2 IClg \u0394 Ho 269 kJ and \u0394S o 113 JK Calculate the equilibrium constant K for

# 2 iclg δ ho 269 kj and δs o 113 jk calculate the

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2 ICl(g) Δ H o = -26.9 kJ and Δ S o = 11.3 J/K. Calculate the equilibrium constant K for this reaction at 298 K. A. 2.3 x 10 3 B. 7.2 x 10 8 C . 2.0 x 10 5 D. 5.2 x 10 17 E. 5.0 x 10 -6 Δ G o = Δ H o - T Δ S o Δ G o = -26900 – (298 x 11.3) = -3.027 x 10 4 J Δ G o = -RTln(K) -3.027 x 10 4 = -8.314 x 298 x ln(K) X = 2.023 x 10 5 = K Chem 162-2012 Exam III review 87 Chem 162-2011 Exam III + Answers Chapter 17 - Thermodynamics Free Energy And Equilibria Calculations 25*. Calculate Δ G for the following reaction at 1200 K and 3.29 x 10 -3 atm of Cl 2 (g). 2AgCl(s) 2Ag(s) + Cl 2 (g) Δ G o = 220 kJ A . 163 kJ B. 190 kJ C. 181 kJ D. 220 kJ E. 277 kJ Δ G = Δ G o + RT ln(Q) Q = P Cl2 Δ G = 220000 + (8.314 x 1200 x (ln(3.29x10 -3 ))) = 1.63 x 10 5 J = 1.63 x 10 2 kJ Chem 162-2012 Exam III review 88 3 Skip Chem 162-2011 Final exam Chapter 17 – Thermodynamics Free energy calculations The enthalpy change of a particular reaction is 15 kJ/mol at 27 °C. This reaction will be spontaneous if the entropy change is: A . > 50J mol -1 K -1 B. < -50J mol -1 K -l C. 15 J mol -l K -l D. -50 J mol -l K -l E. -15 J mol -1 K -1 Δ G = Δ H – T Δ S First determine the value of Δ S when the reaction is at equilibrium. Then determine the value of Δ S when the reaction is spontaneous. 0 = 15000 – (300 x Δ S) Δ S = 50J mol -1 K -1 when the reaction is at equilibrium. If Δ S is greater than 50J mol -1 K -1 , then Δ G would be negative, meaning that the reaction would be spontaneous. Chem 162-2012 Exam III review 89 Skip Chem 162-2011 Exam III + Answers Chapter 17 - Thermodynamics Free Energy And Equilibria Concepts 13. If the numerical value of the equilibrium constant increases as the temperature increases then: A. Δ S° < 0 B. Δ H° < 0 C. Δ G° < 0 D . Δ H° > 0 E. Δ S° > 0 LeChatelier’s principle states that for an endothermic reaction, if the temperature increases then the reaction goes to the right and K increases. (This is supported by the van’t Hoff equation.) Hence, Δ H° > 0. A + B + heat C + D Chem 162-2012 Exam III review 90 38 Skip Chem 162-2011 Final exam Chapter 17 – Thermodynamics Free energy and equilibria concepts Which one of the following statements is true about the equilibrium constant for a reaction if G ˚ for the reaction has a negative value? A. K= 0 B. K = 1 C. K = -1 D. K < 1 E . K > 1 G ˚ = 0 K = 1 G ˚ > 0 K < 1 G ˚ < 0 K > 1 Note that the superscript zero means that everything is being measured under standard state conditions, which means that all reactants and products are at 1M or 1atm. If G ˚ for the reaction is zero, then the reaction is at equilibrium, and K = 1 (i.e., everything is at 1 atm or 1 molar). If G ˚ for the reaction is positive, then the reaction is not spontaneous (or spontaneous to the left), and K<1. If G ˚ for the reaction is negative, then the reaction is spontaneous (or spontaneous to the right), and K > 1. E Chem 162-2012 Exam III review 91 Skip Chem 162-2011 Exam III + Answers Chapter 17 - Thermodynamics Free Energy And Equilibria Calculations 9 . Consider the reaction: Fe 2 O 3 (s) + 3H 2 (g) 2Fe(s) + 3H 2 O(g) Given: Δ H° = 100 kJ and Δ S° = 138 J/K, at what temperature would the equilibrium constant K = 1? A. 465 K B. 685 K C . 725 K D. 575 K E. 845 K Δ G o = Δ H o - T Δ S o Δ G o = -RT ln(K) Δ H o - T Δ S o = -RTln(K) Rearranging terms: T = - Δ H o /(- Δ S o + Rln(K)) T = -100000/(-138 + (8.314 x ln(1))) T = 725K This also could have been solved in a simpler manner.  #### You've reached the end of your free preview.

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