SME
General Chemistry II
Entropy and Gibbs free energy
Use this for “The ionization of ethanoic acid has ΔG° = +27.1 kJ/mol at 25 °C. Find K.”
Diagram
K
=
exp(−ΔG°
÷
RT)
=
1.79 ×
10
-5
:
the
minimum
is
at
ξ
=
K
1 + K
=
1.79 ×
10
-5
,
pressed
against
the
edge:
the
inset
magnifies
it,
and
the
line
below
shows
it
at
log
K.
Drawn for a model A ⇌ B with ideal mixing; G of pure A is 0.
This is an example. Type your own values to start a new problem.
ΔG° (kJ/mol)
=
−R
T (K)
ln
K
ln K
Natural log of K
calculated
Formulas
ΔG°
×
1,000
=
−8.314
×
T
×
(ln
K
)
27.1 × 1,000 = −8.314 × 298.15 × (−10.9326)
K
=
e
ln
K
1.79 ×
10
-5
=
e
−10.9326
Assumptions
•
ΔG° = −RT ln K links the free energy change of the reaction with every species in its standard state to K at the same temperature.
•
R = 8.314 J/(mol·K) and ΔG° is in kJ/mol, so ΔG° is multiplied by 1000 to put it in J/mol.
•
ΔG° < 0 means K > 1 (products favored at equilibrium), not that the reaction is fast.
•
Temperatures are in kelvins (°C + 273.15): 25 °C is 298.15 K.
Step-by-step
WE KNOW
ΔG° = +27.1 kJ/mol, T = 298.15 K
FIND
natural log of K (ln K), equilibrium constant (K)
Step 1 · Find natural log of K (ln K)
Use ΔG° × 1,000 = −8.314 × T × (ln K)
Put ΔG° into J/mol, then divide −ΔG° by RT.
ln
K
=
−ΔG°
×
1,000
8.314 ×
T
ln
K
=
−27.1
×
1,000
8.314 × 298.15
ln
K
=
−27.1
×
1,000
2,478.8191
ln
K
=
−
27,100
2,478.8191
ln
K
=
−10.9326
Step 2 · Find equilibrium constant (K)
Use K = eˡⁿᴷ
Undo
the
natural
log:
raise
e
to
ln
K
.
K
=
e
−10.9326
K
=
1.79 ×
10
-5
→
K
<
1:
reactants
are
favored
at
equilibrium.
Check
Put the numbers back in. Both sides should match.
27.1 × 1,000 = −8.314 × 298.15 × (−10.9326) ✓
1.79 ×
10
-5
=
e
−10.9326
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
More problem types
Free energy change at a reaction quotient
Use this for “A reaction has ΔG° = −12.5 kJ/mol at 310 K. Find ΔG when Q = 25, and say which way the reaction runs.”
Entropy of vaporization
Use this for “Ethanol boils at 351.5 K, and its heat of vaporization is 38.6 kJ/mol. Find its entropy of vaporization.”
Entropy of the universe (second law)
Use this for “For a reaction, ΔH = −92.2 kJ/mol and ΔS = −199 J/(mol·K). Find ΔS of the surroundings and of the universe at 500 K. Is the reaction spontaneous?”