CHM STUDY GUIDE: EXAM III Chapter 17, 18 and 19CHAPTER 17 THERMODYNAMICS1st Law of thermodynamics states:-cannot create or destroy energy-energy can only be transferred/converted∆H stands for: Change in enthalpy-heat of a reactionA positive ∆H =A negative ∆H = Endothermic reaction (needs energy)Exothermic reaction (releases energy)S stands for EntropyEntropy is: Measure of how spread out/diverse the energy of a system is“measure of disorder”Microstate: Possible distribution of atoms, molecules, or energy(W)-a system with fewer microstates (smaller W) among which to spread its energy (small dispersal) has a lower entropy- a system with more microstates(larger W) among which to spread its energy (large dispersal) has a higher entropyS=∆S=S= k*ln(W)W= # of possibilities∆S= k* ln(Wf/Wi)The 2nd law of thermodynamics states that:-Entropy of the universe increases for a spontaneous reaction -entropy of the universe remains unchanged at equilibrium∆Suniverse=(relating the change in entropy to the system/surroundings) ∆Suniverse= ∆Ssystem + ∆Ssurroundings > 0 (spontaneous) -if =0 then reaction is at equilibriumPositive ∆Suniverse =Negative ∆Suniverse=Disorder (spontaneous)More order (non-spontaneous)How to find the ∆S of the reaction: aA +bB → cC + dDaA +bB → cC + dD∆S = ∑∆Sproducts - ∑∆Sreactants-these values are looked up-multiply the value by # of moles-this is the same for ∆HRelationship between ∆S and the # of moles of gas in products vs. reactants-if the reaction produces more moles of gas that it consumes (more moles of gas in products vs. reactants) than the ∆S is positive (creating disorder)- if # of moles of gas diminishes that ∆S is negative (creating order)-if # of moles of gas is same on both sides of the equation than ∆S is +/- but the value is small∆S surroundings = ∆S surr = -∆H sys / TCHM STUDY GUIDE: EXAM III (in relationship to ∆H and the temperature)The third law of thermodynamics states that:-entropy of a crystalline substance is zero at a temperature of 0K G stands for: Gibb’s free energyGibb’s free energy -tells you whether a reaction is spontaneous/non-spontaneous -the energy available to do workGibb’s free energy equation: G=G= H –TSS= J/KH = kJ/molG = kJ or J (must convert to make all the same unit)Change in free energy equation: ∆G=∆G = ∆H - T∆SWhat happens when:∆G > 0 (+)∆G = 0∆G < 0 (-)Non-spontaneous reaction EquilibriumSpontaneous reactionWhat is the standardfree energy of reaction (∆Gorxn)The free energy change for a reaction when it occurs at standard state conditions-get the same way you get ∆S (sum of values for products – sum of reactants)-look up the values (appendix 3)The different effects of ∆S and ∆H on ∆G:Using the equation:∆G = ∆H - T∆S- ∆H is +- ∆S is –- not temp dependent∆G will be + Reaction will be non-spontaneous- ∆H is - - ∆S is +- not temp dependent∆G will be - Reaction will be spontaneous- ∆H is +- ∆S is +- High temp∆G will be –Reaction will be spontaneous-This is because this part of the equation: T∆S will be large at a high temp; subtracting this from ∆H will produce a negative #- ∆H is +- ∆S is +- Low temp∆G will be +Non-spontaneous reaction∆H will be large so subtracting T∆S (which will be small because T is low) from ∆Hwill produce a positive number- ∆H is -- ∆S is -- Low temp∆G will be –Spontaneous reactionT∆S will be a small positive # so subtracting this from a negative number will produce a negative numberCHM STUDY GUIDE: EXAM III - ∆H is - ∆S is -- High temp∆G will be +Non-spontaneous reactionIf a question asks you to determine if a reaction is spontaneous:-Use: ∆G= ∆H - T∆S-If ∆G is negative then the reaction is spontaneous Question asks at what temperature does spontaneity start/stop:Set the above equation equal to 0 so that:0 = ∆H - T∆S-Rearrange equation to set equal to T and solve for TT = ∆H/∆SEquation relating thermodynamics to equilibrium(∆G= …)∆G = ∆G0 + RT lnQR = 8.314 J/mol KQ = [ ] at non-equilibrium conditions ∆Go = free energy at standard states (find by looking up: products – reactants) -need to make sure all values are in same unitsIf reaction is at equilibrium the above equation turns into:0 = ∆Go + RT lnKeqOr…∆Go= -RT ln KeqUsing the equation to find the free energy (above), what is the equationto determine equilibrium constants?Keq = e^ (-∆Go/RT)CHAPTER 18 ELECTROCHEMISTRYBalancing redox equations:- In acidic environment1) Split reaction into two half reactions (place species that look alike in the same half reaction/ separate by what is being reduced or oxidized 2) Make sure the half reactions are atomically balanced (number of atoms are the same)3) Add H2O to account for either side lacking oxygen4) Balance the hydrogens by adding H+ to the appropriate side5) Add up the total charge for the left and the total charge for the right sideof the equation6) Balance the charges by adding a certain amount of electrons to the side that is more positive (multiply by common denominator if needed to make # of electrons the same)7) Combine the half reactions and cancel out any common species - In basic environment-Same steps apply-After combining the half reactions after completing the above steps:-Add OH- atoms wherever there are H+ atoms (add same number as hydrogenCHM STUDY GUIDE: EXAM III ions)-this changes the species to water; then add the same number of OH- to theopposite side of the reaction-Cancel out common species Electrochemical processes:oxidation-reduction reactions in which:• the energy released by a spontaneous reaction is converted to electricity or• electrical energy is used to cause a non-spontaneous reaction to occurOxidation Lose electronsReduction Gain electronsOxidation number: The charge the atom would have in a molecule (or anionic compound) if electrons were completely transferred1. free elements have an ox. # of 02. In monatomic ions, the oxidation number is equal to the charge on the ion.3. The oxidation number of oxygen is usually –2. In H2O2 and O22- it is –1.4. The oxidation number of hydrogen is +1 except when it is bonded to metals in binary compounds. In these cases, its oxidation number is –1.5. Group IA metals are +1, IIA metals are +2 and fluorine is always –1.The sum of the oxidation numbers of all the atoms in a molecule or ion = the charge on the molecule or ionGalvanic
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