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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 + melectron – mneutron = -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. Since 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.

 

Question

Calculate the nuclear binding energy of the He-4 nucleus.

Answer

The theoretical mass of the He-4 nucleus is 2 x (1.007276 + 1.008665) = 4.031882 u, while the experimentally determined mass is 4.000409 u. Therefore,

Ebinding = (4.031882 u – 4.000409 u) x c2 = 0.031473 x 931.5 ≅ 29.3 MeV

 

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Subatomic particle mass

Subatomic particle mass, encompassing the masses of protons, neutrons and electrons, is fundamental to understanding the structure of matter and the behaviour of atoms in chemical reactions.

The mass of a proton can be determined via mass spectrometry in the same way as described in a previous article. The mass of a neutron, mn, however, cannot be measured via mass spectrometry as it lacks a charge.

Particle

 Symbol Relative mass, u

Inertia mass, kg

Proton

 p or H+ 1.007276466879

1.672621898 x 10-27

Neutron

 n 1.00866491588

1.674927471 x 10-27

Electron

 e 5.485799090 x 10-4

9.10938356 x 10-31

From eq9 of the article on mass defect, we have:

m_{D^+}+m_{defect}=m_n+m_p

Hence, the mass of a neutron can be calculated by subtracting the mass of a proton, mp, from the mass of a deuterium nucleus m_{D^+} (both obtained from mass spectrometry), and adding the mass defect of deuterium, mdefect, which can be measured using X-ray diffraction for the gamma ray released when a neutron captures a proton:

n+p\rightarrow D+\gamma

The charge-to-mass ratio of an electron was first estimated by J. J. Thomson, an English physicist, in 1896 using cathode rays. Combining this value with the quantity of charge of a single electron from Robert Millikan’s oil drop experiment, an estimated mass of an electron could be determined. A more precise value, the rest mass of an electron, me, is however calculated from the Rydberg constant, where

m_e=\frac{8\varepsilon _0^{\: 2}h^3cR_{\infty }}{e^4}

 

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Rate laws (chemical kinetics)

The rate of a reaction has been defined in a basic level article as:

rate=\frac{change\, in\, amount\, of \, reactants\, or\, products}{change\, in\, time}\; \; \; \; \; \; \; \; 1

For a simple reaction, A → B at constant volume V1, we can define its rate as

rate=\frac{dn_B}{dt}\; \; \; \; \; \; \; \; 2

where nB is the amount of B.

Since the rate of formation of product must be the same as the rate of consumption of the reactant, and that the rate of a reaction is defined as a positive value, we have:  -\frac{dn_A}{dt}=\frac{dn_B}{dt}. 

Suppose we have another reaction, C → D at a different constant volume V2, with a rate:

rate=\frac{dn_D}{dt}\; \; \; \; \; \; \; \; 3

To have a meaningful comparison of the rates of these two reactions with different reacting volumes, we need a common basis, which can be obtained by dividing eq2 and eq3 by their respective volumes. Therefore, a better definition of the rate of a reaction is:

rate=\frac{d\frac{n}{V_i}}{dt}=\frac{d[n]}{dt}=\frac{change\, in\, concentration\, of\, reactants\, or\, products}{change\, in\, time}\; \; \; \; \; \; \; \; 4

Consider an experiment where HCl is added to CaCO3 to produce CO2 in a rigid vessel with a pressure gauge. The data collected are presented in a CO2 pressure versus time plot. If we use a constant mass of CaCO3 and 3 different concentrations of 20 cm3 of HCl, we have 3 different gradients at the origin, which represent 3 initial rates of reaction (see diagram below) with RA corresponding to the reaction with the highest concentration of HCl and RC relating to the reaction with the lowest concentration of HCl.

The above results imply that the rate of reaction is proportional to the concentration of the reactant used. Therefore, in addition to eq4, we can also expressed the rate of a reaction in the form of an equation known as a rate law:

rate=k\left [ H_3O^+ \right ]\; \; \; \; \; \; \; \; 5

where [H3O+] is the concentration of hydroxonium ions from HCl and k is a proportionality constant called the rate constant. Combining eq4 and eq5,

\frac{d[CO_2]}{dt}=k\left [ H_3O^+ \right ]\; \; \; \; \; \; \; \; 6

Eq6 means that an increase in concentration of hydroxonium ions increases the rate of change in concentration of CO2, which is measured by the rate of change in pressure of CO2, since \frac{n}{V}=\frac{p}{RT}.

 

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Rate laws – differential form

A differential rate law mathematically describes the rate of a reaction in terms of the change in the concentration of a reactant over time.

Eq6 of the previous article is an example of a differential rate law, which can be further analysed by considering the decomposition of dinitrogen pentoxide in a rigid vessel:

N_2O_5(g)\rightleftharpoons 2NO_2(g)+\frac{1}{2}O_2(g)\; \; \; \; \; \; \; \; 7

We monitor the progress of the reaction by measuring the amount of NO2 liberated via spectroscopic methods, with eq1 becoming:

rate=\frac{change\, in\, amount\, of\, NO_2}{change\, in\, time}

Since the magnitude of the overall rate of a reaction is the same regardless of which chemical species is monitored, if we were also able to measure the amount of O2 produced or the amount of N2O5 consumed, we would, in principle, obtained:

\frac{change\, in\, amt\, of\, NO_2}{change\, in\, time}=\frac{change\, in\, amt\, of\, O_2}{change\, in\, time}=-\frac{change\, in\, amt\, of\, N_2O_5}{change\, in\, time}\; \; \; \; \; \; \; \; 8

However, for every two moles of NO2 produced over a certain time, t, half a mole of O2 is liberated and 1 mole of N­2O5 is consumed. This is inconsistent with eq8, as \frac{2}{t}\neq \frac{0.5}{t}\neq -\frac{-1}{t}. Therefore, eq8 has to be modified as follows:

\frac{1}{4}\frac{change\, in\, amt\, of\, NO_2}{change\, in\, time}=\frac{change\, in\, amt\, of\, O_2}{change\, in\, time}=-\frac{1}{2}\frac{change\, in\, amt\, of\, N_2O_5}{change\, in\, time}

Since the volume of the vessel V is constant, we can divide the modified equation throughout by V and express the rate of reaction in terms of the rate of change in concentrations of the species:

\frac{1}{4}\frac{d[NO_2]}{dt}=\frac{d[O_2]}{dt}=-\frac{1}{2}\frac{d[N_2O_5]}{dt}\; \; \; \; \; \; \; \; 9

Eq9 shows that even though the rate of consumption of a species in a reaction may be different from the rate of formation of another species in the same reaction, the magnitude of the overall rate of the reaction must be the same, regardless of the species we are measuring. If we rewrite eq7 as: 2N_2O_5(g)\rightleftharpoons 4NO_2(g)+O_2(g) and compare it with eq9, we see that:

    1. the rate of change of a species in a reaction is multiplied by the reciprocal of its stoichiometric coefficient; and
    2. the rate of change of a reactant is further multiplied by -1

Hence, we can write the differential rate equation for the decomposition of N2O5 as:

\frac{1}{4}\frac{d[NO_2]}{dt}=k[N_2O_5]\; \; or\; \;\frac{d[O_2]}{dt}=k[N_2O_5]\; \; or\; \;-\frac{1}{2}\frac{d[N_2O_5]}{dt}=k[N_2O_5]

In general, for a reaction aA+bB+...\rightarrow pP+qQ+...

rate=\frac{1}{p}\frac{d[P]}{dt}=\frac{1}{q}\frac{d[Q]}{dt}=...=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=...\; \; \; \; \; \; \; \; 10

Comparing the rate equations for the reaction between CaCO3 and HCl (eq6) and that for the decomposition of N2O5, we see that the rates of reaction are directly proportional to the concentration of the respective reactants. This, however, is not always the case. For certain reactions, the rate can be proportional to the square of the reactant or even proportional to the product of the concentrations of two or more reactants. Here are some examples of rate laws:

Stoichiometric equation

Rate law
CH3COOC2H5 + OH– → CH3COO– + C2H5OH rate = k[CH3COOC2H5][OH–]
CH3CHO → CH4 + CO rate = k[CH3CHO]3/2
2N2O5  → 4NO2 + O2 rate = k[N2O5]
NO2 + CO → NO2 + CO2 rate = k[NO2]2

In short, a rate law cannot be predicted from the reaction’s stoichiometric equation. It has to be determined experimentally.

 

Question

The differential rate equation of eq7 can be expressed as \frac{d[N_2O_5]}{dt}=-k[N_2O_5]. If eq7 is written as 2N_2O_5\rightleftharpoons 4NO_2+O_2, the rate equation will be \frac{d[N_2O_5]}{dt}=-2k[N_2O_5]. How do we reconcile the difference?

Answer

For the first rate equation, \frac{d[N_2O_5]}{dt}=-k_1[N_2O_5]

where k1 is the rate constant with reference to the reaction N_2O_5\rightleftharpoons 2NO_2+\frac{1}{2}O_2.

For the second rate equation, \frac{d[N_2O_5]}{dt}=-2k_2[N_2O_5]

where k2 is the rate constant with reference to the reaction 2N_2O_5\rightleftharpoons 4NO_2+O_2.

Comparing the two rate equations, k1 = 2k2.

A reaction can be written in many ways by multiplying or dividing the stoichiometric coefficients on both sides of the reaction by different factors. This implies that many rate equations can be written for a particular reaction. We just need to specify which stoichiometric form of the reaction we are referring to for the rate equation presented.

 

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Order of a reaction

The order of a reaction describes the relationship between the rate of a chemical reaction and the concentration of the species involved in it.

As mentioned in the article on differential forms of rate laws, the rate of a reaction is not always directly proportional to the concentration of a reactant, e.g. for the reaction, NO_2+CO\rightarrow NO+CO_2 :

rate=k[NO_2]^2

The rate equation, except for one that describes an elementary reaction, cannot be predicted from the reaction’s stoichiometric equation and has to be determined experimentally.

In general, for a reaction: aA + bB + cC + dD… → mM + nN + oO + pP… , a proposed but not experimentally verified rate law is:

rate=k[A]^i[B]^j[C]^k[D]^l...

where the exponents i, j, k and l are called the order of the reaction with respect to A, B, C and D  respectively and are not related to the stoichiometric coefficients of the reaction.

Consider the reaction: aA + bB + cC → mM + nN + oO. If experimental results reveal that the rate law is

rate=k[A]^2[C]

we say that the reaction is second order with respect to A, zero order with respect to B and first order with respect to C. We can also say that the reaction is overall third order (2+0+1). The order with respect to a reactant usually ranges from zero to three. It can be a whole number or a fraction.

Next, let’s look at the acid-catalysed hydrolysis of methyl ethanoate, CH3COOCH3 +H2O → CH3COOH + CH3OH, with an overall second order rate law of:

rate=k[CH_3COOCH_3][H_2O]\; \; \; \; \; \; \; \; 11

If water is present in large excess, its concentration remains constant throughout the reaction and eq11 becomes:

rate=k'[CH_3COOCH_3]

where k’ = k[H2O] .

We call such a scenario, where a second (or higher order) rate law reduces to a first order rate law due to the concentration of one (or more) of the reactants being constant throughout the reaction, a pseudo-first order reaction, with the rate law being pseudo-first order.

To determine the order, the rate constant and therefore the rate of a reaction, we need to know how to monitor the progress of a reaction and subsequently analyse the data obtained.

 

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Half-life

Half-life, t1/2, is the time for the amount of a reactant to reduce to half of its initial value. For the first order decomposition of hydrogen peroxide, the rate equation is given by eq13 of the previous article: ln\frac{[H_2O_2]}{[H_2O_2]_0}=-kt.

Assuming that volume of the reacting mixture remains constant, the ratio of the amounts of peroxide, n, at t = t and t = 0 is the same as the ratio of the concentrations of peroxide, \frac{n}{V}, at t = t and t = 0. Hence, hydrogen peroxide’s half-life is the time taken for its initial concentration to fall to half, i.e.:

ln\frac{\frac{1}{2}[H_2O_2]_0}{[H_2O_2]_0}=-kt_{\frac{1}{2}}

t_{\frac{1}{2}}=\frac{ln2}{k}\approx \frac{0.693}{k}\; \; \; \; \; \; \; \; 19

Eq19 shows that the half-life of a species in a first order reaction does not depend on its concentration and only depends on the rate constant, k, of the reaction.

Experimental data for the decomposition of H2O2 with an initial concentration of 8.96×10-2 M is presented in the diagram above. The 1st half-life of H2O2 is about 480s. The 2nd and 3rd half-lives, which are the time taken for \frac{8.96\times10^{-2}}{2} M to fall to \frac{8.96\times10^{-2}}{4} M and \frac{8.96\times10^{-2}}{8} M respectively, are also about 480s each. Hence, the experimental value of the successive half-lives of the first order decomposition of H2O2 is a constant and is consistent with the theoretically derived eq19.

Using eq15 and eq17 from the previous article, the half-lives of a zero order reaction and a second order reaction for a chemical species A,are

t_{\frac{1}{2}}=\frac{[A]_0}{2k}\; \; \; \; \; \; \; \; 20

and

t_{\frac{1}{2}}=\frac{1}{k[A]_0}\; \; \; \; \; \; \; \; 21

respectively.

In summary,

Order

 Simple rate equation Integral rate equation

Half-life
0 rate=k [A]=-kt+[A]_0 t_{\frac{1}{2}}=\frac{[A]_0}{2k}
1 rate=k[A] ln[A]=-kt+[A]_0 t_{\frac{1}{2}}=\frac{ln2}{k}
2 rate=k[A]^2 \frac{1}{[A]}=kt+\frac{1}{[A]_0} t_{\frac{1}{2}}=\frac{1}{k[A]_0}

Next, we shall explore the various methods in monitoring the progress of a reaction and analysing the data obtained.

 

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Monitoring the progress of a reaction

There are many ways to monitor the progress of a reaction in the attempt to determine the rate of the reaction. We can broadly categorise them as:

    1. Real-time monitoring methods; and
    2. Quenching methods

Real-time monitoring involves measuring a physical property of the system while the reaction is in progress. For example, a reaction that produces a gas can be monitored by measuring the volume of gas produced at different times with a syringe, or by noting the pressure of the gas at various time intervals with a pressure gauge (A). Another way to observe the progress of reaction in real-time is to record the change in mass of a reaction mixture over time (B).

The reaction between hydrochloric acid and calcium carbonate to give carbon dioxide is an example that can be monitored by either method A or B.

2H_3O^+(aq)+CaCO_3(s)\rightarrow Ca^{2+}(aq)+CO_2(g)+3H_2O(l)

Colorimetry, an analytical method to determine the concentration of dissolved coloured compounds, is also commonly used to observe the progress of a reaction (C). It begins with radiating the reaction mixture with a specific wavelength of light, which is selected by passing a continuum light source through a monochromatic filter. The monochromatic light is partly absorbed by the coloured compound before it exits the mixture. A detector then captures the exiting light and conveys the data to a computer for analysis. Since the amount of monochromatic light absorbed by the coloured species is proportional to the concentration of the species, the progress of the reaction can be monitored in real-time. The reaction between iodide and hypochlorite to form hypoiodite, which absorbs near 400 nm, is an example of a reaction that can be monitored using colorimetry:

ClO^-(aq)+I^-(aq)\rightarrow IO^-(aq)+Cl^-(aq)

Quenching methods, on the other hand, involve extracting a sample of the reaction mixture called an aliquot, quenching or stopping the reaction in the aliquot, and analysing the quenched mixture using other analytical techniques like titration (D), spectroscopy or chromatography. Some of the ways to quench a reaction mixture include:

    1. Diluting a reaction mixture with water, e.g. for the decomposition of oxalic acid in concentration sulphuric acid.
    2. Cooling a reaction suddenly, e.g. by spraying an aqueous reaction mixture with cold isopentane.
    3. Adding a reagent that combines with one of the reactants to stop the reaction, e.g. adding acid to quench the hydrolysis of ethyl acetate in a basic solution.
    4. Inactivating or removing a catalyst in a reaction mixture, e.g. adding a base in the acid-catalysed iodination of acetone to remove the catalyst:

(CH_3)_2CO(l)+I_2(aq)\; \: \begin{matrix} H^+\\\rightarrow \end{matrix}\: CH_3COCH_2I(l)+HI(aq)

Aliquots of the reaction mixture are extracted over several time intervals and added to excess aqueous sodium hydrogen carbonate to neutralise the acid. The concentration of the remaining iodine is then determined by titration with sodium thiosulphate.

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