SME
General Chemistry II · Problem type
Free energy change at a reaction quotient
Use this for “A reaction has ΔG° = −12.5 kJ/mol at 310 K. Find ΔG when Q = 25, and say which way the reaction runs.”
Diagram
K
=
exp(−ΔG°
÷
RT)
=
128:
the
minimum
is
at
ξ
=
K
1 + K
=
0.992.
ΔG = ΔG° + RT ln Q = −4.2 kJ/mol: Q < K, downhill forward, so it runs forward toward more B.
Drawn for a model A ⇌ B with ideal mixing; G of pure A is 0.
This is an example. Type your own values to start a new problem.
ΔG (kJ/mol)
=
ΔG° (kJ/mol)
+
R
T (K)
ln
Q
Formulas
ΔG
=
ΔG°
+
8.314
×
T
×
ln(
Q
)
1,000
−4.2
=
−12.5
+
8.314
×
310
×
ln(25)
1,000
Assumptions
•
ΔG = ΔG° + RT ln Q gives the free energy change at the mixture’s own reaction quotient Q, at the same temperature as ΔG°.
•
R = 8.314 J/(mol·K), so RT ln Q comes out in J/mol and is divided by 1000 to add it to ΔG° in kJ/mol.
•
ΔG < 0 means the reaction runs forward until Q reaches K; ΔG > 0 means it runs backward.
•
Q is written like K, from the mixture’s concentrations or pressures; solids and pure liquids are left out.
Step-by-step
WE KNOW
ΔG° = −12.5 kJ/mol, T = 310 K, Q = 25
FIND
free energy change (ΔG)
Step 1 · Find free energy change (ΔG)
Use ΔG = ΔG° + 8.314 × T × ln(Q) ÷ 1,000
Find
RT
ln
Q
in
J/mol,
divide
by
1000
for
kJ/mol,
and
add
it
to
ΔG°.
ΔG
=
−12.5
+
8.314
×
310
×
ln(25)
1,000
ΔG
=
−12.5
+
2,577.34
×
3.2189
1,000
ΔG = −12.5 + 8.296
ΔG
=
−4.2
kJ/mol
→
ΔG
<
0:
Q
is
below
K,
so
the
reaction
runs
forward.
Check
Put the numbers back in. Both sides should match.
−4.2
=
−12.5
+
8.314
×
310
×
ln(25)
1,000
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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