SME
General Chemistry I
Intermolecular forces, liquids and solids
Use this for “A liquid’s vapor pressure is 100 torr at 307 K and 400 torr at 336 K. Find ΔHᵥₐₚ and the normal boiling point.”
Diagram
The liquid (triple and critical points sketched): the vapor curve runs from the triple point (261 K, 5.875 torr) to the critical point (421.6 K, 7,882 torr), through (336 K, 400 torr) and (307 K, 100 torr); pressure on a log scale.
Between
neighbouring
points
ln
P
is
a
straight
line
in
1
T
:
Clausius–Clapeyron,
ln(
P₂
P₁
)
=
−(
ΔHᵥₐₚ
R
)(
1
T₂
−
1
T₁
),
with
ΔHᵥₐₚ
constant
on
each
piece.
The melting line is sketched steep (no slope given).
The solid–gas line is sketched dashed, with ΔHsub taken as 1.15 ΔHᵥₐₚ (no point given).
This is an example. Type your own values to start a new problem.
ln(
P₂ (torr)
÷
P₁ (torr)
)
=
−(
ΔHᵥₐₚ (kJ/mol)
÷
R)
1
T₂ (K)
−
1
T₁ (K)
Tb
Normal boiling point
calculated · K
ΔSᵥₐₚ
Entropy of vaporization
worked out (not typed) · J/(mol·K)
Formulas
ln(
P₂
P₁
)
=
−(ΔHᵥₐₚ
×
1,000
8.314
)
×
(
1
T₂
−
1
T₁
)
ln(
400
100
)
=
−(41
×
1,000
8.314
)
×
(
1
336
−
1
307
)
ln(
760
P₁
)
=
−(ΔHᵥₐₚ
×
1,000
8.314
)
×
(
1
Tb
−
1
T₁
)
ln(
760
100
)
=
−(41
×
1,000
8.314
)
×
(
1
351.4
−
1
307
)
ΔSᵥₐₚ
=
ΔHᵥₐₚ
×
1,000
Tb
116.7
=
41
×
1,000
351.4
Assumptions
•
ΔHᵥₐₚ stays the same over the range, so ln P against 1 ÷ T is a straight line with slope −ΔHᵥₐₚ ÷ R (Clausius–Clapeyron), R = 8.314 J/(mol·K).
•
The vapor is an ideal gas and the liquid’s volume is tiny next to it; ΔHᵥₐₚ in kJ/mol is multiplied by 1000 for J/mol.
•
The normal boiling point is where the vapor pressure reaches 760 torr (1 atm); there ΔSᵥₐₚ = ΔHᵥₐₚ ÷ Tb, which Trouton’s rule puts near 88 J/(mol·K).
•
The diagram’s triple and critical points are sketched on the same line, at 0.85 × the lowest and 1.2 × the highest temperature; they are not measured.
Step-by-step
WE KNOW
P₁ = 100 torr, T₁ = 307 K, P₂ = 400 torr, T₂ = 336 K
FIND
enthalpy of vaporization (ΔHᵥₐₚ), normal boiling point (Tb), entropy of vaporization (ΔSᵥₐₚ)
Step 1 · Find enthalpy of vaporization (ΔHᵥₐₚ)
Use ln(P₂ ÷ P₁) = −(ΔHᵥₐₚ × 1,000 ÷ 8.314) × (1 ÷ T₂ − 1 ÷ T₁)
The
slope
of
ln
P
against
1
T
is
−
ΔHᵥₐₚ
R
:
divide
the
change
in
ln
P
by
the
change
in
1
T
,
times
−R,
then
turn
J
into
kJ.
ΔHᵥₐₚ
=
−8.314
×
ln(
P₂
÷
P₁
)
1 ÷
T₂
− 1 ÷
T₁
÷
1,000
ΔHᵥₐₚ
=
−8.314
×
ln(400 ÷ 100)
1 ÷ 336 − 1 ÷ 307
÷
1,000
ΔHᵥₐₚ
=
−8.314
×
1.3863
−2.8114 ×
10
-4
÷
1,000
ΔHᵥₐₚ = 41 kJ/mol
Step 2 · Find normal boiling point (Tb)
Use ln(760 ÷ P₁) = −(ΔHᵥₐₚ × 1,000 ÷ 8.314) × (1 ÷ Tb − 1 ÷ T₁)
The
same
line
reaches
760
torr
at
the
boiling
point:
take
R
ln(760 ÷
P₁
)
ΔHᵥₐₚ
from
1
T₁
,
then
take
the
reciprocal.
Tb
=
1
1 ÷
T₁
− 8.314 × ln(760 ÷
P₁
) ÷ (ΔHᵥₐₚ × 1,000)
Tb
=
1
1 ÷ 307 − 8.314 × ln(760 ÷ 100) ÷ (41 × 1,000)
Tb
=
1
1 ÷ 307 − 8.314 × ln(7.6) ÷ 41,000
Tb
=
1
1 ÷ 307 − 8.314 × 2.028 ÷ 41,000
Tb
=
1
0.003257 − 16.86 ÷ 41,000
Tb
=
1
0.003257 − 0.0004112
Tb
=
1
0.002846
Tb = 351.4 K
Step 3 · Find entropy of vaporization (ΔSᵥₐₚ)
Use ΔSᵥₐₚ = ΔHᵥₐₚ × 1,000 ÷ Tb
At
the
boiling
point
liquid
and
vapor
are
at
equilibrium,
so
ΔG
=
0
and
ΔSᵥₐₚ
=
ΔHᵥₐₚ
Tb
,
in
J/(mol·K).
ΔSᵥₐₚ
=
41
×
1,000
351.4
ΔSᵥₐₚ
=
41,000
351.4
ΔSᵥₐₚ = 116.7 J/(mol·K) → Above Trouton’s rule, about 88 J/(mol·K), as in hydrogen-bonded liquids, whose order is lost on boiling.
Check
Put the numbers back in. Both sides should match.
ln(
400
100
)
=
−(41
×
1,000
8.314
)
×
(
1
336
−
1
307
)
✓
ln(
760
100
)
=
−(41
×
1,000
8.314
)
×
(
1
351.4
−
1
307
)
✓
116.7
=
41
×
1,000
351.4
✓
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