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Chem 161-2009 Chapter 6 - Some Tavss' practice problems

Chem 161-2009 Chapter 6 - Some Tavss' practice problems -...

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CHEM 161-2010 CHAPTER 6 - THERMOCHEMISTRY PRACTICE PROBLEMS DR. ED TAVSS Heat and work Enthalpy Calorim. & Heat capacity (Also, see Chapter 10 - “Phase Changes and Diagrams”) Hess’s Law (Also see Chapter 8 - “Bond Energies & Lengths”) Enthalpy of form’n 1

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HEAT AND WORK 15 Chapter 6 – Thermochemistry Heat and Work What volume change will a system undergo if it absorbs 4.55kJ of heat and ΔU is 3.26kJ and the external pressure is 0.833atm? A. The system will be compressed by 15.3 L B. The system will be compressed 1.55L C. The system will expand by 1.55L D . The system will expand by 15.3 L E. The system will expand by 1.29L ∆E = q + w ∆E = q – P∆V ∆V = (∆E –q)/-P ∆V = (3260J – 4550J)/(-0.833 atm) Convert J into Latm 3260J x (1Latm/101.3J) = 32.18 Latm 4550J x (1Latm/101.3J) = 44.92 Latm ∆V = (32.18 Latm – 44.92 Latm)/(-0.833 atm) = +15.29 L Since ∆ means final minus initial, then +15.29 L means an expansion of 15.29 L. Note: I think that the key, which says “B”, is incorrect. D HILL & PETRUCCI CHAPTER 6 THERMODYNAMICS HEAT AND WORK 10. A gas absorbs 125 J of heat and expands from 2.00 L to 5.00 L under a constant pressure of 1.00 atm. Calculat U for this process. (a) 122 J (b) 429 J (c) -179 J (d) -122 J (e) 347 J ∆E = q – P∆V q = +125J 2
-P∆V = -1.00 atm x (5.00L – 2.00L) = -3.00 Latm Convert to Joules: -3.00 Latm x (101.3 J/Latm) = -304 J ∆E = +125J -304J = -179J 19 Chapter 6 – Thermodynamics Heat and work Assume the sign convention of the text, in which w is positive when work is done on the system. In which of the following processes is the sign of “w” negative? X. A gas is compressed under 2.0 atm pressure Y. A substance undergoes combustion in a bomb calorimeter (constant volume). Z. 1 mole of liquid water vaporizes to 1 mole of gaseous water under atm pressure A. X only B. X and Y only C . Z only D. Y only E. X and Z only X. When a gas is compressed this is work done on the system, and the energy of the system increases; w is positive. Y. When a substance undergoes combustion in a bomb calorimeter, there is no change in volume. Since w = -p∆V, and ∆V = 0, then w = 0, which is neither positive nor negative. Z. When 1 mole of liquid water vaporizes to 1 mole of gaseous water under atm pressure, the gas expands from 0 L to 22.4 L. Expansion is positive work on the environment, and resulting increase in energy of the environment, so it is therefore negative work and negative energy on the system. 3

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37 CHAPTER 6 - THERMODYNAMICS Heat and work Which reaction should have a E>0 under conditions of constant temperature and pressure? This problem is flawed. It should have said, “Which reaction should have a E>0 under conditions of constant pressure and q = 0? A . CO(g) + Cl 2 (g) COCl 2 (g) B. CO(g) + H 2 O(g) H 2 (g) + CO 2 (g) C. H 2 O(s) H 2 O(g) D. Fe 2 O 3 (s) + 3CO(g) 2Fe(s) + 3CO 2 (g) E. 4CuO(s) 2Cu 2 O(s) + O 2 (g) Solving the problem with the corrected version of the problem : E = q + w E = q - P V q = 0 E = - P V E = - P(V f – V i ) V f = the volume of gas in the products. V i = the volume of gas in the reactants.
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Chem 161-2009 Chapter 6 - Some Tavss' practice problems -...

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