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Grade 10 Science · Problem type
The temperature where it turns spontaneous
Use this for “Above what temperature is a reaction with ΔH = 60 kJ/mol and ΔS = 150 J/(mol·K) spontaneous?”
Chart
Zero: T = 400
This is an example. Type your own values to start a new problem.
T (K)
=
1,000
×
ΔH (kJ/mol)
ΔS (J/(mol·K))
Formulas
T
=
1,000
×
ΔH/ΔS
400
=
1,000
×
60
150
Assumptions
•
ΔG = ΔH − TΔS is 0 where the reaction switches: T = ΔH ÷ ΔS, with ΔS changed to kJ.
•
Both positive: spontaneous above that temperature. Both negative: spontaneous below it.
•
With opposite signs ΔG never changes sign, so there is no switch.
Step-by-step
WE KNOW
ΔH = 60 kJ/mol, ΔS = 150 J/(mol·K)
FIND
temperature where ΔG = 0 (T)
Step 1 · Find temperature where ΔG = 0 (T)
Use T = 1,000 × ΔH/ΔS
Set
ΔG
=
0:
then
ΔH
=
TΔS,
so
T
=
ΔH
ΔS
;
ΔS
is
in
J,
so
multiply
by
1,000
to
match
ΔH’s
kJ.
T
=
1,000
×
60
150
T
=
60,000
150
T
=
400
K
(both
positive:
spontaneous
above
400
K)
Check
Put the numbers back in. Both sides should match.
400
=
1,000
×
60
150
✓
Topic check
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