Entropy in Chemistry: Definition, Equation, Formula, Examples

Probably the most native view on entropy is that it measures the amount of disorder or randomness present in a system. It was introduced in the 19th century by Rudolf Clausius as a concept in thermodynamics and statistical mechanics, quantifying the number of possible microscopic configurations that correspond to a given macroscopic state of a thermodynamic system. It can be said to quantify uncertainty in the state of a system. Entropy, associated with natural processes, increases, and this leads to the second law: the total entropy of an isolated system cannot decrease with time.

This Story also Contains

1. Entropy
2. Some Solved Examples
3. Summary

Entropy

It is a thermodynamic state quantity that is used to measure the disorder or randomness of the molecules in a system The disorder or randomness in a system is measured in terms of entropy (S). The absolute value of 'S' is not determined so most change In entropy $\Delta$S is measured.

Randomness $\propto$ Entropy

It is a state function that depends only on the initial and final state of the system that is, it is independent of the path used in going from the initial to the final state.

$\Delta \mathrm{S}=\mathrm{S}_{\text {Final }}-\mathrm{S}_{\text {Initial }}$

For a general chemical reaction at 298K and 1 atm:

$\begin{aligned} & \mathrm{m}_1 \mathrm{P}+\mathrm{m}_2 \mathrm{Q} \rightarrow \mathrm{n}_1 \mathrm{R}+\mathrm{n}_2 \mathrm{~S} \\ & \Delta \mathrm{S}^{\circ}=\left[\left(\mathrm{n}_1 \mathrm{~S}_{\mathrm{R}}^{\circ}+\mathrm{n}_2 \mathrm{~S}_{\mathrm{s}}^{\circ}\right)-\left(\mathrm{m}_1 \mathrm{~S}_{\mathrm{P}}^{\circ}+\mathrm{m}_2 \mathrm{~S}_{\mathrm{Q}}^{\circ}\right)\right] \\ & \Delta \mathrm{S}^{\circ}=\sum \mathrm{S}_{\mathrm{P}}^{\circ}-\sum \mathrm{S}_{\mathrm{R}}^{\circ}\end{aligned}$

It is an extensive property and a state function that depends on state variables like T, P, V, and n which govern the state of a system.

Entropy and Temperature

Entropy increases with the increase of temperature as it is associated with molecular motion which Increases with the increase of temperature due to the increase in the average kinetic energy of the molecules. The entropy of a perfectly ordered Crystalline substance is taken as zero at Zero Kelvin (0K). It is the third law of Thermodynamics. However, in the case of N₂O, NO, Solid CI_2, etc. the value of entropy is not found to be zero at O Kelvin also.

Recommended topic video on (Entropy)

Some Solved Examples

Example 1: The entropy change (in kJK^-1kg^-1) associated with the conversion of 1kg of ice at 273 K to water vapors at 383 K is:( Specific heat of water liquid and water vapour are $4.2 \mathrm{kJK}^{-1} \mathrm{~kg}^{-1}$ and $2.0 \mathrm{kJK}^{-1} \mathrm{~kg}^{-1}$ ; heat of liquid fusion and vapourisation of water are $334 \mathrm{kJk}^{-1}$ and $2491 \mathrm{kJkg}^{-1}$ , respectively ). ( log 273 = 2.436 , log 373=2.572 , log 383 = 2.583)

1)2.64

2)7.9

3)8.49

4) 9.26

Solution

Entropy for phase transition at constant pressure -

$\Delta S=\frac{\Delta H_{\text {Transition }}}{T}$

wherein

Transition Fusion, Vaporisation, Sublimation

$\Delta H \Rightarrow$Enthalpy

$\Delta E \Rightarrow$Internal Energy

$T \Rightarrow$Transitional temperature

Entropy for solid and liquid -

$\Delta S=n C \ln \frac{T_f}{T_i}$

or

$\Delta S=m s \ln \frac{T_f}{T_i}$
wherein

n= no. of moles

C= molar heat capacity

m= mass

s= specific heat capacity

Now, Phase change path -

$\mathrm{H}_2 \mathrm{O}(\mathrm{S}) \xrightarrow{\Delta \mathrm{S}_1} \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \xrightarrow{\Delta \mathrm{S}_2} \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \xrightarrow{\Delta \mathrm{S}_3} \mathrm{H}_2 \mathrm{O}(\mathrm{g}) \xrightarrow{\Delta \mathrm{S}_4} \mathrm{H}_2 \mathrm{O}(\mathrm{g})$

$\begin{array}{lllll}273 \mathrm{~K} & 273 \mathrm{~K} & 373 \mathrm{~K} & 373 \mathrm{~K} & 383 \mathrm{~K}\end{array}$

$\Delta S_1=\frac{\Delta H_{\text {fusion }}}{273}=\frac{334}{273}=1.22$ $\Delta S_2=4.2 \ln \left(\frac{373}{273}\right)=1.31$

$\Delta S_3=\frac{\Delta H_{\text {vap }}}{373}=\frac{2491}{373}=6.67$ $\Delta S_4=2.0 \ln \left(\frac{383}{373}\right)=0.05$

$\Delta S_{\text {Total }}=\Delta S_1+\Delta S_2+\Delta S_3+\Delta S_4=9.26 K_J K^{-1} K^{-1}$

Example 2: The entropy (S^o) in (JK^-1mol^-1) of the following substances are :

$\mathrm{CH}_4(\mathrm{~g}): 186.2 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$

$\mathrm{O}_2(\mathrm{~g}): 205.0 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$

$\mathrm{CO}_2(\mathrm{~g}): 213.6 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$

$\mathrm{H}_2 \mathrm{O}(\mathrm{l}): 69.9 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$

The entropy change $\Delta S^0$ for the reaction $\mathrm{CH}_4(\mathrm{~g})+2 \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}(\mathrm{l})$ is :

1) -312.7

2)-242.8

3)-108.1

4)-37.6

Solution

In the given reaction:

$\mathrm{CH}_{4(\mathrm{~g})}+2 \mathrm{O}_{2(\mathrm{~g})} \rightarrow \mathrm{CO}_{2(\mathrm{~g})}+\mathrm{H}_2 \mathrm{O}_{(\mathrm{l})}$

$\Delta \mathrm{S}=\sum \mathrm{S}^{\circ}($ products $)-\sum \mathrm{S}^{\circ}($ reactants $)$

$\begin{aligned} & \Delta \mathrm{S}=(213.6+69.9)-\{186.2+2(205)\} \\ & \therefore \Delta \mathrm{S}=-312.7 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\end{aligned}$

Hence, the answer is the option (1).

Example 3: A sample of 100g water is slowly heated from 27°C to 87°C. The change in entropy (in J/K) of the water is (the specific heat capacity of water 4200 J/Kg - K).

1) 76.6

2)86.6

3)56.6

4)66.6

Solution

Given:

T₁ = 27^oC = 273 + 27 = 300 K

T₂ = 87^oC = 273 + 87 = 360 K

m = 1000 g = 0.1 kg

S_w = 4200 J/kg-K

Now,

$\begin{aligned} & \Delta Q=m \times s_w \times \Delta T \\ & \Delta S=\frac{\Delta Q}{T} \\ & \Delta S=\frac{m \times s_w \times \Delta T}{T}\end{aligned}$

$\begin{aligned} & \Rightarrow \int_{s 1}^{s 2} d S=\int_{T 1}^{T 2} m s_w \frac{d T}{T} \\ & \Rightarrow S_2-S_1=\left.m s_w \ln T\right|_{T_1} ^{T_2} \\ & \Rightarrow \Delta S=0.1 \times 4200 \times \ln \frac{T_2}{T_1} \\ & \Rightarrow \Delta S=0.1 \times 4200 \times \ln \frac{360}{300} \\ & \Rightarrow \Delta S=0.1 \times 4200 \times \ln 1.2 \\ & \Rightarrow \Delta S=76.6 \mathrm{JK}^{-1}\end{aligned}$

Example 4: Entropy is maximum in the case of

1) Steam

2)Water at 0^oC

3)Water at 4^oC

4)Ice

Solution

The greater the randomness of the molecules, the greater the entropy.

Since gaseous molecules have the most excellent randomness, they have the greatest entropy.

The given options can be arranged in order of their entropy:

Steam $>$ Water at $4^{\circ} \mathrm{C}>$ Water at $0^{\circ} \mathrm{C}>$ Ice

Hence, the answer is the option (1).

Summary

Entropy is one of the fundamental concepts in thermodynamics that describes the state of disorder or randomness of a system. It tells about the quantity of microscopic configurations under the macroscopic state of a system. As introduced by Rudolf Clausius, entropy stands at the core of the second law of thermodynamics, stating that entropy in an isolated system is always increasing with time. In other words, it explains how natural processes are irreversible and the tendency is always toward equilibrium. Entropy is involved in the understanding of the efficiency of engines, the spontaneity of chemical reactions, and the distribution of energy in systems. It connects macroscopic thermodynamic properties to microscopic molecular behavior to acquire insight into the way energy is transformed and distributed.