PROBLEM SOLVING NUMBER 2 Therefore, at 25°C (298.15 K),-AHASsurrrxn-(-1648400 D5528.8 J/K%3D%3DT298.15 KTo get the total entropy of the process (ASuniv),[ASsys +ASurr 1=-549.5 J/K + 5528.8 J/K= 4979.3 J/K- Notice that the change in entropy of the reaction is negative, but the entropy change in the surroundingsis large enough such that the total entropy of the process is positive. This indicates that the oxidation ofiron is a spontaneous process at standard conditions.Activity 2 Problem SolvingDirections: Calculate the total entropy change ASuniv and determine if it is spontaneous or not.1. Calculate the total entropy change for the decomposition of hydrogen peroxide (H,O2). Is the reactionspontaneous or not? Hint: 1kj = 1000JAUm(1)Page | 4

Answered: Image /qna-images/answer/c5b0026d-e07e-49ef-ae25-3ae5e91c3260.jpg

Activity 2 Problem Solving Directions: Calculate the total entropy change ASuniv and determine if it is spontaneous or not. 1. Calculate the total entropy change for the decomposition of hydrogen peroxide (H2O2). Is the reaction spontaneous or not? Hint: 1kj = 1000J %3D 2H.O2 0 → 2H.O + O2

Activity 2 Problem Solving Directions: Calculate the total entropy change ASuniv and determine if it is spontaneous or not. 1. Calculate the total entropy change for the decomposition of hydrogen peroxide (H2O2). Is the reaction spontaneous or not? Hint: 1kj = 1000J %3D 2H.O2 0 → 2H.O + O2

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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Question

100%

PROBLEM SOLVING NUMBER 2

Therefore, at 25°C (298.15 K),
-AH
ASsurr
rxn
-(-1648400 D
5528.8 J/K
%3D
%3D
T
298.15 K
To get the total entropy of the process (ASuniv),
[ASsys +ASurr 1
=-549.5 J/K + 5528.8 J/K
= 4979.3 J/K
- Notice that the change in entropy of the reaction is negative, but the entropy change in the surroundings
is large enough such that the total entropy of the process is positive. This indicates that the oxidation of
iron is a spontaneous process at standard conditions.
Activity 2 Problem Solving
Directions: Calculate the total entropy change ASuniv and determine if it is spontaneous or not.
1. Calculate the total entropy change for the decomposition of hydrogen peroxide (H,O2). Is the reaction
spontaneous or not? Hint: 1kj = 1000J
AUm
(1)
Page | 4

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GENERAL INSTRUCTION: 1. All answers to be entered via short answer are in the numerical format and presented up to the 3rd decimal place. Add trailing zeros if necessary, to complete your answer. NO NEED TO INCLUDE THE UNIT. For uniformity of answers, please be guided by the following examples: If your answer is 5.627594, enter 5.628 If your answer is 3.129934, enter 3.130 If your answer is 8.38, enter 8.380 If your answer is 1.2, enter 1.200 If your answer is 25, enter 25.000 If your answer is less than the value of 1, always place a zero before the decimal point (example: 0.5, enter 0.500) 2. In cases for answers in the exponential format, please be guided by the following examples and format: Do not convert your answer to decimal format If your answer from the calculator is 1.23454566 x 10E-4, enter 1.235x10^-4 (no spaces in between) If your answer from the calculator is 7.379566 x 10E-5, enter 7.38x10^-5 (no spaces in between) 3. For the atomic weights, it should be rounded to…