How to calculate delta G (∆G) of a reaction?

The symbol ∆G is of utmost importance in Chemistry, especially w.r.t thermodynamic systems.

The Greek letter delta (∆) means change, while G stands for Gibbs free energy of a reaction. Thus, ∆G together represents the Gibbs free energy change for a reaction.

In this article, you will learn to calculate ∆G (Delta G) via multiple formulae or equations. So are you ready?

If yes, then dive into the article and give it a good reading!   

What is ∆G (Delta G)?

Gibbs free energy change (∆G) refers to reversible work done on a system at constant temperature and pressure. In other words, ∆G quantifies the energy that is converted to useful work during a process. (Units: J/mol or kJ/mol).

It helps in determining the spontaneity of a chemical reaction and the direction in which the reaction progresses.

For a reversible reaction, A + B ⇌ C + D

• If ∆G < 0: The reaction is thermodynamically favored, thus spontaneous in the forward direction.
• If ∆G > 0: The reaction is non-spontaneous in the forward direction. Some sort of external interference, i.e., energy, is required to propel it forward.
• If ∆G = 0: The reaction stays at equilibrium, where the rate of forward reaction = rate of backward reaction.

Multiple formulae can be applied to determine ∆G under different situations.

Let’s explore when to apply which one with a solved example in each case.

How to calculate ∆G (Delta G) for a general chemical reaction? – Example

A chemical reaction is simply defined as the transformation of reactants into products.

Old reactant bonds break by absorbing a certain amount of energy, while new product bonds are formed by releasing some energy.

Reactants → Products

The Gibbs free energy change (Delta G) can be calculated by the formula given below:

Where,

• ∆G (reaction) = Gibbs free energy change for the reaction
• Σ ∆G_f (products) = Sum of Gibbs free energy change for the formation of products
• Σ ∆G_f (reactants) = Sum of Gibbs free energy change for the formation of reactants

 ∆G_f denotes the Gibbs free energy change of formation, i.e., ∆G when 1 mole of a compound is formed from its constituent elements under standard conditions (273 K temperature and 1 atm pressure).

For example, Find the Gibbs free energy change (∆G) for the reaction given below using the data provided.

2 C₂H_{6 (g)} + 7 O_{2 (g)} → 4 CO_{2 (g)} + 6 H₂O _(l)

• ∆G_f (C₂H₆) = -84.7 kJ/mol
• ∆G_f (CO₂) = -394.4 kJ/mol
• ∆G_f (H₂O) = -273.2 kJ/mol

Solution:

Equation (i) can be re-written as follows w.r.t the reaction given above:

∆G (reaction) = [4 x ∆G_f (CO₂) + 6 x ∆G_f (H₂O)] – [2 x ∆G_f (C₂H₆) + 7 x ∆G_f (O₂)]

You must note that each of the above values is multiplied by the number of moles of each compound in the balanced chemical equation. Also, ∆G_f (O₂) = 0, as oxygen is a molecule and not a compound.

Substituting the other ∆G_f values as given in the question statement:

∆G (reaction) = [4(-394.4) + 6(-273.2)] – [2(-84.7)]

∆G (reaction) = -3216.8 + 169.4

∴ ∆G (reaction) = -3047.4 kJ/mol

Result:  The Gibbs free energy change for the reaction is -3047.4 kJ/mol.

How to calculate Delta G in a thermodynamic system?

For a thermodynamic system in which different types of heat or energy changes are involved, the Gibbs free energy change (∆G) of a reaction can be determined as follows:

Where,

• ∆G = Gibbs free energy change (Units: J/mol or kJ/mol)
• ∆H = Enthalpy change (Units: J/mol or kJ/mol)
• T = Absolute temperature (Unit: K)
• ∆S= Entropy change (Units: J/K.mol or kJ/K.mol)

Enthalpy change (∆H) is the amount of heat exchanged between the system and its surroundings. In contrast, entropy change (∆S) refers to the degree of disorder or randomness in a thermodynamic system.

Now let’s see an example of applying ∆G = ∆H – T∆S, Calculate ∆G for the decomposition of hydrogen peroxide into oxygen and water at 298 K, given that ∆H (reaction) = -98 kJ/mol and ∆S = -0.1 kJ/K.mol.

H₂O_{2 (l)} → H₂O _(l) + ½ O_{2 (g)}

Solution:

Given that,

∆H = -98 kJ/mol

T = 298 K

∆S = -0.1 kJ/mol

∆G =?

Applying the formula and substituting the known values gives us:

⇒ ∆G = ∆H – T ∆S

∆G = -98 – [(298) (-0.1)]

∆G = -98 + 29.8

∴ ∆G = -68.2 kJ/mol

Result: The Gibbs free energy change for the decomposition of hydrogen peroxide is -68.2 kJ/mol, which implies that it is a thermodynamically favored reaction.

How to calculate ∆G using an equilibrium constant?

For a reversible reaction, aA + bB ⇌ cC + dD, the reaction quotient (Q) determines its current state, i.e., whether the equilibrium is more towards the left or towards the right.

∆G is related to Q by the equation shown below:

∆G = ∆G° + RT lnQ……Equation (iii)

Where,

• ∆G = Gibbs free energy change for the reaction
• ∆G° = Standard Gibbs free energy change (at 273 K, 1 atm)
• R = Ideal gas constant (R = 8.314 J/K.mol)
• T = Absolute temperature
• Q = Reaction quotient

At equilibrium, ∆G = 0, and Q = K_eq. Thus, equation (iii) transforms into:

∆G° = -RTln K_eq…Equation (iv)

Under non-standard temperature conditions, equation (iv) can be rewritten as ∆G = -RT lnK_eq.

So, equation (iv) can be used to find the Gibbs free energy change (∆G) for a reversible reaction given its equilibrium constant (K_eq).

For example, The equilibrium constant for the reversible reaction given below is 45.9. Find its Gibbs free energy change (∆G) at 333 K.

H_{2 (g)} + I_{2 (g)} ⇌ 2 HI _(g)

Solution:

Given that,

K_eq = 45.9

T = 333 K

∆G (Delta G) =?

We already know that R = 8.314 J/K.mol

Applying the formula ∆G = -RT lnK_eq to find the unknown value of ∆G:

∆G = – (8.314) (333) (ln 45.9)

∴ ∆G = – 1.06 x 10⁴ J/mol = – 10.6 kJ/mol

Result:  The Gibbs free energy change for the given reaction is -10.6 kJ/mol.  

∆G for a gas-phase reaction can be determined by another variant of equation (iii), i.e.,

∆G = ∆G° + RT ln (P_total/P°)……Equation (v)

Where,

• P_total = Total pressure exerted by the gas
• P^° = Standard pressure (1 atm)

Let’s see an example- The standard Gibbs free energy change for the gas-phase reaction shown below is -33.3 kJ/mol.

N_{2 (g)} + 3 H_{2 (g)} ⇌ 2 NH_{3 (g)}

At 500 K, if the partial pressures of nitrogen, hydrogen, and ammonia gas in the reaction mixture are 1 atm, 2 atm, and 3.5 atm, respectively. Calculate the value of Gibbs free energy change (∆G) for the reaction.

Solution:

Given in the question statement:

• ∆G° = -33.3 kJ/mol = -33,300 J/mol
• T = 500 K
• P_{total =} P(N₂) + P(H₂) + P(NH₃) = 1 + 2 + 3.5 = 6.5 atm
• ∆G =?

We already know that R = 8.314 J/K.mol and P° = 1 atm

Applying the formula:

⇒ ∆G = ∆G° + RT ln (P_total/P°)

Substituting the known values to find the unknown value of ∆G: 

∆G = -33,300 + (8.314) (500) ln (6.5/1)

∆G = -33,300 + 7781.1

∴ ∆G = -2.55 x 10⁴ J/mol =- 25.5 kJ/mol

Result:  The Gibbs free energy change for the given reaction is -25.5 kJ/mol.  

How to calculate ∆G (Delta G) from cell potential?

A redox reaction takes place to generate electricity in an electrochemical cell.

The Gibbs free energy change (∆G) during the redox reaction can be calculated using equation (vi) given below:

Where,

• ∆G = Gibbs free energy change
• n = Number of moles of electrons transferred
• F = Faraday constant (F = 96,485 C/mol.)
• E_cell= Electrical potential difference or cell potential across an electrochemical cell (Unit: V).

The electrical potential difference is determined as follows:

E_cell = E^ϴ (ox) + E^ϴ (red)

Where,

• E^ϴ (red) = Standard cell potential of the reduction half-cell (cathode)
• E^ϴ (ox) = Standard cell potential of the oxidation half-cell (anode)

Red stands for reduction. Reduction is the gain of electrons that takes place at the cathode. Contrarily, ox stands for oxidation, which denotes the loss of electrons taking place at the anode.

The cathode and anode are two half-cells/electrodes that form the electrochemical cell.  The standard cell potentials at these two electrodes can thus be used to find ∆G.

Let us see an example- Find the value of ∆G for the redox reaction given below.

Zn _(s) + CuSO_{4 (aq)} → ZnSO_{4 (aq)} +Cu _(s)

Half-cell reactions:

Oxidation half-cell: Zn_(s) → Zn²⁺ _(aq) + 2e^– (E^ϴ = + 0.76 V)

Reduction half-cell: Cu²⁺ _(aq) + 2e^– → Cu_(s)  (E^ϴ = 0.34 V)

Solution:

Firstly, we need to calculate E_cell as follows:

E_cell = E^ϴ (ox) + E^ϴ (red)

E_cell = 0.76 + 0.34 = 1.10 Volts

As per the balanced chemical equation, n = 2. We also know that F= 96,485 C/mol.

Hence, we can apply the formula ∆G = -nFE_cell to find Delta G:

⇒ ∆G = -nFE_cell

Substituting the known values:

∆G = – (2) (96,485) (1.10)

∆G = – 2.12 x 10⁵ J/mol = – 212 kJ/mol

Result:  The Gibbs free energy change for the given redox reaction is -212 kJ/mol.

 FAQ