For liquid thiacyclobutane, the vapor pressure, in millimeters of mercury, can b
ID: 1041173 • Letter: F
Question
For liquid thiacyclobutane, the vapor pressure, in millimeters of mercury, can be expressed by the equation: Log(P) = 7.01667 - [1321.331 / (t + 224.513)] In which t is in Celsius. Calculate the change in molar enthalpy of vaporization at 298K. Some useful information you may need is 2.302585log(P) = ln(P). For liquid thiacyclobutane, the vapor pressure, in millimeters of mercury, can be expressed by the equation: Log(P) = 7.01667 - [1321.331 / (t + 224.513)] In which t is in Celsius. Calculate the change in molar enthalpy of vaporization at 298K. Some useful information you may need is 2.302585log(P) = ln(P). Log(P) = 7.01667 - [1321.331 / (t + 224.513)] In which t is in Celsius. Calculate the change in molar enthalpy of vaporization at 298K. Some useful information you may need is 2.302585log(P) = ln(P).Explanation / Answer
Answer:
The Clasius Clapeyron equation is
lnP= (-?Hvap/R)(1/T)+C
Where P= pressure, ?H vap= enthalpy of vaporization
T=Kelvin temperature, R=gas xonscons=8.314 J/mol.K
Given expressed equation is log(P)=7.01667-[1321.331/(t+224.513)]
Here t is in (°C) after adding 224.513 then it is converted to Kelvin.
Now we have to convert logP to lnP
Given lnP=2.302585 logP
logP=(1/2.302585)lnP
Therefore the expression rewrite as
(1/2.302585)lnP=-(1321.331)(1/T)+7.01667
lnP=-(3042.4769)(1/T)+16.1565 where T=t+224.513
Comparing the Clausius Clapeyron equation and modified expression,
?Hvap/R=3042.4769
Therefore ?Hvap=3042.4769 x 8.314 =25,295.153 J/mol
?Hvap=25.295 kJ/mol (1kJ=1000J)
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