SME
Physical Chemistry I: Thermodynamics & Kinetics · Problem type
Reaction enthalpy at another temperature (Kirchhoff)
Use this for “A reaction has ΔH = −57.2 kJ/mol at 298.15 K and ΔCₚ = +12.6 J/(mol·K). Find ΔH at 400 K.”
This is an example. Type your own values to start a new problem.
ΔH₂ (kJ/mol)
=
ΔH₁ (kJ/mol)
+
(
ΔCₚ (J/(mol·K))
)
×
(
T₂ (K)
−
T₁ (K)
)
Formulas
ΔH₂
=
ΔH₁
+
ΔCₚ
×
T₂
−
T₁
1,000
−101.3
=
−92.2
+
(−45)
×
500 − 298.15
1,000
Assumptions
•
ΔCₚ = ΣCₚ(products) − ΣCₚ(reactants), each heat capacity times its coefficient in the balanced equation.
•
ΔCₚ stays the same between T₁ and T₂, and no species melts or boils in that range.
•
dΔH ÷ dT = ΔCₚ, so ΔH changes by ΔCₚ(T₂ − T₁); ΔCₚ is in J/(mol·K), so that change is divided by 1000 for kJ/mol.
Step-by-step
WE KNOW
ΔH₁ = −92.2 kJ/mol, ΔCₚ = −45 J/(mol·K), T₁ = 298.15 K, T₂ = 500 K
FIND
reaction enthalpy at T₂ (ΔH₂)
Step 1 · Find reaction enthalpy at T₂ (ΔH₂)
Use ΔH₂ = ΔH₁ + ΔCₚ × (T₂ − T₁) ÷ 1,000
Kirchhoff: ΔH changes by ΔCₚ for each kelvin; divide by 1000 to turn J into kJ.
ΔH₂
=
−92.2
+
(−45)
×
500 − 298.15
1,000
ΔH₂
=
−92.2
+
(−45)
×
201.85
1,000
ΔH₂
=
−92.2
+
−9,083.25
1,000
ΔH₂ = −92.2 + (−9.083)
ΔH₂ = −101.3 kJ/mol → ΔH falls: ΔCₚ < 0, so warming the products takes less heat than warming the reactants.
Check
Put the numbers back in. Both sides should match.
−101.3
=
−92.2
+
(−45)
×
500 − 298.15
1,000
✓
Topic check
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