Page 1 of 1

Homework 9.45

Posted: Mon Jan 25, 2016 9:12 pm
by aristotelis1H
"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?

Re: Homework 9.45

Posted: Mon Jan 25, 2016 9:15 pm
by Dahriel Aron 3A
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

Re: Homework 9.45

Posted: Tue Jan 26, 2016 10:14 pm
by Jacob Afable 3J
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.