SME
Physical Chemistry I: Thermodynamics & Kinetics
Chemical equilibrium
Use this for “For N₂O₄ ⇌ 2NO₂, ΔH° = +57.2 kJ/mol and K = 0.148 at 298.15 K. Find K at 348.15 K.”
Chart
Ringed: the nodes (3.354, −1.911), (2.872, 1.403)
This is an example. Type your own values to start a new problem.
ln
K₂
K₁
=
−(
ΔH° (kJ/mol)
÷
R)
×
1
T₂ (K)
−
1
T₁ (K)
Each point on the graph is one temperature with its K: T₁ with K₁ and T₂ with K₂. ΔH° is the reaction’s standard enthalpy change.
Formulas
ln(
K₂
K₁
)
=
−ΔH°
×
1,000
8.314
×
(
1
T₂
−
1
T₁
)
ln(
4.07
0.148
)
=
−57.2
×
1,000
8.314
×
(
1
348.15
−
1
298.15
)
Assumptions
•
ΔH° stays the same between the two temperatures, so ln K against 1/T is a straight line with slope −ΔH° ÷ R.
•
R = 8.314 J/(mol·K) and ΔH° is in kJ/mol, so ΔH° is multiplied by 1000. Temperatures are in kelvins.
•
An endothermic reaction (ΔH° > 0) has a larger K when hotter; an exothermic one has a smaller K.
•
The graph plots ln K against x = 1000/T so its numbers are easy to read: the slope is then −ΔH° ÷ R with ΔH° in kJ/mol.
Step-by-step
WE KNOW
ΔH° = +57.2 kJ/mol, K₁ = 0.148, T₁ = 298.15 K, T₂ = 348.15 K
FIND
equilibrium constant at T₂ (K₂)
Step 1 · Find equilibrium constant at T₂ (K₂)
Use ln(K₂ ÷ K₁) = −ΔH° × 1,000 ÷ 8.314 × (1 ÷ T₂ − 1 ÷ T₁)
Work
out
how
much
ln
K
changes
from
T₁
to
T₂
,
raise
e
to
it,
and
scale
K₁
by
that.
K₂
=
K₁
×
e^(−ΔH°
×
1,000
÷
8.314
×
(
1
T₂
−
1
T₁
))
K₂
=
0.148
×
e^(−57.2
×
1,000
÷
8.314
×
(
1
348.15
−
1
298.15
))
K₂
=
0.148
×
e
−57.2 × 1,000 ÷ 8.314 × (0.0028723 − 0.003354)
K₂
=
0.148
×
e
−57.2 × 1,000 ÷ 8.314 × (−0.0004817)
K₂
=
0.148
×
e
−57,200 ÷ 8.314 × (−0.0004817)
K₂
=
0.148
×
e
−6,880 × (−0.0004817)
K₂
=
0.148
×
e
3.3141
K₂
=
0.148
×
27.498
K₂
=
4.07
Check
Put the numbers back in. Both sides should match.
ln(
4.07
0.148
)
=
−57.2
×
1,000
8.314
×
(
1
348.15
−
1
298.15
)
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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