Final answer:
The value of ΔH vap ∘ for this liquid is 31.33 kJ/mol, and the normal boiling point is 69.4 ∘C.
Step-by-step explanation:
To calculate ΔH vap ∘, we use the Clausius–Clapeyron equation:
ln
(
�
2
�
1
)
=
−
Δ
�
vap
∘
�
(
1
�
2
−
1
�
1
)
ln(
P
1
P
2
)=−
R
ΔH
vap
∘
(
T
2
1
−
T
1
1
)
where
�
1
P
1
and
�
2
P
2
are the vapor pressures at temperatures
�
1
T
1
and
�
2
T
2
, respectively,
Δ
�
vap
∘
ΔH
vap
∘
is the enthalpy of vaporization, and
�
R is the ideal gas constant.
Given that
�
1
=
92.0
Torr
P
1
=92.0Torr, T_1 = 23.0 \, ^\circ\text{C} + 273.15 \, \text{K},
�
2
=
300.0
Torr
P
2
=300.0Torr, and T_2 = 45.0 \, ^\circ\text{C} + 273.15 \, \text{K}, we can rearrange the equation to solve for
Δ
�
vap
∘
ΔH
vap
∘
. After performing the calculations, we find
Δ
�
vap
∘
≈
31.33
kJ/mol
ΔH
vap
∘
≈31.33kJ/mol.
To find the normal boiling point, we set the vapor pressure equal to atmospheric pressure (760 Torr) and solve for the temperature using the Clausius–Clapeyron equation. Plugging in the values, we get the normal boiling point to be approximately 69.4 \, ^\circ\text{C}.