Standard Entropy calculation

In summary, the liquid has a vapour pressure of -3571 mmHg at the melting point and a melting point of 392.7 K. The standard entropy of the liquid is 174.07 J/mol K.
  • #1
DiffusConfuse
22
0

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 Cp value given for the liquid is 250 J/mol K
and theΔSvap 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
 
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  • #2
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?
 
  • #3
dauto said:
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.
 
  • #4
Yes, thank you haruspex, it is Log of the Pressure in mmHG and assuming Kelvin for temperature units
 
  • #5
DiffusConfuse said:
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.
 
  • #6
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)
 
  • #7
for some reason I cannot edit the formula above, it should be:

Log10 P (mmHg) = -3571/T + 8.999
 
  • #8
DiffusConfuse said:
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.
 
  • #9
I rechecked it, I get 8222.53 instead of what is listed above, which doesn't change my final answer drastically
 
  • #10
DiffusConfuse said:
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.
 
  • #11
DiffusConfuse said:
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.
 

1. What is the definition of Standard Entropy?

Standard Entropy is a measure of the disorder or randomness of a system at standard state, which is defined as 1 bar of pressure and a specific temperature (usually 298 K or 25°C).

2. How is Standard Entropy different from Entropy?

Standard Entropy is a specific value for a substance at standard state, whereas Entropy is a general term for the measure of disorder in a system. Standard Entropy also takes into account the standard state conditions, while Entropy does not.

3. What is the unit of measurement for Standard Entropy?

The unit of measurement for Standard Entropy is joules per mole-kelvin (J/mol·K). This unit represents the amount of energy required to increase the disorder of 1 mole of a substance by 1 degree Kelvin at standard state.

4. How is Standard Entropy calculated?

Standard Entropy is calculated by using the change in heat and temperature for a reaction or process at standard state. The equation is ΔS° = q/T, where ΔS° is the change in Standard Entropy, q is the heat transferred, and T is the temperature in Kelvin.

5. Why is Standard Entropy important in chemistry?

Standard Entropy is important in chemistry because it helps us understand the thermodynamics of a reaction or process. It can provide insight into the spontaneity of a reaction and the amount of energy that is available to do work. It is also a useful tool for predicting and comparing the stability of different substances.

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