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Consider the melting of one mole of water. Use the following data to answer the questions:S ? =( H 2 O, l)=69.9J/K.mol, S ? ( H 2 O, s)=47.8J/K.mol, ?H?( H 2 O, l)=-285.8kJ/mol, and ?H?( H 2...

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Consider the melting of one mole of water. Use the following data to answer the questions:S?=( H2O, l)=69.9J/K.mol, S?( H2O, s)=47.8J/K.mol, ?H?( H2O, l)=-285.8kJ/mol, and ?H?( H2O, s)=-298.8kJ/mol. Assume standard molar entropies and enthalpies of formation are not functions of temperature in this range. a) calculate the entropy change of the system for melting ice at -5?C. b) Calculate the entropy change of the surrounding for melting ice at -5?Cand C) Is this reaction spontaneous at -5?C? Justify.

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The heat capacity is critical in calculating the enthalpy and entropy. Using the data (heat capacity) from book or internet and also?fusH (I2) = 15.52387kJ/mol at 387K and?vap (I2) =41.57kJ/mol at 457K.a) Calculate the heat required to bring one mole of iodine from 0K to 500k. b) calculate the absolute entropyof iodine at 500K
Answered Same Day Dec 22, 2021

Solution

Robert answered on Dec 22 2021
129 Votes
Problem 1
Given data is
 
 
 
 
o 1
f 2
o 1
f 2
o 1 1
2
o 1 1
2
H H O( ) 285.8KJmol
H H O(s) 298.8KJmol
S H O( ) 69.9JK mol
S H O(s) 47.8JK mol


 
 
  
  



The entropy change of the system of ice melting o@ 5 C is given by
   o o osystem 2 2S S H O( ) S H O(s)
69.9 47.8
22.1J
  
 


Now, the change in the entropy of the su
ounding is given by
  
o
systemo
su
ounding
3
3
1 1
H
S
T
285.8 298.8 x10
268.15
13x10
268.15
48.48JK mol 

  
  
 
 
 

Hence, the change in entropy of the universe is
o o o
universe system su
ounding
1 1
1 1
S S S
(22.1 48.48) JK mol
26.38JK mol
 
 
   
 
 
Since ouniverseS 0  , the transition of ice to water
o
@ 5 C (or melting) is not a spontaneous process.
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