HW Key _11 s problems part b

HW Key _11 s problems part b - S40. Using the thermodynamic...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
S40. Using the thermodynamic data from your textbook, calculate Δ S° at 298K for each of the following reactions. Explain the sign of each of the entropy changes by qualitatively comparing the molecular disorder in the reactants and products. To find Δ S°, we need to take the product of the number of moles and the standard molar entropies and sum up those results for both the reactants and the product, then take the difference of the two sums according to the equation Δ S ° = nS m ° (products) nS m ° (reactants) a) Cl 2 (g) 2Cl(g) Δ S ° = (2 moles) × (165.2 J/mole K) (1 mole) × (223.07 J/mole K) =107.33 J/K The sign is positive because chlorine atoms are more randomly distributed (they have a larger number of microstates and hence greater disorder) than the paired atoms or Cl 2 . Therefore going from Cl 2 to 2 Cl increases the entropy of the system. b) H 2 O(l) H 2 O(g) Δ S ° = (1 mole) × (188.83 J/mole K) (1 mole) × (69.91 J/mole K) =118.92 J/K
Background image of page 1

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

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 02/01/2011 for the course CHEM 111 taught by Professor Owens,g during the Fall '08 term at University of Utah.

Page1 / 2

HW Key _11 s problems part b - S40. Using the thermodynamic...

This preview shows document pages 1 - 2. Sign up to view the full document.

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