Chemistry A calculation of enthalpy, entropy, and Gibbs energy of vaporization

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The discussion centers on calculating the enthalpy and entropy of vaporization using the Clausius-Clapeyron equation. The user calculated the enthalpy of vaporization as 17.72 kJ/mol, while the textbook states it as 28.3 kJ/mol. For the standard entropy of vaporization, the user obtained 84.44 J/K·mol, contrasting with the book's value of 91.6 J/K·mol. There is confusion regarding the temperature dependence of enthalpy and entropy in vaporization, as well as the implications of using the Clausius-Clapeyron equation under the assumption of constant enthalpy. The user plans to revisit these calculations for clarity after further reflection.
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Homework Statement
Arsine, ##\mathrm{AsH_3}##, is a highly toxic compound.

Its vapor pressure is 35 Torr at ##-111.95^\circ\text{C}## and 253 Torr at ##-83.6^\circ\text{C}##.

Using these data calculate
Relevant Equations
a) standard enthalpy of vaporization

b) standard entropy of vaporization

c) standard Gibbs free energy of vaporization

d) the normal boiling point of arsine
a) To try to solve this problem I used the Clausius-Clapeyron equation

$$\ln{\frac{P_2}{P_1}}=-\frac{\Delta H^\circ_{vap}}{R}\left (\frac{1}{T_2}-\frac{1}{T_1}\right )$$

with ##P_2=253\ \text{Torr}##, ##P_1=35\ \text{Torr}##, ##T_2=189.55\text{K}##, and ##T_1=161.2\text{K}##.

I obtained ##\Delta H^\circ_{vap}=17.72\mathrm{kJ/mol}##.

The answer at the end of the book says ##+28.3\mathrm{kJ/mol}##.

b) To obtain the standard entropy of vaporization, I used the equation

$$\ln{\frac{P}{P^\circ}}=-\frac{\Delta H^\circ_{vap}-T\Delta S^\circ_{vap}}{RT}$$

which is obtained from

$$\Delta G_{vap}=\Delta G^\circ_{vap}+RT\ln{\frac{P}{P^\circ}}$$

by setting ##\Delta G_{vap}=0## which occurs in the dynamic equilibrium between liquid and gas phases that defined vapor pressure.

There is only one unknown, ##\Delta S^\circ_{vap}## given the ##\Delta H^\circ_{vap}## we solved for in (a).

I get the result of ##84.44\mathrm{\frac{J}{K\cdot mol}}## but the book says ##91.6\mathrm{\frac{J}{K\cdot mol}}##.

Now, one obvious question is why my results seem incorrect.

But more importantly, I am not so sure exactly what is happening in these calculations.

Enthalpy of reaction usually depends on temperature.

In the case of a substance vaporizing at a given temperature, as far as I can tell, the enthalpy of vaporization is the same whatever the temperature.

I am not sure if this is correct, but I am basing this assertion on the equation that says that the relationship between enthalpy and temperature depends on the difference in heat capacities between products and reactants.

In vaporization, the reactant and product are the same substance.

Entropy on the other hand, seems to be dependent of temperature.

$$dH=TdS+VdP$$

and for vaporization at some constant pressure, we have

$$dH=TdS$$

$$\Delta H^\circ_{vap}=T\Delta S^\circ_{vap}$$

But then, enthalpy depends on temperature.

So it is not clear to me what exactly I obtained in my calculations in b). After all, I calculated a ##\Delta S^\circ_{vap}## but for what temperature?

I am sleepy and want to think more about this because I know half of what I wrote above is conceptually incorrect, but I will take this up again when I wake up.
 
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I get the same answers as you. NIST gives a value for ΔHvap of 16.7 kJ/mol, so you are in the right ballpark, and the book is not. I don't know what the book is doing.

ΔH and ΔS are temperature dependent, but often they are approximately constant over a not-too-wide temperature range. The C-C equation in the integral form you used assumes that ΔH is constant.
 
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