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Lecture Notes 4-DELAAT (1).pdf
Lecture Notes 4-DELAAT (1).pdf

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School
University of Guelph
Department
Chemistry
Course
CHEM 1050
Professor
Richard Delaat
Semester
Winter

Description
THERMO - PART 4 SPONTANEITY and GIBBS FREE ENERGY Spontaneous processes tend to minimize energy and maximize disorder. NOTES:  Some processes are " Enthalpy Driven" with S = _______ (e.g. a ball rolling downhill)  Some processes are "Entropy Driven" with H = _______ (e.g. two inert gases mixing). Therefore, for a chemical reaction:  If H Rxnis NEGATIVE and S Rxnis POSITIVE, then the rxn will be _______________ (in the forward direction) as both enthalpy and entropy _________________ the reaction.  If H Rxnis POSITIVE and S Rxnis NEGATIVE, then the rxn will be _________________ as both Enthalpy and Entropy _______________ the reaction (reverse reaction will be spontaneous). What about the cases where (a) H and S are BOTH positive, OR rxn rxn (b) H and S are BOTH negative rxn rxn How do we decide if a reaction is “spontaneous” or not? This balance between energy and entropy can be handled by a new term, the __________________, and it gets us away from having to worry about S Surroundings and hence S Universe We know that: S UniverseS Surroundings SSystem (1) We also know that at constant T and P: S SurroundingsqSurroundings = (2) (this is a key relationship!) Therefore, substituting (2) into equation (1): S = _______________ + S Universe System Multiply both sides by –T: S UniverseH System– TS System ***** We define G = H - TS = –TS Universe, where G is called Gibbs Free Energy **** 42 NOTES: Significance of the Sign of G A reaction at constant temperature and pressure proceeds spontaneously when G rxn< 0, i.e. for a spontaneous process, rxnwill be NEGATIVE.  If G rxnis N EGATIVE, the rxn is _____________________________. (i.e., rxn proceeds in the forward direction from reactants to products).  If G rxnis ZERO, the reaction is _______________________________.  If G rxnis POSITIVE, the reaction is NOT __________________________________________________________. From here on we will use G which includes both energy anndentropy. For a reaction to proceed spontaneously, G must be negative. (Recall that the 2 Law of Thermodynamics says that for a spontaneous process, S > 0.) This means that the direction of spontaneous change it toward lower free energy (i.e., free energy minimization). The equation: G = H – TS can have 4 possibilities Enthalpy Change Entropy Change Spontaneous Reaction? Endothermic (+) Increase (+) If TS > H 1. (H > 0) (S > 0) (______________ driven) Exothermic (–) Increase (+) 2. (H < 0) (S > 0) _______, G < 0 Endothermic (+) Decrease (–) _________, G > 0 3. (H > 0) (S < 0) (But rxn is ______________________.) 4. Exothermic (–) Decrease (–) If TS < H (H < 0) (S < 0) (_____________ driven) 43 NOTES: Note: For 2 and 3, spontaneity depends on size of H, S and T. Some examples: H in kJ, S in J/K) 1. H 2(g)+ Br 2(l) 2HBr (g; H –73 & S +114 2. NH 3(g) + HCl (g)  NH Cl 4 (s; H –176 & S –285 3. NH 4l (s) NH 3(g)+ HCl (g; H +176 & S +285 4. 3O 2(g)  2O 3(g; H +285 & S –137 For each example, state when, if ever, the reaction will be spontaneous: Example 1: Example 2: Example 3: Example 4: For the examples above, we can plot graphs of G vs. temperature: 44 So, for some reactions you might need to check out the possible competing H and NOTES: S requirements (i.e., is the reaction enthalpy or entropy driven). Standard Molar Free Energies of Formation, G f Definition: G fs the G for the reaction in which 1 MOLE of a substance in its STANDARD STATE is FORMED from its ELEMENTS in their STANDARD STATES.  G m,fof any pure element (in its standard state) = _______ (just like ______________) Refer to Appendix C for G valuesf G m,fis a measure of stability with respect to decomposition of a compound into its elements under standard conditions.  If Gm,fis negative at a particular temperature, the elements that make up the compound form the compound spontaneously (i.e., the compound is __________ stable than the elements).  If Gm,fis positive at a particular temperature, the compound decomposes spontaneously into its elements (i.e., the compound is _____________ stable than the elements). Recall Standard States:  pure liquid; pure solid (most stable form at 1 atm and specified T)  gases (1 atm)  solutes (1M concentration) Question: The standard free energy of formation (G ) for CS (gis +66.85 kJ/mol. f 2 What does this say about the molecule CS 2g? G is a STATE FUNCTION (independent of path) We can get G rxnEITHER from: a) Tables of G m,fdata (at a specific temperature, e.g., 25C) i.e., Use:G Rxn = ________________________ NOTE: This equation can ONLY be used for 25C when using G values given for 25C. f OR b) H anf S datm (at any temperature). 45 i.e., For ANY temperature we useG Rxn = __________________ NOTES: This equation assumes that both H and S are independent of temperature (this is OK at our level). Example of calculating G Rxn from H Rxnand S Rxn values Calculate the standard free energy change for the formation of methane at 298 K given H fnd S mor reactants and products. Equation: C(s) + 2H2(g)  CH 4g) Find H Rxn: H Rxn=  H Rxn= Find S Rxn S Rxn= S Rxn= Find G Rxn: G Rxn= H Rxn- T SRxn G Rxn= G Rxn= G f - 50.7 kJ  Reaction goes at 298 K Note: At 1000K, G Rxn= –74.8 kJ – {1000 K × –80.8 J/K × 1 kJ/ 1000 J} 46 G Rxn = G f + 6 kJ NOTES: Reaction is entropically unfavoured at high T. Standard Molar Free Energies of Formation of Aqueous Ions G (aq) f Recall that for Hfof aqueous ions, it was not possible to separate the cation and anion contributions. We got around this by defining H (H ,aq) = _____. f Similarly, S (H , aq) = _____ and G (H , aq) = ____. m m,f Concentration and Free Energy G Rxncan be obtained from: a) G vflues (e.g., at 25C) and G Rxn= nG m,f(Products) - nG m,f(Reactants) b) H afd S vmlues (at any temperature) and G Rxn = H rxn– TS rxn We have a problem however. We RARELY have standard conditions (1 atm, 1M). These conditions are unusual and difficult to set up (except for changes of state at 1 atm or 1 M). Much more useful if we have an expression that we can use with any concentrations. The value of G (NOT G) is much more useful and is easily obtained. For any reactions, Reactants  Products, the sign of GRxntells us if we are going forward, backward or at equilibrium. Complete the following table on your own: G Reaction (,  or ) Rxn negative zero positive Hence G Rxnmust depend on concentrations of reactants and products. You already know this! Think about the REACTION QUOTIENT, Q (material from CHEM*1040) Remember, for a reaction with any concentrations or pressures values: aA + bB  cC + dD Q C QP= 47 NOTES: Units: molarity (solutions) atm (gases) Note: Pure solids and liquids  1 within these expressions. This expression contains ANY values, not just those at equilibrium. If Q = K, then we are at ________________________. If Q < K, the reaction wants to go ________, until Q = K (at equilibrium). i.e, This means that in the expression, the top portion is too small and the bottom portion is too big. To get Q to equal K, the reaction __________________ [C] and [D] and ___________________ [A] and [B] We would say the reaction is ____________________. If Q > K,the top of the expression has values that are too large and the bottom has values that are too small. We must drive the reaction ________________ to _____________ C & D, and __________ A & B. i.e. if Q > K the reaction goes L  R until Q = K. This rxn is _________________ as written, but _________________ in the reverse. How are G rxn and Q/K related? We can write:G rxn= RT ln(Q/K) = 2.303 RT log(Q/K) Q/K ln(Q/K) G rxn Direction Spontaneity? < 1 – – L R > 1 L R 1 L R Spontaneous reactions ALWAYS proceed so as to move TOWARDS a STATE OF EQUILIBRIUM i.e. Q  K i.e. Q/K  1 Thus G rxnvia the value of Q) depends on concentrations of Reactants and Products. We need to think carefully about what we mean brxnnd Gºrxn Using logarithm rules (division within a logarithm is the same as subtracting the logarithm of the top from the bottom): 48 NOTES: G rxn = RT ln(Q/K) = RT lnQ – RT lnK which gives us: G rxn= Grxnis the "standard conditions value" (i.e. value appropriate to unit pressures and concentrations). RT lnQ term is the "correction factor" to account for non–unit concentrations or pressures (i.e., non-standard conditions). N.B. G rxncalculated in this way reflects the ____________________ of a reaction under a particular set of conditions of concentrations or pressures – and as reactants are consumed, G rxnwill change and eventually go to 0 as equilibrium is achieved. G rxnis then a quantitative expression of ______________________. ******************************************************************* Example: Reaction: N 2(g)+ 3H 2(g)  2NH 3(g) At 298K,the partial pressures in a reaction mixture areN2 = 0.25 atm, PH2 = 0.55 atm and P NH3 = 0.95 atm. Detemine G .rxn Problem Solving Strategy: Step 1: Determine the value of G rxnusing tables of Gm,fdata and G Rxn = nG m,fProducts) - nG m,fReactants). Step 2: Determine Q based on the conditions given in the question. Step 3: Solve for G rxn Step 1: G Rxn= nG (frods) - nG (Rfacts) = G Rxn = -33.28 kJ Step 2: Q = [Prod]/[Reactants] = Q = 21.7 Step 3: G rxn= G rxn+ RT ln Q G = -25.7 kJ rxn 49 NOTES: G is negative. Therefore, the reaction is __________________ in the forward rxn direction. An increase in 3H (g) will continue until Q becomes equal to K (the equilibrium constant) and hence G will become zero (i.e., at equilibrium). rxn To see the relationship between Q and Greview the following set of rxn experiments where the partial pressures for the reactants and products are measured for the following reaction2 N (g) +23H (g)  23H (g) where G = -33.28 kJ + RT ln {P 2/ P × P 3} rxn NH3 N2 H2 Note: The equilibrium constant, K, for this reaction at 298K is 6.8×10 . Expt. PN2 PH2 PNH3 Q RT lnQ
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