How Do You Calculate Standard Entropy at the Melting Point?

[DiffusConfuse]

Homework Statement

vapour pressures of a liquid have been measured and fit to the following equation:
Log₁₀ (mmHg) = -3571/T + 6.124
The melting point has been determined to be 392.7 K.
A C_p value given for the liquid is 250 J/mol K
and theΔS_vap is 117 J/mol K

Homework Equations

The Attempt at a Solution

1.
Calculate the standard entropy of the liquid at the melting point.


2. Homework Equations



3. The Attempt at a Solution

T₁= 392.7 K P₁ = 0.8045 mmHg (from the equation, giving the vapour pressure of the LIQUID at the melting point).
T₂= 298.15 K P₂= 1.052E-3 mmHg

From the Clausius Clapeyron:

ln(0.8045/1.052E-3)= ΔH/R*(1/298.15 K - 1/392.7 K)

ΔH/R = 8221.89
ΔH = 68361 J/mol

ΔH/T = ΔS = 68361/392.7 = 174.07 J/mol K

However the answer is about twice this value

 

[dauto]

I'm getting different values for P₁ and P₂, based on the equation Log₁₀ (mmHg) = -3571/T + 6.124.

I find that equation particularly annoying because strictly speaking it makes no sense. What is the logarithm of a unit?

 

[haruspex]

I find that equation particularly annoying because strictly speaking it makes no sense. What is the logarithm of a unit?

It's OK. The value 6.124 is specific to the assumption that the pressure is measured in mmHg. Different units, different constant. Likewise, the 3571 assume degrees Kelvin.

 

[DiffusConfuse]

Yes, thank you haruspex, it is Log of the Pressure in mmHG and assuming Kelvin for temperature units

 

[haruspex]

Yes, thank you haruspex, it is Log of the Pressure in mmHG and assuming Kelvin for temperature units

Good, but I agree with dauto that there's something wrong with your calculation of P1 and P2. If you can't find an error, please post the details of your working.

 

[DiffusConfuse]

Just noticed an error in the formula I gave. The +6.124 is for the pressure in bar, when I gave it in mmHg.
I will correct it to the given data in mmHg (+8.999)

 

[DiffusConfuse]

for some reason I cannot edit the formula above, it should be:

Log10 P (mmHg) = -3571/T + 8.999

 

[haruspex]

for some reason I cannot edit the formula above, it should be:

Log10 P (mmHg) = -3571/T + 8.999

OK, but I don't get your value for ΔH/R from that using your equations. I may well have made a mistake, but perhaps you could check yours.

 

[DiffusConfuse]

I rechecked it, I get 8222.53 instead of what is listed above, which doesn't change my final answer drastically

 

[dauto]

Yes, thank you haruspex, it is Log of the Pressure in mmHG and assuming Kelvin for temperature units

Yes, I know that. Doesn't keep me from feeling annoyed. It's very poor form.

 

[micromass]

for some reason I cannot edit the formula above

That is for a reason. It is discouraged in this forum to edit your posts after you received replies. This for the simple reason that the other posts will look strange and off-topic if you edited out mistakes from your original post. We prefer that you just post the corrections in a new post and leave the original post as it is.