Homework 9.45


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aristotelis1H
Posts: 16
Joined: Fri Sep 25, 2015 3:00 am

Homework 9.45

Postby aristotelis1H » Mon Jan 25, 2016 9:12 pm

"9.45 Use the information in Table 8.3 to calculate the changes in entropy f the surroundings and of the system for (a) the vaporization of 1.00 mol CH4(l) at its normal boiling point; (b) the melting of 1.00 mol C2H5OH(s) at its normal melting point; (c) the freezing of 1.00 mol C2H5OH(l) at its normal freezing point"

When solving for part (c) and using

ΔSsys =

Is the ΔHofus value negative because heat is released from a system when freezing an object?

Dahriel Aron 3A
Posts: 60
Joined: Fri Sep 25, 2015 3:00 am

Re: Homework 9.45

Postby Dahriel Aron 3A » Mon Jan 25, 2016 9:15 pm

for this problem, I was also wondering why they divide the enthalpy of fusion for ethanol by the freezing point and not the melting point to find its melting entropy

Jacob Afable 3J
Posts: 41
Joined: Fri Sep 25, 2015 3:00 am

Re: Homework 9.45

Postby Jacob Afable 3J » Tue Jan 26, 2016 10:14 pm

aristotelis1H wrote:"9.45 Use the information in Table 8.3 to calculate the changes in entropy f the surroundings and of the system for (a) the vaporization of 1.00 mol CH4(l) at its normal boiling point; (b) the melting of 1.00 mol C2H5OH(s) at its normal melting point; (c) the freezing of 1.00 mol C2H5OH(l) at its normal freezing point"

When solving for part (c) and using

ΔSsys =

Is the ΔHofus value negative because heat is released from a system when freezing an object?


ΔHofus indicates the amount of energy it takes to melt C2H5OH(l). So the value would be negative since the process is reversed (freezing). E.g. A -Hofus is used when changing water to ice.


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