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.
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.
To calculate the entropy change when liquid glycol freezes, we can use the equation:
ΔS = ΔH / T
Where:
ΔS is the entropy change
ΔH is the heat of fusion of glycol
T is the temperature in Kelvin
First, we need to convert the freezing temperature from -25 ˚C to Kelvin. The formula to convert Celsius to Kelvin is:
T(K) = T(°C) + 273.15
So,
T = -25 + 273.15 = 248.15 K
Now we can substitute the values into the equation:
ΔS = 2688 cal mol-1 / 248.15 K
Calculating this will give you the entropy change in cal K-1 mol-1.