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

The symbol ∆S represents entropy change, a fundamental concept in Chemistry. The Greek letter delta (∆) means change, while S denotes entropy.

What is ∆S (Delta S)?

Entropy change (∆S) describes the energy distribution in a thermodynamic system.

It refers to the degree of disorder or randomness present in a system as a result of the diverse arrangement of the system’s constituent particles.

If ∆S > 0: The system tends to be disordered.

If ∆S < 0: the Disorder decreases, i.e., the order in the system increases.

In this article, we will teach you how to calculate ∆S through a variety of different formulae that are used under differing situations.

How to calculate ∆S (Delta S) for a general chemical reaction? – Examples

For a general chemical reaction, Reactants → Products, the change in entropy is calculated as:

Where,

• ∆S°(reaction) = Change in entropy of the reaction (Units: J/K.mol or kJ/K.mol)
• ΣS° (products) = Sum of the standard entropies of the products
• ΣS° (reactants) = Sum of the standard entropies of the reactants

The ° sign stands for standard conditions, i.e., a temperature of 298 K and a pressure of 1 atm.

For example, Find the entropy change (∆S) for the reaction (shown below) that takes place at standard conditions.

CH_{4 (g)} + 2 O_{2 (g)} → CO_{2 (g)} + 2 H₂O _(l)

The standard entropies of all the reactants and products are given below:

• S° (CH₄) = 186.3 J/ K.mol
• S° (O₂) = 205.0 J/K.mol
• S° (CO₂) = 213.8 J/K.mol
• S° (H₂O) = 70.0 J/K.mol

Solution:

Modifying equation (i) w.r.t the reaction shown above gives us:

∆S° (reaction) = [S° (CO₂) + 2 S° (H₂O)] – [S° (CH₄) + 2 S° (O₂)]

Important: Do not forget to multiply S° for each of the reactants and products with their respective number of moles as per the balanced chemical equation.

Substituting the known values of S° to calculate unknown ∆S° gives us:

∆S° (reaction) = [(213.8) +2(70.0)]-[186.3+2(205)]

∆S° (reaction) = 353.8 – 596.3

∴ ∆S° (reaction) = -242.5 J/ K.mol

Result:  The entropy change for the combustion of methane is -242.5 J/K.mol.

Further insight: A negative value of ∆S implies that the disorder in the system decreases because of this reaction. Also, the total number of gas molecules present in the system decreases, which in turn means that the stability of the system increases.

Another example is– Calculate ∆S (Delta S) for the reaction given below:

Mg _(s) + 2 HCl _(aq) → MgCl_{2 (aq)} + H_{2 (g)}

The standard entropies of all the above reactants and products are:

• S° (Mg) = 32.69 J/K.mol
• S° (HCl) = 186.9 J/K.mol
• S° (MgCl₂) = 89.88 J/K.mol
• S° (H₂) = 130.59 J/K.mol

Solution:

Modifying equation (i) w.r.t the above reaction:

∆S° (reaction) = [S° (MgCl₂) + S° (H₂)] – [S° (Mg) + 2 S° (HCl)]

Substituting the standard entropy values to find ∆S (Delta S):

∆S° (reaction) = [89.88 + 130.59] – [32.69 + 2 (186.9)]

∆S° (reaction) = 220.47 – 406.49

∴ ∆S° (reaction) = -186.02 J/ K.mol

Result:  The entropy change for the given reaction is -186.02 J/K.mol.

How to calculate ∆S (Delta S) as per the second law of thermodynamics?

As per the 2^nd law of thermodynamics, the entropy of an isolated system always tends to increase over time.

The passage of heat into or out of an isolated system creates a disturbance within the system, which leads to a higher entropy value. Similarly, the entropy of the Universe tends to increase.

∆S = Q_rev/T…Equation (ii)

Where,

• ∆S = Change in entropy of the system
• Q_rev = Heat exchanged under reversible conditions
• T = Absolute temperature

Equation (ii) is mostly used to find the entropy change during the reversible isothermal expansion of an ideal gas.

Let’s see an example- 3600 kJ of heat is supplied to a reversible heat engine at a constant temperature of 300 K. Find the change in entropy (∆S) of the system.

Solution:

As per the question statement:

Q_rev = 3600 kJ = 3600 x 1000 = 3,600,000 J

T= 300 K

∆S (Delta S) =?

Applying the formula:

⇒ ∆S = Q_rev/T

Substituting the known values:

∆S = 3,600,000/300

∴ ∆S = 12000 J/K.mol or 12 kJ/K.mol

Result:  The change in entropy of the reversible heat engine is 12000 J/K.mol.

How to calculate ∆S (Delta S) in a thermodynamic system?

For a thermodynamic system in which different types of heat or energy changes are involved, the entropy change (∆S) of a reaction is determined as per the formula given below:

∆G = ∆H – T∆S….Equation (iii)

Rearranging equation (iii) to make ∆S the subject of the formula:

Where,

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

Let’s solve an example using ∆S = (∆H-∆G)/T, The enthalpy change (∆H) for the endothermic reaction taking place at 298 K (as shown below) is + 580.6 kJ/mol while its Gibbs free energy change (∆G) is 578.4 kJ/mol. Calculate the entropy change (∆S) for the reaction.

PCl_{3 (g)} + 3 HCl _(g) → 3 Cl_{2 (g)} + PH_{3 (g)}

Solution:

Given in the question statement:

• ∆H = 580.6 kJ/mol
• ∆G = 578.4 kJ/mol
• T = 298 K
• ∆S (Delta S) = ?

Applying the formula:

⇒ ∆S = (∆H – ∆G)/T

Substituting all the known values to find the unknown value of ∆S:

∆S = (580.6 – 578.4)/298

∴ ∆S (Delta S) = 7.4 x 10^-3 kJ/K.mol or 7.4 J/K.mol

Result: The entropy change during the endothermic reaction is 7.4 J/K.mol, which means particles gained more order at the end of the reaction.

How to find S as per the Boltzmann equation?

The entropy (S) of a system is directly related to its thermodynamic probability as per the Boltzmann-Planck equation:

S = klnW…..Equation (v)

Where,

• S = Statistical entropy of the system
• k= Boltzmann’s constant per particle (k = 1.3805 x 10^-23 J/K)
• W = no of microstates, also called thermodynamic probability

The greater the possibilities for particles to get arranged in a system (W), the higher the chances of disorder, i.e., entropy.

W = 10¹⁰ in a gaseous system implies that there are 10¹⁰ possibilities for the gas molecules to occupy different positions in the system.  

For example- The thermodynamic probability (W) for a gas at a specific temperature and pressure is 10²⁸ J/K. Calculate the entropy (S) of the system.

Solution:

As per the question statement,

W = 10²⁸ J/K

We already know that k = 1.3805 x 10^-23 J/K.

Therefore substituting the known values into the formula:

⇒ S = k lnW

S = (1.3805 x 10^-23) ln (10²⁸)

∴ S = 8.90 x 10^-22

Result: The entropy of the gas in this example is 8.90 x 10^-22 J/K.mol.

FAQ

How can you find the delta S (∆S) of a reaction?

∆S stands for entropy change, i.e., the degree of disorder in a system. Different formulae that can be used to find ∆S for a system are: ∆S° (reaction) = Σ S° (products) – Σ S° (reactants) Where ∆S° = Entropy change of a chemical reaction, Σ S° = Sum of standard entropies of products and/or reactants multiplied by their stoichiometric coefficients. ∆S = (∆H – ∆G)/T Where, ∆H = Enthalpy change ∆G = Gibbs free energy change T = Absolute temperature ∆S = Qrev/T Where Qrev = Heat exchanged during an isothermal reversible process

When is the formula S = k lnW applied to find the entropy of a reaction?

The formula S= klnW is applied to calculate the statistical entropy of a system. In S= klnW: S = Entropy k = Boltzmann constant (1.3805 x 10-23 J/K) W= number of microstates or thermodynamic probability (total possible arrangements of particles)

What is represented by the formula ∆S = nR ln (Vf/Vi)?

The formula ∆S = nRln (Vf/Vi) is used to find the change in entropy for the expansion of an ideal gas at a constant temperature and pressure where n = number of moles of gas, R = ideal gas constant (8.314 J/K.mol), Vf = final volume of the gas and Vi = initial volume of gas.

What does the formula ∆S = ∆H (fus/vap/sub)/T represents?

The formula ∆S (Delta S) = ∆H/T is used to calculate entropy change for a phase transition at constant temperature and pressure. ∆H denotes enthalpy change, i.e., heat absorbed or released by the system while T stands for the absolute temperature. ∆S = ∆H (fusion)/T → Entropy decreases when heat is released, changing a liquid to a solid. ∆S = ∆H (vaporization)/T → Entropy increases when heat is absorbed, changing a liquid to a gas. ∆S = ∆H (sublimation)/T → Entropy increases when heat is absorbed, changing a solid to a gas.