Chemistry 360 Dr. Jean M. Standard Spring 2015 March 18, 2015 Name __________________________________ Exam 2 – 100 points Note: You must show your work on problems in order to receive full credit for any answers. You must turn in your equation sheet along with this exam in order to receive full credit for the exam. Please turn off cell phones and store them during the exam. 1.) (14 points) Using the values in the table below reported at 25°C, determine ΔGR! at 300ºC for the hydrogenation of acetylene to ethane, C2 H 2 ( g) + 2 H 2 (g) → C2 H 6 (g) . ΔH !f (kJ/mol) C2H2 (g) C2H6 (g) H2 (g) € ΔG !f (kJ/mol) 227.4 209.2 –84.0 –32.0 0 0 2 2.) (14 points) Consider a classroom that is 5 m × 10 m × 3 m in size. Initially, the temperature in the room is 20°C and the pressure is 1 bar. There are 20 people in the class. Each person gives off heat to the room at an average rate of 150 Watts [1 Watt = 1 Joule/second]. Assume that the walls, ceiling, floor, and furniture are perfectly insulating and do not absorb any heat; that is, all the heat from the people in the classroom is absorbed by the air in the room. a.) Determine the heat (in Joules) required to raise the room temperature to 37°C. Assume that the pressure remains constant at 1 bar, the air behaves ideally, and the molar constant pressure heat capacity of air is 7 C p,m = R . 2 b.) How long in minutes will it take for the air in the classroom to reach 37°C? c.) Determine the change in entropy (in J/K) of the air in the classroom. 3 3.) (14 points) Consider the following standard molar entropies at 298 K. ! Sm (J/molK) Compound CO (g) CO2 (g) C2H4 (g) C2H6 (g) 197.7 213.8 € 219.3 229.2 Explain why we are able to obtain absolute molar entropies of substances; that is, what is the theoretical foundation that allows the determination of absolute molar entropies? Discuss and explain any trends observed in the molar entropies of the substances given in the table above. 4 4.) (15 points) True/false, short answer, multiple choice. a.) True or False: The reaction 2 H 2 (g) + O 2 (g) → 2 H 2O ( ℓ) is an example of a formation reaction. b.) True or False: In the statistical definition of entropy, S = kB lnW , the quantity W corresponds to the energy. c.) Short answer The _______________________________ equals H − TS . d.) Short answer The ____________________________________ states that that ΔS ≥ 0 for an isolated system. e.) Multiple Choice: A spontaneous process at constant temperature and pressure must have: 1) ΔU < 0 . 2) ΔH < 0 . 3) ΔA < 0 . 4) ΔG < 0 . 5 5.) (14 points) Determine ΔS for the isothermal compression at 300 K from 100 L to 10 L of two moles of a real gas obeying the equation of state given by € Z = 1 + B , Vm where Z is the compression factor and B is a constant. Assume that B = −0.122 L/mol . € 6 6.) (14 points) The standard molar enthalpy of combustion of graphite is molar enthalpy of combustion of C2H2 (g) is ΔH R! ΔH R! = − 393.51 kJ/mol and the standard = −1299.58 kJ/mol , both at 298 K. Use this information, along with the standard molar enthalpy changes for the following reactions at 298 K, (1) CaC 2 (s) + (2) Ca (s) (3) CaO (s) 2 H 2O (ℓ) 1 2 + + O 2 (g) H 2O (ℓ) → → → Ca(OH) 2 (s) + C 2 H 2 (g) CaO (s) Ca(OH) 2 (s) to determine the standard molar enthalpy of formation of CaC2 (s) at 298 K. € ΔH 1" = − 127.9 kJ/mol ΔH 2" = − 635.1 kJ/mol ΔH 3" = − 65.2 kJ/mol , 7 7.) (15 points) True/false, short answer, multiple choice. a.) True or False: For the reaction 2 CO (g) + O2 (g) → 2 CO2 (g), the entropy change is expected to be negative. !∂ G $ b.) True or False: The expression – # & is equal to the volume V. " ∂ P %T c.) Short answer The ______________________________ is equivalent to the isothermal work under reversible conditions. d.) Short answer The differential dH = T dS + V dP corresponds to one of four _________________________________ . e.) Multiple Choice: The molar constant pressure heat capacity of a crystalline solid at very low temperatures is given by the expression: 1) C p,m = aT 2 . T1 2) C p,m = ∫ 0 aT 2 dT . T 3) C p,m = aT 3. T1 4) C p,m = ∫ 0 aT 3 dT . T 8 PHYSICAL CONSTANTS, CONVERSION FACTORS, AND EQUATIONS R = 0.08206 L atm mol–1 K–1 = 8.314 J mol–1 K–1 = 0.08314 L bar mol–1 K–1 1 atm = 101325 Pa = 1.013 bar = 760 torr 1 bar = 105 Pa = 0.98692 atm 1 L atm = 101.3 J; 1 L bar = 100 J 1 cal = 4.184 J 1 L = 1000 mL = 1000 cm3 = 1×10–3 m3 Fundamental Equations dU = T dS − P dV dH = T dS + V dP dA = − S dT − P dV dG = − S dT + V dP Maxwell€Relations #∂ T & #∂ P & % ( = −% ( $ ∂ V 'S $ ∂ S 'V #∂ T & #∂ V & % ( = % ( $ ∂ P 'S $ ∂ S 'P #∂ S & #∂ P & % ( = % ( $ ∂ V 'T $ ∂ T 'V #∂ S & #∂ V & % ( = −% ( $ ∂ P 'T $ ∂ T 'P €