SME
Principles of Biology I · Problem type
ΔG in the cell
Use this for “With ΔG°′ = +7.5 kJ/mol and products at 1% of reactants, does the reaction run forward at 37 °C?”
Diagram
ΔG°′: Reactants → Products at 1 M, ++7.5 kJ/mol
RT ln Q: Products at 1 M → Products in the cell, −11.87 kJ/mol
ΔG = ΔG°′ + RT ln Q
ΔG = +7.5 + (−11.87) = −4.37 kJ/mol
Down the ladder: ΔG < 0, so it runs forward on its own (exergonic).
This is an example. Type your own values to start a new problem.
ΔG (kJ/mol)
=
ΔG°′ (kJ/mol)
+
R
T (K)
ln
[P] (mM)
[R] (mM)
Q
Reaction quotient
worked out (not typed)
RT ln Q
Shift from the concentrations
calculated · kJ/mol
Formulas
Q
=
[P]
÷
[R]
0.01
=
1
100
RT
ln
Q
=
0.008314
×
T
×
ln(
Q
)
−11.87 = 0.008314 × 310 × ln(0.01)
ΔG
=
ΔG°′
+
RT
ln
Q
−4.37 = 7.5 + (−11.87)
Assumptions
•
ΔG = ΔG°′ + RT ln Q with R = 0.008314 kJ/(mol·K), and Q = [products] ÷ [reactants].
•
ΔG°′ is for 1 M of each at pH 7; 37 °C is 310 K and 25 °C is 298 K.
•
Q below 1 pulls ΔG down: a reaction uphill at 1 M can run forward in the cell.
Step-by-step
WE KNOW
ΔG°′ = +7.5 kJ/mol, [P] = 1 mM, [R] = 100 mM, T = 310 K
FIND
reaction quotient (Q), shift from the concentrations (RT ln Q), free energy change in the cell (ΔG)
Step 1 · Find reaction quotient (Q)
Use Q = [P] ÷ [R]
Products over reactants, in the same unit.
Q
=
1
100
Q
=
0.01
Step 2 · Find shift from the concentrations (RT ln Q)
Use RT ln Q = 0.008314 × T × ln(Q)
RT
in
kJ/mol
times
ln
Q
:
below
0
when
Q
is
below
1.
RT
ln
Q
=
0.008314
×
310
×
ln(0.01)
RT
ln
Q
=
0.008314
×
310
×
(−4.605)
RT
ln
Q
=
2.577
×
(−4.605)
RT
ln
Q
=
−11.87
kJ/mol
Step 3 · Find free energy change in the cell (ΔG)
Use ΔG = ΔG°′ + RT ln Q
Add the concentrations’ shift to ΔG°′.
ΔG = 7.5 + (−11.87)
ΔG = −4.37 kJ/mol → Below 0: it runs forward on its own.
Check
Put the numbers back in. Both sides should match.
0.01
=
1
100
✓
−11.87 = 0.008314 × 310 × ln(0.01) ✓
−4.37 = 7.5 + (−11.87) ✓
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