Chapter 17 Thermodynamics: Directionality of Chemical Reactions

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1 Reactant- & Product-Favored Processes John W. Moore Conrad L. Stanitski Peter C. Jurs Chapter 17 hermodynamics: Directionality of Chemical Reactions Why are equilibria product- or reactant- favored? Why do some reactions occur spontaneously? Why do others require help (heat, spark )? Exothermic reactions are often product favored (but not always). Stephen C. Foster Mississippi State University 2 Al(s) + 3 Br 2 (l) 2 AlBr 3 (l) Product favored; spontaneous Chemical Reactions & Dispersal of Energy Energy will spread out (disperse) unless it is hindered from doing so. Most exothermic reactions are product favored. E is transferred to the surroundings. Bond (potential) E dispersed into many more atoms and molecules. Dispersal of energy is probability driven: E is more likely to spread over many atoms than over just a few. Probability and Dispersal of Energy What are the possible outcomes if 2 excited atoms make contact with 2 more atoms? Excited atoms (A, B) A*B* + C D E packet Unexcited atoms A** B C D A* B* C D A* B C* D A* B C D* A B** C D A B* C* D A B* C D* A B C** D A B C* D* A B C D** E stays with original atoms in only 3 (of 10) cases: 70% chance that E will disperse. More atoms: E concentration is much less likely. Dispersal of Matter E is dispersed when a system expands. Gases expand to fill a container. Quantum theory: E-levels for gas motion get closer together when volume expands. More E-levels become accessible at a given. E will disperse over more levels. vacuum removable barrier Br 2 (g) Measuring Dispersal of E: Entropy In summary: Processes are product favored if E is dispersed from a few levels into many levels. his typically occurs during: exothermic reactions (bond E dispersal). processes in which matter is dispersed. In thermodynamics, nanoscale E dispersal is measured by the Entropy (S) of a system. 1

2 Measuring Dispersal of E: Entropy At constant : ΔS = S final S initial = q rev rev : Only applies to reversible changes. A reversible process: Absolute can be reversed by a slight change in conditions. e.g. ice melting at 0.0 C and 1 atm is reversible. small decrease in will cause it to refreeze. Measuring Dispersal of E: Entropy Calculate ΔS when 25.0 g of Al(s) melts at its normal melting point (660.3 C; 1 atm). ΔH fus (Al)=10.7 kj mol -1 ΔS = q rev = C = K = ΔH fus 25.0 g n Al = = mol g mol -1 ΔS = mol(10.7 kj mol-1 ) K ΔS = 10.6 J K -1 q p = ΔH Reversible if normal melting = kj K -1 Absolute Entropy Values A perfect crystal, at absolute zero (0 K) has: Minimum molecular motion (min. E dispersal). S = 0 (perfect crystal at 0 K) (he 3 rd law of thermodynamics) Measure the heat to change from 0 K room- (reversibly). Gives ΔS and the absolute S at room. ΔS = S final S initial = q rev S room = ΔS S initial = ΔS Absolute Entropy Values Standard molar entropies (S ), are measured by heating 1 mol of substance from 0 K to: A specified (often 25 C = K) At constant P = 1 bar S has J K -1 mol -1 units. S is always positive, for all materials. Standard Molar Entropies ( K) Compound S (J K -1 mol -1 ) Compound S (J K -1 mol -1 ) C (graphite) Cl 2 (g) C(g) Br 2 (l) CH 4 (g) I 2 (s) C 2 H 6 (g) HCl(g) C 3 H 8 (g) H 2 O(g) C 2 H 4 (g) H 2 O(l) CH 3 OH(l) NaCl(s) CO(g) KOH(s) 78.9 CO 2 (g) NaCl(aq) F 2 (g) KOH(aq) 91.6 S gas >> S liquid > S solid Solid molecules vibrate about fixed lattice points. Liquid molecules can slide past each other. Gas molecules have few restrictions on motion. More motion = higher S gas liquid solid S (I 2 ) S (Br 2 ) S (Cl 2 ) (J K -1 mol -1 ) 2

3 Larger, more complex, molecules have larger S: S complex molecule > S simple molecule Species type S (J K -1 mol -1 ) Species mass S (J K -1 mol -1 ) CH 4 alkane 186 Ar C 2 H 6 alkane 230 CO C 3 H 8 alkane 270 C 3 H Larger molecules have more ways to distribute E (more bonds to hold potential and kinetic E). Ionic solids Weaker ionic forces = larger S (Less tightly bound = easier motion) S (J K -1 mol -1 ) Comment MgO and -2 charges NaF and -1 charges NaCl 72 Cl - much larger than F -, less attraction. Solid or liquid dissolving Larger matter dispersal. S usually* increases: Substance S (pure) S (aq) CH 3 COOH(l) NH 4 NO 3 (s) J K -1 mol -1 units *Strong solvation may increase order. ΔS < 0 is possible. Gas dissolving Gas-molecule motion becomes restricted. ΔS < 0. Substance S (gas) S (aq. soln) CO H CH 3 OH J K -1 mol -1 units Predicting Entropy Changes CaCO 3 (s) CaO(s) + CO 2 (g) ΔS = 161 J K -1 mol -1 Large entropy increase: Solid gas, ΔS > 0. 1 molecule 2 molecules, ΔS > 0. H 2 (g) + F 2 (g) 2 HF(g) ΔS = 14 J K -1 mol -1 All gases, S constant. 2 reactant gases, 2 product gases, S constant. Actual ΔS shows a small increase. Predicting Entropy Changes Predict whether S will increase or decrease for: 2 CO(g) + O 2 (g) 2 CO 2 (g) Decrease Increase 3 gas molecules 2 gas molecules. NaCl(s) Na + (aq) + Cl - (aq) H 2 O(l) H 2 O(s) Ions locked in a crystal ions dispersed in water. Decrease Liquid molecules less restricted than solid molecules. 3

4 Calculating Entropy Changes Entropy is a state function (like H). For a reaction: ΔS = (n product S product ) S(n reactant S reactant ) Example Methanol is a common fuel additive. Calculate ΔS at 298 K for its combustion: 2 CH 3 OH(l) + 3 O 2 (g) 2 CO 2 (g) + 4 H 2 O(l) Look up S values (J K -1 mol -1 ): CH 3 OH(l) = O 2 (g) = CO 2 (g) = H 2 O(l) = Calculating Entropy Changes Example: ΔS at 298 K for: 2 CH 3 OH(l) + 3 O 2 (g) 2 CO 2 (g) + 4 H 2 O(l) ΔS = (n product S product ) (n reactant S reactant ) = (2S CO2 + 4S H2O ) (2S CH3OH + 3S O2 ) = [2(213.74) + 4(69.91)] [2(126.8) + 3(205.14)] = = J K -1 mol -1 S decreases as expected: 3 gas + 2 l 2 gas + 4 l Entropy & the 2 nd Law of hermodynamics E dispersal is accompanied by an increase in disorder of a system: Increased disorder = increased S Second Law of hermodynamics he total entropy of the universe is continuously increasing. he universe is slowly becoming more disordered. Entropy & the 2 nd Law of hermodynamics he entropy of a system can increase or decrease. S universe always goes up. S universe = S system + S surr (No subscript? S = S system ). If S system falls, S surr must increase by a larger amount. q rev ΔH surr ΔS surr = = = -ΔH sys Entropy & the 2 nd Law of hermodynamics Calculate ΔS univ for the combustion of octane in air: 2 C 8 H 18 (g) + 25 O 2 (g) 16 CO 2 (g) + 18 H 2 O(l) For the reaction ΔH = -11,024 kj and ΔS = J/K at K. ΔS univ = ΔS sys + ΔS surr given ΔS univ = J/K + (+11,024 x 10 3 J)/ K = J/K kj/k = +35,591 J/K -ΔH sys / Gibbs Free Energy Neither entropy (S), nor enthalpy (H ), alone can predict whether a reaction is product favored. Spontaneous (product favored) reactions can: Be exothermic or endothermic. Increase or decrease the S system. he Gibbs free energy (G), combines H and S. ΔG does predict if a reaction is product favored or not. 4

5 Gibbs Free Energy For a constant change, ΔG is equal to: ΔG = ΔH - ΔS If G: Decreases (ΔG < 0) a reaction is product favored Increases (ΔG > 0) a reaction is reactant favored (at constant P and ). Gibbs Free Energy ΔG can be calculated from ΔH and ΔS values or from Gibbs free energies of formation (ΔG f ). ΔG = (n product ΔG f product ) (n reactant ΔG f reactant ) ΔG f equals: ΔG reaction to make 1 mol of compound from its elements. 0 for an element in its most stable state (like ΔH f ) Gibbs Free Energy Calculate ΔG for the following reaction at 25 C: CO(g) + 2 H 2 (g) CH 3 OH(l) ΔG f values: CO(g) = kj/mol CH 3 OH(l) = kj/mol ΔG = ΔG f CH3OH (2ΔG f H2 + 1ΔG f CO ) Effect of on Reaction Direction Reactions go from reactant to product-favored when: ΔG = 0 = ΔH ΔS or when: ΔH = ΔS or = ΔH ΔS = kj/mol [2(0) + 1( kj/mol)] = kj/mol Product favored element in its standard state Effect of on Reaction Direction Since ΔG = ΔH ΔS Sign of ΔH Sign of ΔS Product Favored? Negative (exothermic) Positive Yes Negative (exothermic) Negative Yes at low ; No at high Positive (endothermic) Positive No at low ; Yes at high Positive (endothermic) Negative No Effect of on Reaction Direction At what will the following reaction be product favored? CH 4 (g) + H 2 O(g) CO(g) + 3 H 2 (g) ΔH f (kj/mol) S (J K -1 mol -1 ) ΔS =[(197.67) +3(130.68)] [ ]J K -1 mol -1 = J K -1 mol -1 ΔH = [( ) + 3(0)] [(-74.81) + ( )] kj mol -1 = kj mol -1 It will switch direction at: ΔH x 10 = = 5 J mol -1 = K ΔS J K -1 mol -1 5

6 Gibbs Free Energy and Equilibrium Constants Gibbs free energy G and the equilibrium constant K both show if a reaction is product-favored and are related. K must be unitless. ΔG = R ln K Each quantity in the K expression is divided by its standard state value: concentrations by 1 mol L -1 (solids and liquids). pressures by 1 bar (gas reactions). Gibbs Free Energy and Equilibrium Constants Evaluate K (298 K) for the reaction: 2 NO 2 (g) N 2 O 4 (g) ΔG = ΔG f (N 2 O 4 ) - 2ΔG f (NO 2 ) = (51.31) kj/mol = kj/mol ln K = ΔG x 10 3 J/mol R = (8.314 J K -1 mol -1 )(298 K) = 1.91 K = 6.75 Gibbs Free Energy & Maximum Work ΔG = maximum useful work that can be done by a reaction on its surroundings (constant and P). Gibbs Free Energy & Maximum Work ΔG also equals the minimum work required to cause a reactant-favored process to occur. ΔG = w system = - w max free energy = available energy For: 2 H 2 (g) + O 2 (g) 2 H 2 O(g) ΔG = kj 2 H 2 O(g) 2 H 2 (g) + O 2 (g) ΔG = kj A minimum of kj of work must be used to produce 2 mol of H 2 from liquid water. Every 2 moles of H 2 consumed can do up to kj of work. hermodynamic and Kinetic Stability hermodynamically stable Reaction is not product-favored. hermodynamically stable: Pd + O 2 PdO 2 ΔG = +326 kj/mol hermodynamically unstable: 4 Al + 3 O 2 2 Al 2 O 3 ΔG = kj/mol Kinetically stable Product-favored, but too slow to be important. Diamond is thermodynamically unstable C diamond C graphite ΔG = -2.9 kj/mol Kinetically stable, too slow to be important. E a is too large 6