Advanced Chemistry Practice Prblems Thermdynamics: Gibbs Free Energy 1. Questin: Is the reactin spntaneus when ΔG < 0? ΔG > 0? Answer: The reactin is spntaneus when ΔG < 0. 2. Questin: Fr a reactin with ΔH = 298 kj/ml and ΔS = 164 J/ml. K, what is the value ΔG at each temperature? Is the reactin spntaneus at that temperature? a. 50 C b. 0 C c. 35 C Answer: When given ΔH and ΔS, ΔG is calculated using Δ G = ΔH TΔS ensuring that ΔH and ΔS are in the same units and temperature is in Kelvin. A reactin will be spntaneus when ΔG is negative. 164 J / ml K Δ G = 298 kj ( 50 + 273) a. 1000 ; spntaneus Δ G = 335 kj / ml b. c. 164 J / ml K Δ G = 298 kj (0 + 273) 1000 ; spntaneus Δ G = 342 kj / ml 164 J / ml K Δ G = 298 kj (35 + 273) 1000 ; spntaneus Δ G = 349 kj / ml
Advanced Chemistry Practice Prblems Thermdynamics: The Third Law Thermdynamics 1. Questin: Withut reerring t tables, rank the llwing xides rm lwest t highest mlar entrpy. Each substance is at the same temperature. a. Na 2 O(s) b. Li 2 O(s) c. K 2 O(s) Answer: Since each substance is a slid and very similar in structure, the mlar mass will determine the ranking. Entrpy increases with mlar mass. Li 2 O(s) < Na 2 O(s) < K 2 O(s) 2. Questin: Withut reerring t tables, predict which the llwing will have the greatest mlar entrpy. Each substance is at the same temperature. a. O 2 (g) b. Na 2 O(s) c. BaO(s) d. H 2 O(l) Answer: Gases have, by ar, the greatest entrpy. The mlecule with the highest entrpy is O 2 (g). 3. Questin: Withut reerring t tables, predict which the llwing will have the greatest mlar entrpy. Each substance is at the same temperature. a. N 2 (g) b. O 2 (g) c. NO(g) d. Ar(g) Answer: The mlar masses each substance are very similar. The dierences in entrpy result rm mlecular cmplexity. NO, as a cmpund, is mre cmplex than the elements (mnatmic and diatmic, alike) and therere will have the highest entrpy.
Advanced Chemistry Practice Prblems Thermdynamics: Calculating Standard Entrpy Change 1. Questin: Cnsider the llwing reactin. Predict the sign r DS rxn. Then, given the table standard mlar entrpies at 25 C, calculate the standard entrpy change (DS ) r the reactin. H 2 (g) + I 2 (g) 2 HI(g) Substance S (J/ml. K) H 2 (g) 130.7 I 2 (g) 260.7 HCl(g) 206.6 Answer: The number mles gas n each side the equatin is equal (2 mles gas n each side). The change in entrpy is due t a change in the mlecular cmplexity. It is expected that there will be an increase in the entrpy as the elements are cnverted t cmpunds. T calculate the standard entrpy change, the llwing equatin will be used: rxn Δ S = ns (prducts) ns (reactants ) ΔS rxn = 2 ml S (HI) [1 ml S (H 2 ) + 1 ml S (I 2 )] ΔSrxn = 2 ml 206.6 J/ml. K [1 ml 130.7 J/ml. K + 1 ml 260.7 J/ml. K] ΔSrxn = 21.8 J/K
Advanced Chemistry Practice Prblems 2. Questin: The standard mlar entrpies (S ) N 2 and Mg 3 N 2 are 191.6 J/ml. K and 88 J/ml. K, respectively. Calculate the standard mlar entrpy Mg(s) given the llwing reactin. 3Mg(s) + N 2 (g) Mg 3 N 2 (s) DS = -201.7 J/K Answer: Use the equatin: rxn Δ S = ns (prducts) ns (reactants ) ΔS rxn = 1 ml S (Mg 3 N 2 ) [3 ml S (Mg) + 1 ml S (N 2 )] 201.7 J/K = 1 ml 88 J/ml. K [3 ml S (Mg) + 1 ml 191.6 J/ml. K] 207.1 J/K = 88 J/K 3 ml S (Mg) 191.6 J/K 201.7 J/K 88 J/K + 191.6 J/K = -3 ml S (Mg) 98.1 J/K = 3 ml S (Mg) 33 J/ml. K = S (Mg)
Advanced Chemistry Practice Prblems Thermdynamics: Calculating Standard Free Energy Change 1. Questin: Write the reactin assciated with the standard ree energy rmatin ( Δ G ) r each substance listed belw. a. CaO(s) = -603.3 kj/ml b. C 6 H 12 O 6 (s) = -910.4 kj/ml Answer: Free energy rmatin is the ree energy when ne mle a cmpund is rmed rm the elements in their mst stable rm. The cmpund is listed n the prduct side (it is being rmed) and the equatin must be balanced with ne mle this cmpund. a. The stable rm elemental calcium is Ca(s). The stable rm xygen is O 2 (g). Ca(s) + ½ O 2 (g) CaO(s) b. The stable rm elemental carbn is graphite, C(graphite), elemental hydrgen is H 2 (g) and xygen is O 2 (g). 6 C(graphite) + 6 H 2 (g) + 3 O 2 (g) C 6 H 12 O 6 (s) Did yu knw that graphite was the stable rm? Yu may have nly put C(s), but C(graphite) is mre accurate. 2. Questin: Determine the standard ree energy change (DG ) r the llwing reactin, given the table belw. 6 CO 2 (g) + 6 H 2 O(l) C 6 H 12 O 6 (s) Substance (kj/ml) CO 2 (g) -394.4 H 2 O(l) -237.1 C 6 H 12 O 6 (s) -910.4
Advanced Chemistry Practice Prblems Answer: The llwing equatin is used t determine the DG rxn when DG are prvided: Δ G rxn = nδg (prducts) nδg (reactants ) Δ G rxn ΔGrxn ΔGrxn = 1ml ΔG (C6H12O6) [6 ml ΔG (CO2) + 6 ml ΔG (H2O)] = 1 ml ( 910.4 kj/ml) [6 ml ( 394.4 kj/ml) + 6 ml( 237.1 kj/ml)] = 2878.6 kj 3. Questin: Calculate the standard ree energy rmatin ( ) r Br(g) given the llwing: 2 Br(g) Br 2 (g) ΔGrxn = -12.6 kj/ml Answer: One way t apprach this prblem is t cnsider the reactin that ΔG Br(g) represents. The rmatin ne mle Br(g) rm elemental Br 2 (l) is given by: ½ Br 2 (l) Br(g) The abve reactin, i reversed and dubled will give the reactin prvided in the prblem. rxn 2 Br(g) Br 2 (l) ΔG = 12.6 kj/ml Therere, t btain ΔG rxn r the tp reactin equatin (the rmatin ne mle Br(g) rm its elements), reverse the sign and divide by 2 t give = 6.3 kj/ml. Anther way t apprach the prblem is t use the equatin: Δ G rxn = nδg (prducts) nδg (reactants )
Advanced Chemistry Practice Prblems ΔGrxn = 1ml ΔG (Br2 ) 2 ml ΔG ( Br) Substitute in the values that are given: 12.6 kj/ml = 1ml 0 kj/ml 2ml (Br 2 ) Then, slve r G Δ (Br 2 ). 6.3 kj/ml = (Br 2 ) 4. Questin: Determine the Δ G rxn r the llwing reactin H 2 (g) + I 2 (s) 2 HI(g) Given the llwing: 2 H(g) H 2 (g) DG = -406.6 kj 2 I(g) I 2 (g) DG = -121.1 kj I 2 (g) I 2 (s) DG = -19.3 kj H(g) + I(g) HI(g) DG = -271.8 kj Answer: The equatins will be manipulated as llws t btain the equatin r which DG is needed. The irst three equatins are used t btain the reactants and prducts needed (in red belw). The rth reactin is used t cancel the substances nt needed. H 2 (g) 2 H(g) DG = 406.6 kj I 2 (s) I 2 (g) DG = 19.3 kj 2 H(g) + 2 I(g) 2 HI(g) DG = 2( 271.8 kj) = 543.6 kj I 2 (g) 2 I(g) DG = 121.1 kj Veriy that the reactins d add t btain the reactin needed. Then, add the ΔG values. H 2 (g) + I 2 (s) 2 HI(g) ΔG = 3.4 kj