Standard entropy of a liquid at melting temperature

In summary: There might be additional steps, for example: 3. solid at 377 and equilibrium sublimation pressure at 377 --> gas at 3774. gas at 377 --> liquid at 392.7 and 1 atm5. liquid at 392.7 and 1 atm --> gas at 298 and 1 atmIn summary, the data suggests that the heat of vaporization is not constant, and there is no value given for the heat of fusion.
  • #1
DiffusConfuse
22
0
I am presented a review of data which gives:
vapour pressures of a liquid have been measured and fit to the following equation:
Log10 (mmHg) = -3571/T + 8.999
The melting point has been determined to be 392.7 K.
A Cp value given for the liquid is 250 J/mol K
and the ΔSvap is 117 J/mol K

The standard entropy S°(392.7) for the liquid is calculated, although I am unable to recalculate it correctly for verification.


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 K-1
ΔH = 68361 J/mol

ΔH/T = ΔS = 68361/298.15 = 229.29 J/mol K

However the answer is S°(392.7) = 350.60 J/mol K.

I tried adding the ΔSvap to the derived value (229.29+117.2) , however that only gives me 346.5 J/mol K
 
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  • #2
Is the standard reference state of the material a vapor at 298K and a hypothetical pressure of 1 atm?

Chet
 
  • #3
at 298K it is a solid
 
  • #4
DiffusConfuse said:
at 298K it is a solid
Is 298K and 1 atm (solid) the standard reference state for this material? The vapor pressure relationship suggests that the heat of vaporization is constant, but yet there is a heat capacity given, implying that the heat of vaporization is not constant? Is there any value given for the heat of fusion? The entropy of vaporization...is that from liquid to vapor? At what temperature? There doesn't seem to be enough information given to solve this problem.

Chet
 
  • #5
Yes this is the standard reference state of the material. Heat of vaporization is not constant.
The heat of fusion is unknown (no measured data).

The entropy of vaporization is from liquid to vapour at 460 K.

There is also vapour pressure data for the solid-gas; Log10 (mmHg) = -3571/T + 18.279, with entropy of sublimation ΔSsub = 295 J/ mol K at 377 K.
 
  • #6
DiffusConfuse said:
Yes this is the standard reference state of the material. Heat of vaporization is not constant.
The heat of fusion is unknown (no measured data).

The entropy of vaporization is from liquid to vapour at 460 K.

There is also vapour pressure data for the solid-gas; Log10 (mmHg) = -3571/T + 18.279, with entropy of sublimation ΔSsub = 295 J/ mol K at 377 K.
Is there any heat capacity data on the solid between 298 and 377?
Chet
 
  • #7
yes, it is 155 + 0.086T - 1.8[itex]\times[/itex]106 T-2 (J/mol K)
 
  • #8
OK. You're starting out with solid at 298 and 1 atm, and your final state is liquid at 392.7 and 1 atm. You need to dream up a sequence of reversible processes that takes you from state 1 to state 2, and you need to calculate the change in entropy for each of these steps. The first two steps might be:

1. solid at 298 and 1 atm ---> solid at 377 and 1 atm.
2. solid at 377 and 1 atm ----> solid at 377 and equilibrium sublimation pressure at 377

You need to be able to get the change in entropy over each step of the entire sequence.

Can you think of the additional steps to complete the sequence?

Chet
 

Related to Standard entropy of a liquid at melting temperature

What is the standard entropy of a liquid at its melting temperature?

The standard entropy of a liquid at its melting temperature is a measure of the amount of disorder or randomness in the molecules of the liquid at that specific temperature. It is denoted by the symbol S° and its unit is joules per mole-kelvin (J/mol•K).

How is the standard entropy of a liquid at its melting temperature calculated?

The standard entropy of a liquid at its melting temperature can be calculated using the formula ΔS° = ΔH°/T, where ΔH° is the standard enthalpy of fusion and T is the melting temperature in Kelvin. Alternatively, it can also be calculated by integrating the heat capacity of the liquid with respect to temperature.

Why is the standard entropy of a liquid at its melting temperature important?

The standard entropy of a liquid at its melting temperature is important because it provides valuable information about the state of the liquid at that temperature. It can also be used to predict the behavior of the liquid, such as its melting and freezing points, as well as its stability and reactivity.

How does the standard entropy of a liquid at its melting temperature differ from its standard entropy at other temperatures?

The standard entropy of a liquid at its melting temperature is different from its standard entropy at other temperatures because it is a specific measurement of the disorder of the liquid molecules at that particular temperature. As the temperature changes, the standard entropy of the liquid will also change.

What factors can affect the standard entropy of a liquid at its melting temperature?

The standard entropy of a liquid at its melting temperature can be affected by various factors such as changes in pressure, changes in the molecular structure of the liquid, and the presence of impurities. It can also be influenced by the temperature and pressure at which the liquid was formed.

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