SME
Organic Chemistry II · Problem type
How much hydrate or enol at equilibrium
Use this for “Methanal’s hydrate has K = 2.3 × 10³ in water at 25 °C. What percent is hydrate, and what is ΔG°?”
This is an example. Type your own values to start a new problem.
% (%)
=
100
×
1
+
ΔG° (kJ/mol)
=
−R
T (K)
ln
K
Formulas
%
=
100
×
K
1 +
K
0.139804
=
100
×
0.0014
1 + 0.0014
ΔG°
=
−8.314
×
T
×
ln(
K
)
1,000
+16.3
=
−8.314
×
298.15
×
ln(0.0014)
1,000
Assumptions
•
K is the added form over the carbonyl form; a hydrate’s K folds in water’s concentration, so the two forms’ ratio is K.
•
The percent in the added form is 100K ÷ (1 + K): over half when K > 1, almost none when K is tiny.
•
ΔG° = −RT ln K with R = 8.314 J/(mol·K) gives J/mol, divided by 1000 for kJ/mol; 25 °C is 298.15 K.
•
Bulky or electron-giving groups on the C=O carbon lower K; electron-pulling groups such as CCl₃ raise it.
Step-by-step
WE KNOW
K = 0.0014, T = 298.15 K
FIND
percent in the added form (%), standard free energy change (ΔG°)
Step 1 · Find percent in the added form (%)
Use % = 100 × K ÷ (1 + K)
The
added
form
is
K
parts
for
each
1
part
of
carbonyl,
so
it
is
K
parts
of
1
+
K
.
%
=
100
×
0.0014
1 + 0.0014
%
=
100
×
0.0014
1.0014
%
=
0.14
1.0014
% = 0.139804%
Step 2 · Find standard free energy change (ΔG°)
Use ΔG° = −8.314 × T × ln(K) ÷ 1,000
Multiply
−R
by
T
and
ln
K
for
J/mol,
then
divide
by
1000
for
kJ/mol.
ΔG°
=
−8.314
×
298.15
×
ln(0.0014)
1,000
ΔG°
=
−8.314
×
298.15
×
−6.571
1,000
ΔG°
=
−2,478.8191
×
−6.571
1,000
ΔG°
=
16,290
1,000
ΔG°
=
+16.3
kJ/mol
→
K
<
1:
ΔG°
>
0,
so
most
of
it
stays
as
the
carbonyl.
Check
Put the numbers back in. Both sides should match.
0.139804
=
100
×
0.0014
1 + 0.0014
✓
+16.289
=
−8.314
×
298.15
×
ln(0.0014)
1,000
✓
Topic check
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