Entropy of the universe (Basics)

In chemical thermodynamics, the universe is composed of a system and its surroundings. The system is a part of the universe that is being studied, while its surroundings is the remaining part of the universe, with the two regions separated by a boundary. For example, if we are studying a gaseous reaction in a closed vessel in thermal equilibrium with its surroundings, the system is the gas in the vessel, while its surroundings is the space outside the vessel, and the boundary is the thermally conducting vessel walls (see diagram below).

The change in entropy of the universe is given by the sum of the change in entropy of the system and the change in entropy of the surroundings:

\Delta S_{uni}=\Delta S_{sys}+\Delta S_{sur}\; \; \; \; \; \; \; \; 2

or under standard conditions:

\Delta S_{uni}^o=\Delta S_{sys}^o+\Delta S_{sur}^o\; \; \; \; \; \; \; \; 3

The change in entropy of the universe is sometimes called the total entropy change ΔStoto. However, we are mostly concerned about chemical reactions that are usually defined as occurring within a system, and especially whether such reactions are spontaneous. To determine the spontaneity of a reaction using eq3, we need to know the following facts:

This means that we require, in addition to ΔSsyso, the value of ΔSsuro, which may not be easily available. To circumvent this problem, we seek a new measure, which can be developed as follows:  

For a reaction carried out under standard conditions, ΔSsyso is given by eq1 of the previous article, i.e., the standard reaction entropy of the system, ΔSro, while

\Delta S_{sur}^o=\frac{\Delta H_{sur}^o}{T}\; \; \; \; \; \; \; \; 4

where T is the temperature of the surroundings, which is equal to the temperature of the system, as the system is at thermal equilibrium with its surroundings.

Although the derivation of eq4 requires knowledge of chemical thermodynamics at the advanced level, we can interpret it by, firstly, noting that the change in enthalpy of the surroundings at constant pressure is equivalent to the amount of heat transferred from the system to the surroundings. We also mentioned in the previous article that an increase in entropy of a system is associated with an increase in number of moles of substances. Rearranging eq4, we get

T=\frac{\Delta H_{sur}^o}{\Delta S_{sur}^o}\; \; \; \; \; \; \; \; 5

We can say that the ‘spread’ of a given value of ΔHsuro (for a reaction occurring in the surroundings) over a larger increase in the number of moles of substances (i.e. a higher value of ΔSsuro) leads to lower energy per particle and hence, lower T.

Having interpreted eq4, let’s consider an exothermic reaction occurring in the system that releases thermal energy that is measured as the change in enthalpy of the system, ΔHsyso. Since the system’s vessel walls are thermally conducting, by the theory of conservation of energy, energy transferred from the system to its surroundings increases the energy of the surroundings and decreases the energy of the system, i.e.,

\Delta H_{sys}^o+\Delta H_{sur}^o=0\; \; \; \; \; \; \; \; 6

Substituting eq6 in eq4,

\Delta S_{sur}^o=-\frac{\Delta H_{sys}^o}{T}\; \; \; \; \; \; \; \; 7

Substituting eq7 in eq3,

\Delta S_{uni}^o=\Delta S_{sys}^o-\frac{\Delta H_{sys}^o}{T}\; \; \; \; \; \; \; \; 8

From the previous article, we know that a reaction occurring in a system can result in either positive or negative ΔSsyso. Furthermore, we mentioned earlier that:

    • ΔSunio > 0 for a spontaneous reaction
    • ΔSunio = 0 when a reaction reaches equilibrium, where there is no net transfer of thermal energy between a system and its surroundings.

Therefore, with reference to eq8, if the reaction is exothermic and ΔHsyso is large and negative to the extent that ΔSunio > 0, the reaction is spontaneous. For an endothermic reaction to be spontaneous, ΔSsyso must be positive enough to compensate for the term -\frac{\Delta H_{sys}^o}{T}.

Substituting eq1 from the previous article and eq7 in eq3,

\Delta S_{uni}^o=\sum_Pv_P(S_m^o)_P-\sum_Rv_R(S_m^o)_R-\frac{\Delta H_{sys}^o}{T}\; \; \; \; \; \; \; \; 9

 

Question

Calculate ΔSuni for the reaction BaCO3(s) → BaO(s) + CO2(g) at 25°C and 1500°C given Smo[BaCO3(s)] = 112.1 JK-1mol-1, Smo[BaO(s)] = 70.4 JK-1mol-1, Smo[CO2(g)] = 213.6 JK-1mol-1 and ΔHsyso = 269.0 kJmol-1.

Answer

At 25°C, using eq9,

\Delta S_{uni}^o=70.4+213.6-112.1-\frac{269000}{298.15}=-730.3\,JK^{-1}mol^{-1}

Since ΔSuni < 0, the reaction is not spontaneous. This is consistent with the fact that barium carbonate is thermodynamically stable at room temperature.

At  1500°C, if we assume ΔHsyso and the standard absolute molar entropies remain unchanged,

\Delta S_{uni}(1798.15K)=70.4+213.6-112.1-\frac{269000}{1798.15}=22.3\, JK^{-1}mol^{-1}

The decomposition of barium carbonate becomes spontaneous at this elevated temperature.

 

Eq8 is a better way to measure the spontaneity of a reaction, as it only requires the calculation of thermodynamic properties of the system. In fact, we can generate a new measure (a new function) in place of ΔSunio to determine the spontaneity of a reaction, starting from:

\Delta S_{sys}^o-\frac{\Delta H_{sys}^o}{T}>0\; \; \; \; \; \; \; \; 10

Multiplying eq10 throughout by T,

T\Delta S_{sys}^o-\Delta H_{sys}^o>0

\Delta H_{sys}^o-T\Delta S_{sys}^o<0\; \; \; \; \; \; \; \; 11

or

\Delta G_{sys}^o<0\; \; \; \; \; \; spontaneous\, reaction\; \; \; \; \; \; \; \; 12

where ΔGsyso = ΔHsyso  TΔSsyso.

We call this function, ΔGsyso, the Gibbs energy of the system.

Finally, just as  ΔSunio  = 0 when a reaction reaches equilibrium, where there is no net transfer of thermal energy between a system and its surroundings, ΔGsyso = 0 for a reaction at equilibrium.

 

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Standard reaction entropy

Just as the standard reaction enthalpy is

\Delta H_r^{\; o}=\sum_Pv_P\left (\Delta H_f^{\; o}\right )_P-\sum_Rv_R\left (\Delta H_f^{\; o}\right )_R

the standard reaction entropy of a system is

\Delta S_r^{\; o}=\sum_Pv_P\left (S_m^{\; o}\right )_P-\sum_Rv_R\left (S_m^{\; o}\right )_R\; \; \; \; \; \; \; \; 1

where vP and vR are the stoichiometric coefficients of the products and reactants respectively.

The difference between ΔHfo and Smo is that the former is the relative energy between a substance and the constituents of the substance in their reference states, while the latter is the absolute energy of the substance. This is why ΔHfo[C(graphite)] = 0 but Smo[C(graphite)] ≠ 0. Despite the difference, the quantities of ΔHro and ΔSro are comparable, as the zero-enthalpies of reference states in \sum_Pv_P\left ( \Delta H_f^{\; o} \right )_p and \sum_Rv_R\left ( \Delta H_f^{\; o} \right )_R cancel out when we compute ΔHro (see articles on standard enthalpy change of formation and Hess Law for details).

 

Question

Calculate the standard reaction entropy of C(s)+\frac{1}{2}O_2(g)\rightarrow CO(g) given S_m^{\; o}[C(s)]=5.7\,JK^{-1}mol^{-1}, S_m^{\; o}[O_2(g)]=205.0\,JK^{-1}mol^{-1} and S_m^{\; o}[CO(g)]=197.6\,JK^{-1}mol^{-1}.

Answer

Using eq1,

\Delta S_r^{\; o}=197.6-[5.7+\frac{1}{2}(205.0)]=+89.4\, JK^{-1}mol^{-1}

 

The positive change in standard reaction entropy of the incomplete combustion of carbon illustrates the dependency of the change in entropy of a system on the change in the number of moles of gas in a reaction. In the above example, the double in number of moles of gas outweighs the removal of a mole of carbon solid in the system, as the increase in the number of moles of gas leads to the dispersal of energy over a greater number of translational energy states of the system. This implies that ΔSro can be negative when we have a reaction that results in a decrease the number of moles of gas.

 

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Entropy

Entropy, denoted by the letter S, is a measure of energy dispersal at a specific temperature.

From the articles on chemical energetics, we learn that a reaction’s enthalpy change indicates whether it is exothermic or endothermic. Scientists previously thought that only exothermic reactions occur spontaneously, as the energy of systems are lowered during such reactions in a way similar to the lowering of an object’s gravitational potential energy when it falls from a height to the ground. However, the change in enthalpy of a reaction does not predict whether the reaction is spontaneous, because both exothermic and endothermic reactions can occur spontaneously. For example, both NaCl and NaOH dissolve spontaneously in water but the enthalpy change of solution of NaCl is positive, while that of NaOH is negative. 

One common aspect of the solution of NaCl and the solution of NaOH is the change of the ions from an ordered solid lattice structure to a random distribution of solvated ions throughout the vessel (see diagram above). Similarly, when ice dissolves spontaneously to form water at a temperature above 0°C, the regular arrangement of H2O molecules in ice becomes a disordered distribution of H2O molecules.

This suggests that chemical reactions occur spontaneously in a direction that increases their energy dispersal, which is sometimes inaccurately termed as ‘disorderliness‘. To quantify the change in the energy dispersal of a system, a measure called entropy is developed.

 

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Covalent and molecular solids

A covalent solid consists of atoms held together by covalent bonds, creating a network that extends throughout the crystal, while a molecular solid is formed when molecules are held together by van der Waals forces, hydrogen bonds, or a combination of the two types of intermolecular forces.

Diamond is an example of a covalent solid with a structure similar to that of zinc blende. The structure involves carbon atoms (orange in diagram above) occupying four tetrahedral holes within a face-centred cubic structure formed by carbon atoms (blue). Like zinc blende, the unit cell of diamond is face-centred cubic. Other elements, such as silicon and germanium in group 14, also have the same structure.

An example of a molecular solid is ice. Water molecules, when cooled below 273.15 K at standard atmospheric pressure, are held together in a chair configuration (see diagram I above) by hydrogen bonds. When viewed from the top, hexagonal rings extend throughout the crystal, with the unit cell being hexagonal (demarcated by red lines in diagram II).

The structure of graphite, on the other hand, features characteristics of both a covalent solid and a molecular solid. The diagram above shows carbon atoms covalently bonded to form giant planar sheets of hexagonal rings that are stacked in an XYXY… manner. Each sheet is called graphene. Every carbon in graphite is sp2 hybridised, leaving an electron in the orbital that is free to form a π-bond with one of three neighbouring orbitals. One may expect alternating single and double bonds throughout each sheet. However, x-ray diffraction data show that all covalent bonds in the sheets have a similar length of 0.141 nm, implying that the orbitals overlap to form a conjugated -electron system where the electrons are delocalised. This is the reason why graphite conducts electricity. Experiments also show that graphite is only electrically conductive in the direction parallel to the sheets, but not perpendicular to them, suggesting that the sheets are not held together by the overlap of pz orbitals in the perpendicular direction. Layers of graphene are instead loosely stabilised by van der Waals forces and can slide against each other, making graphite a good lubricant. The thermodynamically stable form of graphite, as shown in the diagram above, has a hexagonal unit cell containing four carbon atoms.

 

Question

Why does the unit cell of graphite contain four carbon atoms?

Answer

The edges of the hexagonal unit cell of graphite extend through three layers of graphene (see above diagram). The top sheet has four corner atoms, each being shared by six unit cells, and one other atom shared by two unit cells \left ( \frac{4}{6}+\frac{1}{2} \right ). The same occurs for the bottom layer. For the middle layer of graphene, four peripheral atoms are shared by six unit cells but one atom remains entirely within the unit cell \left ( \frac{4}{6}+1\right ). Therefore, there are a total of four atoms in a single hexagonal unit cell of graphite.

 

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Radius ratio

The radius ratio for an ionic solid is defined as the ratio of the radius of the smaller ion, rs, to the radius of the larger ion rl. An ionic solid with a 8:8 coordination structure (e.g. CsCl) has the following radius ratio:

a=2r_1\; \; \; \; \; \; \; \; 1

where a is the length of a side of the cube. Using Pythagoras’ theorem, the body diagonal of the cube, d, is:

d=a\sqrt{3}\; \; \; \; \; \; \; \; 2

Substituting eq1 in eq2

d=2r_l\sqrt{3}\; \; \; \; \; \; \; \; 3

The body diagonal is also given by:

d=2r_l+2r_s\; \; \; \; \; \; \; \; 4

Substituting eq3 in eq4 and rearranging,

\frac{r_s}{r_l}=\sqrt{3}-1\approx 0.732

The radius ratio of 0.732 is the minimum ratio required for a 8:8 coordination compound. If the radius of the cation is smaller, it will not be in contact with the anions, resulting in a compound that has a coordination of less than 8:8, i.e. a 6:6 coordination compound.

The diagram above shows one of the six surfaces of an ionic solid with a 6:6 coordination structure (e.g. NaCl). Its radius ratio is calculated as follows:

a=2r_l+2r_s\; \; \; \; \; \; \; 5

f=a\sqrt{2}\; \; \; \; \; \; \; \; 6

where f is the face diagonal. Substituting eq5 in eq6,

f=2\sqrt{2}(r_l+r_s)\; \; \; \; \; \; \; \; 7

The face diagonal is also given by:

f=4r_l\; \; \; \; \; \; \; \; 8

Combining eq7 and eq8 and simplifying,

\frac{r_S}{r_l}=\sqrt{2}-1\approx 0.414

Again, the radius ratio of 0.414 is the minimum ratio required for a 6:6 coordination compound. If the radius of the cation is smaller, it will not be in contact with the anions, resulting in a compound that has a coordination of less than 6:6, i.e. a 4:4 coordination compound.

Finally, for a 4:4 coordination structure like zinc blende (see diagram above), its radius ratio is calculated by selecting the appropriate ions (diagram IIf):

\frac{f}{2}=2r_l\; \; \; \; \; \; \; \; 10

Since the tetrahedral angle is 109.5°,

sin\frac{109.5^o}{2}=\frac{\frac{f}{2}/2}{r_l+r_s}\; \; \; \; \; \; \; \; 11

Substituting eq10 in eq11 and simplifying,

\frac{r_s}{r_l}=\frac{1}{sin\frac{109.5^o}{2}}-1\approx 0.225

Once again, the radius ratio of 0.225 is the minimum ratio required for a 4:4 coordination compound.

Therefore, assuming that ions in crystal are rigid spheres and that cations are in contact with anions, we can use the three calculated radius ratios to predict structures of ionic compounds as follows:-

Radius Ratio

0.225<\frac{r_s}{r_l}<0.414 0.414<\frac{r_s}{r_l}<0.732 \frac{r_s}{r_l}>0.732

Coordination

4:4 6:6

8:8

Structure

Zinc blende type NaCl type

CsCl type

 

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Ionic solids

Ionic solids are characterised by robust lattice structures formed through the electrostatic attraction between positively and negatively charged ions.

We can, as we did for metallic solids, use the concept of packing spheres to study ionic solids. We also need to consider the relative sizes of the oppositely charged ions and the charges of the ions to determine the structure.

Consider two layers of closely packed anions, with the second layer (orange) sitting in the notches of the bottom layer (blue), forming two types of space (or holes) between the layers (see diagram I above). As the volume of a cation is usually smaller than that of an anion, a cation can fit into either hole, depending on the relative sizes of cations and anions in the crystal. A relatively larger cation occupies an octahedral hole, while a relatively smaller cation lodges in a tetrahedral hole (diagram II), forming ionic compounds with two different arrangements. For each of the arrangements, the anion-cation-anion layered structure then repeats itself with the third layer of anions having the same arrangement as the first layer of anions.

For the ionic compound where octahedral holes are filled, each cation has six anions in its vicinity and each anion has six cations in its vicinity. Therefore, we also call such a structure a 6-6 coordination structure. For the ionic compound where tetrahedral holes are filled, each cation has four anions in its vicinity and each anion has four cations in its vicinity. Therefore, we also call such a structure a 4-4 coordination structure.

If the cation-anion radius ratio is even larger (but still less than 1), the layer of anions must move apart to accommodate the cations (diagram III), such that both cations and anions are in the same layer with every cation in contact with four anions (diagram IV). A second layer has the anions sitting in the notches formed by the anions of the bottom layer (diagram V), with each second-layer cation again in contact with four second-layer anions. The third layer of anions are directly above those in the first layer, giving the crystal an XYXY… structure. In such a structure, each cation has eight anions in its vicinity and each anion has eight cations in its vicinity. Therefore, we also call such a structure a 8-8 coordination structure.

Having seen how cations and anions are packed in a crystal, let’s find the respective unit cells. Firstly, we depict a crystal that has a packing pattern like diagram V in the lattice form (diagram Va), and rotate it for a better view (diagram Vb). Note that some anions are coloured black for contrast.

The space lattice Vb is formed by replicating two interpenetrating cells, a cationic simple cubic unit cell (pink) and an anionic simple cubic unit cell (blue; see diagram Vc). By pairing a cation in the middle of an anionic simple cubic unit cell with one of the anions of that cell, and using that as a basis (diagram Vd), we get the unit cell of V, which is again a simple cubic unit cell (diagram Ve). To maintain charge neutrality within the crystal, the total charges for cations and anions must be equal and opposite. In short, a compound with a relatively high radius ratio like caesium chloride adopts this structure.

A compound with a lower radius ratio (e.g. NaCl) has cations occupying octahedral holes that are formed by layers of anions (see diagram II and IIa below). The result is a space lattice consisting of replications of two interpenetrating cells, a cationic face-centred cubic unit cell and an anionic face-centred cubic unit cell (diagram IIb). By pairing a cation with an anion and using the pair as a basis, we get a face-centred cubic unit cell for NaCl.

For a compound with even lower radius ratio (e.g. ZnS, zinc blende), cations occupy tetrahedral holes formed by layers of anions (see diagram II and IId below). By pairing a cation with an anion and using the pair as a basis, we once again get a face-centred cubic unit cell for zinc blende (diagram IIe).

 

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Metallic solids

The crystal structures of metallic solids are usually composed of three types of unit cells that are formed due to the way metal atoms are packed in a crystal.

Consider the atoms as hard spheres touching each other to form a closely packed single layer (see diagram above). To maintain the close packed arrangement, a second layer (orange) is placed above the bottom layer (blue) with each second-layer atom sitting in the notches of the bottom layer. There are two ways to position atoms for the third layer (yellow), starting with notch A or notch B. If the first atom in the third layer starts at notch A, all atoms of the third layer are directly above atoms in the first layer, with the layers having an XYXYXY… pattern, giving the crystal (e.g. magnesium) a structure called hexagonal close-packed (hcp). However, if the first atom in the third layer starts at notch B, all atoms of the third layer are staggered relative to atoms in the first layer, resulting in the layers having an XYZXYZXYZ… pattern. A crystal with such an arrangement of layers (e.g. copper) has a structure called cubic close-packed (ccp).

Another common crystal structure of metals, shown in the diagram below, is the body centred cubic (bcc). The bottom layer (blue) is less closely packed than hcp and ccp, with each atom touching only four neighbouring atoms in the same layer instead of six. The second layer of atoms (orange) sits in the notches of the bottom layer and the third layer fits into the notches of the second layer, giving an XYXYXY… pattern. Iron and potassium have bcc structures.

Having seen how metal atoms are packed in a crystal, let’s find the respective unit cells. A crystal with a ccp structure has a face-centred cubic unit cell (see below diagram) where a = b = c and α = β = γ = 90°. There is one lattice point at each corner of the cube and one lattice point in the middle of each surface plane. Each lattice point meets the criteria of having the same environment by having six and eight neighbouring atoms at distances a and (√2/2)a respectively (it is easier to visualise the neighbouring atoms by replicating the unit cell around itself).

A crystal with a bcc structure has a unit cell of the same name (see below diagram) where a = b = c and α = β = γ = 90°. There is one lattice point at each corner of the cube and one lattice point in the middle of the cube. Each lattice point meets the criteria of having the same environment by having six and eight neighbouring atoms at distances a and (√3/2)a respectively (using Pythagoras’ theorem).

To map out the unit cell of a hcp structure, we let a pair of atoms be a lattice point (denoted by a yellow oval in the 3rd diagram below). The atom pair or basis consists of an atom from the bottom layer (blue) and one from the second layer (orange). Connecting the centres of four of the ovals (numbered 1 to 4), we get a rhombus with two opposite angles of 120°; and linking two adjacent rhombi, we obtain the unit cell called the hexagonal unit cell. The reason it is called a hexagonal unit cell is the combination of three such unit cells form a hexagon.

Finally, the different forms of packing and hence, unit cells, are a result of the different electron densities of atoms and molecules and their interactions to produce energetically stable arrangements in the solid state.

 

Question

Calculate the packing fraction of Cu (i.e. the fraction of volume occupied by Cu spheres).

Answer

Cu has a fcc unit cell and the volume of an fcc unit cell is a3. A fcc unit cell contains 4 Cu spheres since one-eighth of a sphere resides in each corner of the unit cell and half of a sphere remains on each surface of the unit cell. The length of a diagonal across a surface of the unit cell is determined using Pythagoras’ theorem as a√2 and is equivalent to exactly four radii of a Cu sphere. Hence, the radius of a Cu sphere is \frac{a\sqrt{2}}{4} and the volume of 4 Cu spheres is \frac{4(4)}{3}\pi \left (\frac{a\sqrt{2}}{4} \right )^3. Therefore, the packing fraction of Cu is \frac{4(4)}{3}\pi \left (\frac{a\sqrt{2}}{4} \right )^3/a^3 , which computes to 0.740.

 

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The unit cell

The unit cell is a parallelepiped that is the simplest repeating unit of a crystal (diagram I).

Diagram II shows a crystal that is composed of unit cells where a = b = c and α = β = γ = 90°. Each blue point, known as a lattice point, must have identical surroundings (or environment) in the crystal. An infinite three-dimensional array of lattice points forms a space lattice. The unit cell in II that has eight lattice points at all corners of a cube is called a simple cubic unit cell.

Consider a crystal structure containing two types of atoms (diagram III). As each lattice point must have the same surroundings, it is, in this case, a point consisting of a pair of blue and red atoms (diagram IV). Therefore, each lattice point, in general, may represent an atom, a molecule or a collection of atoms or molecules. Using the pair of atoms as a basis in forming the space lattice, we find that the unit cell for compound III is also a simple cubic unit cell.

Finally, a unit cell must fill all space of the crystal when replicated. As a result, spherical unit cells do not exist.

 

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Nuclear binding energy per nucleon

The nuclear binding energy per nucleon of a nucleus is the total binding energy of the nucleus divided by the total number of nucleons.

56Fe has the highest nuclear binding energy per nucleon and is therefore the most stable isotope (see graph above). The negative gradient of the curve beyond 56Fe implies that electrostatic forces of repulsion increase per nucleon for isotopes with mass number greater than 56.

 

Question

With reference to the data below and using the concept of nuclear binding energy per nucleon, explain why isotopes with atomic numbers lower than that of 12C have relative masses that are slightly higher than their respective mass numbers, while those with atomic numbers that are higher than 12C have relative masses that are slightly lower than their respective mass numbers.

Atomic no. (Z)

Mass no. (A) Symbol

Relative isotopic mass*

1

1 1H

1.007825

2

2H

2.014104

2

3 3He

3.016029

4

4He

4.002603

3

6 6Li

6.015122

4

9 9Be

9.012182

5

10

10B

10.012937

11

11B

11.009305

6

12 12C

12.000000

8

16

16O

15.994915

17 17O

16.999132

18

18O

17.999160

9

19 19F

18.998403

10

20

20Ne

19.992440

21 21Ne

20.993847

22

22Ne

21.991386

Answer

Let’s rewrite eq9 of the previous article as:

m_{isotope}+m_{binding}=Z\left ( m_{proton}+m_{electron} \right )+(A-Z)m_{neutron}

Dividing the above equation throughout by A and rearranging,

\frac{m_{isotope}}{A}+\frac{m_{binding}}{A}=\frac{Z}{A}\left ( m_{proton}+m_{electron}-m_{neutron}\right )+m_{neutron}\; \; \; \; \; \; \; \; 11

where \frac{m_{binding}}{A} is the binding energy per nucleon.

Using data in a previous article, mproton + melectronmneutron = -0.000839869u. Furthermore, with the exception of 1H, \frac{Z}{A}<1. Therefore, the RHS of eq11 approximately equals to mneutron:

\frac{m_{binding}}{A}\approx m_{neutron}-\frac{m_{isotope}}{A}

For 12C, m_{isotope}=A=12, and so \frac{m_{isotope}}{A}=1. If \left (\frac{m_{binding}}{A}\right )_{isotope}<\left (\frac{m_{binding}}{A}\right )_{^{12}C} , then \frac{m_{isotope}}{A}>1. By the same logic, if \left (\frac{m_{binding}}{A}\right )_{isotope}>\left (\frac{m_{binding}}{A}\right )_{^{12}C} , then \frac{m_{isotope}}{A}<1.

 

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Mass defect

Mass defect is the difference in the combined masses of an atom’s components and the measured mass of the atom.

An atom is composed of electrons and nucleons (protons and neutrons). The constituents of its nucleus, the neutrons and positively charged protons, are held together by an energy called the nuclear binding energy, Ebinding. In other words, the nuclear binding energy is the energy that is released when a stable nucleus is formed from its components. The total energy of the atom can be expressed as:

This equation represents the balance of energies within the atom, where Ebinding accounts for the energy required to hold the nucleus together. As energy and mass are related by Einstein’s formula of E = mc2, we can rewrite eq8 as:

Eq9 shows that the mass of an atom is always less than the sum of the masses of its isolated components, with the exception of 1H where mneutrons = 0 and thus mbinding = 0. As this defect in mass (for atoms other than 1H) is due to the atom’s nuclear binding energy, we refer the mass equivalent of the nuclear binding energy as the mass defect of the atom:

m_{binding}=m_{defect}

Hence, mass defect is the difference in the combined masses of an atom’s components and the measured mass of the atom.

m_{defect}=m_{protons}+m_{neutrons}+m_{electrons}-m_{atom}\; \; \; \; \; \; \; \; 10

The heavier the atom, that is the more protons and neutrons it has, the greater the mass defect, since a larger amount of binding energy is needed to hold a greater number of positively charged protons and the neutrons together.

As an illustration, if we simply add the subatomic particles of a deuterium atom, we have:

m_{D,calculated}=m_{protons}+m_{neutrons}+m_{electrons}

Substituting the values of subatomic masses from the previous article in the above equation, we have: mD, calculated = 2.01648996 u. The value of the relative mass of deuterium that is experimentally determined via mass spectrometry is 2.014104 u. The difference of about 0.00238596 u is due to the mass defect of deuterium.

 

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