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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 behavior 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 is 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 can 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 the chemical species monitored, if we were able to also measure the amount of O2 produced or the amount of N2O5 consumed, we would have:

\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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Analysing rate data (chemical kinetics)

The aim of analysing rate data of an experiment is to determine the order with respect to different reactants, the rate constant and hence the rate law.

Although various approaches can be employed to achieve that, depending on the type of reaction being investigated, most experimental procedures are structured in a way that narrow the approaches down to one or two.

The commonly used methods for analysing rate data are:

    1. the half-life method
    2. the differential method
    3. the integral method
    4. the initial rate method; and
    5. the isolation method

Some experiments may require a combination of these methods to derive the rate equation. We shall look at them individually and see how they can be used.

 

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Half-life method

The half-life method for analysing rate data is useful when the rate law involves a single reactant, i.e.

rate=k[A]^i\; \; \; \; \; \; \; \; 22

To determine the values of k and i, we let the reaction proceed, measure the appropriate physical property and plot the data in a concentration versus time graph.

If the curve is a straight line (see graph above), the gradient or the change in concentration versus the change in time (i.e. rate of reaction) is a constant. This implies that i = 0 in eq22, i.e. a zero order reaction. We can further confirm this by finding the first three successive half-lives from the graph (there is no need to consider the subsequent half-lives) and substitute them, along with the initial concentration of the reactant, in eq20 from the article on half-life: t_{\frac{1}{2}}=\frac{[A]_0}{2k}. The three substitutions should give three different values of k that should be close to one another. The average of these values of k is then taken to give the rate equation: rate = kave.

If the curve is concave (see diagram above), it represents a higher order reaction. Similarly, the first three successive half-lives is determined from the graph. If the values are relatively constant, the graph corresponds to a first order reaction, with i = 1. The value of the rate constant is then found using eq19 from the article on half-life: t_{\frac{1}{2}}=\frac{ln2}{k}, resulting in the rate equation: rate = k[A].

If the half-lives are very different from one another, they could be those of a second order reaction, i = 2. We can verify this by substituting the three half-life values, along with the initial concentration of the reactant, in eq21 from the article on half-life: t_{\frac{1}{2}}=\frac{1}{k[A]_0}, which should give three different values of k that may be close to one another. The average value of the rate constant can be found and hence, the rate equation: rate = kave[A]2.

 

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Differential method (rate laws)

The differential method of determining the rate of a reaction makes use of the differential form of a rate law.

Just like the half-life method, the differential method is useful for reactions involving a single reactant, e.g.

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

where we can write the rate equation as

-\frac{d[N_2O_5]}{dt}=k[N_2O_5]^i\; \; \; \; \; \; \; \; 23

Taking the natural logarithm on both sides of eq23, we have

ln\left [ -\frac{d[N_2O_5]}{dt} \right ]=iln[N_2O_5]+lnk\; \; \; \; \; \; \; \; 24

To find the rate constant, k, and the order, i:

    1. Run the experiment in a rigid vessel and record the concentrations of N2O5 using spectroscopic methods at various times.
    2. Plot a graph of [N2O5] versus time.
    3. Find the gradients to the curve at selected concentrations; that is, find \frac{d[N_2O_5]}{dt} at various [N2O5].
    4. With reference to eq24, plot a graph of ln\left [- \frac{d[N_2O_5]}{dt} \right ] versus ln[N2O5] using the values determined in step 3.

The gradient and the vertical intercept of the line in the second graph give the values of i and lnk respectively. The differential method is commonly used together with the initial rate method and the isolation method in determining the unknowns of a rate equation with multiple reactants.

 

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