What does Delta G less than zero physically mean?

A process with negative Delta G can proceed spontaneously and can, in principle, do useful non-expansion work equal to minus Delta G. Real processes deliver less because they are not reversible.

Is a reaction with positive Delta G ever possible?

Yes, if energy is supplied externally - electrolysis, photosynthesis, or coupled reactions. Positive Delta G means the reaction is not spontaneous under the given conditions but can be driven by external energy input.

What is the significance of Delta G standard = -RT ln K?

It links Gibbs energy directly with equilibrium composition through the equilibrium constant K. It lets us predict K for any reaction from tabulated Delta H formation and standard entropy values, without doing the experiment.

How is Delta G related to cell EMF?

Delta G = -nFE cell where n is the number of electrons transferred and F is Faraday's constant. For a spontaneous cell E is positive so Delta G is negative. At standard conditions, Delta G standard = -nFE standard cell.

What is the physical meaning of Delta G formation of an element being zero?

It is a convention so that Delta G reaction can be computed from a table of Delta G formation values. Setting the reference element in its standard state to zero makes all differences work out correctly regardless of true absolute G.

Gibbs Energy Change and Equilibrium

Q: Why do we need Gibbs free energy when we already have entropy of the universe?

Delta S universe predicts spontaneity but requires knowing Delta S of surroundings. Delta G = Delta H - T Delta S uses only system quantities and works at constant T and P (common lab conditions). Since Delta G = -T Delta S universe, they are equivalent.

Q: Why does temperature favor different reactions?

Because Delta G = Delta H - T Delta S: at high T the T Delta S term dominates. Positive Delta S makes Delta G more negative at high T (favors reaction); negative Delta S makes it more positive at high T (disfavors). This is why ice melts above 0 C and freezes below.

Q: Does Delta G tell us anything about reaction rate?

No. Delta G is thermodynamic and predicts whether a reaction is spontaneous, not how fast. Rate is a kinetics question governed by activation energy. Diamond has positive Delta G relative to graphite but converts so slowly that it seems permanent.

ChemistryChemical ThermodynamicsFor JEE aspirants

Using $Δ S_{u ni v er se}$ to judge spontaneity is technically correct but often inconvenient - it requires knowing the surroundings' behavior. J. W. Gibbs introduced a purely system-based quantity, the Gibbs free energy $G = H - T S$ , whose sign directly indicates spontaneity at constant temperature and pressure. This concept covers the definition of $G$ , the criterion $Δ G < 0$ for spontaneous processes, the effect of temperature via the four sign-combinations of $Δ H$ and $Δ S$ , the crucial relation $Δ G^{\circ} = - R T ln K$ between free energy and the equilibrium constant, the Gibbs-Helmholtz equation, and the connection to cell EMF via $Δ G = - n F E_{ce l l}$ .

Key Formulas - Quick Reference

Definition: $G = H - T S$ ; extensive state function.
At constant $T$ : $Δ G = Δ H - T Δ S$
Fundamental relation: $Δ G_{sy s t e m} = - T Δ S_{u ni v er se}$ (at constant $T$ , $P$ )
Spontaneity criterion (at constant $T$ , $P$ ): $Δ G < 0$ spontaneous, $Δ G = 0$ equilibrium, $Δ G > 0$ non-spontaneous
Non-standard states: $Δ G = Δ G^{\circ} + R T ln Q$
At equilibrium ( $Q = K$ ): $Δ G^{\circ} = - R T ln K = - 2.303 R T lo g K$
Gibbs energy from formation: $Δ G_{r x n}^{\circ} = \sum ν_{i} Δ G_{f}^{\circ} (products) - \sum ν_{i} Δ G_{f}^{\circ} (reactants)$
Cell EMF: $Δ G = - n F E_{ce l l}$ ; $Δ G^{\circ} = - n F E_{ce l l}^{\circ}$
Gibbs-Helmholtz: $(\frac{\partial G}{\partial T})_{P} = - S$

1. From $Δ S_{u ni v er se}$ to $Δ G$

The second law tells us a process is spontaneous when $Δ S_{u ni v er se} > 0$ . But $Δ S_{u ni v er se} = Δ S_{sy s} + Δ S_{s u r r}$ requires knowing $Δ S_{s u r r}$ , which is inconvenient. At constant $T$ and $P$ :

$Δ S_{s u r r} = - \frac{q _{sy s}}{T} = - \frac{Δ H _{sy s}}{T}$

Substituting:

$Δ S_{u ni v er se} = Δ S_{sy s} - \frac{Δ H _{sy s}}{T}$

Multiplying by $- T$ :

- T Δ S_{u ni v er se} = Δ H_{sy s} - T Δ S_{sy s} \equiv Δ G_{sy s}

Define the Gibbs free energy $G = H - T S$ ; then $Δ G_{sy s} = - T Δ S_{u ni v er se}$ at constant $T, P$ . Since $T > 0$ :

Δ S_{u ni v er se} > 0 ⟺ Δ G_{sy s} < 0

So $Δ G < 0$ (system's free energy decreases) $\Leftrightarrow$ process is spontaneous. This is the practical spontaneity criterion for chemistry, since it involves only the system.

2. Gibbs free energy: definition and properties

G = H - T S

$G$ is a state function (since $H, T, S$ are).
$G$ is extensive (proportional to amount of substance).
At constant $T$ : $Δ G = Δ H - T Δ S$ .
Physical meaning: $- Δ G$ equals the maximum non-expansion (useful) work obtainable from a process at constant $T$ and $P$ .

3. Spontaneity criterion at constant T and P

Δ G < 0 \Rightarrow

process is spontaneous (forward direction favored)

Δ G = 0 \Rightarrow

process is at equilibrium (no net direction)

Δ G > 0 \Rightarrow

process is non-spontaneous (reverse direction favored)

4. Effect of temperature: the four sign combinations

The spontaneity of a process depends on the signs of $Δ H$ and $Δ S$ and the temperature. There are four cases:

Figure: Sign combinations of

Δ H

and

Δ S

and their effect on spontaneity (

Δ G

sign). Crossover temperature for mixed-sign cases is

T = Δ H /Δ S

$Δ H$	$Δ S$	$Δ G = Δ H - T Δ S$	Spontaneity	Example
$< 0$	$> 0$	always $< 0$	Spontaneous at all $T$	Combustion, $2 H_{2} O_{2} \to 2 H_{2} O + O_{2}$
$< 0$	$< 0$	$< 0$ if $T$ small; $> 0$ if $T$ large	Spontaneous only at low $T$	Freezing of water (below $0^{\circ}$ C)
$> 0$	$> 0$	$< 0$ if $T$ large; $> 0$ if $T$ small	Spontaneous only at high $T$	Melting of ice (above $0^{\circ}$ C); decomposition of CaCO $_{3}$ at high $T$
$> 0$	$< 0$	always $> 0$	Non-spontaneous at all $T$	Reverse of combustion

Crossover temperature

For mixed-sign cases, the temperature at which $Δ G$ changes sign is:

T_{crossover} = \frac{Δ H}{Δ S}

Solved Example 1

Determine the temperature at which the reaction

FeO (s) + C(graphite) \to Fe (s) + CO (g)

becomes spontaneous. Given:

Δ H_{f}^{\circ} (FeO) = - 266.3

kJ/mol,

Δ H_{f}^{\circ} (CO) = - 110.5

kJ/mol;

S^{\circ}

: FeO

= 57.49

, C(graphite)

= 5.74

, Fe

= 27.28

, CO

= 197.6

J K

^{- 1}

mol

^{- 1}

Solution:

$Δ H_{r x n}^{\circ} = [Δ H_{f}^{\circ} (Fe) + Δ H_{f}^{\circ} (CO)] - [Δ H_{f}^{\circ} (FeO) + Δ H_{f}^{\circ} (C)]$

$= [0 + (- 110.5)] - [(- 266.3) + 0] = + 155.8 kJ$ .

$Δ S_{r x n}^{\circ} = [S^{\circ} (Fe) + S^{\circ} (CO)] - [S^{\circ} (FeO) + S^{\circ} (C)]$

$= [27.28 + 197.6] - [57.49 + 5.74] = 161.65 J K^{- 1}$ .

For spontaneity, $Δ G < 0$ :

$T > \frac{Δ H}{Δ S} = \frac{155800 J}{161.65 J K ^{- 1}} \approx 964 K$ .

Below this temperature the reaction is non-spontaneous; above, it is spontaneous. This is why iron ore reduction (blast furnace) requires very high temperatures.

5. Standard Gibbs energy of formation

By analogy with $Δ H_{f}^{\circ}$ : the standard Gibbs energy of formation $Δ G_{f}^{\circ}$ of a compound is the change in Gibbs energy when 1 mole of the compound is formed from its elements in standard states.

Convention:

Δ G_{f}^{\circ}

(element in its stable standard state)

= 0

Gibbs energy of a reaction:

Δ G_{r x n}^{\circ} = \sum ν_{i} Δ G_{f}^{\circ} (products) - \sum ν_{i} Δ G_{f}^{\circ} (reactants)

6. Gibbs energy and non-standard states: $Δ G = Δ G^{\circ} + R T ln Q$

For a general chemical reaction proceeding at non-standard concentrations, the Gibbs energy change is:

Δ G = Δ G^{\circ} + R T ln Q

where $Q$ is the reaction quotient (product concentrations over reactant concentrations, raised to stoichiometric powers).

Cases

$Q < K$ : $Δ G < 0$ , forward reaction is favored.
$Q = K$ : $Δ G = 0$ , system is at equilibrium.
$Q > K$ : $Δ G > 0$ , reverse reaction is favored.

7. Relation between $Δ G^{\circ}$ and $K$ - the key formula

At equilibrium, $Δ G = 0$ and $Q = K$ . Substituting into $Δ G = Δ G^{\circ} + R T ln Q$ :

Δ G^{\circ} = - R T ln K = - 2.303 R T lo g K

This is one of the most important relations in physical chemistry:

$Δ G^{\circ} < 0 \Rightarrow K > 1$ (products favored at equilibrium).
$Δ G^{\circ} = 0 \Rightarrow K = 1$ .
$Δ G^{\circ} > 0 \Rightarrow K < 1$ (reactants favored).

Solved Example 2

Calculate

Δ G^{\circ}

for the conversion of oxygen to ozone:

\frac{3}{2} O_{2} (g) \to O_{3} (g)

298 K

, given

K_{p} = 2.47 \times 1 0^{- 29}

Solution:

$Δ G^{\circ} = - 2.303 R T lo g K_{p}$

$= - 2.303 \times 8.314 \times 298 \times lo g (2.47 \times 1 0^{- 29})$

$= - 5705 \times (- 28.607)$

$\approx + 163200 J mol^{- 1} = + 163.2 kJ mol^{- 1}$ .

The large positive $Δ G^{\circ}$ means ozone formation is highly non-spontaneous under standard conditions - $K_{p}$ is extremely small.

Solved Example 3

K_{a}

for CH

_{3}

COOH at

2 5^{\circ}

C is

1.754 \times 1 0^{- 5}

; at

5 0^{\circ}

C it is

1.633 \times 1 0^{- 5}

. Calculate

Δ H^{\circ}

and

Δ S^{\circ}

for the ionization of acetic acid.

Solution:

$Δ G_{298}^{\circ} = - 2.303 \times 8.314 \times 298 \times lo g (1.754 \times 1 0^{- 5}) \approx + 27194 J mol^{- 1}$ .

$Δ G_{323}^{\circ} = - 2.303 \times 8.314 \times 323 \times lo g (1.633 \times 1 0^{- 5}) \approx + 29605 J mol^{- 1}$ .

Using $Δ G^{\circ} = Δ H^{\circ} - T Δ S^{\circ}$ at both temperatures:

$27194 = Δ H^{\circ} - 298 Δ S^{\circ}$

$29605 = Δ H^{\circ} - 323 Δ S^{\circ}$

Subtracting: $- 2411 = 25 Δ S^{\circ} \Rightarrow Δ S^{\circ} = - 96.44 J K^{- 1} mol^{- 1}$ .

Substituting back: $Δ H^{\circ} = 27194 + 298 \times (- 96.44) = 27194 - 28739 = - 1545 J mol^{- 1} \approx - 1.55 kJ mol^{- 1}$ .

Ionization of acetic acid is slightly exothermic but entropy-disfavoured (ions organize water molecules around them, reducing entropy).

8. Gibbs energy and cell EMF

For a galvanic cell, the maximum electrical work obtainable equals $- Δ G$ :

Δ G = - n F E_{ce l l}

where $n$ is the number of moles of electrons transferred and $F$ is Faraday's constant ( $96485 C mol^{- 1}$ ).

At standard conditions:

Δ G^{\circ} = - n F E_{ce l l}^{\circ}

Combining with $Δ G^{\circ} = - R T ln K$ :

E_{ce l l}^{\circ} = \frac{R T}{n F} ln K = \frac{0.0592}{n} lo g K

298 K

Solved Example 4

Calculate free energy change when one mole of NaCl is dissolved in water at

2 5^{\circ}

C. Lattice energy

= 700 kJ mol^{- 1}

;

Δ S = + 26.5 J K^{- 1} mol^{- 1}

; enthalpy of hydration

= - 696 kJ mol^{- 1}

Solution:

$Δ H_{so l n} = Δ H_{l a tt i ce} + Δ H_{h y d r a t i o n} = + 700 + (- 696) = + 4 kJ mol^{- 1}$ .

$Δ G = Δ H - T Δ S = 4000 - 298 \times 26.5 = 4000 - 7897 = - 3897 J mol^{- 1} \approx - 3.9 kJ mol^{- 1}$ .

Slightly negative, so NaCl dissolution is spontaneous - even though the enthalpy change is slightly positive (endothermic), the entropy increase from mixing more than compensates.

9. The Gibbs-Helmholtz equation

Differentiating $G = H - T S$ gives $d G = d H - T d S - S d T$ . Combining with $d H = d U + P d V + V d P$ and the first/second laws, one obtains:

d G = V d P - S d T

Two important consequences:

At constant $T$ : $d G = V d P$ . For an ideal gas, integrating gives $G_{2} - G_{1} = n R T ln (P_{2} / P_{1})$ .
At constant $P$ : $(\frac{\partial G}{\partial T})_{P} = - S$ . Higher entropy means Gibbs energy drops faster with rising $T$ - which is why high $T$ favors high-entropy products.

10. Third law revisited: absolute entropy in $Δ G^{\circ}$

The third law lets us tabulate absolute standard molar entropies $S_{m}^{\circ}$ , which then feed into:

Δ S_{r x n}^{\circ} = \sum ν_{i} S_{m}^{\circ} (products) - \sum ν_{i} S_{m}^{\circ} (reactants)

Combined with $Δ H_{r x n}^{\circ}$ from formation enthalpies, we get $Δ G_{r x n}^{\circ} = Δ H_{r x n}^{\circ} - T Δ S_{r x n}^{\circ}$ , giving the equilibrium constant $K$ from purely thermodynamic tables - no experiment on the reaction itself required.

11. Common Mistakes to Avoid

Watch out

Assuming $Δ G < 0$ means fast reaction. $Δ G$ predicts spontaneity, not rate. A reaction with $Δ G ≪ 0$ may still be very slow because of a high activation barrier.
Using $Δ G$ instead of $Δ G^{\circ}$ in $Δ G^{\circ} = - R T ln K$ . The relation $Δ G^{\circ} = - R T ln K$ uses standard Gibbs energy, not the actual $Δ G$ under working conditions. At equilibrium the actual $Δ G = 0$ , not $Δ G^{\circ}$ .
Confusing units of $R$ . When computing $Δ G$ in kJ, use $R = 8.314 \times 1 0^{- 3} kJ mol^{- 1} K^{- 1}$ . When in J, use $R = 8.314 J mol^{- 1} K^{- 1}$ . Mixing kJ and J gives errors of $1000 \times$ .
Forgetting that $Δ S^{\circ}$ is in J K $^{- 1}$ but $Δ H^{\circ}$ often in kJ. Always convert to consistent units before combining in $Δ G = Δ H - T Δ S$ .
Ignoring the sign of $lo g K$ when $K < 1$ . If $K$ is very small (e.g. $1 0^{- 29}$ ), $lo g K$ is a large negative number, giving a large positive $Δ G^{\circ}$ .
Using $Δ G^{\circ} = - n F E_{ce l l}$ instead of $- n F E_{ce l l}^{\circ}$ . Distinguish standard vs non-standard: $Δ G = - n F E_{ce l l}$ under any conditions; $Δ G^{\circ} = - n F E_{ce l l}^{\circ}$ only at standard.
Assuming the third law lets us compute absolute $H$ or $G$ . The third law gives absolute entropy, but $H$ and $G$ still have no absolute reference. Only $Δ H$ , $Δ G$ are meaningful.

Frequently Asked Questions

Q1. Why do we need Gibbs free energy when we already have entropy of the universe?

$Δ S_{u ni v er se} > 0$ correctly predicts spontaneity but requires knowing $Δ S_{s u r r o u n d in g s}$ - which depends on external conditions. $Δ G = Δ H - T Δ S$ uses only system quantities and works at constant $T$ and $P$ (common lab conditions). Since $Δ G = - T Δ S_{u ni v er se}$ , the two criteria are equivalent.

Q2. What does $Δ G < 0$ physically mean?

A process with $Δ G < 0$ can proceed spontaneously and can, in principle, do useful (non-expansion) work equal to $- Δ G$ . If $Δ G = - 100 kJ$ , the process can do up to $100 kJ$ of useful work. In practice actual work is less because real processes aren't reversible.

Q3. Is a reaction with $Δ G > 0$ ever possible?

Yes - if energy is supplied from outside (electrolysis, photosynthesis, coupled reactions). $Δ G > 0$ means the reaction won't proceed spontaneously at the given conditions, but it can be driven by external energy. This is why plants use sunlight to drive endergonic reactions and batteries drive electrolysis.

Q4. What is the significance of $Δ G^{\circ} = - R T ln K$ ?

It links thermodynamics (Gibbs energy) with equilibrium composition (equilibrium constant $K$ ) via one compact formula. It lets us predict the equilibrium constant of any reaction without doing the experiment - given the tabulated $Δ H_{f}^{\circ}$ and $S^{\circ}$ values.

Q5. Why does temperature favor different reactions?

Because $Δ G = Δ H - T Δ S$ : at high $T$ the $T Δ S$ term dominates. If $Δ S > 0$ , high $T$ makes $Δ G$ more negative (favors reaction). If $Δ S < 0$ , high $T$ makes $Δ G$ more positive (disfavors reaction). This is why ice melts above $0^{\circ}$ C but freezes below.

Q6. How is $Δ G$ related to cell EMF?

$Δ G = - n F E_{ce l l}$ where $n$ is the number of electrons transferred and $F$ is Faraday's constant. For a spontaneous cell ( $E_{ce l l} > 0$ ), $Δ G < 0$ . At standard conditions, $Δ G^{\circ} = - n F E_{ce l l}^{\circ}$ .

Q7. Does $Δ G$ tell us anything about reaction rate?

No. $Δ G$ is a thermodynamic quantity that predicts whether a reaction is spontaneous, not how fast. Rate is a kinetics question, governed by activation energy. Diamond has $Δ G > 0$ relative to graphite but converts to graphite so slowly that diamonds seem permanent.

Q8. What is the physical meaning of $Δ G_{f}^{\circ}$ of an element being zero?

It is a convention chosen so that $Δ G_{r x n}^{\circ}$ can be computed from a table of $Δ G_{f}^{\circ}$ values. Since we set the reference element in its standard state to zero, all differences work out correctly - regardless of the true absolute value of $G$ (which is unknowable).

Previous year questions on Gibbs Energy Change and Equilibrium

21 questions from past papers, each with a step-by-step solution.

Show all 21 questions

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1. From ΔSuniverse​ to ΔG

2. Gibbs free energy: definition and properties

3. Spontaneity criterion at constant T and P

4. Effect of temperature: the four sign combinations

Crossover temperature

5. Standard Gibbs energy of formation

6. Gibbs energy and non-standard states: ΔG=ΔG∘+RTlnQ

Cases

7. Relation between ΔG∘ and K - the key formula

8. Gibbs energy and cell EMF

9. The Gibbs-Helmholtz equation

10. Third law revisited: absolute entropy in ΔG∘

11. Common Mistakes to Avoid

Frequently Asked Questions

Q1. Why do we need Gibbs free energy when we already have entropy of the universe?

Q2. What does ΔG<0 physically mean?

Q3. Is a reaction with ΔG>0 ever possible?

Q4. What is the significance of ΔG∘=−RTlnK?

Q5. Why does temperature favor different reactions?

Q6. How is ΔG related to cell EMF?

Q7. Does ΔG tell us anything about reaction rate?

Q8. What is the physical meaning of ΔGf∘​ of an element being zero?

Previous year questions on Gibbs Energy Change and Equilibrium

Ready to master Chemical Thermodynamics?

1. From $Δ S_{u ni v er se}$ to $Δ G$

6. Gibbs energy and non-standard states: $Δ G = Δ G^{\circ} + R T ln Q$

7. Relation between $Δ G^{\circ}$ and $K$ - the key formula

10. Third law revisited: absolute entropy in $Δ G^{\circ}$

Q2. What does $Δ G < 0$ physically mean?

Q3. Is a reaction with $Δ G > 0$ ever possible?

Q4. What is the significance of $Δ G^{\circ} = - R T ln K$ ?

Q6. How is $Δ G$ related to cell EMF?

Q7. Does $Δ G$ tell us anything about reaction rate?

Q8. What is the physical meaning of $Δ G_{f}^{\circ}$ of an element being zero?