 Second Law of Thermodynamics: Introduction to Entropy Chemistry Tutorial

Key Concepts

• Thermodynamics is the study of the changes of energy, or transformations of energy or energy conversions, which accompany physical and chemical changes in matter.
• Thermodynamics describes the behaviour of macroscopic systems.
• Entropy of a chemical system(1) can be thought of as the amount of disorder in the system.(2)

amount of disorder relative entropy
highly disordered
high entropy
highly ordered low entropy

• Entropy can also be thought of in terms of how energy is distributed in the chemical system.(3)

amount of disorder relative distribution of energy relative entropy
highly disordered high high entropy
highly ordered low low entropy

• The Second Law of Thermodynamics says that the entropy of the universe is constantly increasing.(4)

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Entropy and States of Matter

Cconsider the particles (atoms or molecules) making up a solid held inside a container. If we repesent each particle (an atom or a molecule) as a circle, we could represent the solid as shown in the diagram below:

The atoms or molecules are highly ordered so the entropy of the system is said to be low.
The movement of the atoms or molecules (their kinetic energy) within the solid is severely restricted so the energy distribution within the solid is low, so, the entropy of the system is low.

Now, consider the particles (atoms or molecules) making up a gas held inside a container as shown below:

Each gas particle is in constant motion. The particles bump into each other and the walls of the container, constantly changing direction so the particles in the system are highly disordered. We say that the entropy of the system is high.
Because the particles are constantly moving they have greater kinetic energy than they did when they formed part of a solid, and because they are being scattered in different directions the distribution of energy in this system is higher than it is for the solid system , therefore the entropy of the system is higher.

If you consider the particles making up a liquid, we would say that they have more kinetic energy than those in the solid, but less kinetic energy than those in the gas.
The particles in a liquid can roll over each, so the particles in a liquid are more disordered than those in the solid, but the particles in the liquid cannot move around as freely as in a gas, so the particles in the liquid system are not as disordered as they are in the gas system.
The amount of disorder in a liquid system is greater than that of the solid, but less than that of the gas.
Because the moving particles in the liquid are free to move within the body of liquid the energy distribution within the liquid is greater than that of the solid. Because the moving particles in the liquid are confined and not free to move outside the body of the liquid, the energy distribution for the liquid system is not as great as that for the gas system.
The entropy of the liquid system is greater than that of the solid but less than that of a gas.

For the states of matter:

 lower entropy → → → → higher entropy solid liquid gas

We can see that one way to increase the entropy of a system is to increase the energy of the particles in the system, for example, by heating the system.
Similarly, one way to reduce the entropy of a system is to decrease the energy of the particles in the system, for example, by cooling the system.

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Entropy and Solutions

Let's think about what happens when you make an aqueous solution of sodium chloride (NaCl(aq)).

First, you would weigh out some solid sodium chloride (NaCl(s)) which is the solute.
Solid sodium chloride is in a low entropy state. All the ions are held together by ionic bonds forming a 3-dimensional lattice of ions (an ionic solid).

Then you would add some liquid water (the solvent).
The solid sodium chloride breaks up into sodium ions (Na+(aq)) and chloride ions (Cl-(aq)) which can move freely within the body of liquid water, making an aqueous solution of ions.
The sodium chloride system has gone from a highly ordered, low entropy state, to a less ordered, higher entropy state.
The distribution of energy in the solid sodium chloride lattice is lower than it is in the aqueous solution where the ions are constantly moving, so the entropy of the sodium chloride system has increased.

We could summarise this as shown below:

 NaCl(s) ⇋ Na+(aq) + Cl-(aq) low entropy ⇋ higher entropy

There are other types of solutions.
Consider what happens when a gas like oxygen (O2(g)) dissolves in water.
In the gaseous state, the molecules of oxygen have kinetic energy and move freely so the entropy of the system is high.
Once these molecules of oxygen are surrounded by water (dissolved), they are confined to move within the body of liquid water, so their entropy has decreased.

 O2(g) ⇋ O2(aq) higher entropy ⇋ lower entropy

And you can mix one gas with another to make a gaseous solution.
Imagine you have a 1 L jar of argon gas (Ar(g)) at 25°C and 100 kPa, and another 1 L jar containing neon gas (Ne(g)) also at 25°C and 100 kPa.
Both gas are now released into a 2 L container so that the temperature and pressure of the gaseous solution remain constant (25°C, 100 kPa).
Has the entropy of the system changed?
Yes, because the energy of the gas particles is now distributed over a greater volume so the entropy of the system has increased.

And if you mix one liquid with another liquid, the entropy of the system is increased as the two liquids mix together in a greater volume to form a solution.

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Entropy and Chemical Reactions

We saw above how entropy changes during physical changes like changes of state or preparing solutions.
Now we will turn our attention to what happens to entropy during chemical reactions.
We will only be considering chemical reactions in which there is at least one gas present.

Consider the thermal decomposition of solid calcium carbonate (CaCO3(s)) to produce solid calcium oxide (CaO(s)) and carbon dioxide gas (CO2(g)) as shown in the balanced chemical equation below:

CaCO3(s) ⇋ CaO(s) + CO2(g)

Has the entropy of the chemical system increased?
Yes it has!
The solid calcium carbonate (CaCO3(s)) is a highly ordered arrangement of ions so the distribution of energy in the system is low, hence, the entropy of the system is low.
Once decomposed, we have a highly ordered calcium oxide solid, but, we have also produced a highly disordered gas in which the energy distribution will be large because the gas molecules have greater kinetic energy and are more mobile.
The entropy of this system has increased from low entropy to higher entropy.

Consider now the production of 2 moles of gaseous ammonia (NH3(g)) from 1 mole of gaseous nitrogen (N2(g)) and 3 moles of gaseous hydrogen (H2(g)), as shown by the balanced chemical equation below:

N2(g) + 3H2(g) ⇋ 2NH3(g)

Has the entropy of this system increased?
On the left hand side of the equation (the reactants) we have 4 moles of gas and a very disordered system in which the energy of the system is greatly dispersed.
On the right hand side of the equation (the products) we have only 2 moles of gas so the system is less disordered and the distribution of energy in the system is less.
For this chemical reaction the entropy of the system has decreased, it has gone from higher entropy to lower entropy.

For gaseous chemical reactions, the more particles of gas there are in the system, the higher the entropy of the system.

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Worked Example of Entropy Changes in a Chemical System

(Using the StoPGoPS approach to problem solving)

 Question: 1 L of colourless 0.1 mol L-1 aqueous solution of lead(II) nitrate (Pb(NO3)2(aq)) is added to 1 L of colourless 0.2 mol L-1 aqueous solution of potassium iodide (KI(aq)). A bright yellow precipitate of lead(II) iodide (PbI2(s)) is produced. Has the entropy of the system increased or decreased?

STOP STOP! State the Question.
What is the question asking you to do?
Determine whether the entropy of the chemical system has increased or decreased.
PAUSE PAUSE to Prepare a Game Plan
(1) What information (data) have you been given in the question?

(i) Reactants:

concentration = c(Pb(NO3)2(aq)) = 0.1 mol L-1
volume = V(Pb(NO3)2(aq)) = 1 L

potassium iodide solution: KI(aq)
concentration = c(KI(aq)) = 0.2 mol L-1
volume = V(KI(aq)) = 1 L

(ii) Product:

(2) What is the relationship between what you know and what you need to find out?

(i) Ions in solution have higher entropy than ions bound up in a solid.

(ii) Having more particles in solution results in higher entropy.

(iii) Number of particles in solution is determined by the moles of particles in solution.
moles = concentration (mol L-1) × volume (L)

GO GO with the Game Plan

(i) Write a balanced chemical equation for this reaction.
NOTE: we want to see all the particles so we we will write an ionic equation:

 Molecular Equation: Ionic Equation: Pb(NO3)2(aq) + 2KI(aq) ⇋ PbI2(s) + 2NO3-(aq) + 2K+(aq) Pb2+(aq) + 2NO3-(aq) + 2K+(aq) + 2I-(aq) ⇋ PbI2(s) + 2NO3-(aq) + 2K+(aq)

(ii) Consider the number of particles (moles) in solution on the left hand side of the equation compared to the right hand side of the equation:

Calculate moles of Pb(NO3)2(aq) = 0.1 mol L-1 × 1 L = 0.1 mol
Since Pb(NO3)2 dissociates in water: Pb(NO3)2 → Pb2+(aq) + 2NO3-(aq)
We use the mole ratio (stoichiometric ratio) Pb(NO3)2 : Pb2+ (1:1) and Pb(NO3)2 : NO3- (1:2) to calculate the moles of each ion in the solution:
moles of Pb2+(aq) = moles of Pb(NO3)2 = 0.1 mol
moles of NO3-(aq) = 2 × moles of Pb(NO3)2 = 2 × 0.1 mol = 0.2 mol

Calculate moles of KI(aq) = 0.2 mol L-1 × 1 L = 0.2 mol
Since KI dissociates in solution: KI → K+(aq) + I-(aq)
moles of K+(aq) = moles KI = 0.2 mol
moles of I-(aq) = moles of KI = 0.2 mol

Calculate moles of lead(II) iodide precipitate that forms:
mole ratio (stoichiometric ratio) Pb(NO3)2(aq) : PbI2(s) is 1:1
moles PbI2(s) = moles Pb(NO3)2(aq) = 0.1 mol

Note that the mole ratio (stoichiometric ratio) Pb(NO3)2 : KI is 1 :2 and we have this same ratio, that is, 0.1 mol : 0.2 mol, so there is no limiting reagent for this reaction.

Potassium ions (K+(aq)) and nitrate ions (NO3-) are spectator ions, they are not used to make the precipitate so they are present on both sides of the equation in the same numbers as before the reaction occurred.

Now, compare the moles of ions in solution on the left hand side and right hand side of the chemical equation:

 Ionic Equation: Moles of each species in solution Total Moles in solution on each side of equation Pb2+(aq) + 2NO3-(aq) + 2K+(aq) + 2I-(aq) ⇋ PbI2(s) + 2NO3-(aq) + 2K+(aq) 0.1 + 0.2 + 0.2 + 0.2 ⇋ 0 + 0.2 + 0.2 0.7 ⇋ 0.4

Since the number of ions in solution has decreased from 0.7 moles to 0.4 moles, there are fewer ions moving around in the aqueous system so the system has become less disordered, and, since there are fewer moving particles in solution the dispersal of energy through the system has decreased. Therefore, the entropy of the system has decreased.

PAUSE PAUSE to Ponder Plausibility

Consider the reverse reaction, solid lead(II) iodide (PbI2(s)) dissolving in water to form an aqueous solution of lead(II) ions (Pb2+(aq)) and iodide ions (I-(aq)) as shown in the balanced chemical equation below:

Pb(NO3)2(s) ⇋ Pb2+(aq) + 2I-(aq)

The entropy of a solid is less than the entropy of the aqueous solution of ions, therefore there has been an increase in entropy:

 Pb(NO3)2(s) ⇋ Pb2+(aq) + 2I-(aq) low entropy ⇋ higher entropy

Therefore, for the reaction in which Pb2+(aq) reacts with I-(aq) to produce solid PbI2(s) there will be a decrease in entropy:

 Pb2+(aq) + 2I-(aq) ⇋ Pb(NO3)2(s) higher entropy ⇋ low entropy

Since this agrees with the answer we got above, we are reasonably confident that our answer is plausible.

STOP STOP! State the Solution
Entropy of the system has decreased

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Footnotes

(1) The chemical system must be defined as reversible. Thermodynamics can only define entropy changes for reversible processes.

(2) This is an introduction to entropy. For a high school chemistry course the concept of entropy as the degree of disorder is useful.
We can equate "entropy" with "disorder" in the chemical systems being studied because these systems are all:
(i) molecular-level systems
(ii) capable of exchanging thermal energy with the surroundings.
For S = k ln W
S = entropy, k = Boltzmann's constant (1.38 × 10-23 JK-1), W = thermodynamic probability
W for a chemical system in a given state is then the number of alternative ways the particles in the chemical system can be arranged to constitute the state.
W will be high when randomness, or disorder, is high.

(3) For a process that occurs at constant temperature, the entropy change (ΔS) of the surroundings depends only on the
(i) amount of heat absorbed by the surroundings from the system undergoing change (-q, at constant pressure -q = -ΔHsystem)
(ii) temperature at which the heat is transferred (T)
That is, ΔSsurrounds = -ΔHsystem ÷ T
For a high school chemistry course, it is generally acceptable to relate the distribution of energy in the system (-ΔHsystem) term to the number of particles in the system available to distribute the energy.
Increasing temperature increases entropy change and disorder in a reversible chemical system. A more disordered chemical system therfore has a greater distribution of energy.

(4) Rudolf Clausius (1822-1888) summarised the first and second laws of thermodynamics as:
First Law: the energy of the universe is constant.
Second Law: the entropy of the world is constantly increasing.
Note that entropy and energy are both state functions, that is, they both depend on the state of a system and are independent of the past history of the system.