SME
Chemical Engineering Thermodynamics
Reaction equilibria
Use this for “The water–gas shift reaction has ΔG° = −28.6 kJ/mol at 298.15 K. Find its equilibrium constant K.”
Diagram
K
=
exp(−ΔG°
÷
RT)
=
581,000:
the
minimum
is
at
ξ
=
K
1 + K
=
1
−
1.72 ×
10
-6
,
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
K
=
e
Formulas
ΔG°
×
1,000
=
−8.314
×
T
×
(ln
K
)
−32.9 × 1,000 = −8.314 × 298.15 × 13.27
K
=
e
ln
K
5.81 ×
10
5
=
e
13.27
Assumptions
•
Standard state: each gas pure at 1 bar, so K is written with fugacities (ideal: pressures) in bar.
•
ΔG° belongs to the reaction as written: doubling every coefficient doubles ΔG° and squares K.
•
ΔG° = −RT ln K with R = 8.314 J/(mol·K); ΔG° in kJ/mol is multiplied by 1000.
Step-by-step
WE KNOW
ΔG° = −32.9 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
=
−(−32.9)
×
1,000
8.314 × 298.15
ln
K
=
−(−32.9)
×
1,000
2,478.8191
ln
K
=
32.9
×
1,000
2,478.8191
ln
K
=
32,900
2,478.8191
ln
K
=
13.27
Step 2 · Find equilibrium constant (K)
Use K = eˡⁿᴷ
Undo
the
natural
log:
raise
e
to
ln
K
.
K
=
e
13.272449
K
=
5.81 ×
10
5
→
K
>
1:
products
are
favored
at
equilibrium.
Check
Put the numbers back in. Both sides should match.
−32.9 × 1,000 = −8.314 × 298.15 × 13.27 ✓
5.81 ×
10
5
=
e
13.272449
✓
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Equilibrium constant at another temperature
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