Exam+review+24Nov09

Exam+review+24Nov09 - 1 Under which temperature conditions...

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1. Under which temperature conditions will the following reaction be spontaneous? A(aq)+B(aq) C(s)+D(aq) ΔH°=-34.0kJ/mol ΔS° = -57.0 J/mol•K A) above 273°C B) below 323°C C) above 552°C D) below 705°C E) At all temperatures. When ΔG° is negative, the reaction is spontaneous at that temperature. At 274 0 C (above 273°C) ΔG° = ΔH° - TΔS° ΔG° = -34,000 J/mol – 547 K(-57.0 J/mol•K) ΔG° = -2821 J/mol At 322°C (below 323°C) ΔG° = -34,000 J/mol – 595 K(-57.0 J/mol•K) ΔG° = -85 J/mol At 553°C (above 552°C) ΔG° = -34,000 J/mol – 826 K(-57.0 J/mol•K) ΔG° = 13,082 J/mol At 704°C (below 705°C) ΔG° = -34,000 J/mol – 977 K(-57.0 J/mol•K) ΔG° = 21,689 J/mol 2. What is the value of ΔG° (in kJ/mol) for this reaction at 25°C? A (g) B (g) ΔH° = -42.4 kJ/mol ΔS°= -56 J/mol•K A) -136 B) +236 C) -26 D) -59 E) none of the above At 25 0 C ΔG° = ΔH° - TΔS° ΔG° = -42,400 J/mol – 298 K(-56 J/mol•K)
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ΔG° = -25,712 J/mol ΔG° = -25.7 kJ/mol 3. Consider a pure, crystalline solid being heated from absolute zero to some very high temperature. Which one of the following processes produces the greatest increase in the entropy of the substance? A) Melting the solid B) Heating the liquid C) Heating the gas D) Heating the solid E) Boiling the liquid Because you are converting liquid to gas. The gas phase has the greatest entropy among 3 states of matter. 4. The equilibrium constant for a reaction is 0.50 at 30°C. What is the value of ΔG° (kJ/mol) at this temperature? A) -4.1 B) 4.1 C) 1.4 × 102 D) 1.7 E) More information is needed. ΔG° = -RT ln K ΔG° = - 8.314 J/mol K (303 K) ln 0.5 ΔG° = 1,746.14 J/mol ΔG° = 1.75 kJ/mol 5. The equilibrium constant for the following reaction is 5.0 × 108 at 25°C. N 2 (g) + 3H 2 (g) 2NH3 (g) The value of ΔG° for this reaction is __________ kJ/mol. A) -4.2 B) -22 C) 22 D) -50 E) -25 ΔG° = -RT ln K ΔG° = - 8.314 J/mol K (298 K) ln 5.0 × 10 8 ΔG° = -49,626.06 J/mol ΔG° = -49.6 kJ/mol 6. Given the thermodynamic data in the table below, calculate the equilibrium constant for the reaction:
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2SO2 (g) + O2 (g) 2SO3 (g) Substance ΔHf° (kJ/mol) ΔS° (J/mol U K) SO2 -297 249 O2 0 205 SO3 -395 256 A) 3.82 × 1023 B) 1.06 C) 1.95 D) 32 × 1024 E) More data are needed. Assuming temperature is 25 0 C: H 0 = Σ n H 0 f of products - Σ n H 0 f of reactants H 0 = [2(-395)] – [1(0) + 2(-297)] H 0 = -196 kJ/mol S 0 reaction = Σ n S 0 of products - Σ n S 0 of reactants S 0 = [2(256)] – [1(205) + 2(249)] S
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This note was uploaded on 02/06/2011 for the course CHEM 115L taught by Professor Nakkiew,pichaya during the Spring '10 term at Grand Canyon.

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Exam+review+24Nov09 - 1 Under which temperature conditions...

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