Answer Key - CHEM111 Week 7 Discussion questions Fall 2014 1 Predict the higher entropy based on the following scenarios Why A B C D E 2 PCl3(g vs CO(g

Answer Key - CHEM111 Week 7 Discussion questions Fall 2014...

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Unformatted text preview: CHEM111 Week 7 Discussion questions Fall 2014 1. Predict the higher entropy based on the following scenarios? Why? A. B. C. D. E. 2. PCl3(g) vs CO(g) Co(s) @25oC vs Co(ℓ) @ 1495 oC(s) NaCl(s) (1mole) vs NaCl(s) (2 moles) O2(g) (1mol) vs O3(g) (1mol) HCl(g) vs B(s) A. PCl3, higher molar mass and more vibrational microstates B. Co(ℓ) @ 1495oC; higher temperature, liquid phase (more microstates) C. NaCl(s) 2 mol; more moles, more vibrational microstates D. O3(g) 1 mol; more complex, higher molar mass E. HCl(g) ; more atoms, gaseous (has vibrational, rotational, translational microstates) Predict the sign of ΔSsys from the following reactions A. PCl3(g) → PCl3(l) B. Co(s) → Co(l) C. Co(s) @ 25oC → Co(s) @ 100oC D. NaHCO3(s) + CH3COOH(l) → NaCH3COO(aq) + H2O(l) + CO2(g) E. Cl2(g) @ 1 atm → Cl2(g) @ 3 atm A. Negative B. Positive C. Positive D. Positive E. Negative 3. Care must be taken when dissolving solid pellets of sodium hydroxide (NaOH) in water as the temperature of the water can rise quickly. Taking NaOH as the system, what can you deduce about the signs of ∆Ssys and ∆Ssurr? Is the ∆Suniv >0 or =0? ΔSsys is positive; dissolving of solid to aqueous ions. ΔSsurr is positive; heat is absorbed by surroundings ΔSuniv is positive 4) a) For the reaction Cu(s) + Hg22+(aq) ⇌ Cu2+(aq) + 2Hg(l) compoute the standard state enthalpy change and the standard state entropy change at 25oC. The following values are given. ΔHfo (kJ/mol) So (J/molK) Cu(s) 0 33.15 2+ Hg2 (aq) 172.4 84.5 Cu2+(aq) 64.77 -­‐99.6 CHEM111 Hg(l) Week 7 Discussion questions 0 Fall 2014 76.02 Cu (c) + Hg22+ (aq) ! Cu2+ (aq) + 2 Hg (l) ΔHf° = 0 172.4 64.77 0 S° = 33.15 84.5 –99.6 2(76.02) ΔrH° = Hf°(Cu2+ (aq)) – Hf°(Hg22+ (aq)) ⇒ 64.77 – 172.4 = –107.63 kJ/mol ΔrS° = S°(Cu2+ (aq)) + 2S°(Hg (l)) – S°(Cu (c)) – S°(Hg22+ (aq)) ΔrS° = –99.6 + 2(76.02) – 33.15 – 84.5 = –65.21 J/K⋅mol b) For this reaction what are the signs of the following quantities: ΔSo, ΔSsurro, and ΔSTotalo. Why? With ΔS is negative, ΔSsurr must be positive if the reaction is spontaneous as written (ΔStot positive). From the following combined 1st and 2nd law statement, ΔStot will be positive: ΔStot = ΔS – ΔH/T at constant temperature and pressure. For one mole: ΔStot = – 65.21 J/K – (–107.63 kJ/298 K) = 296 J/K 5) Consider the reaction SiO (g) + POCl3 (g) ! SiO2 (c) + PCl3 (g) a) Estimate ΔrS° for the above reaction based on Campbell’s rule. Net loss of 1 mole of gas suggests an estimate of –140 J/K•mol for this reaction. b) This reaction is spontaneous as written at room temperature and pressure. Using your estimate of ΔrS°, assign a plus (positive value), minus (negative), or a zero for each of the following changes: ΔS ΔSsurr ΔStotal ΔH ΔEsurr ΔEtotal – + + – + 0 There is one answer we should know without considering anything else -­‐ ΔEtotal is always zero (First Law). From “spontaneous as written” ΔStotal must be positive (Second Law). Our estimate of ΔrS° tells us that ΔSσ is negative (which again is the key, because if ΔS were positive, this would not be a solvable problem without additional information). So, we should recognize ΔEtotal, ΔStotal, and ΔS very quickly, and then we can determine everything else. From the second law, ΔStotal = ΔS + ΔSsur, so from the information we already have, ΔSθ must be positive. ΔSsur = ΔEsur/T so ΔEsur must be positive. ΔSsur = – ΔH/T, so ΔH must be negative. CHEM111 Week 7 Discussion questions Fall 2014 6)(a) One method for synthesizing methanol (CH3OH) involves reacting carbon monoxide and hydrogen gases: CO(g) + 2H2(g) ! CH3OH(l). (a) Predict the signs of the following values if you know the reaction is exothermic and spontaneous. ΔHo ΔSθo ΔSσo ΔSToto Go ΔETot - + - + - 0 6 points (b) Calculate ΔG at 25oC for this reaction when carbon monoxide gas has a pressure of 5.0 atm and hydrogen gas has a pressure of 3.0 atm are converted to liquid methanol. ΔGfo(CO(g))= -137.17 kJ/mol ΔGfo (H2(g))= 0.0 kJ/mol ΔGfo (CH3OH(l))= 166.27 kJ/mol ΔG=ΔGo+RTlnQ 2 pts. ΔGo= ΣνPGfo(P)-­‐ ΣνRGfo(R)=-­‐166.27kJ/mol–(-­‐137.17 kJ/mol)-­‐2(0.0)=-­‐29.1kJ/mol 2pts. 1 1 = = !" !!! 5.0 ∙ 3.0! 2 pts. 29100J 8.314J 1 ΔG = ΔG! − = − + 298K ∙ ln = −38.5kJ/mol mol molK 5.0 ∙ 3.0! 6pts. 12 points total (c) Is the reaction spontaneous at this temperature and these pressures? How do you know? Spontaneous because ΔG is negitive 4pts (d) What is the equilibrium constant for this expression? ΔGo=-RTlnK K=e- Go/RT=5.6x106 4 pts. Δ ...
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