ΔG vs ΔG°: Using the Reaction Quotient Q

By Aniket Bhardwaj · 13 October 2026 · Updated 13 October 2026 · Thermodynamics · IIT-JAM / GATE / CSIR-NET

Students often use ΔG and ΔG° as if they were the same thing. They are not, and the difference is one of the most useful ideas in chemical thermodynamics. ΔG° is a fixed number for a reaction at standard conditions. ΔG tells you what happens in the flask you actually have. The bridge between them is the reaction quotient Q. This page shows the relation, three worked calculations, and the errors that cost marks in JAM, GATE and CSIR-NET style questions.

A number line shows ΔG negative at Q = 0.10, zero at K = 0.144, and positive at Q = 2.0. — ΔG changes sign as Q crosses K: forward below equilibrium, reverse above.

The key relation

ΔG = ΔG° + RT ln Q
At equilibrium: ΔG = 0 and Q = K, so ΔG° = −RT ln K
Combining: ΔG = RT ln (Q/K)
Base-10 form at 298.15 K: ΔG° = −5.708 log₁₀K kJ/mol (since RT ln 10 = 5.708 kJ/mol)

What each symbol means

ΔG° is the standard Gibbs energy change: reactants and products each in their standard states (1 bar for gases; 1 mol/L for solutes in most courses). It depends only on temperature.
ΔG is the Gibbs energy change for the mixture you have now. It changes as the reaction proceeds and reaches zero at equilibrium.
Q is the reaction quotient, calculated with the present pressures (in bar) or concentrations (in mol/L). Pure solids and liquids are left out.
K is the equilibrium constant at the same temperature.
R = 8.314 J/(mol K). Use J with J, or convert to kJ. T is in kelvin.

Edition note: some books define the standard pressure as 1 atm and older ones used 1 atm for K_p; the current IUPAC standard is 1 bar. The difference is small for most problems, but follow the convention your question or textbook states.

What the sign of ΔG tells you

Comparison	ΔG	Direction
Q < K	negative	forward reaction is spontaneous
Q = K	zero	equilibrium
Q > K	positive	reverse reaction is spontaneous

This is the point many students miss: a reaction with a positive ΔG° is not "impossible". It simply has K < 1, and it will still run forward if Q is small enough that ΔG comes out negative.

Worked examples

Example 1: N₂O₄(g) ⇌ 2NO₂(g) at 298.15 K. Take ΔG° = +4.8 kJ/mol (a typical data-book value; use the value given in your question).
Step 1: K = exp(−ΔG°/RT). RT = 8.314 × 298.15 = 2478.8 J/mol. ΔG°/RT = 4800 / 2478.8 = 1.936.
Step 2: K = e^−1.936 = 0.144.
Step 3: Now take Q = 0.10 (pressures in bar). ΔG = 4800 + 2478.8 × ln(0.10) = 4800 + 2478.8 × (−2.3026) = 4800 − 5707.7 = −907.7 J/mol.
Step 4: ΔG is negative, so the forward reaction proceeds. This matches Q = 0.10 < K = 0.144.
Answer: K = 0.144 and ΔG = −0.91 kJ/mol at Q = 0.10. Even though ΔG° is positive, the forward reaction runs because Q < K.

Example 2: same reaction with Q = 2.0.
Step 1: ΔG = 4800 + 2478.8 × ln 2.0 = 4800 + 2478.8 × 0.6931 = 4800 + 1718.2 = +6518 J/mol.
Step 2: ΔG is positive and Q (2.0) is greater than K (0.144).
Answer: ΔG = +6.52 kJ/mol. The reverse reaction is spontaneous: NO₂ turns back into N₂O₄.

Example 3: a reaction with ΔG° = −30.0 kJ/mol at 298.15 K.
Step 1: ln K = −ΔG°/RT = 30000 / 2478.8 = 12.103. log₁₀K = 12.103 / 2.3026 = 5.256, so K = 1.80 × 10⁵. Cross-check with the shortcut: 30.0 / 5.708 = 5.256 ✓.
Step 2: At Q = 1.0 × 10³: ΔG = −30000 + 2478.8 × ln(1000) = −30000 + 2478.8 × 6.9078 = −30000 + 17123 = −12877 J/mol, which is −12.9 kJ/mol. Still forward.
Step 3: At Q = 1.0 × 10⁶: ΔG = −30000 + 2478.8 × 13.8155 = −30000 + 34246 = +4246 J/mol, which is +4.25 kJ/mol. Now the reverse is spontaneous, because Q has passed K = 1.80 × 10⁵.
Answer: K = 1.8 × 10⁵. ΔG is −12.9 kJ/mol at Q = 10³ and +4.25 kJ/mol at Q = 10⁶. ΔG changes sign exactly when Q crosses K.

Example 4: finding ΔG° from a measured K.
Step 1: Suppose K = 1.0 × 10⁻⁵ at 298.15 K.
Step 2: ΔG° = −RT ln K = −2478.8 × ln(1.0 × 10⁻⁵) = −2478.8 × (−11.513) = +28538 J/mol.
Step 3: Shortcut: −5.708 × (−5) = +28.54 kJ/mol ✓.
Answer: ΔG° = +28.5 kJ/mol. A small K corresponds to a positive ΔG°.

Quick feel for the numbers

Because RT ln 10 is about 5.7 kJ/mol, every factor of 10 in K changes ΔG° by about 5.7 kJ/mol at 298 K. So ΔG° = −17 kJ/mol means K near 10³, and ΔG° = −34 kJ/mol means K near 10⁶.

Treating ΔG° as the criterion for direction in the flask. Direction depends on ΔG, that is, on Q. ΔG° only fixes K.
Using ΔG = ΔG° − RT ln Q. The sign is plus. The wrong sign gives the opposite direction.
Mixing J and kJ. RT is in J/mol. Convert ΔG° to J before adding, or convert RT to kJ.
Using °C for T. Always use kelvin.
Putting solids or pure liquids in Q. They are left out.
Ignoring standard states. Q should use pressures in bar for gases and mol/L for solutes. Using mmHg or atm without conversion gives the wrong Q.
Claiming ΔG° > 0 means "no reaction". It means K < 1, not K = 0.

Exam relevance

Question type	Method
Find K from ΔG°	K = exp(−ΔG°/RT)
Find ΔG at given pressures or concentrations	Compute Q, then ΔG = ΔG° + RT ln Q
"Is the reaction spontaneous under these conditions?"	Sign of ΔG, or compare Q with K
Find the Q at which direction reverses	Set ΔG = 0, so Q = K

Check the current official syllabus for your exam to see how this topic is tested. Do not assume a fixed number of marks.

Work the logarithms by hand for practice, then check ΔG for your own numbers with the calculator.

Open the Gibbs Free Energy Calculator →