Gibbs Free Energy

Gibbs free energy is the maximum useful work obtainable at constant temperature and pressure. It is denoted by G and is widely used in chemistry to understand energy changes in reactions and processes. It takes into account two important factors: enthalpy (heat content) and entropy (degree of disorder) of a system.

Equation of Gibbs Free Energy

This equation shows that Gibbs free energy depends on both the enthalpy and the entropy of a system. It helps in understanding how these two factors together determine the energy available in a system.

G = H - TS

where:

• S is the entropy of the system,
• H is Enthalpy, and
• T is the Temperature.

Change in Gibbs Free Energy

The change in Gibbs free energy, denoted by ΔG, represents the difference in free energy between the final and initial states of a system during a process. It helps in understanding the direction and feasibility of a reaction at constant temperature and pressure.

It depends on both enthalpy (ΔH) and entropy (ΔS) of the system, which together determine how the energy of the system changes.

ΔG = ΔH - T ΔS

Where:

• ΔG = Change in Gibbs free energy
• ΔH = Change in enthalpy
• T = Temperature (in Kelvin)
• ΔS = Change in entropy

Gibbs Energy and Spontaneity

The concept connects enthalpy (ΔH) and entropy (ΔS) to predict the direction of a process. Spontaneity means a process can occur on its own without external energy. Gibbs free energy gives a clear criterion for this.

• If ΔG is negative, the system has enough energy to proceed on its own, so the process occurs naturally.
• If ΔG is positive, energy is required to make the process occur, so it is not spontaneous.
• If ΔG is zero, there is no net change, and the system is in balance (equilibrium).

ΔG = ΔH − TΔS

Where:

• ΔG = Change in Gibbs free energy
• ΔH = Change in enthalpy
• T = Absolute temperature (in Kelvin)
• ΔS = Change in entropy

Factors affecting Gibbs Free Energy

Gibbs free energy depends on both enthalpy (ΔH) and entropy (ΔS). Thus, the factors that affect Gibbs free energy are:

1. Enthalpy Change (ΔH)

• If ΔH is negative (exothermic) , ΔG becomes more negative, process becomes more spontaneous.
• If ΔH is positive (endothermic), ΔG increases, process becomes less spontaneous.

Example: Burning of fuel releases heat , ΔH is negative , favors spontaneity.

2. Entropy Change (ΔS)

• If ΔS is positive (increase in disorder) , ΔG decreases , favors spontaneity.
• If ΔS is negative (decrease in disorder), ΔG increases, opposes spontaneity.

Example: Solid → Liquid → Gas, entropy increases , ΔG becomes more negative.

3. Temperature (T)

• Temperature plays a key role in the term TΔS.
• If ΔS is positive: Increasing temperature makes ΔG more negative, increases spontaneity.
• If ΔS is negative: Increasing temperature makes ΔG more positive, decreases spontaneity.

Standard Gibbs Free Energy (ΔG°)

Standard Gibbs free energy, denoted as ΔG°, is the change in Gibbs free energy when all reactants and products are in their standard states at a specified temperature (usually 298 K) and pressure (1 bar).

ΔG° = ΔH° − TΔS°

Where:

• ΔG° = Standard Gibbs free energy change
• ΔH° = Standard enthalpy change
• ΔS° = Standard entropy change
• T = Temperature in Kelvin

ΔG and Equilibrium constant

This equation shows the link between standard Gibbs free energy (ΔG°) and the equilibrium constant (K). It tells us how thermodynamics (ΔG°) is related to equilibrium (K).

• If K > 1, ΔG° is negative
• If K < 1, ΔG° is positive
• If K = 1, ΔG° = 0, system is at equilibrium

ΔG° = −RT ln K

Where:

• ΔG° = Standard Gibbs free energy change
• R = Gas constant
• T = Temperature in Kelvin
• K = Equilibrium constant

Solved Examples

Problem 1: Determine whether the reaction is spontaneous or non-spontaneous for the given value of ΔH and ΔS. Also, state whether they are exothermic or endothermic.

• ΔH = - 40 kJ and ΔS = +135 JK^-1 at 300K
• ΔH = - 60 kJ and ΔS = -160 JK^-1 at 400K

Solution:

ΔG = ΔH - T ΔS

 ΔH = - 40 kJ,  ΔS = +135 J K^-1 = 0.135 kJ K^-1 and T = 300K

 ΔG = -40 (kJ) - 0.135(kJ K^-1) × 300(K)

ΔG = –80.5 kJ

Because ΔG is negative, the reaction is spontaneous. The negative ΔH value indicates that the reaction is exothermic.

• ΔG = ΔH - T ΔS

ΔH = - 60 kJ,  ΔS = - 160 J K^-1 = - 0.160 kJ K^-1 and T = 400K

 ΔG = −60 −[ 400 ( − 0.160 ) ]

 = -60 kJ + 64 kJ = 4kJ

The reaction is non-spontaneous because ΔG is positive and exothermic as ΔH is negative.

Problem 2: For a certain reaction ΔH = -25kJ and ΔS = -40J K^-1. At what temperature will it change from spontaneous to non-spontaneous.

Solution:

T = ΔH / ΔS

 ΔH = - 25 kJ,  ΔS = -40 J K^-1  = -0.04 kJ K-1 

Hence, T = -25(kJ) / -0.04 kJ K^-1  = 625K

Because both ΔH and ΔS are negative, the reaction will occur spontaneously at lower temperatures. As a result, the reaction will be spontaneous below 625K and non-spontaneous beyond 625K.

At 625K, the transition from spontaneous to non-spontaneous occurs.

Problem 3: Determine  ΔS_total and decide whether the following reaction is spontaneous at 298K.

ΔH° = -24.8 kJ, ΔS° = 15 J K^-1

Solution: 

The heat evolved in the reaction is 24.8 kJ. The same quantity of heat is absorbed by the surroundings. 

Hence, Entropy change of the surrounding will be,

ΔS = −ΔH / T

= - [(24800 (J)) / 298 (K)]

= + 83.2 J K^-1

ΔS_total = ΔS_System + ΔS_Surr 

ΔS_Sys = ΔS° = 15 J K^-1

 = 15(J K^-1) + 83.2 (J K^-1)

= 98.2 J K^-1

As ΔS_total is positive, the reaction is spontaneous at 298 K.

Problem 4: Determine whether the reaction,

N₂​O₄ ​(g)⟶2NO₂​ (g)

is spontaneous at 298 K from the following data.

Δ_fH° (N₂O₄) = 9.16 kJ mol^-1, Δ_fH° (NO₂) = 33.2 kJ mol^-1

ΔS° = 175.8 × 10 ⁻³

Solution: 

ΔH^∘ = ∑ Δ_f​H^∘ (products) − ∑Δ_f​H^∘ (reactants)

= 2 × Δ_fH° (NO₂) - Δ_fH° (N₂O₄)

= 2(mol) × 33.2(kJ mol^-1) -1(mol) × 9.16(kJ mol^-1)

= +57.24 kJ

ΔG° = ΔH° - TΔS°

57.24(kJ) - 298(K) × 175.8 × 10^-3 (kJ K^-1)

= +4.85 kJ.

Because ΔG° is positive, the reaction is non-spontaneous at 298 K. 

The temperature at which the reaction changes from spontaneous to non-spontaneous is given by,

T = ΔH° / ΔS°

= 57.24(kJ) / 0.1758(kJ K^-1)

= 325.6 K

Because ΔH° and ΔS° are both positive, the reaction will be spontaneous at high temperature.

The reaction will be spontaneous above 325.6 K.

Problem 5: Determine K_p for the reaction,

2SO₂​(g) + O₂​(g) ⟶ 2SO_3​(g)

is 7.1 × 10²⁴ at 298 K. Calculate ΔG° for the reaction (R = 8.314 JK^-1 mol^-1).

Solution:

ΔG° = -2.303RT log₁₀ K_p

K_p = 7.1 × 10²⁴ 

R = 8.314 JK^-1 mol^-1

T = 298K

Hence,

  ΔG°  = -2.303 × 8.314 × 10^-3 (kJ K^-1 mol^-1) ×  log₁₀ (7.1 ×  10²⁴)

          = -141.8 kJ mol^-1

Problem 6: Calculate K_p for the reaction at 513 K,

2NOCl (g) ⟶ 2NO(g) + Cl₂​(g)

with ΔG° = 17.38 kJ mol^-1.

Solution:

ΔG° = -2.303 RT log₁₀ K_p

ΔG° = 17.38 kJ mol^-1

R = 8.314 J K^-1 mol^-1

T = 513K

Hence,

log₁₀ k_p = - ΔG° / 2.303 RT

             = - (17380(J mol^-1) / 2.303 × 8.314 ( J K^-1 mol^-1) × 513(K))

             = -1.769

Hence, K_p = antilog(-1.769)

                = 0.017          

Related Articles:

• Spontaneity
• Second Law of Thermodynamics