2  P 4 O 10 Using these enthalpies for reactions: P 4 + 3O 2  P 4 O 6 ΔH = kJ P 4 + 5O 2  P 4 O 10 ΔH = 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. P 4 O 6  P 4 + 3O 2 ΔH = 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 kJ kJ = kJ

L AWS 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 = E final - E initial If the ∆E is positive then energy is gained from the surroundings. If the ∆E is negative then energy is released into the surroundings

H-C-C-C-C-H + 13 O=O  8 O=C=O + 10 H-O-H H H H H HH H H 2 20(413 kJ/mol) + 3(348) + 13(498) = kJ/ mol 16(799) + 20(463)= kJ/ mol = kJ/mol **Calculate the ΔH for this reaction in kJ/mol.

E NTHALPY 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)

E NTHALPIES OF F ORMATION 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.

C ALORIMETRY 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

q=mc(ΔT) (20g)(65 o C)(4.18 J/g o C)= 5434 J/g (5434 J/g)(18 g/mol)(.001 kJ/J)= kJ/mol “Random Mike Thing” -q fuel =q net · q water ( J)(12.25g/mol)=(1250J/K)( ) + m( )(4.18J/g°C) m= g that is a lot of agua! 20g of water is heated up from 20 o C to 85 o C. Calculate the ΔH in kJ/mol for this process. The point of that was, -q rxn = q cal

T HERMODYNAMICS DONE RIGHT WITH M OSES AND M IKE