"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?
Homework 9.45
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Re: Homework 9.45
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
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Re: Homework 9.45
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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