I'm a little confused about the question because you show the specific heats but don't list a starting T. So I will ignore those specific heats.
Then dG = dH - TdS
dG at freezing is an equilibrium and = 0 so
0 = dH - TdS
dH is 2688
T is -25. Convert to K.
Solve for dS.
What is the entropy change of liquid glycol (in cal K-1 mol-1
) when it freezes at -25
˚C?the heat of fusion of glycol is 2688 cal mol-1
,
and the specific heats of liquid and solid glycol are 33.8 and 20.1 cal K-1 mol-1
,
respectively.
1 answer