- #1
zenterix
- 705
- 84
- 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.
$$\ln{\frac{P_2}{P_1}}=-\frac{\Delta H^\circ_{vap}}{R}\left (\frac{1}{T_2}-\frac{1}{T_1}\right )$$
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.