SME
Physical Chemistry I: Thermodynamics & Kinetics · Problem type
Enthalpy and entropy of heating
Use this for “2.50 mol of liquid ethanol (Cₚ = 112.3 J/(mol·K)) warms at constant pressure from 293.15 K to 333.15 K. Find ΔH and ΔS.”
This is an example. Type your own values to start a new problem.
ΔH (J)
=
n (mol)
×
Cₚ (J/(mol·K))
×
(
T₂ (K)
−
T₁ (K)
)
ΔS (J/K)
=
n (mol)
×
Cₚ (J/(mol·K))
×
ln
T₂ (K)
T₁ (K)
Formulas
ΔH
=
n
×
Cₚ
×
(
T₂
−
T₁
)
+3,765 = 1 × 75.3 × (348.15 − 298.15)
ΔS
=
n
×
Cₚ
×
ln(
T₂
T₁
)
+11.67
=
1
×
75.3
×
ln(
348.15
298.15
)
Assumptions
•
Constant pressure, so the heat taken in is ΔH = nCpΔT.
•
Cₚ stays the same over the range, and nothing melts or boils between T₁ and T₂.
•
Reversible heating, so dS = nCₚ dT ÷ T, which integrates to ΔS = nCₚ ln(T₂ ÷ T₁).
•
Temperatures are in kelvins (°C + 273.15).
Step-by-step
WE KNOW
n = 1 mol, Cₚ = 75.3 J/(mol·K), T₁ = 298.15 K, T₂ = 348.15 K
FIND
enthalpy change (ΔH), entropy change (ΔS)
Step 1 · Find enthalpy change (ΔH)
Use ΔH = n × Cₚ × (T₂ − T₁)
At constant pressure the heat taken in is ΔH: the moles times Cₚ times the rise.
ΔH = 1 × 75.3 × (348.15 − 298.15)
ΔH = 75.3 × 50
ΔH = +3,765 J
Step 2 · Find entropy change (ΔS)
Use ΔS = n × Cₚ × ln(T₂ ÷ T₁)
Add
up
dS
=
nCₚ
dT
T
from
T₁
to
T₂
:
the
integral
is
nCₚ
ln(
T₂
T₁
).
ΔS
=
1
×
75.3
×
ln(
348.15
298.15
)
ΔS = 75.3 × ln(1.1677)
ΔS = 75.3 × 0.15504
ΔS = +11.67 J/K → ΔS > 0: the warmer sample has its energy spread over more ways.
Check
Put the numbers back in. Both sides should match.
+3,765 = 1 × 75.3 × (348.15 − 298.15) ✓
+11.6743
=
1
×
75.3
×
ln(
348.15
298.15
)
✓
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Reversible adiabatic expansion
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Reaction enthalpy at another temperature (Kirchhoff)
Topic check
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Related lessons
Laws of thermodynamics
Main lesson
Reversible adiabatic expansion
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Reaction enthalpy at another temperature (Kirchhoff)
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State or path function?
Use this for “A gas goes from state 1 to state 2 by two different routes. Which of q, w, ΔU and ΔS are the same for both?”