Ch. 17 - Chemical ThermodynamicsWorksheetSee all chapters
All Chapters
Ch.1 - Intro to General Chemistry
Ch.2 - Atoms & Elements
Ch.3 - Chemical Reactions
BONUS: Lab Techniques and Procedures
BONUS: Mathematical Operations and Functions
Ch.4 - Chemical Quantities & Aqueous Reactions
Ch.5 - Gases
Ch.6 - Thermochemistry
Ch.7 - Quantum Mechanics
Ch.8 - Periodic Properties of the Elements
Ch.9 - Bonding & Molecular Structure
Ch.10 - Molecular Shapes & Valence Bond Theory
Ch.11 - Liquids, Solids & Intermolecular Forces
Ch.12 - Solutions
Ch.13 - Chemical Kinetics
Ch.14 - Chemical Equilibrium
Ch.15 - Acid and Base Equilibrium
Ch.16 - Aqueous Equilibrium
Ch. 17 - Chemical Thermodynamics
Ch.18 - Electrochemistry
Ch.19 - Nuclear Chemistry
Ch.20 - Organic Chemistry
Ch.22 - Chemistry of the Nonmetals
Ch.23 - Transition Metals and Coordination Compounds

Solution: Consider the following reaction at constant pressure. Use the information here to determine the value of ∆Ssurr at room temperature. Predict whether or not this reaction will be spontaneous at this temperature. N2 (g) + 2 O2 (g) → 2 NO2 (g)              ∆H° = + 66.4 kJ ∆Ssurr = - 223J/K, not spontaneous ∆Ssurr = + 265 J/K, not spontaneous ∆Ssurr = + 223 J/K, is spontaneous ∆Ssurr = - 66.4 J/K, not spontaneous ∆Ssurr = - 66.4 J/K, not possible to predict the spontaneity 

Problem

Consider the following reaction at constant pressure. Use the information here to determine the value of ∆Ssurr at room temperature. Predict whether or not this reaction will be spontaneous at this temperature.

N(g) + 2 O(g) 2 NO(g)              ∆H° = + 66.4 kJ

  1. ∆Ssurr = - 223J/K, not spontaneous
  2. ∆Ssurr = + 265 J/K, not spontaneous
  3. ∆Ssurr = + 223 J/K, is spontaneous
  4. ∆Ssurr = - 66.4 J/K, not spontaneous
  5. ∆Ssurr = - 66.4 J/K, not possible to predict the spontaneity