In freezing a mole of liquid water at the freezing point, the enthalpy of fusion is 6.01 KiloJoules per mole. Answer the questions that follow. (A). What is the entropy change of this process (in Joules per Kelvin)? Express answer in THREE SIGNIFICANT FIGURES. (B). What is the entropy change of the surroundings for this process (in Joules per Kelvin)? Express answer in THREE SIGNIFICANT FIGURES. (C). What is the total entropy change (or the entropy change of the universe) for this process (in Joules per Kelvin)? Express answer in THREE SIGNIFICANT FIGURES.

Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
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In freezing a mole of liquid water at the freezing point, the enthalpy of fusion is 6.01
KiloJoules per mole. Answer the questions that follow.
(A). What is the entropy change of this process (in Joules per Kelvin)? Express answer in
THREE SIGNIFICANT FIGURES.
(B). What is the entropy change of the surroundings for this process (in Joules per Kelvin)?
Express answer in THREE SIGNIFICANT FIGURES.
(C). What is the total entropy change (or the entropy change of the universe) for this process
(in Joules per Kelvin)? Express answer in THREE SIGNIFICANT FIGURES.
Transcribed Image Text:In freezing a mole of liquid water at the freezing point, the enthalpy of fusion is 6.01 KiloJoules per mole. Answer the questions that follow. (A). What is the entropy change of this process (in Joules per Kelvin)? Express answer in THREE SIGNIFICANT FIGURES. (B). What is the entropy change of the surroundings for this process (in Joules per Kelvin)? Express answer in THREE SIGNIFICANT FIGURES. (C). What is the total entropy change (or the entropy change of the universe) for this process (in Joules per Kelvin)? Express answer in THREE SIGNIFICANT FIGURES.
Expert Solution
Step 1

Given,

Hf=6.01 KJ/mol, n=1 mole, 

as given freezing point=273 K

(a) dS=nHfT=1×6.01×1000273=22.014 J/k

(b) Entropy change of surrounding,

dSsurrounding=-nHfT=-1×6.01×1000273=-22.014 J/k 

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