: The heat of dissociation per mole of gaseous water at 18 °C and 1 atm is 241750 J/mol. Calculate its value at 68 °C. H,0(g) → H2(9) + 0-(9) Data given are Cp(H20) = 33.56 JK-ªmol¬1 Cp(H2) = 28.83 JK-!mol-1 С- (0) %3D 29.12 Jк--mol-1 %3D
: The heat of dissociation per mole of gaseous water at 18 °C and 1 atm is 241750 J/mol. Calculate its value at 68 °C. H,0(g) → H2(9) + 0-(9) Data given are Cp(H20) = 33.56 JK-ªmol¬1 Cp(H2) = 28.83 JK-!mol-1 С- (0) %3D 29.12 Jк--mol-1 %3D
Chemistry by OpenStax (2015-05-04)
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Author:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark Blaser
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Chapter5: Thermochemistry
Section: Chapter Questions
Problem 17E: Would the amount of heat absorbed by the dissolution in Example 5.6 appear greater, lesser, or...
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Question
![: The heat of dissociation per mole of gaseous water at 18 °C and 1 atm is 241750
J/mol. Calculate its value at 68 °C.
H,0(g) → H2(9) + 0-(9)
Data given are
Cp(H20) = 33.56 JK-ªmol¬1
Cp(H2) = 28.83 JK-!mol-1
С- (0) %3D 29.12 Jк--mol-1
%3D](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7ec88b1b-0b88-41d6-8038-33af9680a315%2F8e087994-dae4-486b-ae23-d6d040592f05%2Foaiyabl.png&w=3840&q=75)
Transcribed Image Text:: The heat of dissociation per mole of gaseous water at 18 °C and 1 atm is 241750
J/mol. Calculate its value at 68 °C.
H,0(g) → H2(9) + 0-(9)
Data given are
Cp(H20) = 33.56 JK-ªmol¬1
Cp(H2) = 28.83 JK-!mol-1
С- (0) %3D 29.12 Jк--mol-1
%3D
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