PROBLEM SOLVING NUMBER 1 The decomposition of CaCO3 gives a positive AS, which means that the system is in a more random state undèrstandard conditions than when CaCO3 is formed (reverse reaction)- As a rule, a reaction that produces more number of moles of gas results in an increase in its entropy. A reactionthat produces less number of moles of gas will result in a decrease in entropy.Sample Problem 2 The Haber process for the industrial synthesis of ammonia is another process that involves gaseousspeciesN2 ( + 3H2 )2NH3 (2)NH3 (2)SubstanceS (J/mol-K)130.6192.3191.5The AS is obtained using the following equation.[(H) SE + (N) s1-[(HN) S7] =SV= [(2 mol)(192.3 J/mol-K)]-[(lmol)(191.5 J/mol-K) + (3mol)(130.6 J/mol-K)%3D%3D= - 198.7 J/K- The negative AS° value results from the decrease in the number of moles of gas.Activity 1. Problem SolvingDirections: Calculate the AS for each given reaction. Show your complete solution.1.) 2H2 + 02+ 02 (g(3) 0'H7SubstanceS (J/mol-K)205.0188.7130.6Lesson 3 Second Law of ThermodynamicsV The spontaneity of a process depends on the total change in entropy.Page | 3

Change in entropy ΔS0 for a reaction is given by, ΔS0rxn = Σ S0products - Σ S0reactants

activity 1. Problem Solving Directions: Calculate the AS for each given reaction. Show your complete solution. 1.) 2H2 (g) + 02@ + O2 (g) → 2H20 (g) H2 (2) O2 (e) (g) 188.7 Substance S (J/mol-K) 130.6 205.0

activity 1. Problem Solving Directions: Calculate the AS for each given reaction. Show your complete solution. 1.) 2H2 (g) + 02@ + O2 (g) → 2H20 (g) H2 (2) O2 (e) (g) 188.7 Substance S (J/mol-K) 130.6 205.0

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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