SME
General Chemistry II · Problem type
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?”
This is an example. Type your own values to start a new problem.
ΔSₛᵤᵣᵣ (J/(mol·K))
=
−
(
ΔHₛᵧₛ (kJ/mol)
)
÷
T (K)
ΔSᵤₙᵢᵥ (J/(mol·K))
=
ΔSₛᵧₛ (J/(mol·K))
+
ΔSₛᵤᵣᵣ (J/(mol·K))
Formulas
ΔSₛᵤᵣᵣ
=
−ΔHₛᵧₛ
×
1,000
T
+184.4
=
−(−92.2)
×
1,000
500
ΔSᵤₙᵢᵥ = ΔSₛᵧₛ + ΔSₛᵤᵣᵣ
−14.6 = −199 + 184.4
Assumptions
•
The second law: a process is spontaneous when the entropy of the universe rises, ΔSᵤₙᵢᵥ = ΔSₛᵧₛ + ΔSₛᵤᵣᵣ > 0.
•
The surroundings are large and stay at temperature T, so the heat the system gives off, −ΔHₛᵧₛ at constant pressure, raises their entropy by −ΔHₛᵧₛ ÷ T.
•
ΔHₛᵧₛ is in kJ/mol and entropies are in J/(mol·K), so ΔHₛᵧₛ is multiplied by 1000 to put it in J/mol.
•
Temperatures are in kelvins (°C + 273.15): −10 °C is 263.15 K.
Step-by-step
WE KNOW
ΔSₛᵧₛ = −199 J/(mol·K), ΔHₛᵧₛ = −92.2 kJ/mol, T = 500 K
FIND
entropy change of the surroundings (ΔSₛᵤᵣᵣ), entropy change of the universe (ΔSᵤₙᵢᵥ)
Step 1 · Find entropy change of the surroundings (ΔSₛᵤᵣᵣ)
Use ΔSₛᵤᵣᵣ = −ΔHₛᵧₛ × 1,000 ÷ T
Put
ΔHₛᵧₛ
into
J/mol;
the
surroundings
take
in
−ΔHₛᵧₛ
as
heat,
so
divide
it
by
T
.
ΔSₛᵤᵣᵣ
=
−(−92.2)
×
1,000
500
ΔSₛᵤᵣᵣ
=
92.2
×
1,000
500
ΔSₛᵤᵣᵣ
=
92,200
500
ΔSₛᵤᵣᵣ = +184.4 J/(mol·K)
Step 2 · Find entropy change of the universe (ΔSᵤₙᵢᵥ)
Use ΔSᵤₙᵢᵥ = ΔSₛᵧₛ + ΔSₛᵤᵣᵣ
Add the entropy changes of the system and the surroundings.
ΔSᵤₙᵢᵥ = −199 + 184.4
ΔSᵤₙᵢᵥ = −14.6 J/(mol·K) → ΔSᵤₙᵢᵥ < 0: the process is not spontaneous; the reverse one is.
Check
Put the numbers back in. Both sides should match.
+184.4
=
−(−92.2)
×
1,000
500
✓
−14.6 = −199 + 184.4 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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