chapter_2018

chapter_2018 - CHAPTER 18 THERMODYNAMICS 18.6 The...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
CHAPTER 18 THERMODYNAMICS 18.6 The probability ( P ) of finding all the molecules in the same bulb becomes progressively smaller as the number of molecules increases. Based on the information given in the problem, we can come up with an equation relating the probability to the number of molecules present. 1 2 ⎛⎞ = ⎜⎟ ⎝⎠ N P where, N is the total number of molecules present. Using the above equation, we calculate the probability of finding 100 molecules in the same bulb to be: 100 1 2 == 31 81 0 P × Extending the calculation to Avogadro’s number of molecules would result in such a small number that for all practical purposes, there is zero probability that all molecules would be found in the same bulb. 18.9 (a) This is easy. The liquid form of any substance always has greater entropy (more microstates). (b) This is hard. At first glance there may seem to be no apparent difference between the two substances that might affect the entropy (molecular formulas identical). However, the first has the O H structural feature which allows it to participate in hydrogen bonding with other molecules. This allows a more ordered arrangement of molecules in the liquid state. The standard entropy of CH 3 OCH 3 is larger. (c) This is also difficult. Both are monatomic species. However, the Xe atom has a greater molar mass than Ar. Xenon has the higher standard entropy. (d) Same argument as part (c). Carbon dioxide gas has the higher standard entropy (see Appendix 2). (e) O 3 has a greater molar mass than O 2 and thus has the higher standard entropy. (f) Using the same argument as part (c), one mole of N 2 O 4 has a larger standard entropy than one mole of NO 2 . Compare values in Appendix 2. Use the data in Appendix 2 to compare the standard entropy of one mole of N 2 O 4 with that of two moles of NO 2 . In this situation the number of atoms is the same for both. Which is higher and why? 18.10 In order of increasing entropy per mole at 25 ° C: (c) < (d) < (e) < (a) < (b) (c) Na( s ): ordered, crystalline material. (d) NaCl( s ): ordered crystalline material, but with more particles per mole than Na(s). (e) H 2 : a diatomic gas, hence of higher entropy than a solid. (a) Ne( g ): a monatomic gas of higher molar mass than H 2 . (b) SO 2 ( g ): a polyatomic gas of higher molar mass than Ne.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
CHAPTER 18: THERMODYNAMICS 424 18.11 Using Equation (18.7) of the text to calculate rxn Δ S o (a) rxn 2 2 ( SO) [ (O) ( S ) ] Δ= ° ° + ° SS S S o rxn (1)(248.5 J/K mol) (1)(205.0 J/K mol) (1)(31.88 J/K mol) Δ = ⋅− = 11.6 J/K mol S o (b) rxn 2 3 (MgO) (CO ) (MgCO ) ° + ° ° S S o rxn (1)(26.78 J/K mol) (1)(213.6 J/K mol) (1)(65.69 J/K mol) Δ = ⋅+ = 174.7 J/K mol S o 18.12 Strategy: To calculate the standard entropy change of a reaction, we look up the standard entropies of reactants and products in Appendix 2 of the text and apply Equation (18.7). As in the calculation of enthalpy of reaction, the stoichiometric coefficients have no units, so rxn Δ S o is expressed in units of J/K mol.
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 24

chapter_2018 - CHAPTER 18 THERMODYNAMICS 18.6 The...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online